ra MI 
 
 \ Irving 
 
 2MOEIAM 
 Stringham 
 
GEOMETRY FOR BEGINNERS. 
 
 BY 
 
 G. A. HILL, A.M. 
 
 BOSTON: 
 
 PUBLISHED BY GINN, HEATH, & CO. 
 1884. 
 
VV 
 
 Entered according to Act of Congress, in the year 1880, by 
 
 GEORGE A. HILL, 
 in the office of Jie Librarian of Congress, at Washington. 
 
 J. S. CUSHING & Co., PKINTERS, 101 PKAKL STREET, BOSTON. 
 
PREFACE. 
 
 ' I **HE formal method of teaching Geometry, as we find it in Euclid 
 JL and in the common text-books, whatever may be its merits, is 
 a very bad method for beginners. It makes the study of the subject 
 unnecessarily dry, tedious, and difficult ; and it ignores, in most cases, 
 the great law of mental development, that clear perceptions and intui- 
 tions must precede the intelligent use of the faculties of comparison 
 and reasoning. The usual consequence is that this method exercises 
 the memory more than the intelligence, and utterly fails to impart any 
 habits of thought that are useful in after-life. 
 
 In the present work, a method is employed which seems to the author 
 much better suited to promote the natural growth of the mental powers. 
 The method will speak for itself to those who will take the trouble to 
 examine it. A marked feature of it consists in the numerous exercises 
 which are to be worked by the learner. By this kind of work, if the 
 exercises are well chosen and faithfully studied, the inventive faculty is 
 called into play, and the habit of seizing quickly the true relations of 
 things is acquired. And it is just here, in its power to form this 
 habit of mind, that the value of the study of Geometry, as a prepar- 
 ation for the varied duties and labors of life, can hardly be over-esti- 
 mated. Does not the possession or want of this habit constitute, in 
 most instances, the chief intellectual difference between a man who 
 succeeds in the world and a man who does not ? 
 
 In Germany the true worth of Geometry as an educational means is 
 better recognized than elsewhere; and the simpler parts of the subject, 
 
 800542 
 
IV PREFACE. 
 
 treated much as in this work, have been introduced with the happiest 
 results into the common schools. The author, while residing in Ger- 
 many in 1877 and 1878, made himself familiar with their methods and 
 text-books, and he has freely used the knowledge thus acquired in 
 writing the present work. 
 
 As regards the subjects treated in the last foui 1 chapters, the author 
 was not satisfied with what he found, even in the best German text- 
 books, and he has planned and written these chapters according to his 
 own ideas. 
 
 To make the study both more interesting and more useful, much 
 attention has been given to the practical uses of Geometry ; such, for 
 example, as measuring inaccessible distances, and computing lengths, 
 areas, and volumes. 
 
 In a few cases the method of Limits has been used. By this means 
 rigor of proof has been secured ; and when the method is presented in 
 its simplest form (as on page 230) , the author believes that it is quite 
 within the comprehension of the average boy fifteen years old. 
 
 If answers to the exercises are generally desired they will soon be 
 published. 
 
 The author desires to express his obligations to Prof. G. A. WENT- 
 WORTH, who has kindly read the proof-sheets, and furnished him with 
 many valuable suggestions. 
 
 G. A. HILL. 
 CAMBRIDGE, June i, 1880. 
 
CONTENTS. 
 
 PAGE. 
 
 CHAPTER I. INTRODUCTION 1 
 
 I. Space ( i). II. The -Cube ( 2-5). III. The Cylinder ( 6). 
 IV. Bodies, Surfaces, Lines, and Points ( 7-15). V. Generation of 
 Space Magnitudes ( 16). VI. Straight and Curved Lines ( 17). 
 
 VII. Plane and Curved Surfaces ( 18). VIII. Cornered and Curved 
 Bodies ($ 19). IX. Geometry ( 20). 
 
 CHAPTER II. STRAIGHT LINES 17 
 
 I. Direction of One Line ($ 21-25). H. Change of Direction. The 
 Circle ( 26-28). III. Directions of Two Lines ( 29, 30). IV. Di- 
 rections of Three Lines ( 31). V. Length of a Line ( 32-36). VI. Ra- 
 tio of Two Lines ( 37, 38). VII. Units of Length ( 39-41). 
 
 VIII. Measuring a Line ( 42, 43). 
 
 CHAPTER III. ANGLES 51 
 
 I. Definition of an Angle ( 44). II. Magnitude of an Angle ( 45, 
 46). III. Magnitudes of Particular Angles ( 47-49). IV. Measure 
 of an Angle ( 50-52). V. Angles made by Two Lines ( 53, 54). 
 VI. Angles made by Three Lines ( 55-58). 
 
 CHAPTER IV. TRIANGLES 73 
 
 I. Sides of a Triangle ($$ 59-63). II. Angles of a Triangle ( 64-66). 
 III. Similarity, Equivalence, and Equality ($ 67, 68). IV. Equal Tri- 
 angles ( 69-79). V. Some Consequences of the Equality of Triangles 
 ( 80-93). VI. Applications ( 94-98). 
 
 CHAPTER V. QUADRILATERALS Ill 
 
 I. Sides and Angles of a Quadrilateral ( 99, 100). II. Different Kinds 
 of Quadrilaterals ( 101-105). III. Construction of Quadrilaterals 
 ( 106-111). IV. Subdivision of a Line ( 112-114). 
 
VI CONTENTS. 
 
 PAGE. 
 
 CHAPTER VI. POLYGONS 123 
 
 I. Sides and Angles of a Polygon ($ 115-117). II. Regular Polygons 
 ( 118-122). III. Symmetrical Figures ( 123). 
 
 CHAPTER VII. AREAS 136 
 
 I. Units of Area ( 124, 125). II. The Areas of Polygons ( 126-135). 
 
 III. Practical Exercises and Applications ( 136-141). IV. Theorem 
 of Pythagoras ($ 142, 143). V. Transformation of Figures ( 144-150). 
 
 VI. Partition of Figures ( 151-155). 
 
 CHAPTER VIII. SIMILAR FIGURES 171 
 
 I. Proportional Lines and Figures ($ 156-158). II. Similar Triangles 
 ( 159-168). III. Problems and Applications ( 169-176). IV. Simi- 
 lar Polygons ( 177-179). 
 
 CHAPTER IX. THE CIRCLE 199 
 
 I. Sectors, Angles at the Centre, Chords, Segments ( 180-184). II. In- 
 scribed Angles ( 185-188). III. Secants and Tangents ( 189-192). 
 
 IV. Two Circles ($ 193, 194). V. Inscribed and Circumscribed Fig- 
 ures ( 195-207). VI. Length of a Circumference ( 208-211). 
 
 VII. Area of a Circle ( 212-218). 
 
 CHAPTER X. THE ELLIPSE 24O 
 
 I. The Properties of the Ellipse ( 219-221). II. Construction of 
 Ellipses ( 222-224). 
 
 CHAPTER XL PLANES 248 
 
 I. Straight Lines in Space ( 225-227). II. A Plane ( 228-230). 
 III. A Plane and a Straight Line ( 231-235). IV. Two Planes 
 ( 236-239). V. Three Planes ( 240). VI. Solid Angles ( 241). 
 
 CHAPTER XII. GEOMETRICAL BODIES 263 
 
 I. Polyhedrons ( 242, 243). II. The Prism, Cylinder, Pyramid, and 
 Cone ( 244-250). III. The Sphere ( 251, 252). 
 
 CHAPTER XIII. SURFACES AND VOLUMES OF BODIES . 279 
 I. Surfaces of Bodies ($ 253-258). II. Volumes of Bodies ( 259- 
 274). III. Exercises and Applications ( 275-279). 
 
GEOMETRY FOR BEGINNERS. 
 
 CHAPTER I. 
 INTRODUCTION. 
 
 CONTENTS. I. Space ($ i). II. The Cube ($ 2-5). III. The Cylinder ( 6). 
 IV. Bodies, Surfaces, Lines, and Points ( 7-15). V. Generation of Space 
 Magnitudes ( 16). VI. Straight and Curved Lines ($ 17). VII. Plane and 
 Curved Surfaces ( 18). VIII. Cornered and Curved Bodies ( 19). IX. Geom- 
 etry ( 20). 
 
 L Space. 
 
 1. We cannot define space. We know, however, that all 
 things exist in space. I lay a book on the table : it has now a 
 definite position in space on the table. The table is in the space 
 occupied by the room ; the room is in the space enclosed by the 
 house ; the house rests upon the earth ; and the earth is always 
 moving swiftly through space. Thus all things are in space ; but 
 what space is, where it begins, or where it ends, these are questions 
 which no man can answer. 
 
 Space contains all things, and extends in all directions, without 
 beginning and without end. 
 
 II. The Cube. 1 
 
 2. The Cube {fig. /) occupies a portion of space which is 
 limited on all sides. A limited portion of space is called a BODY 
 (see 7) . Therefore the cube is a body. 
 
 1 A model of the cube is supposed to rest on the table or other support, with 
 one face turned towards the eye of the learner. 
 
2 
 
 GEOMETRY FOR BEGINNERS. 
 
 [is- 
 
 The cube is extended in space in three chief directions : from 
 right 'to left, from front to ^^/(% from <?^z;<? 
 downwards. We express this fact by say- 
 ing that the cube has three DIMENSIONS, 
 and we name them Length, Breadth, and 
 Thickness (Height or Depth). Either one 
 of the dimensions may be termed length 
 or breadth or thickness, but the terms 
 height and depth are applied only to the 
 dimension from above downwards. 
 
 Fig. i. 
 
 The three dimensions of the cube are equal in magnitude. 
 
 Exercises. 1. Here are several cubes. Have they all the same shape? 
 the same size or magnitude? Are they all composed of the same substance 
 or material ? Why are they called bodies? Why are they called cubes? 
 
 2. Mention objects which are cubical in shape, or resemble a cube. 
 (Lumps of sugar, pieces of soap, tea-chests, some boxes, etc.) 
 
 3. The cube is limited or bounded on all sides by a SURFACE. 
 When we handle the cube it is its surface alone which we touch. 
 If the cube is not transparent, it is its surface alone which we 
 are able to see. 
 
 The entire surface consists of six parts or partial surfaces, called 
 its Faces : the upper, lower, front, back, right, and left faces. 
 Each face has two dimensions, length and breadth (height). 
 Thus the upper face has length (extending from right to left) and 
 breadth .(extending from front to back) ; the front face has length 
 and height, etc. 
 
 The faces of the cube are flat or plane surfaces. 
 
 Exercises. 1. Name the faces of the cube, and likewise the dimensions 
 of each face. 
 
 2. Have the faces of the cube the same shape or form? the same size? 
 How can you test whether they have the same size or not? 
 
 Answer. Lay the cube on a piece of paper, mark around its base with a pen- 
 cil, then apply to the surface thus marked out the other faces of the cube, 
 
4.] CHAPTER I. INTRODUCTION. 3 
 
 3. Give examples of plane surfaces both like and unlike the faces of the 
 cube in form. 
 
 4. Do the faces of one cube have the same shape as those of another cube? 
 the same size ? 
 
 4. The faces of the cube are bounded by its Edges. These 
 edges are LINES, and we see, (a] that each face is bounded by four 
 lines or edges, (V) that each edge is also the place where two faces 
 meet. 
 
 The edges of the cube have only one dimension : length. 
 
 The edges of the cube are straight lines. 
 
 Exercises. 1. How many edges has the cube in all? 
 
 2. If the cube has 6 faces, and each face has 4 edges, why has not the 
 cube in all 6 X 4= 24 edges? 
 
 3. Have the edges of the cube the same form? the same size (in other 
 words, the same length") ? Test whether they have the same length. 
 
 4. Of these cubes, which has the longest edges? Which the shortest? 
 
 5. The edges of the cube are limited by its Corners. These 
 corners are POINTS, and we see, (a) that each edge is limited by 
 two corners, (b) that each corner is the place where three edges 
 meet. 
 
 The corners of the cube have no dimensions, neither length, 
 breadth, nor thickness. 
 
 How many corners are there on the cube ? 
 
 Exercises. 
 
 1. This -body {Fig. 2, I.) is a three-sided (or triangular) Prism : 
 () Point out and name its dimensions (height, length, and breadth). 
 
 NOTE. The front edge gives the height of the prism. The other two dimen- 
 sions are, (i) the distance from this edge to the opposite face, and (2) the upper or 
 lower edge of this face ; either of these may be taken as the length, and then the 
 other will be the breadth. 
 
 (]}) How many faces, edges, corners, are there ? Which are surfaces, 
 which lines, which points? (In prisms the upper and lower faces are called 
 the bases, the other faces the lateral faces.) 
 
 r Point out and name the dimensions of each face. 
 
GEOMETRY FOR BEGINNERS. 
 
 [ 6. 
 
 (X) By how many edges (lines) is each face bounded? 
 (e) How many faces meet in each edge? 
 (/) How many edges meet at each corner? 
 
 (g) Have the faces the same shape (form) ? the same size (magnitude) ? 
 (h) Have the edges the same form? the same magnitude? 
 (t) Does shape or size or material determine whether a body is a three-sided 
 Prism? 
 
 2. Examine in the same way the four-sided (square) Prism (Fig. 2, II.). 
 
 3. Examine in the same way the six-sided (hexagonal) Prism (Fig. 2, III.). 
 
 4. Examine in the same way the four-sided (rectangular) Pyramid (Fig. 2, 
 
 Fig. 2. 
 
 III. The Cylinder.^ 
 
 6. The Cylinder (Fig. j) is also, like the cube, a limited 
 portion of space : in other words, a body. 
 Its three dimensions are usually called 
 length, breadth, and height. The height 
 (the straight dotted line in the figure) may 
 have any value : so may the length and 
 breadth, but they must be always equal to 
 each other. 
 
 The cylinder is bounded by three sur- 
 faces : two plane surfaces called its Bases, 
 and a curved surface. The bases of the cylinder are Circles. 
 
 1 Several models of right cylinders are supposed to be placed on the teacher's 
 desk before the eyes of the pupils. 
 
7-] 
 
 CHAPTER I. INTRODUCTION. 
 
 The cylinder has only two edges ; these form the boundaries of 
 the bases, and are called Circumferences. They are not straight 
 lines, but curved lines. 
 
 The cylinder has no corners. 
 
 Exercises. 1. Here are several cylinders. How do they differ from 
 one another? In what do they agree? 
 
 2. Give examples of objects which have the shape of cylinders (a pencil, 
 stove-funnel, etc.). 
 
 3. Examine, as regards dimensions, surface, etc., the Cone (Fig. 4, I.). 
 4 Examine in like manner the Truncated 1 Cone (Fig. 4, II.). 
 
 5* Examine in like manner the Sphere (Fig. 4, III.). 
 
 IV. Bodies, Surfaces, Lines, and Points. 
 
 7. A body, in common language, is a limited portion of 
 space filled with matter, which we can see, touch, handle, etc. : 
 for example, a pencil, a book, the table, a tree, the earth, the sun. 
 But in Geometry we pay no regard to the matter of which a body 
 is composed ; we study simply its shape and its size : in other 
 words, we regard it simply as a limited portion of space. 
 
 Definition. A GEOMETRICAL BODY, or SOLID, is a limited por-^ 
 tion of space. 
 
 I Truncated means cut off. The body is also called the Frustum of a cone. 
 
6 GEOMETRY FOR BEGINNERS. [ 8. 
 
 A body has three dimensions, denoted by the terms length, breadth, 
 and thickness (height and depth). 
 
 NOTE. Either dimension may be called length, or breadth, or thickness. A 
 dimension estimated from below upwards, is often called height, as the height of 
 a monument ; a dimension estimated from above downwards, is often called depth, 
 as the depth of a well. Width is also used in the sense of breadth. 
 
 8. The limits or boundaries of a body are surfaces. Surfaces 
 exist not only on the exterior of a body, they may also be thought 
 of as existing in its interior. When we cut an apple into two parts 
 and then hold the parts together, the common boundary between 
 the two parts is a surface. But without actually cutting through a 
 body we may imagine a surface extending in any direction com- 
 pletely through the body. This surface would divide the body 
 into two parts and form the common boundary between them. 
 Every body is divisible into parts, and the common boundary 
 between two adjacent parts is a surface. 
 
 Definition. A SURFACE is the limit of a body, or common 
 boundary between two adjacent parts of a body. 
 
 A surface has two dimensions, length and breadth (height). 
 
 9. The limits of a surface are lines. Lines exist not only as 
 the limits or boundaries of surfaces : we can also think of them 
 as existing anywhere in a surface, where they form the common 
 boundaries between adjacent parts of the surface. 
 
 Definition. A LINE is the limit of a surface, or the common 
 boundary between two adjacent parts of a surface. 
 
 A line has one dimension, length. 
 
 10. The limits or ends of a line are points. Points exist not 
 only at the ends of lines, but also between the ends, where they 
 form the common boundaries between adjacent parts of the line. 
 
 Definition. A POINT is the limit of a line, or common bound- 
 ary between two adjacent parts of a line. 
 
 A point has no dimensions. 
 
II.] CHAPTER I. INTRODUCTION. 7 
 
 11. Bodies, surfaces, lines, and points are four different kinds 
 of things relating to space which we can think and reason about. 
 
 Definition. Bodies, surfaces, lines, and points are called 
 SPACE CONCEPTIONS. 
 
 The first three : bodies, surfaces, and lines, are extended in space, 
 and hence possess shape {form} and size {magnitude}. 
 
 Definition. With reference to form, bodies, surfaces, and 
 lines are called SPACE FIGURES ; with reference to magnitude, they 
 are called SPACE MAGNITUDES. 
 
 The point has no extension, therefore neither form nor magni- 
 tude. 
 
 12. We have seen that two faces of the cube, prism, etc., 
 meet in an edge, that is to say, in a line. Whenever two surfaces 
 meet they are said to cut or intersect each other, and the place of 
 meeting is a line called their Intersection. This line lies at once 
 in both surfaces, or is common to both surfaces. 
 
 Again, as we have seen, the edges of the cube, etc., meet at the 
 corners ; that is to say, in points. Whenever two lines meet, the 
 place of meeting is a point common to both lines, and is called 
 their intersection. 
 
 Definition. The place where two surfaces or two lines meet 
 is called their INTERSECTION. 
 
 The intersection of two surfaces is a line. 
 
 The intersection of two lines is a point. 
 
 13. Divide a body (for example, an apple) into any number 
 of parts ; each part is also a body, smaller of course than the 
 whole body, because a part of a thing is less than the whole. 
 
 Grind a lump of salt into the finest particles you can, so fine 
 that you cannot distinguish them by the eye. Under a micro- 
 scope, each particle is seen to be what it really is, a little lump 
 of salt, having length, breadth, and thickness : in other words, a 
 
8 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 
 
 body. In short, subdivide a body in thought as far as you please, 
 the smallest conceivable part is also a body, occupying space, and 
 having length, breadth, and thickness. We could never arrive at 
 a part which was a surface, a line, or a point. 
 
 In like manner, every part of a surface, however small, is also a 
 surface ; and every part of a line is also a line. 
 
 T/ie parts of a body are bodies, the parts of a surface are sur- 
 faces, and the parts of a line are lines. 
 
 14. A surface, therefore, is not a part of a body. Lay, in 
 thought, any number of surfaces one upon the other : in this way 
 you would never obtain a body, but always again a surface. The 
 common boundary which separates water 
 from oil resting on it (-Fig*5), is neither 
 water nor oil nor any other substance : it 
 is not a body, but a surface. A surface is 
 no more a part of a body than a shadow 
 is a part of the wall on which it rests. 
 
 A line is no part either of a surface or 
 of a body. Lay any number of lines side 
 by side, the result is neither a surface nor 
 a body, but always a line. The line which 
 forms the common boundary between a 
 
 Fig. 5- 
 
 red surface and a blue surface is itself neither red nor blue, nor 
 has it any other color. 
 
 Lastly, a point is no part of a line, a surface, or a body. In 
 conversation we often hear the word "point" used in the sense of 
 a surface of a certain extent. Thus, we speak of the " point of 
 crossing" of two or more roads. Substitute, in thought, for these 
 roads, lines absolutely without breadth or thickness ; then their in- 
 tersection would be a point, as this word is understood in Geom- 
 etry. 
 
i5-3 
 
 CHAPTER I. INTRODUCTION. 
 
 Fig. 6. 
 
 15. Except in thought, surfaces, lines, and points cannot 
 exist apart from bodies. In thought, however, we can detach 
 them, so to speak, from bodies, and study their forms and proper- 
 ties apart from the bodies to which they belong. 
 
 To aid the mind in reasoning about space conceptions, we rep- 
 resent them to the eye as follows : 
 
 A point is represented by a fine dot made with a pencil on paper 
 {Fig. 6), or with a crayon on the 
 blackboard : and is named by means 
 of a letter (A, B, C, etc.). 
 
 A line is represented by the fine 
 continuous trace of a pencil on pa- 
 per, or of a crayon on the black- 
 board, and named either by two let- 
 ters, placed one at each of its ends 
 (see Fig. 7), or by a single small 
 letter (a, b, c, etc.). 
 
 A surface is represented and named by means of the lines which 
 form its boundaries. 
 
 A body is represented and named by means of the surfaces 
 which form its boundaries. 
 
 Exercises. 1. Give examples of bodies. 
 
 2. Is the space occupied by the room a body? Why? 
 
 3. Name the largest body you can; also the smallest. 
 
 4. Point out and name the dimensions of the room; this book; this pen- 
 cil; this piece of rubber. 
 
 5. Name the dimensions of a board; a tower; a ditch ; a well. 
 
 6. How many partial surfaces has the Voom? Point out and name the di- 
 mensions of each. 
 
 7. How many edges has the room? How many corners? Of what are 
 the edges the intersections? the corners? 
 
 8. How many surfaces has this piece of rubber? Name their dimensions. 
 Which are equal in magnitude? 
 
 9. A glass tumbler has an exterior (outer) and an interior (inner) sur- 
 face. Give other examples of such bodies. 
 
10 GEOMETRY FOR BEGINNERS. [ 16. 
 
 10. How many edges and corners has a pane of glass ? 
 
 11. Cut twice through this body (say, an apple) so as to divide it (if pos- 
 sible) into two parts; three parts; four parts; five parts. 
 
 12. Are most natural objects (for example, stones, trees, leaves, flowers, 
 animals) simple (like the cube, prism, etc.) or complex in shape ? 
 
 13. When we use such an expression as "a shapeless mass," do we mean 
 that the mass has no shape at all ? What do we really mean ? 
 
 14. In the process of beating gold, leaves are obtained not over g-cjuVfftf 
 of an inch thick. Are such leaves surfaces or bodies? 
 
 15. In conversation, we often hear a string, a telegraph wire, etc., spoken 
 of as lines. What are they, strictly speaking? Why? 
 
 16. Is the trace of a pencil-point on paper, which we call a line, really a 
 line? What is it? Why? 
 
 17. When we make a point on paper with the end of a pencil, is it really 
 a point, or does it stand for or represent a point? What is it, strictly speak- 
 ing? Why? 
 
 V. Generation of Space Magnitudes. 
 
 16. When we move the point of a pencil on paper, it leaves 
 behind a trace which we call a line : reduce, in thought, the point 
 of the pencil to a point as understood in Geometry, then its trace 
 or path would be a geometrical line. Again, every one has ob- 
 served that the path of the red-hot end of an iron rod, when 
 moved rapidly in the dark, presents the appearance of a luminous 
 line. 
 
 The path of a moving point is always a line : in other words, 
 when a point moves in space it describes or generates a line. 
 
 When a line moves in space, it usually generates a surface. 
 This fact may be illustrated by moving a crayon over the surface 
 of the blackboard, with its longest side pressed against the board. 
 
 When a surface moves in space, it usually generates a solid. 
 We may illustrate this truth by taking a thin piece of board (or a 
 slate) to represent a surface, and pressing it down through a soft 
 substance, like snow : the hole thus made is the path of the sur- 
 face, and is a geometrical body or solid. 
 
1 7-1 
 
 CHAPTER' I. INTRODUCTION. 
 
 11 
 
 We have then, in all, three general truths or laws : 
 
 1. The path of a moving point is a line. 
 
 II. The path of a moving line is a surface, 
 
 III. The path of a moving surface is a solid, 
 
 NOTE. Practically speaking, in Law I. material points are always meant; that 
 is, bodies which seem to be points, or are for the time regarded as points. The stars, 
 to our eyes, are material points ; in reality, they are bodies of enormous size. 
 
 Exercises. 1. What exception is there to Law II.? Illustrate with a 
 pencil. 
 
 2. What exception is there to Law III.? Illustrate with a slate. 
 3t What is the path of a moving solid? 
 
 VI. Straight and Curved Lines. 
 
 17. If a point move always in the same direction, its path 
 is a straight line ; if, on the contrary, the direction of its motion 
 is continually changing, its path is a curved line or curve. In 
 Fig. 7, A B is a straight line, CD is a curve. 
 
 Definitions. I. A STRAIGHT LINE is a line which has every- 
 where the same direction. 
 
 II. A CURVED LINE, or CURVE, is a line whose direction is con- 
 tinually changing. 
 
 A line like E F (Fig. 7) , composed A 
 of several straight lines, is called a / 
 broken line : and a line like G H, c 
 composed of straight and curved 
 lines, is called a composite line. 
 
 The figures on carpets and wall- 
 papers, and the ornamental designs Q ~ H 
 employed on articles of furniture and 
 
 Fig. 7. 
 
 in architecture, furnish examples of broken and composite lines 
 in endless variety. 
 
 The word line, when used alone, usually means a straight line. 
 
12 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 
 
 Exercises. 1. On the blackboard are lines a, b, c, d, etc. What kind of 
 a line is a? Why? b ? c? d? etc. Give the reason in each case. 
 
 2. Draw and name with letters two lines of each kind. In what two ways 
 may you name or denote a line ? 
 
 3. What kind of lines are the edges of the cube? the prism? the pyra- 
 mid? the cylinder? this table? this ruler? this rubber? the room? also the 
 dm of a tumbler? the spokes of a carriage-wheel? its tire? a telegraph wire ? 
 the letters M, N, O, U, V, Z? 
 
 Fig. 8. 
 
 4. What sort of a line does a flash of lightning often resemble? (Fig. 5.) 
 
 5. What is the path of a falling apple? a ball thrown into the air? the end 
 of a watch-hand? the end of a pendulum? a drop of rain? a flying bird? a 
 point on the tire of a carriage-wheel? 
 
 6. Give other examples of straight lines and curves. 
 
 7. In these exercises have we been using the word "line "in its stricl 
 geometrical sense? Why? 
 
 VIL Plane and Curved Surfaces. 
 
 18. The faces of the cube are flat or plane : the side of ^ 
 cylinder is round or curved. How can we define these two classes 
 of surfaces? 
 
19.] CHAPTER I. INTRODUCTION. l3 
 
 Definitions. LA PLANE SURFACE, or PLANE, is a surface on 
 which straight lines can be drawn in all directions. 
 
 II. A CURVED SURFACE is a surface on which straight lines 
 cannot be drawn in all directions. 
 
 A straight line which moves in space, with one end fixed and 
 the other end sliding along another straight line, generates a plane 
 surface ; but if the second line is a curve, the surface generated is 
 in general a curved surface. 
 
 If, for example, the second line is the circumference of a circle, 
 the surface generated will be that of the cone (Fig. 4), except 
 when the moving line is in the plane of the circle : in this case the 
 surface generated will be a plane surface. 
 
 Test of a Plane Surface. Apply to the surface, in different 
 directions, the straight edge of a ruler ; if the surface is plane, the 
 edge of the ruler will always touch the surface from end to end. 
 
 Exercises. 1. On a tin pail point out a plane surface; a curved surface. 
 
 2. What kind of surfaces are the faces of a pyramid? the side of a cone? 
 its base ? the surface of the sphere ? 
 
 3. Are the walls of the room plane or curved? the surface of the table? 
 of a lamp-shade? of a hat ? of a mirror? of a wash-basin? of the water in 
 the basin? of a flower? of the ocean? 
 
 4. Give other examples of plane and curved surfaces. 
 
 5. Can you draw a straight line on the side of a cylinder? on the side of 
 a cone ? on the surface of a sphere ? 
 
 6. Test whether the surface of your desk is a perfect plane. 
 
 VIII. Cornered and Curved Bodies. 
 
 19. The cube and the bodies shown in Fig. 2 are examples of 
 cornered bodies. The cylinder and the bodies shown in Fig. 4 
 are instances of curved bodies. 
 
 Definitions. I. A CORNERED BODY, or POLYHEDRON, is a body 
 bounded entirely by plane surfaces. 
 
 II. A CURVED BODY is a body bounded partly or wholly by 
 curved surfaces. 
 
14 GEOMETRY FOR BEGINNERS. [ 2O. 
 
 Exercises. 1. Give examples () of polyhedrons, (<$) of curved bodies. 
 2. To which class does the earth on which we live belong? 
 
 IX. Geometry. 
 
 20. The Science of space conceptions is called GEOMETRY. 
 The aim of Geometry is threefold : 
 
 I. To discover the various properties of space figures arising 
 from their having different forms . 
 
 II. To show how to draw or construct space figures. 
 
 III. To show how space magnitudes may be measured. 
 Geometry is usually divided into Plane Geometry and Solid 
 
 Geometry. 
 
 PLANE GEOMETRY treats of space conceptions confined to one and 
 the same plane. 
 
 SOLID GEOMETRY treats of space conceptions not confined to one 
 plane {for example, solids and curved surfaces) . 
 
 Chapters II. X. of this work treat of Plane Geometry; the re- 
 maining chapters, of Solid Geometry. 
 
 NOTES. I. The word Geometry comes from two Greek words meaning land- 
 measuring. In ancient times, when the science was in its infancy, it consisted en- 
 tirely of a few practical rules employed in measuring land, and used especially by the 
 Egyptians in settling disputes about boundary lines destroyed by the annual over- 
 flow of the river Nile. Hence the name Geometry (or land-measuring) was given 
 to the science by the Greeks. 
 
 2. No science excels Geometry in the number and importance of its practical 
 applications. This fact is strikingly seen in the power which it gives us of measur- 
 ing space magnitudes. If one asks : How much more will grow on this field than 
 on that one ? how much must this ship be loaded to sink ten feet into water ? how to 
 find the right way across the pathless ocean ? how maps of the land and sea are 
 made ? how the distance of the moon can be found ? at precisely what instant an 
 eclipse of the sun will take place ? when a comet will again be seen ? how a cannon 
 must be pointed in order that the ball may hit a certain mark? these and similar 
 questions, only a man skilled in Geometry can answer. 
 
CHAPTER I. INTRODUCTION. 15 
 
 REVIEW OF CHAPTER I. 
 QUESTIONS. 
 
 1. How many dimensions has the cube, and what are they called? 
 
 2. How many faces, edges, and corners has the cube? 
 
 3. How many faces meet in an edge? how many edges in a corner? 
 
 4. Are the faces equal or unequal? the edges? 
 
 5. Answer the preceding questions for each of the bodies in Fig. 2. 
 
 6. What account can you give of the surfaces and edges of the bodies in 
 Figs. 3 and 4? 
 
 7. Name a body which has no corners; also one which has neither edges 
 nor corners. 
 
 8. Define a solid, a surface, a line, a point; how many dimensions has 
 each, and what are they called? 
 
 9. Define the intersection of surfaces or of lines. What is the intersection 
 of a surface? of a line? 
 
 10. What are space conceptions ? space magnitudes ? space figures ? 
 
 11. What are the parts of a body? the parts of a surface? the parts of a line? 
 
 12. Illustrate the facts, that a surface is no part of a body, and that a line is 
 no part of a surface. 
 
 13. Illustrate what is meant by a point in Geometry. 
 
 14. How are space conceptions represented and named? 
 
 15. Illustrate how space magnitudes are generated by motion. What are 
 the general laws? the exceptions to them? 
 
 16. Define the four kinds of lines, and give examples. 
 
 17. Define plane and curved surfaces. How is a plane surface tested? 
 
 18. Show how a line must move to generate a plane surface; a curved sur- 
 face. 
 
 19. Define polyhedrons and curved bodies, and give examples. 
 
 20. What is Geometry ? its chief aims ? its chief divisions ? 
 
 EXERCISES. 
 
 1. Give examples of bodies shaped like a prism, a pyramid, a cone, a sphere. 
 
 2. Name the dimensions of a telegraph wire, a monument, a river, a well. 
 
 3. Name the dimensions of a cylinder, (a] when standing on its base, () 
 when resting on its side. 
 
 4. How can you cut a prism in two so that each part will also be a prism? 
 
16 GEOMETRY FOR BEGINNERS. 
 
 5. How can you place two cylinders having equal bases together so as to 
 form another cylinder? 
 
 6. Can you place two cubes together so as to form a new cube? 
 
 7. Here are several equal cubes: put them together so as to form anew 
 cube. How many cubes does it take? How much larger is the new 
 cube than one of its parts? its face than the face of one of its parts? 
 its edge than the edge of one of its parts? 
 
 8. Mention bodies of regular shape, but differing in shape, from those thus 
 far examined. 
 
 9. What kind of space magnitude is the human skin? 
 
 10. Give an instance where more than two lines intersect in the same point. 
 
 11. Can more than two surfaces intersect in the same line? Can you give an 
 illustration? 
 
 12. When will a large body appear very small? Give examples. 
 
 13. Draw (if possible) a line with no ends ; with one end ; with two ends ; 
 with more than two ends. 
 
 14. If you slide a straight line along the two rails of a railroad, what kind 
 of a surface would be generated? 
 
 15. If you move a straight line around the tires of two carriage-wheels 
 on the same axle (always keeping it in contact with both tires), what 
 kind of a surface would be generated? 
 
2!.] 
 
 CHAPTER II. STRAIGHT LINES. 
 
 17 
 
 CHAPTER II. 
 STRAIGHT LINES. 
 
 CONTENTS. I. Direction of One Line ( 21-25). H. Change of Direction. 
 The Circle ( 26-28). III. Directions of Two Lines ( 29, 30). IV. Direc- 
 tions of Three Lines ( 31). V. Length of a Line ( 32-36). VI. Ratio of 
 Two Lines ($$ 37, 38). VII. Units of Length ($ 39-41). VIII. Measuring a 
 Line (42,43). 
 
 L Direction of One Line. 
 
 21. A straight line, A B (Fig. 9) , may be regarded as hav- 
 ing either the direction from A towards A B 
 
 B, or the opposite direction from B to- ^ 
 
 wards A; in the first case, we should 
 call it the line AJ3, in the second case, 
 
 Fig. 9. 
 
 the line BA. Sometimes (as, for instance, on a guide-post) an 
 arrow-head or a hand is used to point out which direction is in- 
 tended ; but, in general, we may assign to the line either direc- 
 tion. 
 
 22. If we know only the direction of a straight line, we can- 
 not determine its position or locate it ; 
 for, through different points, A, B, C, 
 D, (Fig. id), different straight lines 
 may be drawn, all having the same 
 direction. 
 
 Likewise, if we know only that a 
 straight line passes through a known or 
 given point, A, the line is not known ; Fi &' I0 ' 
 
 for, through a point any number of straight lines may be drawn. 
 
18 GEOMETRY FOR BEGINNERS. [ 2 3- 
 
 But through the point A, in a given direction, as that of the 
 arrow, only one straight line can be drawn. And, also, through 
 any two points, as A and M, but one straight line can be drawn. 
 
 These two truths may be stated as follows, in a general form, 
 without reference to Fig. 10, or to any particular case : 
 
 1. Through a given point in a given direction only one straight 
 line can be drawn. 
 
 II. Throttgh two given points only one straight line can be 
 drawn. 
 
 Hence, if one point and the direction of a line are given, or if 
 two points of the line are given, the line is determined in position. 
 
 23. For drawing straight lines on paper, or on the blackboard, 
 a RULER (Fig. 7/) 1 with a straight edge is employed. 
 
 In order to draw a line on paper through two given points, lay 
 the ruler on the paper so that its edge 
 just touches the two points ; then draw 
 the line with a hard, well-sharpened 
 pencil, holding the pencil nearly vertical, and always in contact 
 with the edge of the ruler. 
 
 When we draw a line without the aid of any instrument to 
 guide the hand, we are said to draw it FREE-HAND. 
 
 Exercises. 1. Draw four lines through a point, and name them with letters. 
 
 2, Make two points, and then join them by a straight line. 
 
 NOTE. If A and B are two points, the expression join AB \slo be understood 
 as meaning draw a straight line from A to B. 
 
 3* Make three points not in the same line; then join them by straight lines. 
 How many straight lines in all can be drawn? 
 
 4. How many straight lines in this way can be drawn between four points? 
 five points? six points? 
 
 5. Ttrzw, free-hand, a straight line; then correct or rectify it with the aid 
 of a ruler. Repeat the exercise several times. 
 
 1 The divided rule (see $ 42), which the pupil requires for other purposes, 
 will also serve for drawing straight lines. 
 
24.] CHAPTER II. STRAIGHT LINES. 19 
 
 6. How can you test whether the edge of a ruler is straight? 
 
 Ans. Draw a line with the ruler through any two points; -then turn the ruler 
 end for end, and, along the same edge, draw a line again through the same two 
 points. If the edge is straight, the two lines will coincide and form a single line. 
 (See 22, II.).' 
 
 7. Test the edge of your ruler or divided rule. 
 
 8. Test the edge of the ruler at the blackboard. 
 
 9. A farmer has a four-sided field, and wishes to run a straight fence be- 
 tween two opposite corners. Explain how, with a person to assist him, he 
 can find points on the proposed line of the fence between the corners. 
 
 Suggestion. If the farmer stands a little way behind one corner, and places 
 his eye so that the line of sight passes through both corners, this line will coincide 
 with the proposed line of the fence. 
 
 24. Among the directions which a straight line may have, 
 there are two of more practical importance than the 
 others. 
 
 Fasten a weight to one end of a cord, and then hold 
 the other end at rest in the hand : this forms what is 
 called a plumb line. When the cord is at rest its di- 
 rection is said to be VERTICAL. On the other hand, 
 a pencil, stick, or similar object, when floating on 
 the surface of still water, is said to have a HORI- ^ 
 ZONTAL direction. Fi s- I2 - 
 
 Definitions. LA VERTICAL LINE is a line which has the 
 direction of a plumb line. 
 
 II. A HORIZONTAL LINE is a line which. has the direction of any 
 line in the surface of still water. 
 
 III. Lines neither vertical nor horizontal are called INCLINED 
 LINES. 
 
 Plane surfaces, also, are either vertical, horizontal, or inclined. 
 Every plane in which a vertical line can be drawn, is a vertical 
 plane ; the surface of still water, or any plane similarly placed 
 with reference to the earth's surface, is a horizontal plane ; planes 
 neither vertical nor horizontal are inclined planes. 
 
 The sides of the room, for example, are vertical planes ; the 
 
20 GEOMETRY FOR BEGINNERS. [ 24. 
 
 floor and ceiling are horizontal planes ; the roof of the house is 
 an inclined plane. 
 
 NOTES. i. The surface of water, if of considerable extent (the ocean, for in- 
 stance), is sensibly a curved surface, because the earth is round. This curvature 
 makes itself evident on the seashore when we watch a ship sailing out of sight on 
 the horizon : first the hull sinks out of sight ; then the sails and lower parts of the 
 masts ; last of all, the tops of the masts. But on a pond three or four miles in 
 length, the curvature is so small that it cannot be observed ; hence, for all common 
 purposes, the surface of the pond is regarded as a plane. 
 
 2. On paper draw vertical lines towards or from you ; horizontal lines from right 
 to left, or left to right. 
 
 Exercises. 1. Of the lines on the blackboard, which are vertical? which 
 horizontal? and which inclined? 
 
 2. Draw a line on paper; then hold the paper so that the line shall be 
 vertical. Is now the plane of the paper also vertical? 
 
 3. Hold a pencil vertical; horizontal; inclined. 
 
 4. Hold a book vertical; horizontal; inclined. 
 
 5. Draw three lines of each kind. 
 
 6. Draw a vertical line, mark five points on it, and through these points 
 draw horizontal lines. 
 
 7. On the cube, point out vertical edges and faces; also horizontal edges 
 and faces. 
 
 8. What directions have the edges of the prism? those of the pyramid? 
 those of the cylinder? 
 
 9. What direction has a ladder when leaning against a wall? when lying 
 on the ground? when suspended from one end .by a rope? What direction 
 have the rounds of the ladder in these three cases? 
 
 10. What direction has the mast of a ship? the path of a falling apple? 
 the beam of a balance at rest? the surface of a lake? the side of a house? 
 its roof? 
 
 11. At what time of the year are the sun's rays most nearly vertical ? 
 When are the sun's rays horizontal? 
 
 12. Give other examples of the vertical and horizontal lines and planes. 
 
 13. When is the minute-hand of a clock vertical? when horizontal? 
 Answer the same questions for the hour-hand. 
 
 14. When two vertical planes (for example, the two sides of a room) in- 
 tersect each other, what kind of a line is their intersection? 
 
 15. When a vertical plane intersects a horizontal plane what kind of a 
 line is their intersection? Give an example. 
 
 16. Through a point how many vertical lines can be drawn? how many 
 horizontal lines? how many inclined lines? how many planes of each kind? 
 
25.] CHAPTER II. STRAIGHT LINES. 21 
 
 17. In a vertical plane how many vertical lines can be drawn? how many 
 horizontal lines? 
 
 18. In a horizontal plane how many vertical lines can be drawn? how 
 many horizontal lines? 
 
 25. Applications. I. In order to test whether a line or a 
 plane (as a wooden beam, the wall of a house, etc.) is ver- 
 tical, a PLUMB-RULE (Fig. 13) is often used. It consists of 
 a piece of wood having two straight edges everywhere equally 
 distant from each other. Near one end, exactly half way 
 between the edges, a plumb-line is attached. Near the 
 other end, also exactly half way between the edges, a mark 
 is made. If we place one edge of the plumb-rule against a 
 surface which is vertical, the line, when at rest, will cover 
 the mark ; if the surface is not vertical, the line will pass on 
 one side of the mark. It is well to apply the test twice, 
 using first one then the other edge of the plumb-rule ; in 
 both cases, the line ought to cover the mark. 
 
 II. To test whether a line or a plane surface is horizontal (or, 
 as workmen say, level), instruments called LEVELS are used. The 
 simplest kind of level is the Plumb-line Level (Fig. 14) . It con- 
 sists of three pieces of wood put together in the shape of the let- 
 ter A, with a plumb-line suspended from the top. In order to 
 adjust the instrument for use, a mark must be made on the cross- 
 piece, where the plumb-line passes, when the feet of the level are 
 
 Fig. 14. Fig. 15. 
 
 at the same height, or horizontal. This is done as follows : Place 
 the feet of the level on any two points, and mark on the cross-bar 
 
22 
 
 GEOMETRY FOR BEGINNERS. 
 
 [25. 
 
 the place of the plumb-line ; then turn the instrument end for 
 end, rest it on the same points, and again mark the place of the 
 plumb-line. The point midway between the two marks is the 
 right one. 
 
 Fig. is shows the instrument in use to test whether a wooden 
 beam is horizontal. As we see in the figure, the plumb-line passes 
 a little to the left of the mark on the cross-bar. Which end of the 
 beam is the highest? 
 
 In testing whether a surface (as that of a table) is horizontal, 
 we must apply the level in at least two different directions, because 
 a line in one direction on the surface might be horizontal while 
 lines in other directions were inclined. But if two lines on the 
 surface in different directions are horizontal, then the surface must 
 be horizontal. 
 
 NOTES. i. A common Spirit-level is shown in Fig. 16. It is more accurate 
 
 than the plumb-line level, but (when properly 
 made) more expensive. The essential part of 
 the spirit-level is a glass tube, slightly curved, 
 and filled with alcohol, except the space oc- 
 cupied by a bubble of air, which always seeks 
 Fig. 16. the highest part of the tube. 
 
 2. Surveyors and engineers, in establishing horizontal lines on the ground, or 
 prolonging them, use levels mounted either on 
 a staff which is driven into the ground, or on 
 a three-legged stand (tripod) which rests on the 
 ground. One may make a simple (but not very 
 accurate) plumb-line level for use on the ground, 
 as follows : Fasten a common carpenter's square 
 {Fig. 77) in a slit in the top of a staff by means 
 of a screw, and then tie a plumb-line at the 
 angle so that it may hang beside one arm. 
 When it has been made to do so by turning 
 the square, the other arm will be horizontal. 
 Fig. 77. 
 
 Exercises. 1. Choose some object in the room, and test whether it is 
 vertical. 
 
 2. Choose some object in the room, and test whether it is horizontal. 
 
 3. Draw on the blackboard (with the aid of a plumb-rule) a vertical line. 
 
 4. Draw on the blackboard (with the aid of the plumb-line level) a hori- 
 zontal line. 
 
26.] 
 
 CHAPTER II. STRAIGHT LINES* 
 
 II. Change of Direction. The Circle. 
 
 26. The direction of a straight line is susceptible of change. 
 The hands of a clock, for instance, may be regarded as two lines 
 which are constantly changing in direction. The minute-hand, in 
 the course of an hour, points in all possible directions contained 
 in the plane of the face of the clock. 
 
 When a straight line, A B (Fig. 18) , moves without leaving the 
 point A, until it has another direction, 
 as A C, it is said to turn or ROTATE 
 about the point A. If the motion 
 continues, the rotating line will at 
 length return to its first position, A B; 
 it is then said to have made one REV- 
 OLUTION. 
 
 Fig. 18. 
 
 27. If, during the motion about A, the length of the line re- 
 mains unchanged, its other end, B, will 
 describe a curved line which returns into 
 itself, and has the property that all its 
 points are at the same distance from A. 
 This curved line (Fig. 19) is called a 
 CIRCUMFERENCE ; the point A is called 
 its CENTRE. Of all curved lines, the 
 circumference is the simplest and the 
 most important. 
 
 Definitions. I. A ciirved line which is everywhere equally 
 distant from a point is called a CIRCUMFERENCE. 
 
 II. The point is called the CENTRE. 
 
 III. A part of a circumference is called an ARC (for example, 
 B C or CD, Fig. 19). 
 
GEOMETRY FOR BEGINNERS. 
 
 [ 28. 
 
 IV. A straight line joining the centre to the circumference is 
 called a RADIUS 1 (for example, AB or A C, Fig. 19). 
 
 V. The plane surface bounded by a circumference is called a 
 CIRCLE. 
 
 NOTE. The word circle is also used in certain expressions in the sense of cir- 
 cumference ; thus, in the expressions,"^ draw a circle" "arcs of circles" strictly 
 speaking, not the circle but the circumference or boundary of the circle is meant. 
 
 From definitions I. and IV. it follows that 
 All radii of the same circle are equal in length. 
 
 Exercises. 1. Name objects on which circles and circumferences present 
 themselves (ends of a cylinder, rim of a tea-cup, a hoop, etc.). 
 
 2. What does the tire of a carriage-wheel represent? the spokes? the 
 axle ? the part of the tire between two spokes ? 
 
 3. In Fig. 18, as A B rotates about A, what kind of paths do points be- 
 tween A and B describe? 
 
 28. Circles (or, speaking more exactly, circumferences and 
 arcs) are drawn on paper, etc., by the aid of an instrument called 
 the. COMPASSES or DIVIDERS 2 (Fig. 20). 
 
 The compasses (for use on paper) consist of two metal legs 
 connected together by a pivot about which 
 they can turn. The legs have fine, hard 
 steel points. The lower part of one leg 
 can be removed (by turning a screw on 
 the side) and its place supplied by a pencil- 
 leg, or a leg provided with a steel pen. 
 In Fig. 20 the pencil-leg is shown. 
 
 To draw a circle with a given radius, say 
 two inches, fit the pencil-leg to the com- 
 Fig. 20. passes, open the legs till the distance apart 
 
 of their points, or opening, is just two inches, and then proceed 
 as shown in Fig. 20. 
 
 1 Plural, radii. 
 
 2 So called from its use in dividing lines and angles into equal parts. 
 
28.] 
 
 CHAPTER II. STRAIGHT LINES. 
 
 25 
 
 On the ground, circles are often described by the aid of a cord. 
 When a gardener, for instance, wishes to trace a circumference, he 
 proceeds as illustrated in Fig. 21. 
 
 
 Fig. 21. 
 
 Exercises. 1. Describe how to draw a circle of given radius on paper. 
 
 2. Describe how a gardener traces a circle on the ground. 
 
 3. Draw (on paper) a circle with a radius equal to the breadth of your 
 ruler or divided rule. 
 
 4. Draw a circle with the large compasses on the blackboard. 
 
 5. Can you draw a circle on the blackboard with a radius greater than the 
 greatest opening of the compasses? 
 
 6 Draw several arcs of circles. 
 
 7. Draw, free-hand, several circles. Which is drawn the best? 
 
 8. Draw three circles all having the same centre, or concentric circles, as 
 they are called. The waves caused by throwing a stone into water are exam- 
 ples of concentric circles ; what is their common centre ? 
 
 9. Draw (a) any circle; () a circle Math a point A as centre, and any 
 radius; (<:) a circle with a radius of one foot, and in any position (that is, hav- 
 ing any point as centre) ; (of) a circle with the point A as centre, and a ra- 
 dius of one foot. 
 
 10. What two things completely determine the position and magnitude of 
 a circle? 
 
26 GEOMETRY FOR BEGINNERS. 
 
 III. Directions of Two Lines. 
 
 29. Two straight lines in the same plane must have either 
 (a) the same or (^) different directions. 
 Case I. When two different lines, as AB and CD (Fig. 
 
 A B 22 }, have the same direction, they are 
 
 called PARALLEL lines. Bearing in 
 
 mind what was said in 21, we may 
 
 also regard one of them as having one 
 
 G ~~ H direction, and the other the opposite 
 
 Fig. 22. 
 
 direction. It is also evident that sev- 
 eral lines, as A B, CD, F F, G H, may be parallel to one another. 
 
 Definition. PARALLEL LINES are lines which have the same 
 direction. 
 
 That a line AB is parallel to a line CD is sometimes ex- 
 pressed, for the sake of brevity, by a sign ; thus : A B || CD. 
 
 The two rails of a railroad furnish a familiar illustration of par- 
 allel lines ; substitute for the rails geometrical lines, and these would 
 be parallel lines as understood in Geometry. The opposite edges 
 of a table are another example of parallel lines. 
 
 Fig. 23. 
 
 Every one has observed^that the rails of a railroad course along 
 without either approaching or separating. We may suppose them 
 
29-] 
 
 CHAPTER II. STRAIGHT LINES. 
 
 27 
 
 Fig. 24 gives 
 
 to be prolonged as far as we please, without meeting. The same 
 is true of the edges of the table and of all lines parallel to each 
 other. In fact, if two parallel lines could meet when prolonged, 
 they would then be two straight lines passing through the same 
 point (the point of meeting), yet not coinciding; but this is im- 
 possible, for all straight lines drawn through the same point in the 
 same direction must coincide and form only one line (see 22, 1.) . 
 Therefore, 
 
 Parallel lines can never meet, however far prolonged. 
 
 But two lines which would not meet if prolonged are not neces- 
 sarily parallel. For instance : an edge of the table and one of the 
 legs on the opposite side would never meet if prolonged, yet they 
 are not parallel. The lines, in order to be parallel, must lie in the 
 same plane. 
 
 Two curves, also, may be parallel to each other, 
 an example. The curves would not meet if pro- 
 longed, and, at corresponding points, as A and B, 
 or C and D, they have the same direction. When a 
 railroad makes a curve, the rails do not cease to 
 be parallel. Concentric circles supply another ex- 
 ample of parallel curves. (See Fig. 190, p. 212.) 
 
 Two planes, or a plane and a straight line, are 
 parallel, if they would not meet however far they 
 were extended. The floor and ceiling of the 
 room is an instance of parallel planes ; the floor 
 and an edge of the ceiling is an example of a plane and straight 
 line parallel to each other. 
 
 Case II. Far more common is the case where the two lines 
 have different directions. Two 
 such lines, as A B and C D Q-'~-~^. 
 (Fig. 25} , either actually meet, 
 or, if prolonged in one or the 
 other direction, will approach 
 
 2 4> 
 
28 GEOMETRY FOR BEGINNERS. [ 3- 
 
 and at last meet in a point, (9, called their intersection ( 12). 
 Two edges of a table which meet at a corner are an example 
 of lines having different directions ; the corner is their intersec- 
 tion. 
 
 Exercises. 1. Which lines on the blackboard are parallel? which not 
 parallel? Prolong the latter until they meet. 
 
 2. Can two lines, not parallel, intersect in more than one point? 
 
 3. How do you read the expression AB || CD? 
 
 4. Hold two pencils (#) parallel; (b~) so that they would intersect; (V) 
 so that they are neither parallel nor would they intersect. 
 
 5. Hold two books (a) parallel ; (U} not parallel. 
 
 6. Hold a pencil and a book (#) parallel ; (b*} not parallel. 
 
 7. Point out on the cube (a) parallel edges ; () intersecting edges : (^ 
 edges neither parallel nor intersecting. 
 
 8. Examine, in the same way, the edges of the prism and the pyramid. 
 
 9. Which edges of the room are parallel to the floor? 
 
 10. Give another instance of parallel lines; of parallel planes ; of a line 
 parallel to a plane. 
 
 11. Are two vertical lines parallel to each other? 
 
 Ans. Strictly speaking, No; because, owing to the round shape of the earth, 
 two vertical lines intersect at a point near the centre of the earth. But the dis- 
 tance to the earth's centre (nearly four thousand miles) is so great compared with 
 the distance apart of two vertical lines, such as the edges of a house, etc., that the 
 latter are always regarded as parallel. 
 
 12. Are horizontal lines always parallel to each other? 
 
 13. Can two vertical planes intersect each other? 
 
 14. Can two horizontal planes intersect each other? 
 
 15. Give an instance of parallel lines which are vertical ; horizontal ; 
 inclined. 
 
 30. The simplest way to draw lines parallel to each other 
 (especially when several are desired) is illustrated in Fig. 26. 
 Besides a ruler, a piece of wood with three smooth, straight edges 
 called a SQUARE, is needed. Keep the ruler at rest, slide the 
 square along the edge of the ruler, and, in different positions of 
 the square, draw lines along its edge ; these lines will he parallel 
 to each other. To draw a line through the point J/, parallel to the 
 
3'-] 
 
 CHAFl'ER II. STRAIGHT LINES. 
 
 29 
 
 line P Q, place the ruler and square as shown in the figure, then 
 
 slide the square along till its 
 
 edge just touches M \ the 
 
 line drawn through M t along 
 
 the edge of the square, will 
 
 be the parallel required. 
 
 NOTE. How parallel lines are 
 drawn with ruler and compasses will 
 be explained later. (See $ 78). 
 
 Exercises. 1. Drawastraight 
 line ; and then, through a point not 
 on the line, a parallel to it. Pig- 26. 
 
 2. Draw a straight line, and, through a point on the line, a parallel. 
 
 3. Draw five parallel lines about equally distant from each other. 
 
 4. Draw free-hand several parallel lines. Test and correct them with 
 ruler and square. 
 
 IV. Directions of Three Lines. 
 
 31. Three straight lines in the same plane present, as regards 
 direction, four cases, 
 
 (a) The three lines are all parallel ; 
 
 (b) Two of the lines are parallel ; 
 
 (c) No two lines are parallel, and all intersect in one point ; 
 (//) No two lines are parallel, and they do not all intersect in a 
 
 point. 
 
 These cases are all illustrated in Fig. 27. When necessary, the 
 
 \ 
 
 \ 
 
 (a) 
 
 \ 
 
 Fig. 27. 
 
 lines must be prolonged until they intersect. In case (//) the 
 lines (prolonged, if necessary) intersect in three points, and com- 
 
30 GEOMETRY FOR BEGINNERS. [ 3 2 - 
 
 pletely enclose a portion of a plane surface, which is called a 
 TRIANGLE. 
 
 NOTE. The properties of the triangle will be studied in Chapter IV. 
 
 Exercises. 1. Draw any three lines, and then prolong them (by dotted 
 lines) till they intersect. 
 
 2. In how many points can four lines intersect one another? (a) when all 
 four lines are parallel ? () when three are parallel ? (r) when two are par- 
 allel ? (</) when no two are parallel ? 
 
 3, Examine five lines in the same way. 
 
 V. Length of a Line. 
 
 32. The straight line which passes through two points, A and 
 B (fig. 28) , is shorter than any other line 
 passing through the points, and hence its 
 length is taken as the DISTANCE of the 
 points from each other. 
 
 We all learn as soon as we begin to 
 - 28 - walk and to move about from place to 
 
 place, that a straight line is the shortest distance between two 
 points. 
 
 Illustration. We may illustrate this truth as follows : Fasten 
 one end of a string to a nail which is fixed in a wall, and pass the 
 other end through a ring also attached to the wall. Now pull the 
 
 Fig. 2 9 . 
 
 free end with one hand, and the part of the string between the 
 wall and the nail will diminish in length until it is straight ; and 
 then it cannot be made any shorter. 
 
33-] CHAPTER II. STRAIGHT LINES. 31 
 
 Application. Sign-painters, carpenters, etc., make use of 
 this property when they wish to trace straight lines on wood. They 
 chalk a cord and stretch it tightly between the points through which 
 the line is to pass. Then seizing the cord by the middle, they 
 
 Fig. so. 
 
 draw it away a little from the wood and then let it go. It springs 
 back, strikes the wood a sharp blow, and leaves on it a white trace, 
 which is a straight line. 
 
 33. As regards length, two lines must be either equal or 
 unequal. 
 
 Definition. Two straight lines are EQUAL, if, when laid one 
 upon the other, their ends or extremities can be made to coincide. 
 If this cannot be done the lines are UNEQUAL. 
 
 Expression of Equality. If two lines, AB and CD, are 
 
 equal, we express the fact thus : A B = CD. A B 
 
 This expression is called an EQUATION, and is c D 
 
 read : the line AB\s> equal to the line CD. E F 
 
 The sign = is called the SIGN OF EQUALITY. Fig. 31. 
 
 Expression of Inequality. The lines AB and E F {Fig. 
 ji) are unequal lines. AB is greater than E F, or EF is less 
 than- A B. In place of the words in italics, we often employ the 
 signs > and <, and write : AB>EF, and EF<AB. 
 
 Exercise. Read the following: a = l>; c>a; x<y; AJ3<PQ; 
 
32 GEOMETRY FOR BEGINNERS. [ 34. 
 
 34. In order to test the equality or inequality of two lines, it 
 is not necessary to place one line upon the other, as the definition 
 of the last section might seem to require. We find it usually more 
 convenient to compare both lines with a third line, which can be 
 readily placed upon or by the side of each of them. 
 
 In constructing geometrical figures and diagrams, we make this 
 comparison (especially when great accuracy is required) by means 
 of the dividers (compasses). Suppose, for example, that we wish 
 to test whether the lines A B and CD (Fig. ji) are equal. Open 
 the dividers, and place one point on A, the other on B ; this is 
 called taking the distance A B between the points of the dividers. 
 Then, taking care not to alter the opening of the dividers, place 
 one point on C and observe whether the other point will fall on 
 D; if it does fall on D, we know that CD = A B. 
 
 NOTE. In general, lines are compared in length by measuring them, and ex- 
 pressing their lengths in feet, miles, metres, etc., as will be explained in Parts VII. 
 and VIII. of this chapter. 
 
 Exercises. 1. Draw two lines, and then test whether they are equal. 
 
 2. Draw a horizontal line ; then draw a vertical line equal to it. 
 
 3. Explain how to make a line equal to a given line. 
 
 4. Draw a line equal to one edge of this cube. 
 
 5. Draw lines equal to the different edges of this prism. 
 
 6. Draw four lines, a, b, c, d, as follows : a = b, c > a, d< b. 
 
 7. Draw any four lines, m, n, o, p ; then test their lengths, and write the 
 results with the proper signs (=, or >, or <). 
 
 8. Try to draw, free-hand, two equal lines, one horizontal, the other ver- 
 tical ; then test their equality. 
 
 9. Draw two lines, each longer than the greatest opening of the dividers. 
 How can you test whether they are equal or not? 
 
 35. In testing, with the dividers, the lengths of two lines, the 
 third line with which we compare each of them is the distance be- 
 tween the points of the dividers. We assume that if the two lines 
 are each equal to this distance, they must be equal to each other. 
 This assumption is so very obvious that it almost escapes notice. 
 
36.] CHAPTER II. STRAIGHT LINES. 33 
 
 If two lines or any two magnitudes are each equal to a third, 
 of course they are equal to each other. In fact, tjis is a truth so 
 simple in its nature that it cannot be made clearer by trying to 
 prove it. Truths of this kind that is to say, setf-evitftnttniths 
 are called AXIOMS. 
 
 Besides the axiom here under consideration, there are four 
 others often useful in Geometry. We shall now state them all, and 
 shall refer to them hereafter as here numbered. 
 
 Axioms. I. If two magnitudes are each equal to a third, 
 they are equal to each other. 
 
 II. If equals are added to equals, the sums are equal. 
 
 III. If equals are subtracted from equals, the remainders are 
 equal. 
 
 IV. If equals are multiplied by equals, the products are equal. 
 
 V. If equals are divided by equals, the quotients are equal. 
 
 36. The length of a line (like its direction) is susceptible of 
 change. We may suppose it either to increase or to diminish. 
 
 Prolong the line AB (Fig. 32} to any point C ; then the line 
 
 A C is equal to the sum of the lines < 
 
 A B and B C ; here we may employ A *~~ ~~ 
 
 to advantage the sign + of addition, 
 
 and write, 
 
 A C= AB + B C. 
 
 Conversely, the line A B is the difference between the lines A C 
 and B C ; or we may use the sign of subtraction, and write, 
 
 AB = AC-BC. 
 
 Prolong a line A B (Fig. 33} by adding to it repeatedly (by 
 means of the dividers) lines B C, CD, D E, etc., each equal to 
 A B. Then A C is twice as long as A B, A D is three times as 
 long, A E four times as long, etc. That is, we have multiplied the 
 
34 GEOMETRY FOR BEGINNERS. [ 36. 
 
 line A B by 2, 3, 4, etc. It is usual to omit the sign x of multi- 
 plication, and write the results thus : AC=2AB,AD=$AB, 
 A E = 4 AB, etc. Also 
 
 
 Conversely, a line can always be divided into equal parts. In 
 the above line, for example, A B is one-half of A C, one third of 
 A D, one-fourth of A E, etc. ; in other words, 
 
 234 3 
 
 These results may be summed up in the following general truth : 
 Lines, like numbers, may be added, subtracted, multiplied^ and di- 
 vided. 
 
 Remark. Observe, however, that we have only multiplied and 
 divided a line by a number, and that the result has always been 
 another line. If, for instance, we multiply the line AB by the 
 number 10, we simply repeat the line a certain number of times, 
 and the result is the line A L ; or, writing the operation in the 
 form of an equation, 
 
 10 x AB = AL. 
 
 In the converse operation of division there are two cases : - 
 
 (a) We may divide the line ALty the number 10, obtaining 
 as the quotient the line A B. In this case we divide the line into 
 a certain number of equal parts. 
 
 (b) Or, we may divide the line A L by the .line AB, and now 
 our quotient will be the number 10. In this case we find how 
 many times one line is contained in the other line. In this sense 
 division is the comparison, as regards magnitude, of two things of 
 the same kind, and the quotient is their relative magnitude or ratio. 
 
37-] CHAPTER II. STRAIGHT LINES. 35 
 
 Exercises. 1. In Fig. 33, what line is equal to BD + DE? toAE 
 A D ? to three times the sum of A B and B C ? to four times the difference 
 of A Cand CD? 
 
 NOTE. Perform the following exercises with ruler and dividers, and not with 
 the aid of the divisions of a divided rule. 
 
 2. Draw two lines ; then draw lines equal (a) to their sum, (3) to their 
 difference. 
 
 3. Drw A = CD, and MNPQ; then draw lines equal to A B + 
 MN, CD + PQ,AB MN, CD PQ. Why are the first "two of these 
 lines equal? Why are the last two also equal? 
 
 4. Draw a line three times as long as your pencil. 
 
 5. Draw a line, A B, and prolong it to a point, P, so that AP= $A B. 
 
 6. Explain how to repeat or multiply a line any number of times. 
 
 7. Draw a line equal in length to the sum of the edges which bound one 
 face of this cube. 
 
 8. Draw a line ; then divide it into two, four, and eight equal parts. 
 NOTE. This is to be done here by repeated trial with the dividers. 
 
 9. Divide a line, free-hand, into five equal parts. Test and correct with 
 the dividers. Repeat this exercise several times. 
 
 VI. Ratio of Two Lines. 
 
 37. Definitions. I. If a line is contained in another line 
 one or more times without a remainder, the first line is called a 
 MEASURE, or ALIQUOT PART, of the second ; and the second line is 
 called a MULTIPLE of the first. 
 
 II. A COMMON MEASURE of two or more lines is a line which is 
 exactly contained in each of them. 
 
 III. The RATIO of two numbers is the quotient obtained by di- 
 viding one number by the other ; in other words, it expresses how 
 often one of the numbers will contain the other. 
 
 IV. The RATIO of frvo lines is the number of times one of the 
 lines will contain the other ; it is equal to the ratio of the numbers 
 which express how often a common measure is contained in each 
 of the lines. 
 
 V. Of the two terms of a ratio, the first is called the ANTECEDENT, 
 the second is called the CONSEQUENT. 
 
36 GEOMETRY FOR BEGINNERS. [ 38. 
 
 For example : in Fig. 34, where AB = B C = CD D E E F, 
 the line AB is a measure of AB, of A C, of CF, in short, of 
 every line represented in the figure ; therefore A B is a common 
 measure of all these lines. On the other hand, all the lines in the 
 figure are multiples of A B. 
 
 Fig- 34- 
 
 If we compare the two lines A B and A F, we see that their 
 common measure A B is contained once in A B and five times in 
 A F; hence, the lines A B and A F are to each other as the num- 
 bers i and 5 ; that is, they have the ratio 1:5. By expressing this 
 ratio as a fraction, we obtain the equation, 
 
 AB _ i X A B _ i 
 ~AF ~ 5 x AB ~ 5 ' 
 
 The reciprocal ratio of AF to AB is that of 5 to i, or simply 5. 
 
 NOTE. In common language, instead of using the term ratio, we say, A B is 
 one-fifth as long as AF; or, reciprocally, A F"\s five times as long as A B. 
 
 In like manner, it is easy to see from Fig. 34 that 
 The lines A B and A C have the ratio 1:2; 
 
 2:5; 
 
 AE " " 3:4; etc. 
 In the form of fractions, these ratios would be written, 
 
 ^J?_i- ^_*_ BD ! - 2 ^_1 etc 
 A C~ 2' Z^~7~ ' AF~ $ ' AE~ 4' 
 
 Exercise. What is the ratio of A D and A B (Fig. 34) ? AD and A C? 
 A D and AD? A D and A E ? AD and A F? A Cand A E? A E and . / F? 
 Write the values both with dots (: ) and as fractions. In these ratios, which 
 term is the antecedent? which the consequent? 
 
 38. In order to find the ratio of two lines, apply the shorter 
 line to the longer line as often as possible ; if the shorter line is a 
 
' 3 8.] CHAPTER II. STRAIGHT LINES. 37 
 
 measure of the longer, if, for example, it is contained in the longer 
 exactly five times, then i : 5 is the ratio of the shorter line to the 
 longer. 
 
 If, however, the shorter line is not a measure of the longer, the 
 problem becomes more difficult. After we have applied the shorter 
 line to the other line as many times as possible, there is a remainder 
 less than the shorter line. For example : in the case of the lines 
 MNand PQ (Fig. 35), when MNhas been applied twice from 
 P towards Q, there remains the portion R Q of the longer line. 
 To meet this difficulty, apply R Q to M ' N as often as possible ; 
 then apply the new remainder SjVto R Q, which contains it ex- 
 actly twice. 
 
 This series of operations is precisely like the process of finding 
 the greatest common divisor of two numbers ; it is, in fact, the 
 method of finding the greatest common measure of two lines. 
 
 P R 
 
 Fig- 35- 
 
 From what precedes we see that 
 JR Q = 2 SN; 
 
 MN= 3 R Q + SN= 6 SN + SN= 7 
 P Q = 2MN+ RQ = I46W + 2 SN 
 
 hence P Q - l6 SN - l6 and MN ^ SN 
 
 ---> [= 
 
 S N is the greatest common measure of the two lines ; the 
 longer contains it sixteen times, the other seven times ; the ratio 
 of the lines, therefore, is that of the numbers 16 and 7, either 
 16 : 7 or 7 : 16. 
 
38 GEOMETRY FOR BEGINNERS. [ 38. 
 
 It may happen that we cannot find a common measure of two 
 lines by this process, for the reason that the lines have no common 
 measure ; such lines are called INCOMMENSURABLE LINES. 
 
 Remark. The exact ratio of two incommensurable lines can- 
 not be expressed by numbers ; but by dividing one of the lines 
 into a great number of equal parts, and applying one of these parts 
 to the other line as often as possible, and neglecting the remainder 
 (which will be very small compared with either of the lines) we 
 are able to approximate very nearly to the true value of the ratio. 
 
 If, for instance, we divide one of two lines into 1000 equal 
 parts, and find that the other line contains more than 600, but less 
 than 60 1, of these parts, we know that the true value of the ratio 
 lies between $$#, and $$j\j. For nearly all purposes we should 
 assume that the ratio was r 6 o(yo> or I \ m other words, we should 
 assume that one line was three-fifths as long as the other. 
 
 Exercises. 1. In Fig. 36, of what lines is MN a common measure? 
 How many times is it contained in each of them? Of what lines is PQ 
 (=MN) a common measure? How many times is it contained in each? 
 
 2. Express the values of the other lines in terms of P Q. 
 
 D 
 
 Fig. 36. 
 
 3. How many times is i P Q contained in each line in Fig. 36 ? Is it a 
 common measure of all of them? If we divide P Q into three, four, five, etc., 
 equal parts, will one of the parts be a common measure of all the lines? 
 
 4. Of what lines is 2 PQ a common measure? 2 MN? 
 
 5. What is the greatest common measure of MN and A B? of A B and 
 CD? of MN, ABvn&CD? of MN, A B, CD and 
 
39-] CHAPTER II. STRAIGHT LINES. 39 
 
 6. Find the ratio of P Q to each of the other lines in Fig. 36, and write 
 the results both fractionally and with dots. 
 
 7. Find the ratio of M ' N to all the other lines, and write the results 
 both fractionally and with dots. 
 
 8. Find the ratio of A B to the other lines, and write out the results. 
 
 9. What is the ratio of CD to E F? What is the reciprocal ratio? 
 
 10. Draw any two lines, and then find their ratio. 
 
 VII. Units of Length. 
 
 39. In comparing the lengths of lines with one another a great 
 amount of time and trouble is saved by employing units of length. 
 
 Definitions. LA UNIT of length is a known length which 
 men agree to employ as a common term of comparison, or common 
 measure for all lines. 
 
 Units of length receive individual names, as the foot, the yard, 
 the meter, the mile, etc. 
 
 11. To MEASURE a line is to find how often it will contain a 
 unit of length. 
 
 In other words, to measure a line is to find its ratio to the unit 
 of length. 
 
 III. The number of times a line contains a unit of length 
 is the NUMERICAL VALUE of the line with reference to that unit; 
 joined to the name of the unit, the whole expression is called the 
 LENGTH of the line. 
 
 Examples. 6 feet, 9 meters, 12 miles. 
 
 NOTE. In olden times, units of length were usually taken from some part of 
 the human body. The word foot indicates its origin. The cubit was the length of the 
 forearm, from the elbow to the end of the middle finger. The first attempt to es- 
 tablish a standard yard in England was in 1120, in the reign of King Henry I., who 
 ordered that the yard, the ancient ell (Latin ulna, elbow, arm), should be of the 
 exact length of his own arm. At the present day, horses are described as so many 
 " hands high," and a place is sometimes said to be so many "paces distant." The 
 span, the palm, and \htfathom all derive their origin from the human body; the 
 fathom was the distance between the extreme points of the outstretched arms. But 
 the units of length now recognized by law among civilized nations are defined 
 by the State with the utmost precision, by reference to a standard made of plati- 
 num or other hard metal, which suffers no sensible change from age to age. This 
 standard is kept at the capital of the nation, and carefully made copies of it ar? 
 preserved in public buildings in the chief cities where the unit is employed. 
 
40 GEOMETRY FOR BEGINNERS. [ 40 
 
 40. The units of length established by law, and used for mosl 
 purposes in the United States and in Great Britain, are the INCH, 
 the FOOT, the YARD, the ROD, the MILE. 
 
 12 inches (in.) = i foot (ft.). 5^ yards = i rod. 
 
 3 feet = i yard (yd.). 320 rods = i mile. 
 
 NOTE. The geographical or sea mile is one-sixtieth of a degree of a meridian 
 of the earth and is equal to 1.15 common or statute miles, very nearly. 
 
 Exercise. How many inches in one yard ? in one rod ? in one mile ? 
 
 41. The units of length now employed by the people of most 
 civilized nations, and by the men of science of all nations, are 
 called METRIC units, from the name of the fundamental unit, the 
 meter. 1 They form, taken together, a part of the Metric System 
 of Weights and Measures, first adopted by the French just after 
 the great revolution of 1792. 
 
 Definition. The METER the fundamental unit from which 
 all other units in the metric system are derived is defined by law 
 as the length of a certain platinum bar kept at Paris. 
 
 NOTES. i. The nations which have adopted the metric system take from 
 the Paris meter the most perfect copies which can be made, and in the capital of 
 each nation are preserved a set of metric measures and weights with which those 
 used by the people must agree. 
 
 2. The founders of the metric system defined the meter as one ten-millionth 
 part of the distance from the Equator to the North Pole. Extraordinary efforts were 
 made to construct a platinum bar which should have this exact value. An arc of 
 a meridian passing through France and Spain was measured with extreme care, 
 and by combining the results with other known facts the distance from the Equator 
 to the Pole was calculated. The platinum bar based on these results was supposed 
 at the time to be exactly what it was denned to be. But not long afterwards errors 
 in the measurements were discovered, and it is now known that the Paris meter 
 (the platinum bar) is less than its intended value by an amount not exceeding ^ 
 of the whole. 
 
 The other units of length are multiples and divisions of the 
 meter as follows : 
 
 Ten meters are called one DEKAMETER. 
 
 One hundred meters are called one HEKTOMETER. 
 
 1 The word meter comes from a Greek word meaning measure, and is pro- 
 nounced me6-ter. The French write the word mttre. 
 
41.] CHAPTER II. STRAIGHT LINES. 41 
 
 One thousand meters are called one KILOMETER. 
 One tenth of a meter is called one DECIMETER. 
 One hundredth of a meter is called a CENTIMETER. 
 One thousandth of a meter is called a MILLIMETER. 
 
 NOTES. i. Ten thousand meters is also called a MYRIAMETER.. 
 
 2. The names of the multiples de&ameter, hectometer, kilometer, myn'ameter are 
 formed by prefixing to the word meter the Greek numerals deka (io) P hecto (100), 
 kilo (1000), and myria (10000) ; for the divisions of the metre, the Latin numep 
 als, deci, centi, and milli are used. 
 
 Abbreviations. Meter = m ; dekameter = dkm ; hecto- 
 meter = hm ; kilometer = km ; decimeter = dm ; centimeter =* 
 cm ; millimeter mm. 
 
 TABULAR VIEW. 
 
 km. hm. dkm. m. din. cm. mm. 
 
 I = 10 = IOO = IOOO = IOOOO =IOOOOO = IOOOOOO 
 
 I = 10 = IOO = IOOO = IOOOO = IOOOOO 
 
 I = IO = IOO = IOOO = IOOOO 
 
 I = 10 = IOO = IOOO 
 
 I ' = IO = IOO 
 
 I IO 
 
 This table shows that a given number of units of one value are 
 reduced to units of the next lower value by multiplying by 10, and 
 to units of the next higher value by dividing by 10. 
 
 Examples. 1. Reduce 32 meters to decimeters, and also to dekameters. 
 
 Ans. 32O dm ; 3.2 dkm - 
 
 2. Reduce 0.764 of a kilometer to meters. 
 
 Solution. Since we wish to reduce kilometers to the third lower unit, we must 
 multiply by 10 three times. As in decimal fractions, this is done by moving the 
 decimal point three places to the right. Ans. 764- 
 
 3. Reduce 376 millimeters to kilometers. 
 
 Solution. Here we have to reduce millimeters to the sixth higher unit; there- 
 fore, we must divide by 10 six times. That is, move the decimal point six places 
 to the left. Ans. 0.000376^", 
 
 Remarks. i. The units most used are, the kilometer (for 
 long distances), the meter, and the centimeter and millimeter (for 
 lengths less than one meter) . 
 
42 GEOMETRY FOR BEGINNERS. [41. 
 
 2. In practice, only one unit is employed to express a length. 
 We say 76" or 76o mm (not y dm 6 cm ) ; 332 (not 3 hm 3 dkm 2 m , 
 and not 33 dkm 2 m ), etc. 
 
 3. In reading infraction of a unit, the word signifying the num- 
 ber of decimal places may be omitted ; thus, 4.5 may be read : 
 four meters five; 4.5 2 m may be read : four meters fifty-two, etc. 
 This is analogous to the common way of reading money; $2.75, 
 for instance, is read : two dollars seventy-five (or, yet more sim- 
 ply : two seventy-five) . In the case of money, the omitted word 
 is cents ; with metric units it is tenths, hundredths, thousandths, 
 etc., according to the number of decimal places. 
 
 4. In adding, subtracting, etc., metric units, they must first all 
 be reduced to a common unit, as, for example, the meter. Thus, 
 
 6.4 m -f- 27 cm = 6.4 m f 0.27 = 6.67 m - 
 
 COMPARISON OF METRIC AND ENGLISH UNITS. 
 
 kilometer = 0.62137 miles (about |- of a mile), 
 
 meter = 1.0936 yards (about 39-3- inches), 
 
 centimeter = 0.3937 inch (about f- of an inch), 
 
 millimeter = 0.0394 inch (about ^V of an inch). 
 
 mile =1609 meters (about if- kilometers), 
 
 yard = 0.9144 meter (about -JT meter), 
 
 foot = 0.3048 meter (about 30 centimeters). 
 
 inch = 2.54 centimeters (about 2^ centimeters). 
 
 NOTE. It might be well for the pupil to learn by heart the approximate values 
 (not the others) just given in parentheses; but the best way to become familiar 
 with the values of the meter and its divisions is by comparing a meterstick with 
 a yardstick, by comparing the divisions on a divided rule, and by solving exercises 
 like the following. 
 
 Exercises. 1. Give the metric units of lengths thus: 10 millimeters 
 make I centimeter, 10 centimeters make I decimeter, etc. 
 
 2. Write a tabular view of the metric units of length. 
 
 3. Write with abreviations : 8 meters; 72 kilometers; 324.6 dekameters; 
 $7.34 hectometers; 9.27 centimeters; 4.2 decimeters; 13 millimeters. 
 
4I-] CHAPTER II. STRAIGHT LINES. 43 
 
 4. Read the following : 37.5 ; 947 ; 846.7 cm ; l8 dm ; 9.O4 km ; 
 45.2o6 hm ; 77-96 dkm . 
 
 5. Reduce SS44 mm to meters and to kitometers. 
 
 6. Reduce 3-27 km to meters and to millimeters. 
 
 7. Reduce I2.4 hm to all the lower units. 
 
 8. Reduce i8o cm to all the higher units. 
 
 9. What part of a kilometer is i m ? i cm ? i mm ? 
 
 10. 25 cm are what part of a meter? also what part of a kilometer? 
 
 11. 24.32 m + 6 dm 4- i8 dm -f- 26o cm = how many meters? 
 
 12. 4 dkm 4- 4 m 4- 4 dm 4" 4 cm 4- 4 mm = how many meters? 
 
 13. 6 km 4- 6 hm 4- 6 dkm 4- 6 m 4- 6 dm = how many kilometers? 
 
 14. A shopkeeper sells ribbon as follows : to the first customer 8 m , to 
 the second 7O Cm , to the third 4.5, to the fourth 2 dm , to the fifth 8o cm ; how 
 many meters does he sell in all? 
 
 15. What is- the difference in length of two nails, one of which is 4 cm , the 
 other 28 mm long? 
 
 16. From io m take io mm . 
 
 17. From ioo km take ioo mm . 
 
 18. 87 cm X 36 = how many meters? 
 
 19. 42.64 m X 80 = how many kilometers? 
 
 20. If the post-office is 900 from my house, and I go to it twice a day 
 for letters, how many kilometers do I walk in a week? 
 
 21. I receive an order for 20 rulers, each to be 3O cm long; what length 
 of wood is required? 
 
 22. How many rails each 7.5 m long are required to build a track 3oo km 
 long? 
 
 23. How long would it take an express train going at the average rate of 
 4O km an hour to travel from Boston to San Francisco, a distance of 55OO km ? 
 
 24. If i m of silk costs $3.00, what costs i dm ? i cm ? i mm ? i dkm ? i hm ? i km ? 
 
 25. Reduce the following values to English equivalents : velocity of an ex- 
 press train = 6o km an hour, height of Mt. Blanc = 481 5, usual height of 
 a barometer = j6 cm . 
 
 26. Reduce the following values to metric equivalents : velocity of light 
 = 190,000 miles a second; distance from Boston to New York = 236 miles; 
 average height of a man = 5 feet 8 inches. 
 
 27. Which is greater, 3 km or 2 miles? 3 dm or I foot? 
 
 28. What unit would you use to express the distance from New York to 
 Chicago? the height of a mountain? the length of a room? the dimensions 
 of a sheet of paper? the thickness of a soap-bubble? 
 
44 GEOMETRY FOR BEGINNERS. [42. 
 
 29. Why is the Metric system also called a decimal system? 
 
 30. Which do you consider preferable, metric units or the foot, yard, etc.? 
 
 VIII. Measuring a Line. 
 
 42. In order to measure a line various instruments are em- 
 ployed. Among them we may mention the Divided Rule, the 
 Yardstick (or Meterstick), the Tape-Rule or Roulette, and the 
 Surveyor's Chain. 
 
 Lines on paper are generally measured by placing beside them 
 a divided rule. A very useful kind of divided rule is a rule one 
 foot long, divided on one side into inches and parts of an inch, 
 on the other into decimeters, centimeters, and millimeters. On 
 the blackboard a yardstick or meterstick is usually employed. 
 
 Longer lines, such as the length of a room or of a garden, can 
 be more rapidly measured by the help of a tape-rule or roulttte. 
 These are made of cloth or flexible steel, wound around an axis, 
 and enclosed in a small case suitable for the pocket. They are 
 made of various lengths up to one hundred feet. 
 
 In measuring lines on the ground surveyors employ a chain 
 sixty-six feet long, containing one hundred links, or a steel tape one 
 hundred feet long. French surveyors use a chain one dekameter 
 long, containing fifty links. If a chain or steel tape is not at hand, 
 a rope divided by knots into yards or meters may be employed. 
 
 In the absence of any better means of measuring a line, pacing 
 must be resorted to. The average value of a pace must be found 
 by counting how many paces are made in walking a certain known 
 distance, as ten meters. Pacing is not a very exact way of meas- 
 uring a line, but it is a rapid way, and is always at our disposal. 
 
 In many cases a line is not measured directly, but its length is 
 computed from lines of known length by methods which will be 
 described hereafter. Indeed, it often happens that the direct 
 measurement of a line is impossible ; for example, the breadth of 
 a river, the height of a mountain, the diameter of the earth, and, in 
 
43 -J CHAPTER II. STRAIGHT LINES. 45 
 
 general, the distance of an inaccessible object (as the sun or the 
 moon) . 
 
 Exercises. 1. Make a line 3 dm long, and divide it into centimeters. 
 
 2. Draw lines with lengths as follows : i6 cm ; o.6 dm ; I75 mm . 
 
 3. Draw lines with lengths as follows : j6 cm ; I4 dm ; i.36 m . 
 
 4. Draw, free-hand, a line 3 dm long. Measure it with the divided rule, 
 and find what error you have made. Repeat the exercise three times. 
 
 5. Measure, in metric units, with the divided rule, the following lines, and 
 write out the results: () Length, breadth, and thickness of a book; (<) 
 length and breadth of your desk; (<r) length of your lead-pencil ; (</) length 
 of your middle finger. 
 
 6. Estimate first, by the eye, then measure with the meterstick, the height 
 of the door. 
 
 7. Measure, with a roulette, the length and breadth of the room; also, 
 the length and breadth of the play-ground. 
 
 8. Find the value of your pace by pacing twenty meters. 
 
 9. Measure, by pacing, the distance from the schoolhouse to your home. 
 
 43. If we look on a map of a large country like the United 
 States, we see that distances which amount to hundreds of miles 
 are reduced in length and represented by lines at most a few inches 
 long. It is clear that, in constructing maps, plans 1 of estates, etc., 
 the actual lengths of lines must be very much reduced ; for, other- 
 wise, the map or plan would need to be as large" as the country or 
 the estate which it represents. 
 
 A divided line, or other means for enabling us to reduce lengths 
 measured on the ground in a given constant ratio, is called a RE- 
 DUCING SCALE ; and the operation of drawing on paper lines whose 
 length shall be one-half, one-fourth, one-tenth, or any other frac- 
 tional part of the lines measured on the ground, is called DRAWING 
 TO SCALE. 
 
 1 Every one knows what a map is ; a plan is a portion of a horizontal surface, 
 or surface assumed as horizontal, represented on paper together with its subdivis- 
 ions and other details of importance which it contains. A diagram is any set of 
 lines drawn on paper, etc., to illustrate a statement or aid in a demonstration ; the 
 so-called Figures of Geometry are diagrams. 
 
46 GEOMETRY FOR BEGINNERS. [43- 
 
 Suppose, for example, that we take as a reducing scale the di- 
 visions and subdivisions of a divided rule, and that lines are to be 
 drawn to the scale of i meter to 500 meters ; in other words, that 
 the ratio or scale of reduction is i : 500. Then i meter would be 
 represented on the paper by a line -^V" 2 mm l n g> a distance 
 of 48 meters by a line -^ m = 9.6 cm long ; a length of 66.5 meters 
 by a line |^(f m = 13.3'" long, etc. And, in general, the reduced 
 length would be found by dividing the real length by 500. 
 
 Conversely, the real length of a line would be found from the 
 reduced length by multiplying the reduced length by 500. Thus, 
 & line 8 centimeters long on the paper, would represent a line which 
 was in reality 8 cm X 500 = 4ooo cm 40 long. 
 
 Scales are named by expressing the ratio of reduction, either 
 fractionally, thus, T ^<j, -5-^, TOIJFU-J e tc., or with two dots, thus, 
 i : 100, i : 500, i : 10000, etc. 
 
 A reducing scale is sometimes printed or drawn by the side of the 
 map or diagram, and its divisions numbered in terms of the real 
 lengths of the lines represented on the paper. We can then easily 
 ascertain with the dividers the real length represented by a line on 
 the paper. How? 
 
 The scale of reduction nmst be the same for all lines on the 
 same map, plan, or diagram. If this rule were not followed, the 
 appearance of two lines on the paper would give a false idea of 
 their actual relative lengths ; and, besides, the shape of the map or 
 diagram would differ from the shape of what it professed to rep- 
 resent ; in other words, the map or diagram would be distorted. 
 
 NOTES. i. The choice of a scale depends on the real lengths of the lines 
 compared with the size of the paper on which they are to be represented. The 
 scale chosen should be such as will enable us to represent all the lines on the paper, 
 and also leave a fair margin around the edge of the paper. 
 
 2. The following are examples of scales actually employed: United-States 
 Coast Survey, charts of small harbors, etc., i : 5000 down to i : 60000 ; charts of 
 bays, sounds, etc., 1:80000 down to 1:200000; general charts, usually, 1:400000; 
 Ordnance Survey of Great Britain, partly i inch to I mile (i : 63360), and partly 6 
 inches to i mile (i : 10560) ; Government Survey of France, original scale, i : 20000. 
 scale of copies, i : 40000 and i : 80000. 
 3. For ordinary purposes, a reducing scale need not be made on the paper ; we 
 
43-] CHAPTER II. STRAIGHT LINES. 47 
 
 choose the scale, then compute the reduced lengths, and then lay off these lengths 
 on the paper with the aid of a divided rule. But when great accuracy is demanded, 
 it is necessary to have on the paper a reducing scale by means of which we can de- 
 termine the length of a line with extreme precision. Such a reducing scale is de- 
 scribed in Chapter VIII. (See p. i84-) 
 
 Exercises. 1. Draw five parallel lines, and take on them to the scale 
 I : 200 the distances 30, 2O m , 9 m , 48, I5 m . 
 
 2. Draw any three lines, and find what lengths they represent to the scale 
 I : 200. 
 
 3. What is the value of the scale if i m is represented by 4 cm ? 
 
 4. If 5 cm represent ioo m natural length, what length will represent i m ? 
 
 2Q ? jdkm ? i2^m p 
 
 5. In what ratio are natural lengths reduced if the scale is i inch to the 
 foot? i inch to the mile? 2 millimeters to the kilometer? 
 
 Ans. i : 48 ; i : 126720 ; 1 : 500000. 
 
 6. The distance between two towns, drawn on paper to the scale i : 40000, 
 is represented by a line i dm long; find the distance between the towns. 
 
 Ans. 4 km . 
 
 7. In a railroad survey the total length is eighty miles. What scale will 
 enable us to represent a plan of the road on a piece of paper just one foot 
 in length? 
 
 8. Draw on the blackboard three horizontal lines. Then represent them 
 on paper, choosing a suitable scale for this purpose. Write the value of the 
 scale on the paper under the lines. 
 
 REVIEW OF CHAPTER II. 
 QUESTIONS. 
 
 1. How many directions has a straight line? 
 
 2. How is a straight line determined in position? 
 
 3. How is a straight line drawn? 
 
 4. What is free-hand drawing? 
 5* What is a phimb line? 
 
 6. Define vertical, horizontal, and inclined lines, and give examples. 
 
 7. Define vertical, horizontal, and inclined planes, and give examples. 
 
 8. How can you test whether a line or a plane is vertical? 
 9 How can you test whether a line or a plane is horizontal? 
 
48 GEOMETRY FOR BEGINNERS. 
 
 10. When is a line said to rotate about a point? When does the rotation 
 amount to one revolution ? 
 
 11. Define a circumference ; \\s, centre ; an arc ; a radius; a circle. 
 
 12. What is true of radii of the same circle? 
 
 13. Describe how circles (arcs, etc.) are drawn. 
 
 14. In considering the directions of two lines, what two cases arise? 
 
 15. Define parallel lines, and give examples. 
 
 16. Show that parallel lines can never meet. 
 
 17. Are lines which would never meet always parallel? Give an example. 
 
 18. Give examples of parallel curves ; of parallel planes ; of a line and a 
 plane parallel to each other. 
 
 19. How are parallel lines drawn? 
 
 20. What four cases arise when we consider the directions of three lines? 
 Represent the four cases by figures. 
 
 21. How can we illustrate the truth that a straight line is the shortest dis- 
 tance between two points? Give an application of the same truth. 
 
 22. Define equal and unequal lines. 
 
 23. Explain how the equality and the inequality of two lines are expressed. 
 
 24. How is the equality of two lines tested? 
 
 25. What is an Axiom ? Give Axioms I., II., III., IV. and V. 
 
 26. Explain how lines are added, subtracted, multiplied, and divided. 
 
 27. In what two senses of the word can division be performed on a line? 
 
 28. Define a 1 measure of a line; a multiple of a line; a common measure of 
 two or more lines; the ratio of two numbers and of two lines. What 
 are the terms of a ratio called? 
 
 29. In what two ways can a ratio be expressed or written? 
 
 30. How is the ratio of two lines found? 
 
 31. What are incommensurable lines? 
 
 32. How can we approximate to the value of their ratio? 
 
 33. Define a unit of length; measuring a line; the numerical value and the 
 length of a line. 
 
 34. What is the ratio of a foot to an inch? of a yard to a foot? of a rod to a 
 yard? of a mile to afoot? What are the corresponding reciprocal ratios? 
 
 35. Define the meter. 
 
 36. Name and define the multiples and divisions of the meter. 
 
 37. How are metric units reduced from one denomination to another? 
 
 38. Which of the metric units are most used? 
 
 39. In expressing a length, how many units should be used? 
 
 40. How would you read a length containing a fraction of a unit? 
 
CHAPTER II. STRAIGHT LINES. 49 
 
 41.- In adding, subtracting, etc., metric units, what rule must be observed? 
 
 42. What is the value of one meter in inches? 
 
 43. How are lines on paper usually measured? longer lines, such as the 
 length of a room? lines on the ground? 
 
 44. What are the advantages and disadvantages of pacing as a means of 
 measuring a line? 
 
 45. Give examples of lines which cannot be directly measured. 
 
 46. What is a reducing scale, and what is drawing to scale ? 
 
 47. I low are scales of reduction named and written? 
 
 48. Why must the ratio or scale of reduction be constant for all lines belong- 
 ing to the same map* plan, or diagram ? 
 
 EXERCISES. 
 
 1. How many straight lines can be drawn through the North and South 
 poles of the earth? 
 
 2. How many straight lines can be drawn through n points (it being any 
 number), each line connecting two points? 
 
 Solution. From the first point we can draw lines through all the other points; 
 therefore, in all, n I lines. The same number may be drawn from each of the 
 other points, and since there are n points, this would make n (n i) lines in all. 
 But, in this way, two lines would be drawn in every case through the same two 
 points ; these two lines, however, would coincide and form but one line by $ 22, II. 
 Therefore, the total number of different straight lines which can be drawn is 
 n (n i) 
 
 2, 
 
 3. Find the greatest possible number of straight lines which can be drawn 
 through ten points. Ans. 45. 
 
 4. Through a vertical line how many vertical planes can be passed in 
 thought (that is, how many different vertical planes can be imagined, 
 all of which shall contain the vertical line) ? how many horizontal 
 planes? how many inclined planes? 
 
 5. Through a horizontal line how many vertical planes can be passed? 
 how many horizontal planes? how many inclined planes? 
 
 6. Through an inclined line how many vertical planes can be passed? how 
 many horizontal planes? how many inclined planes? 
 
 7. How many centres can a circle have? how many circles can have the 
 same centre? 
 
 8. Can two curved surfaces be parallel to each other? a curved surface and 
 a plane surface? 
 
50 . GEOMETRY FOR BEGINNERS. 
 
 9. Can you draw a line parallel to a given line with the aid only of the 
 dividers ? 
 
 10. Find in how many points six lines can intersect one another. 
 
 11. Can you illustrate with numbers Axioms II.-V.? 
 
 12. If you multiply the ratio of two lines by their reciprocal ratio, what is 
 the product always equal to? 
 
 13. Draw any two lines, and find their ratio. 
 
 14. Choose two lines in the room ; estimate their ratio by the eye, then find 
 it by measuring them. 
 
 15. i8 m -r-4 dm 64 cm -|- 268 mm = how many meters? 
 
 16. 9 km 220 m -4- 84 dkm 72 hm = how many kilometers ? 
 
 17. The sum of two lines is 24 feet, and their difference is 9 feet ; find the 
 lines. 
 
 18. Two lines are to each other as 5 : 3, and their sum is 48 m ; find the lines. 
 
 Ans. i X 48 m = 30, and f X 48 m i8 m . 
 
 19. The sum of two lines is 84, and their ratio is 5:16; find the lines. 
 
 20. The difference of two lines is 20, and their ratio is 4:9; find the lines. 
 
 Ans. | X 20 m = i6 m , and | X 20 = 36. 
 
 21. I read in a newspaper that "the river Theiss has fallen 3o cm "; how 
 much is this in inches? 
 
 22. Among the scales prescribed for the United-States engineer service are : 
 general plans of buildings, I inch to 10 feet; maps i-i miles square, 2 
 feet to i mile ; do., 3 miles square, I foot to I mile ; do., from 4 to 8 
 miles square, 6 inches to I mile ; do., 9 miles square, 4 inches to I mile ; 
 do., not exceeding 24 miles square, 2 inches to I mile ; do., 50 miles 
 square, I inch to I mile. Express these scales as fractions, with unity 
 for the numerator. 
 
 Ans. T 1 -, 2FIOJ 52 l gO> TU5F<J> T5&4'S> 3T68tf> F3&60- 
 
 23. Which scale reduces lines the most : a scale of I inch to i mile, or a scale 
 of I centimeter to I kilometer? 
 
 24. If the scale to which a map is made is not known, how can it be found? 
 
440 
 
 CHAPTER III. ANGLES. 
 
 51 
 
 CHAPTER III. 
 ANGLES. 
 
 CONTENTS. I. .Definition of an Angle (44). II. Magnitude of an Angle 
 (45,46). III. Magnitudes of Particular Angles ( 47-49). IV. Measure 
 of an Angle ($ 50-52). V. Angles made by Two Lines ( 53, 54). VI. 
 Angles made by Three Lines ( 55-58). 
 
 L Definition of an Angle. 
 
 44. When we open the legs of the compasses (Fig. J/) they 
 point in different directions : they 
 are now said to" make an ANGLE 
 with each other. Replace, in 
 thought, the legs by geometrical 
 lines A B and A C, and we have 
 an angle as understood in Geome- 
 try. 
 
 Definition. An ANGLE is the 
 difference in direction of two lines. 
 
 The two lines are called the SIDES of the angle, and the point 
 where they meet is called the VERTEX.! If the lines do not 
 actually meet (see Fig. 25, p. 27), the point where they would 
 meet if prolonged, is the vertex of the angle. 
 
 An angle is named in one of three ways : 
 
 (1) By a small letter placed just inside the vertex; 
 
 (2) By a large letter placed just outside the vertex ; 
 
 (3) By three letters, one at the vertex, the others on the sides. 
 
 1 Plural, vertices. 
 
52 GEOMETRY FOR BEGINNERS. [45- 
 
 In the last case, in reading or writing the angle, the letter at 
 the vertex must stand between the others. 
 
 For example : the angle of the lines A B and A C (Fig. 37) is 
 called either the angle d, or the angle A, or the angle B A C or 
 CAB. In general, the word angle need not be written ; the let- 
 ters d, or A, or A C, or CAB are sufficient. 
 
 / Exercises. 1. Name the angles drawn on the black- 
 
 / board. 
 
 2. Draw four angles ; name them in different ways. 
 3* If (as in Fig. j<?) several angles have the same 
 p. r, vertex, which method of naming an angle cannot be 
 
 used? Why not? 
 
 II. Magnitude of an Angle. 
 
 45. Angles differ in magnitude. We shall get a clearer no- 
 tion of the magnitude of an angle if we regard the angle as gen- 
 erated or described by the rotation of a line. 
 
 Hold one leg of the compasses at rest, and make the other rotate 
 about the pivot in the direction shown by the arrow (Fig. 37) ; 
 then the angle made by the leg will increase in magnitude. On 
 the other hand, if we rotate the legs in the opposite direction, the 
 angle will diminish ; and when the legs 'are brought together, it will 
 vanish entirely. 
 
 In like manner, we may regard the angle made by any two lines, 
 A B and A C (fig. 3f) 9 as described by the rotation of one of the 
 lines about their intersection A, until it coincides in direction with 
 the other line. 
 
 It is obvious that the size or magnitude of an angle depends 
 entirely on the amount of rotation required to describe it, and not 
 at all on the lengths of its sides. 
 
 Definition. Two angles are EQUAL if they can be placed one 
 upon the other so that their vertices coincide in position, and their 
 sides coincide in direction ; otherwise the angles arc UNEQUAL. 
 
46.] 
 
 CHAPTER III. ANGLES. 
 
 53 
 
 We shall soon see that the equality of two angles (like that of 
 two lines) can be tested without actually placing one upon the 
 other. 
 
 Exercises. 1. Of the two angles on the blackboard, which has the 
 longer sides? which is the greater angle? 
 
 2. Must the two sides of an angle have the same length? 
 
 46. If, in the angle B s ^ ^ ./. jp), the 
 side A C is made to rotate about the vertex 
 A away from A B, until it comes into the 
 position A D, it is clear that the angle BAD 
 is equal to the sum of the angles B A C and 
 CAD: that is, 
 
 BAD=BA C+ CAD. 
 
 39- 
 
 D 
 
 If, on the other hand, the side AD of the angle BAD be 
 made to rotate towards A B until it 
 arrives at the position A C, there re- 
 mains the angle B A C, which is equal 
 to the difference of the angles BAD 
 and CAD: in other words, 
 
 BAC=BAD- CAD. 
 
 If the angles AOB,BOC,C OD, 
 D OE, E OF (Fig. 40), are equal 
 to one another, then A OC = 2 A OB, 
 AOD=$AOB, A OE= 4 A OB,AOF= 
 
 Conversely, it is evident that, 
 
 A OB = $A O C= A OD = A OE = I A OF. 
 
 Angles, in the same sense as lines, may be added, subtracted, 
 multiplied, and divided. 
 
 Definitions. I. To BISECT an angle (or a line) is to divide it 
 into two equal parts. 
 
 II. A line which bisects an angle is called a BISECTOR. 
 
54 
 
 GEOMETRY FOR BEGINNERS. 
 
 [47- 
 
 Exercises. 1. In Fig. 40 what angle is equal to the sum B C-\- COE? 
 to the difference A O F C O F? 
 
 2. Draw freehand, three angles, of which one is twice, the other five times 
 as large as the first. 
 
 3. Bisect freehand, a given angle. 
 
 4. How many bisectors can an angle have? 
 
 5. Divide an angle, freehand, into three, four, five, six equal parts. 
 
 6. In Fig. 40 what is the ratio of the angles A O B and A OF? A C 
 and A OE? 
 
 III. Magnitudes of Particular Angles. 
 
 47o Angles receive different names according to their mag- 
 nitude. 
 
 If we conceive a line to revolve 
 round O (Fig. 41) , starting from the 
 initial position O A and revolving in 
 the direction of the arrow, it will 
 describe an angle of constantly in- 
 creasing magnitude. 
 
 When the line has made exactly 
 one-fourth of a complete revolution, 
 the angle described, A O C, is called 
 a RIGHT ANGLE ; it is sometimes de- 
 noted by the letter R. 
 
 Fig. 41. 
 
 When the line has made one-half of a revolution, it is in the 
 position OE ; the angle described, A OE, is equal to two right 
 angles, and is called a STRAIGHT ANGLE, because its sides O A and 
 O E, having opposite directions, form one straight line. 
 
 When the line has made three-fourths of a revolution, it is in 
 the position O G. The angle which has been described, A O G, 
 is equal to three right angles. 
 
 Finally, when the line has returned to the initial position O A, 
 it has made a complete revolution, and has described four right 
 angles. 
 
47-] CHAPTER III. ANGLES. 55 
 
 Definitions. I. A STRAIGHT ANGLE is the angle described by 
 a line which makes one -half of a revolution. 
 
 II. An angle less than a straight angle is called a CONCAVE 
 ANGLE ; an angle greater than a straight angle is called a CONVEX 
 ANGLE. 
 
 III. A RIGHT ANGLE is one-half of a straight angle. 
 
 IV. A concave angle, if less than a right angle, is called an 
 ACUTE ANGLE ; if greater, it is called an OBTUSE ANGLE. 
 
 In Fig. 41, A OB, A O C, A O D are concave angles ; A OF, 
 A O G, A O H are convex angles ; A O B is an acute, A O D an 
 obtuse angle. 
 
 NOTES. i. Acute and obtuse angles are sometimes called oblique angles, in 
 distinction from the right angle. 
 
 2. Whenever two lines meet, there is formed both a concave and a convex angle ; 
 but, when not otherwise stated, the concave angle is to be understood. 
 
 Since the sides of a straight angle form a straight line, and one 
 straight line can always be placed on another straight line so as to 
 coincide with it in direction, it follows that two straight angles sat- 
 isfy the test of equal angles ( 45) ; in other words, 
 
 All straight angles are equal to one another. 
 
 And since a right angle is half of a straight angle, it follows 
 (by Axiom III.) that 
 
 All right angles are equal to one another. 
 
 Exercises. 1. Open the legs of the compasses so that they make aright 
 angle ; a straight angle ; three right angles j an acute angle ; an obtuse 
 angle ; a concave an'gle ; a convex angle. 
 
 2. On the blackboard are several angles : what kind is each, and why? 
 
 3 What kind of angles are found on the cube? the prism? the pyramid? 
 
 4. Point out angles on various objects (doors, windows, tables, walls of 
 the room, etc.), and name the kind in each case. What kind of angle occurs 
 most frequently? 
 
 5. Draw, free-hand, a right angle; an acute angle; an obtuse angle; a 
 concave angle; a convex angle. " 
 
 6. At 12 o'clock what angle do the hands of a watch make with one an- 
 other? at 3 o'clock? at 6 o'clock? at 9 o'clock. 
 
56 
 
 GEOMETRY FOR BEGINNERS. 
 
 [48. 
 
 7. Mention a time of the day when the angle between the hands of a 
 watch is an acute angle? a right angle? an obtuse angle? 
 
 8. What part of one revolution does the minute-hand describe in 5 min- 
 utes? (A MS. -/Q =iV-) I n I0 > 20 > 3> 40. 5 minutes? 
 
 9. How long does it take the minute-hand to describe a right angle? the 
 hour-hand? 
 
 10. What kind of an angle does the hour-hand describe in I hour? in 3 
 hours? in 5 hours? in 6 hours? in 8 hours? in 9 hours? in 12 hours? 
 
 11. What kind of an angle does a vertical line make with a horizontal 
 line? 
 
 12. Draw a right angle with one side vertical ; with one side horizontal; 
 with one side inclined. 
 
 48. Definitions. I. A straight line is said to be PERPEN 
 DICULAR to another straight line when it makes a right angle with z>, 
 
 II. Two lines not perpendicular to each other are said to be 
 INCLINED to each other. 
 
 Examples. The lines AE and C G (Fig. 41}, the sign -f 
 
 in arithmetic, a common cross, 
 the slats of a window (Fig. 
 42) , are examples of perpen- 
 dicular lines ; the lines A B 
 and A C (Fig. 39) are inclined 
 lines. 
 
 The sign 1 is sometimes used 
 for the word perpendicular; 
 thus : A EL CG is read : the 
 line A E is perpendicular to the 
 line C G. 
 
 Exercises. 1. Give examples 
 of perpendicular lines ; of inclined 
 lines. 
 
 2. What is the difference be- 
 tween a perpendicular line and a 
 Fig. 42. ' vertical line ? 
 
49-] CHAPTER III. ANGLES. 57 
 
 3. If two lines are perpendicular to each other, and one of them is ver- 
 tical, must the other be horizontal? 
 
 4. If two lines are perpendicular to each other, and one of them is hori- 
 zontal, must the other be vertical ? 
 
 5. Give an example of two perpendiculars, one of which is vertical, the 
 other horizontal. 
 
 49. Among the most common problems which the surveyor, 
 carpenter, draughtsman, etc., have to solve, are 
 
 (/.) To erect a. perpendicular at a given point of a given line. 
 
 (//.) To let fall a perpendicular from a given point to a given line. 
 
 On the ground, surveyors use various instruments, the simplest 
 of which is the Surveyors' Cross {Fig. 43). 
 
 It consists of a block of wood having two saw-cuts made very 
 precisely at right angles to each other. This block 
 is fixed on a pointed staff on which it can turn 
 freely. To erect a perpendicular, set the cross at 
 the point of a line where a perpendicular is wanted, 
 then turn the block till on looking through one saw- 
 cut you see the ends of the line ; then the other 
 saw-cut will point out the direction of the perpen- 
 dicular. To let fall a perpendicular to a line from 
 some object (as the corner of a field, a tree, etc.), 
 set the cross at a point of the line which seems to i &' 43 ~ 
 
 the eye about right, then note how far from the object the perpen- 
 dicular at this point strikes, and move the cross that distance ; re- 
 peat this operation till the correct point is found. 
 
 On paper, or the blackboard, both of the above problems are 
 readily solved by means of the Ruler and the Square. Explain 
 how. 
 
 NOTE. The method of solving these problems on paper with the compasses 
 will be shown in Chapter IV. 
 
 Exercises. 1. Draw a straight line, and then, free-hand, (ci) erect a 
 perpendicular at a point on the line, () let fall a perpendicular from a point 
 not on the line. Test and correct the results by the aid of the ruler and 
 the square. 
 
58 GEOMETRY FOR BEGINNERS. [ 50. 
 
 2. How many perpendiculars can "be erected at a given point of a given 
 line? How many can be let fall from a given point to a given line? 
 
 3. Draw a horizontal line, and, at five points on it, erect perpendiculars. 
 What kind of lines are they? 
 
 4. Draw a vertical line, and, at five points on it, erect perpendiculars. 
 What kind of lines are they? 
 
 5. Draw an inclined line, and, at five points on it, erect perpendiculars. 
 What kind of lines are they? 
 
 6. Draw a horizontal line, erect a perpendicular, and at a point of this 
 perpendicular erect also a perpendicular. What direction has the second per- 
 pendicular? 
 
 7. Erect a perpendicular at a point of a line on the ground. 
 
 8. Choose a point in the yard (playground or field), and let fall a perpen- 
 dicular to the nearest side of the yard. 
 
 IV. Measure of an Angle. 
 
 50. Angles (like lines) are measured by choosing a unit, and 
 then comparing other angles with this unit. The unit for measur- 
 ing angles is obtained by dividing the right angle (which would be 
 too large a unit) into ninety equal parts, and calling one of these 
 parts a degree. The degree is subdivided into minutes and seconds. 
 One ninetieth part of a right angle is called a DEGREE. 
 One sixtieth part of a degree is called a MINUTE. 
 One sixtieth part of a minute is called a SECOND. 
 Abbreviations. Degree = ; minute = ' ; second = ". 
 Thus, 24 1 7' 8" is read : 24 degrees 1 7 minutes 8 seconds. 
 The values in degrees of the angles defined in 47 are as fol- 
 lows : 
 
 A right angle = 90. An acute angle < 90. 
 
 A straight angle =180. An obtuse angle > 90. 
 
 Three right angles 2 70. A concave angle < 1 80. 
 
 Four right angles = 360. A convex angle > 180. 
 
 Exercises. 1. How many seconds in i? in io? in 45? in 360? 
 
 2. Reduce 48 54' 36" to seconds. 
 
 3. Reduce 120000" to degrees. 
 
51-] 
 
 CHAPTER III. ANGLES. 
 
 59 
 
 4. Add 37 48' 35", 28 39', and 78 9' 55". 
 
 5. From 128 15' 31" take 69 42' 18". 
 
 6. Multiply 1 8 35' by 2, 3, 4, and 5. 
 
 7. Divide 72 27' by 2, 3, 4, and 5. 
 
 8. Reduce to degrees 5 R (5 right angles); 6 R ; 8 A 1 / 12 A 5 ; % R ; 
 R; % R; & A'/ i AV i R; % R; \ R; T \> R; T V A. 
 
 9. How many right angles in 540? 900? 225? 30? 5? 
 10. Find the ratio between 10 and three right angles. 
 
 51. In measuring angles we take advantage of a simple rela- 
 tion between angles and the arcs intercepted between their sides, 
 and described with the same radius from their vertices as centres. 
 
 In Fig. 44 compare the angles A O B, A O C, A O D, with the 
 arcs A B, AC, AD, intercepted by 
 their sides. We see that the greater 
 the angle the greater the corresponding 
 arc, or arc intercepted by its sides. 
 This conclusion is general : in the 
 same circle, the greater the angle at the 
 centre the greater the corresponding 
 arc, and the less the angle the less the 
 arc. 
 
 Again, if the two angles A OB and 
 FOG are equal, the corresponding 
 arcs A B and FG are also equal. For the angles, being equal, 
 may be placed one on the other so that they will coincide ; and 
 then the arcs A B and F G must also coincide, since all the points 
 of one arc are at the same distance from the centre O as the points 
 of the other arc. 
 
 In like manner it can be shown that if the arcs A B and F G 
 are equal, the corresponding angles at the centre, AOB and FOG, 
 will also be equal. 
 
 The same reasoning may be applied with the same result to any 
 angles and their corresponding arcs ; hence, in general : 
 
 Fig. 44. 
 
60 GEOMETRY FOR BEGINNERS. [ 51. 
 
 I. Eqtial angles at the centre of a circle intercept on the cir- 
 cumference equal arcs. 
 
 II. Equal arcs on the circumference correspond to equal 
 angles at the centre. 
 
 Accordingly, the circumference of a circle is divided into 360 
 equal arcs, each corresponding to an angle of i at the centre. 
 These arcs are also called degrees, and are subdivided like the angu- 
 lar degree into minutes and seconds. And, in practice, an angle is 
 measured by finding how many degrees, minutes, and seconds 
 there are in the corresponding arc, it being obvious that the num- 
 ber of angular degrees, minutes, and seconds in an angle is the 
 same as the number of arc degrees, minutes, and seconds in the 
 arc described with any radius from the vertex of the angle as a 
 centre, and intercepted between its sides. 
 
 We have just said that the arc employed to measure an angle 
 
 may be described with any ra- 
 dius. With a longer radius we 
 have, it is true, a longer arc 
 (Fig. 45), but each division of 
 the arc is also correspondingly 
 increased in length, so that the 
 number of divisions (degrees, 
 etc.) in the entire arc remains 
 the same as before. 
 
 We are now prepared to an- 
 swer the question : How can the equality of two angles be tested 
 without placing one of them upon the other ? We reply : two an- 
 gles are equal, if, when their corresponding arcs are measured (as 
 shown in the next section), these arcs are found to contain the 
 same number of degrees, minutes, and seconds. 
 
 NOTE. How an arc can be divided into degrees, etc., is a question which must 
 be deferred until we know more about the properties of the circle. 
 
52.] CHAPTER ill. ANGLES. 61 
 
 Exercises. 1. Find the angle described by the hour-hand of a watch in 
 I hour; 2 hours; 5 hours; 12 hours. 
 
 2. Find the angle described by the minute-hand of a watch in i, 5, 10, 15, 
 20, 30, 45 minutes. 
 
 3. In how many minutes does the minute-hand describe an angle of i? 
 60 ? 225? 300? 
 
 4. What angle do the hands of a watch make with each other at I o'clock? 
 at 2, 3, etc., up to 12 o'clock? 
 
 5. The earth turns on its axis in 24 hours ; what angle does a point on the 
 surface describe in I hour? in 6, 12, 15 hours? 52 hours? 
 
 52. On paper or the blackboard, when great accuracy is not 
 required, angles are measured with an instrument called a PRO- 
 TRACTOR (Fig. 46). It is a semicircle, made of paper, horn, 
 
 Fig. 46. 
 
 brass, or silver, the circular edge of which is divided into 180 equal 
 parts or degrees. To measure an angle with the protractor, place 
 the centre of the instrument on the vertex of the angle, and its 
 zero line on one side ; then read off on the edge of the protractor 
 the division through which the second side of the angle passes. 
 On the ground, surveyors and engineers employ for measuring 
 
62 
 
 GEOMETRY FOR BEGINNERS. 
 
 [52. 
 
 angles, costly instruments called THEODOLITES. A cheap substitute 
 for a theodolite is shown in Fig. 4.7. It consists of two pieces of 
 wood shaped like rulers mounted on a vertical axis, by a pin driven 
 
 Fig. 47. 
 
 through their exact centres. The vertical needles inserted near 
 the end of the rulers are used for sighting. In place of the needle 
 nearest the eye, it is better to employ a thin strip of wood, A t 
 having a fine vertical slit ; and in place of the other needle, a ver- 
 tical wire fixed in a light frame, B. 
 
 By the help of this instrument, arid a protractor, one can mea*. 
 ure with considerable accuracy an angle on the ground for in- 
 stance, the angle M ON (Fig. 48). 
 
53-] CHAPTER III. ANGLES. 63 
 
 Exercises. 1. On the blackboard are several angles. Measure each 
 with the protractor, and write the result between its sides. 
 
 2. Draw five different angles. Estimate the value of the first in degrees; 
 then measure it with a protractor. Proceed in like manner with each of the 
 other angles. Give the results in 3 columns: in column I, the estimated val- 
 ues ; in column 2, the measured values ; in column 3, the differences between 
 the estimated and the measured values. 
 
 3. From a point on a straight line draw two lines, both on the same side 
 of the given line ; measure each of the three angles thus formed, andadd the 
 results. What is the sum? What ought the sum to be? 
 
 4. Through a point draw three lines ; measure the six angles thus formed 
 about the point ; add the results. What is the sum? What ought it to be? 
 
 5. Open the legs of the compasses so that they make angles of 90; 60; 
 
 45; 30. 
 
 6. Imagine two lines drawn from your eye, one to the right-hand upper 
 corner of the room, the other to the left-hand upper corner ; what angle 
 would the lines make? 
 
 7 . With the aid of the protractor make an angle equal to a given angle ? 
 
 8. Draw angles equal to 20; 30; 40; 60; 90; 100; 150; 180; 
 
 15; 45; 79; 81; 142. 
 
 9. Explain how the instrument in Fig. 4.7 is constructed, and how you 
 would proceed to measure by means of it the angle M O N'\n Fig. 48. 
 
 V. Angles made by Two Lines. 
 
 53. The angles A O Cand C OB (Fig. 49) are called adjacent 
 angles. 
 
 Definition I. Two angles are 
 ADJACENT when they have the vertex 
 and one side common, and the other 
 sides are opposite parts of the same 
 straight line. _ 
 
 We see from the figure that 
 
 Since this equation would hold true in whatever direction the line 
 O C was drawn, we conclude that in all cases 
 
 The sum of two adjacent angles is two right angles, or 180. 
 
64 GEOMETRY FOR BEGINNERS. [ 54. 
 
 Definition II. If the sum of two angles is 180, each is 
 
 called the SUPPLEMENT of the other. 
 
 Adjacent angles are always supplements of each other. 
 
 Exercises. 1. If one of two adjacent angles is a right angle, what is 
 the other? 
 
 2. If one of two adjacent angles is acute, what is the other? 
 
 3. If one of two adjacent angles is obtuse, what is the other? 
 
 4. If one of two adjacent angles is known, how is the other found? 
 
 5. How is the supplement of a given angle found? 
 
 6. Find the adjacent angles of the following angles: 10; 30; 45; 
 75; 99; 100; 179; 15 48'; 79 13' 52". What is the supplement of 
 each of these angles? 
 
 54. Whenever two lines cross one another there are formed 
 about the intersection of the lines four 
 angles, a, b, c, d (Fig. jo). Of these 
 angles the pairs a and b, b and c, c and d> 
 d and a are adjacent angles ; the pairs 
 a and c 9 b and d are called vertical 
 angles. 
 
 Definition. Two angles are called 
 
 Fig. co. 
 
 VERTICAL angles if they have a common 
 
 vertex, and the sides of one angle have opposite directions to the sides 
 of the other angle. 
 
 If we measure two vertical angles, as a and c (Fig. 50) , either 
 with a protractor or by cutting them out on a piece of cardboard 
 and laying them one upon the other, we shall find that they are 
 equal. But without measuring them we can prove that they must 
 be equal, if we bear in mind what has been said about adjacent 
 angles in the last section. Since b is an adjacent angle to both 
 a and c, therefore, 
 
 and 
 
55-] CHAPTER III. ANGLES. 65 
 
 Hence it follows from Axiom I. (state the axiom) that, 
 
 a+b=b+c 
 subtract b = b 
 
 There remains a = c 
 
 For, if equals are taken from equals, the remainders are equal 
 (Axiom III.). 
 
 Since this reasoning holds good, however the intersecting lines 
 cut each other, we come to the general conclusion, that 
 
 Two vertical angles are always equal to each other. 
 
 Remark. In the above reasoning we have made use of the 
 general truth shown in the last section ; namely, that the sum of 
 two adjacent angles is 180 ; and also of Axioms I. and III. By 
 reasoning upon these truths we have proved or demonstrated a new 
 truth ; namely, that two vertical angles are equal. Geometrical 
 truths which are capable of proof by reasoning from known truths 
 are called THEOREMS. 
 
 Exercises. 1. Draw adjacent and vertical angles and name them. 
 
 2. What is the sum of the angles a and b (Fig. 50)! Why? 
 
 3. What is the sum of the angles a, b, c, and d (Fig. jo~) ? Why? 
 
 4. One of the angles formed by two intersecting lines is 36; find the 
 others. 
 
 5. One of the angles formed by two intersecting lines is 90 ; find the others. 
 
 6. Draw any two intersecting lines, and find the values of the angles 
 formed at the intersection. 
 
 ? If one of the angles formed by two intersecting lines is known, how 
 are the others found? 
 
 8. Each of two vertical angles is 45; find the value of each angle in the 
 other pair of vertical angles. 
 
 9. Prove that b = d (Fig. 50) by reasoning like that used above to show 
 that a=c. 
 
 VI. Angles made by Three Lines. 
 55. Thus far the angles considered have had a common ver- 
 tex j let us now consider angles formed about two different ver- 
 
66 GEOMETRY FOR BEGINNERS. [55- 
 
 tices. Such angles are formed when two straight lines are cut by 
 a third line. 
 
 The most important case, and the only one which we shall exam- 
 ine, is that in which two parallel lines, 
 AB and CD (Fig. jz), are cut by a 
 -% third line, E F. There are formed 
 about the two points of intersection 
 _ eight angles. These angles receive 
 D special names. 
 
 The four angles, c, d, n> m, which 
 lie between the parallel lines, are called 
 
 INTERNAL ANGLES ; the other four, a, b, 0, /, are called EXTERNAL 
 ANGLES. 
 
 An external angle and an internal angle, as a and m, having dif- 
 ferent vertices, and lying on the same side of the intersecting line, 
 are called CORRESPONDING ANGLES. 1 
 
 An external angle and an internal angle, as a and , having dif- 
 ferent vertices, and lying on different sides of the intersecting line, 
 are called CONJUGATE ANGLES. 
 
 Two external angles, as a and p, or two internal angles, as d and 
 m, having different vertices, and lying on the same side of the in- 
 tersecting line, are called OPPOSITE ANGLES. 
 
 Two external angles, as a and o, or two internal angles, as d and 
 n, having different vertices, and lying on different sides of the in- 
 tersecting line, are called ALTERNATE ANGLES. 
 
 Exercises. 1. Name the four pairs of corresponding angles in Fig. 51 ; 
 the four pairs of conjugate angles ; the four pairs of opposite angles ; the 
 four pairs of alternate angles. 
 
 2. Which are external, which internal opposite angles? which are ex- 
 ternal, which internal alternate angles? 
 
 8 Which is the angle corresponding to o? conjugate to p? opposite to 
 m ? alternate to n ? 
 
 _ 1 _ . __ . 
 
 1 They are also called external-internal angles. 
 
56.] CHAPTER III. ANGLES. 67 
 
 4. Draw two parallel lines, and a third line intersecting them. Letter the 
 eight angles e,f,g, h, and/, r, s, t. Then write in six vertical columns (a) 
 the external angles ; (li) the internal angles ; (c) the corresponding angles; 
 (c/) the conjugate angles ; (i) the opposite angles ; (/) the alternate 
 angles. 
 
 56. If we conceive the line AB to move along the line E F, 
 remaining always parallel to its first 
 position, then, since its direction does 
 not change, the four angles a, l>, c, d ; 
 which it makes with EF do not 
 change. When A B reaches the par- 
 allel line CD, it will coincide with it 
 in direction, and the four angles a, b, 
 c, d will coincide respectively with the 
 
 corresponding angles m, ;/, o, p. Moreover, each pair of conjugate 
 angles, as a and ;/, become now adjacent angles ; therefore their 
 sum is 1 80 ( 52) : the same is true of each pair of opposite an- 
 gles, as a and p ; lastly, each pair of alternate angles, as a and o, 
 become now vertical angles, so that they are equal ( 53). 
 Hence, 
 
 I. 
 
 2. 
 
 3- 
 
 4- 
 
 a = m. 
 
 a + n= 2 R. 
 
 a + p = 2 R. 
 
 a= o. 
 
 b = n. 
 
 b +771=2 R. 
 
 b + 0=2 R. 
 
 b = p. 
 
 C =0. 
 
 c + / = 2 R. 
 
 c -f- n 2 R. 
 
 c = m. 
 
 d = p. 
 
 d+ o = 2 R. 
 
 d-\-m= 2 R. 
 
 d= n. 
 
 Putting these results into words, we obtain the following theo- 
 rem : 
 
 Theorem. If two parallel lines are cut by a third line, 
 /. The corresponding angles are equal ; 
 
 2. The sum of two conjugate angles is two right angles ; 
 
 3. TJic sum of two opposite angles is two right angles ; 
 4.. The alternate angles are equal. 
 
68 GEOMETRY FOR BEGINNERS. [' 57. 
 
 liemark. Theorems often involve consequences which are 
 
 easily deduced from the theorems : 
 such consequences are called in 
 _ Geometry COROLLARIES. The pres- 
 ent theorem furnishes an example. 
 Corollary. If a = 90 (that is, 
 
 D ,toowstatw = 90 
 
 (that is, that CD-lEF) ; or, to 
 Fig. 53. state the corollary in general 
 
 terms, 
 
 If a straight line is perpendicular to one of two parallels, it is 
 also perpendicular to the other. 
 
 Exercises. 1. Give the proof separately for each of the last three 
 (equations in each of the above four series of equations. 
 
 Models. b = n, because when A B coincides with CD, b coincides with n. 
 
 b -J- m = 2 R, because when A B coincides with C D, b becomes adjacent to m, 
 and the sum of two adjacent angles is 180. 
 
 2. Find the other seven angles when (7 = 112; when a = 90. 
 
 3. Draw two parallel lines, and a line intersecting them; then find the 
 eight angles thus formed, using a protractor as little as possible. 
 
 4. If either one of the four parts in the theorem of this section is not 
 true, what can we conclude as to the lines A B and CD? 
 
 57. Theorem. If two straight lines are cut by a third line 
 .so that two corresponding angles are equal, then the lines must be 
 parallel. 
 
 If, for example, a = m (Fig. 52), then is AB \\ CD. For, if 
 we move A B towards C D, keeping it parallel to its first position, 
 during the motion the angle a does not change because the direc- 
 tion of A B does not change. When the intersection of A B and 
 E F coincides with the intersection of CD and F F, AB must 
 coincide in direction with CD, since a = m. Therefore, AB, in 
 its first position, must be parallel to CD. 
 
 Corollary. By making in this theorem a = m = 90, we ob- 
 tain the following corollary : 
 
58.] CHAPTER III. ANGLES. 69 
 
 If two straight lines are both perpendicular to a third line, they 
 are parallel. (Fig. 54.) 
 
 Remark. All the four properties enumerated in the theorem 
 of the last section are so related that if either one of them is 
 true the other three must also be true, and the two intersected 
 lines must be parallel. 
 
 Hence arises the importance of corresponding, conjugate, oppo- 
 site, and alternate angles. Before we can assert with absolute cer- 
 tainty that two lines are parallel, we ought to show that the lines, 
 when prolonged farther than even the imagination can reach, would 
 not meet. Such an actual extension of the lines P R 
 is of course impossible ; it is, however, quite un- 
 necessary, for the parallelism of two lines is very 
 simply tested by the angles which are formed 
 when the lines are cut by a third line. 
 
 Exercises. 1. Prove the corollary, independently 
 of the theorem, with lines drawn and lettered as in 
 
 &% 54- 
 
 2 When more than two lines are perpendicular to a r^ TJ i TV 
 
 third, are they all parallel to each other? 
 
 Fig. 54- 
 
 58. In Fig. 55, we have drawn AB II CD and EF \\ G H. 
 If we compare the angle a with either of the angles x, y, *, we see 
 that their sides are respec- 
 tively parallel. But there 
 are these differences : (Y.)the 
 sides of x have the same di- 
 rection respectively as those 
 of a ; (it.) the sides of z 
 have opposite directions ; -^ 
 (///.) the sides of y have, one 
 the same direction, the other 
 the opposite direction, ~. . 
 
 fj s 55* 
 
70 GEOMETRY FOR BEGINNERS. 
 
 In case (/.), a = b (corresponding angles on the parallels EF 
 and G H}, and x = b (corresponding angles on the parallels C D~ 
 and A B) : therefore a = x (Axiom I.). 
 
 Theorem I. If two angles have their sides respectively par- 
 allel, and directed the same way from the vertex, the angles are 
 equal. 
 
 In case (.), x = z (vertical angles). And a = x (Theorem I.). 
 Therefore a = z (Axiom I.). 
 
 Theorem II. If two angles have their sides respectively par- 
 allel, and directed opposite ways from the vertex, the angles are equal. 
 
 In case (in.), x + y = 2R (adjacent angles). But a = x 
 (Theorem L). Therefore a -f- y = 2 R. 
 
 Theorem III. If two angles have their sides respectively par- 
 allel, and directed, the one pair the same way, the other pair opposite 
 ways ', from the vertex, the sum of the angles is two right angles. 
 
 Exercises. 1. In Fig. 55, if a = 115, what is the value of x? of y ? ofz? 
 
 2. Prove Theorem I., using in the proof the angle c in Fig. 55, instead of 
 the angle b. 
 
 3. What relation exists between a and s {Fig. 55) ? Why? 
 
 4. Prove Theorems I., II., and III., with a new figure, drawn and lettered 
 differently from Fig. 33. 
 
 REVIEW OF CHAPTER III. 
 QUESTIONS. 
 
 1. Define an angle; its sides ; its vertex. 
 
 2. How is an angle named? 
 
 3. How may an angle be conceived as produced by motion? 
 
 4. On what does the magnitude of an angle depend? 
 
 5. Define equal angles. 
 
 6. Explain (by figures) what is meant by adding, subtracting, multiplying, 
 and dividing angles. 
 
CHAPTER III. ANGLES. 71 
 
 7 * What is bisecting an angle or a line ? What is the bisector of an angle ? 
 . 8. Define a straight angle ; a concave angle ; a convex angle ; a right 
 
 angle ; an acute angle ; an obtuse angle. 
 
 9. Show that all straight angles are equal ; also, that all right angles are 
 equal. 
 
 10. When are two lines perpendicular to each other? when are they inclined? 
 
 11. W T hat are two of the commonest problems in surveying and drawing? 
 
 12. Describe the surveyor's cross, and how to use it. 
 
 13. How are perpendiculars to a given line drawn on paper? 
 
 14. Define a degree, a minute, and a second. 
 
 15. W 7 hat are the values in degrees of the angles mentioned in Question 8? 
 
 16. Explain why an angle can be measured by means of the arc of a circle 
 intercepted between its sides. 
 
 17. Why is the measure of an angle the same for any radius of the arc? 
 
 18. How is the equality of two angles tested without placing one of them 
 on the other? 
 
 19. Describe the protractor, and how to use it. 
 
 20. Describe a simple substitute for a theodolite, and how an angle on the 
 ground is measured by means of it. 
 
 21. Define adjacent angles ; the supplement of an angle. 
 
 22. Show that the sum of two adjacent angles must be 180. 
 
 23. Define vertical angles. 
 
 24. Two vertical angles are equal. 1 
 
 25. What is a theorem ? 
 
 26. Define external angles ; internal angles ; corresponding angles ; conju- 
 gate angles ; opposite angles ; alternate angles. 
 
 27. If a straight line intersect two parallel lines, the corresponding angles 
 are equal, the alternate angles are equal, and the sum of two conjugate 
 angles, or of two opposite angles, is 180. 
 
 28. What corollary follows from this theorem? 
 
 29. What is a corollary ? 
 
 30. If two lines are cut by a third line so that two corresponding angles are 
 equal, the two lines are parallel. 
 
 31. Give a corollary of this theorem. 
 
 32. How can you test if two straight lines are parallel? 
 
 1 In the review we shall always (as here) simply state the theorem to be proved 
 or problem to be solved The learner is to understand that the proof (or solution) 
 is required. 
 
72 GEOMETRY FOR BEGINNERS. 
 
 83. Two angles have their sides respectively parallel; then, 
 
 (1) If these sides are directed the same way from the vertex, or if they 
 are directed opposite ways, the angles are equal. 
 
 (2) If the sides are directed, the one pair the same way, the other pair 
 opposite ways, the sum of the angles is 180 
 
 EXERCISES. 
 
 1. What is the ratio of one right angle to three right angles? of one-third 
 of a right angle to four right angles? 
 
 2. How can you test whether two intersecting lines are perpendicular to 
 each other? 
 
 3. How can you test whether the two edges of a square are truly at right 
 angles to each other? 
 
 Suggestion. How was the straight edge of a ruler tested ( 23, Exercise 6)? 
 
 4. Make any angle, and then, by the aid of a protractor, make an angle 
 four times as large ; also, an angle one-fourth as large. 
 
 5. Mark three points, join them by lines, then measure the three angles 
 which the lines make with one another. What is their sum? 
 
 6. Can you make three angles with two lines? 
 
 7. If two lines intersect, and one of the four angles is a right angle, prove 
 that the other three are also right angles. 
 
 8. Draw five lines meeting at a point ; how many angles are formed? what 
 is their sum? If the angles are equal, find the value of each in degrees. 
 
 9. Ten lines meet at a point so as to form a regular ten-rayed star ; find the 
 value of the angle between any two rays. 
 
 10. What is the supplement of i? of 179? of 180? 
 
 11. Of two adjacent angles the greater is twice the less ; find the values of 
 the angles in degrees. What is the ratio of each to four right angles? 
 
 12. The bisectors of adjacent angles are at right angles to each other. 
 18. The bisector of one of two vertical angles bisects the other also. 
 
 14. If two parallel lines are cut by a third line, then (a) the bisectors of two 
 alternate angles are parallel to each other; (U} the bisectors of two op- 
 posite angles are perpendicular to each other. 
 
 15. What is the difference between a theorem and an axiom? between a 
 theorem and a corollary ? 
 
59-] 
 
 CHAPTER iv. TRIANGLES. 
 
 73 
 
 CHAPTER IV. 
 TRIANGLES. 
 
 CONTENTS. I. Sides of a Triangle ( 59-63)- II. Angles of a Triangle ($$ 
 64-66). III. Similarity, Equivalence, and Equality ( 67, 68). IV. Equal Tri- 
 angles ($ 69-79). V. Some Consequences of the Equality of Triangles ( 
 80-93). VI. Applications ( 94-98). 
 
 L Sides of a Triangle. 
 
 59. In Fig. 56 are several figures enclosed on all sides by 
 lines. Each is a portion of a plane surface (the surface of the 
 paper) , and is called a Plane Figure. 
 
 Fig. 56. 
 
 Definition I. A PLANE FIGURE is a portion of a plane sur- 
 face bounded on all sides by lines. 
 
 Many surfaces with which we are familiar, for example, the out- 
 side of our houses, the floor and ceiling of a room, the leaf of a 
 book, and, in many cases, gardens, fields, parks, etc., furnish in- 
 
74 GEOMETRY FOR BEGINNERS. [ 60. 
 
 stances of plane figures bounded by straight lines. Such figures 
 are called Polygons. 
 
 Definition II. A POLYGON is a plane figure bounded by 
 straight lines. 
 
 The bounding lines are called the SIDES of the polygon, and 
 their sum is called the PERIMETER of the polygon. 
 
 Exercises. 1. In Fig, 56 which figures are polygons and which are 
 not polygons? 
 
 2. How many sides and how many angles has each polygon in Fig. 56 ? 
 
 3. Draw a polygon of five sides ; six sides ; eight sides ; ten sides. How 
 many angles has each? 
 
 60. Polygons receive different names according to the num- 
 ber of their sides. 
 
 At least three straight lines are required to enclose completely a 
 part of a plane surface : the two sides of an angle are not sufficient. 
 Therefore a polygon cannot have less than three sides. 
 
 Definition. A polygon of three sides is called a TRIANGLE 
 
 (Fig- 57)- 
 
 A triangle is usually denoted by three letters standing at the 
 vertices of its angles. Instead of writing the word triangle, the 
 sign A is sometimes employed. Thus the triangle in Fig. 57 is 
 called the triangle A B C, or A A B C. 
 
 Every triangle has six PARTS, three 
 sides, AB, AC, B C (Fig. 57), and 
 three angles, A, B, and C. 
 
 Each side, as A B, has two adja- 
 cent angles, A and B, and one oppo- 
 site angle, C ; each angle, as A, is included between two sides, 
 A B and A C, and is opposite to the third side, B C.* 
 
 Exercises. 1. What angles are adjacent to the side A C (Fig. 57) ? to 
 the side B C ? What angle is opposite to B C? 
 
 2. By what sides is the angle B included? the angle C? What side is 
 opposite to B? to C? 
 
6i.] 
 
 CHAPTER IV. -^TRIANGLES. 
 
 61. If we prolong one side of a triangle, the prolongation 
 makes with the adjoining side an an- 
 gle called an EXTERIOR ANGLE of the 
 triangle, and the three angles of the 
 triangle are termed, by way of dis- 
 tinction, INTERIOR ANGLES. 
 
 In Pig. j8 m is an exterior angle, Fi * 5 8 - 
 
 b is the adjacent interior angle ; a and c are the opposite interior 
 angles. 
 
 Exercises. 1. Draw a triangle, and then prolong each side in both direc- 
 tions. How many exterior angles are formed? For each exterior angle point 
 out the adjacent interior angle, and also the opposite interior angles. 
 
 2. If an exterior angle of a triangle is 75, find the adjacent interior angle. 
 
 62. Theorem. The sum of two sides of a triangle is al- 
 ways greater than the third side. 
 
 For the straight line AB (Pig. 57) is the shortest line from 
 AtoB ( 32), therefore less than the broken line A C B ; that is, 
 A C + CB> AB. For the same reason it follows that A C + 
 AB>CB, and that A B + B C> A C. 
 
 Triangles are divided into three classes, according to the lengths 
 of their sides : the EQUILATERAL (Pig. 59, I.), in which the three 
 
 III. 
 
 Fig- 59- 
 
 sides are equal, the ISOSCELES (Pig. 59, II.), in which two of the 
 sides are equal, and the SCALENE (Pig. 59, IH.)> in which the sides 
 are all unequal. 
 
GEOMETRY FOR BEGINNERS. 
 
 [63. 
 
 Exercises. 1. Can a triangle have for its sides. 2 m , 3, and 6 m ? 8 m , 4, 
 and 2 m ? i m , 2 m , and 3? 3, 5, and y m ? 
 
 2. If the perimeter of a triangle is 2 m , what is the greatest value which one 
 side can have? 
 
 3. Draw, free-hand (#), an equilateral; (), an isosceles ; (c), a scalene 
 triangle. 
 
 4. The perimeter of an equilateral triangle is 1.5; find each side. 
 
 5 Can you draw accurately with the dividers an equilateral triangle of 
 given side, say 6o cm ? 
 
 63. We may regard a triangle as resting on one of its sides 
 as a base ; this side is then called the BASE of the triangle ; the 
 vertex of the angle opposite to the base is called the VERTEX of 
 the triangle ; and the perpendicular line drawn from the vertex to 
 the base (or base prolonged) is called the ALTITUDE of the tri- 
 angle. If the triangle has a horizontal side, this is usually taken 
 for the base. 
 
 Thus in the triangle ABC (Fig. 60) we would naturally take 
 
 the side A B for the base ; 
 then C is the vertex, and CD 
 is the altitude. If A C is 
 taken as the base, B is the 
 vertex, and B E the altitude. 
 If B C is taken as the base, 
 A is the vertex, and A F the 
 altitude. 
 
 D 
 
 N 
 
 M 
 Fig. 60. 
 
 Notice that the three altitudes intersect in one point. 
 
 The altitude may be outside the triangle ; this is the case in 
 the triangle MNO, in which MJVis the base, and OP the altitude. 
 The altitude must fall outside if the triangle has an obtuse angle, 
 and one of the sides of this angle is taken as the base. 
 
 In an isosceles triangle, the side unequal to the other sides is 
 always taken as the base, and the equal sides are usually referred 
 to as the sides of the triangle. 
 
64.] 
 
 CHAPTER IV. TRIANGLES. 
 
 77 
 
 Exercises. 1. Draw a triangle having all its angles acute, and then 
 draw the three altitudes. 
 
 2. Draw a triangle having an obtuse angle, and then draw the three altiv 
 tudes. How many of the altitudes lie outside the triangle? 
 
 3. Draw an isosceles triangle and its altitude. 
 
 4. On a given line as base, how many isosceles triangles can be con. 
 structed? How many equilateral triangles? 
 
 II. Angles of a, Triangle. 
 
 64. In order to find the sum of the angles of a triangle ABC 
 (Fig. 61), let us bring them all to a common vertex. For this 
 purpose, suppose a line D E drawn through the vertex C of the tri- 
 angle parallel to the base A B : we thereby make the angles ;;/ 
 and ;z. The angles m and a are equal because they are alternate 
 angles ( 56), and the angles n and b are equal for the same 
 
 Fig. 61. 
 
 reason. Hence, a + b-\-c=m-\-n + c (Axiom II.). But 
 m -f- n -h c = 1 80, or a straight angle. Therefore, a + b + c = 
 1 80 (Axiom I.). 
 
 This proof applies equally well to any triangle ; therefore, to all 
 triangles ; hence : 
 
 Theorem. The sum of the angles of a triangle is eqital to 
 two right angles, or 180. 
 
 Illustration. The truth of this theorem may be shown to 
 the eye as follows : Cut out of paper a triangle, ABC (Fig. 62), 
 and draw its altitude CD. Then fold over the corners on the 
 
78 GEOMETRY FOR BEGINNERS. [65. 
 
 dotted lines as edges. This will bring the vertices of the three 
 angles to the same point D, and we shall see that the three angles 
 together just make a straight angle. 
 
 In the same way we might test one triangle after another, find- 
 ing in each case that the sum of the angles was 180 ; after a time 
 we should become convinced that all triangles were subject to this 
 law. 
 
 But in the proof given above a single case is sufficient ; for a 
 little reflection shows that the same reasoning holds good for all 
 cases. 
 
 65. From the preceding important theorem several corolla- 
 ries follow. 
 
 Corollaries. I. The sum of two angles of a triangle must 
 always be less than 180. 
 
 2. A triangle can have only one right, or one obtuse, angle. 
 
 In other words, two angles of a triangle must be acute ; the 
 third may be either acute, right, or obtuse. If it is acute, the tri- 
 angle is called an ACUTE TRIANGLE (Fig. 6j, I.) ; if it is right, the 
 
 I. II. 
 
 Fig. 63. 
 
 triangle is called a RIGHT TRIANGLE (Fig. 63, II.) ; if it is obtuse, 
 the triangle is called an OBTUSE TRIANGLE (Fig. 63, III.). 
 
 In a right triangle the side opposite to the right angle is called 
 the HYPOTENUSE, and the other sides are called the LEGS. 
 
 3. If two angles of a triangle are known, the third angle will 
 be found by subtracting the sum of the two known angles from 
 1 80. 
 
66.] CHAPTER IV. TRIANGLES. 79 
 
 4. If two angles of one triangle are equal respectively to two 
 angles of another triangle, then the third angle of the one is equal 
 to the third angle of the other. 
 
 5. In a right triangle the sum of the acute angles is equal to a 
 right angle or 90. 
 
 Hence, if one of the acute angles in a right triangle is known, 
 the other can be found by subtracting the first from 90. 
 
 Definition. If the sum of two angles is 90, each is called 
 the COMPLEMENT of the other. 
 
 The acute angles of a right triangle are always complements of 
 each other. 
 
 Exercises. 1. Can a triangle have one right aw^/one obtuse angle? 
 
 2. If one leg of a right triangle is taken as base, what is the altitude? 
 
 3. Two angles of a triangle are, 
 
 O) 37 and 71; (V) 45 32' 18" and 62 50' 57"; 
 
 () 82 and 48; O) 64 47' 33" and 77 18' 41"; 
 
 (<) 40 28' and 18 57'; (/) 179 o' 54" and o 59' 5"; 
 
 find the third angle in each case. 
 
 4. If one acute angle of a right triangle is (a) 30; () 45; (c] 63; 
 (</) 27 15'; 0) 58 12' 48", find in each case the other acute angle. 
 
 5. What is the complement of 30? 45? 60 ? 89? 90? 
 
 66. If we subtract the angle b (Fig. 64) from 180, the re- 
 mainder is equal to a + c ( 64). 
 But this remainder is also equal to the 
 exterior angle, m; why ( 53)? 
 Therefore, m = a -f- c (Axiom I.). 
 That is, 
 
 Theorem. An exterior angle of Fi g- &* 
 
 a triangle is equal to the sum of the two opposite interior angles. 
 
 Exercises. 1. Find the exterior angle of a triangle if the opposite inte- 
 rior angles are 38 35' and 69 46'. 
 
 2. An exterior angle of a triangle is 60, and one of the opposite interior 
 angles is 30; find the other. 
 
 3. An exterior angle of a triangle is 90. What kind of a triangle is it? 
 
 4. Prove that the sum of the six exterior angles of a triangle is 720. 
 
80 GEOMETRY FOR BEGINNERS. [ 67. 
 
 III. Similarity, Equivalence, and Equality. 
 
 67. We may compare two triangles with respect either to their 
 shape (form) or to their size {magnitude}. 
 
 There are four cases. Either (a) , the triangles have the same 
 shape; or (), they have the same size; or (c), they agree in 
 both shape and size ; or (d) , they differ in both shapeand size. 
 For example : in Fig. 65 the triangles ABC and P O Q have 
 the same shape ; ABC and D E F 
 have the same size ; A B C and M ' N O 
 agree in both shape and size; DEF 
 and P O Q differ in both shape and 
 size. 
 
 If two triangles have the same shape, 
 they are called SIMILAR TRIANGLES ; if 
 they have the same size, they are called 
 EQUIVALENT TRIANGLES ; if they agree 
 both in shape and size that is, if they 
 are both similar and equivalent they are called EQUAL TRI- 
 ANGLES. 
 
 Exercises. 1. To which class of triangles do P O Q and M N be- 
 long? M N and D E F? 
 
 2. Draw (free-hand) two triangles of each kind. 
 
 68. What has just been said about triangles may be extended 
 to space magnitudes in general. 
 
 Two straight lines always have the same shape whatever be their 
 lengths ; so have two circles, two cubes, or two spheres, however 
 much they differ in size. v 
 
 Again, a curved line may have the same length as a straight 
 line ; a field bounded by curved lines may enclose the same extent 
 of surface as a field bounded by straight lines ; a cubical vessel 
 may hold the same quantity of water as a vessel shaped like a cyl- 
 
69.] CHAPTER IV. TRIANGLES. 81 
 
 cylinder or a sphere. In all these cases the size is the same, the 
 shape different. 
 
 Lastly, two magnitudes, whether lines, surfaces, or solids, may 
 have at once the same size and the same shape. 
 
 Definitions. I. Two magnitudes which have the same shape 
 are called SIMILAR MAGNITUDES. 
 
 II. Two magnitudes which have the same size are called 
 EQUIVALENT MAGNITUDES. 
 
 III. Two magnitudes which agree both in shape and in size are 
 called EQUAL MAGNITUDES. 
 
 Between two similar magnitudes we place the sign ^ ; between 
 two equivalent magnitudes the sign = ; between two equal mag- 
 nitudes both signs ^ ; or, if the shape is not taken into account, 
 simply the sign =. 
 
 Of these three kinds of agreement between space magnitudes, 
 that of equality is the simplest, and ought therefore to be studied 
 first. 
 
 IV. Equal Triangles. 
 
 69. Since two equal triangles agree both in size and in shape, 
 they can differ from each other only in their position in space, and 
 must, when one is laid on the other, coincide in all their six parts ; 
 in other words, their three sides and three angles must be equal 
 each to each. 
 
 Compare, for example, the equal triangles ABC and M ' N O 
 
 Of the sides, A B = MN; A C = MO; B C = N O. 
 
 Of the angles, C = O; B = N; A = M. 
 
 In two equal triangles eqtial sides are opposite to equal angles, 
 and equal angles are opposite to equal sides. 
 
 In any two triangles by corresponding sides (angles) are meant 
 those which are opposite to equal angles (sides). 
 
82 GEOMETRY FOR BEGINNERS. [ 'JO. 
 
 70. There would be nothing more to say about equal triangles 
 were it not for the fact that, when in two triangles a certain number 
 of the six parts are equal each to each, the remaining parts must 
 also be equal, and the triangles must be equal. This fact is one 
 of the corner-stones in the science of Geometry ; upon it are based 
 the proofs of many important theorems, and it is likewise a fact 
 of great practical value in the various uses to which Geometry can 
 be applied. (See Part VI. of the present chapter.) 
 
 Let us therefore proceed to find, once for all, how many and 
 what parts in two triangles must be equal in order to make the tri- 
 angles equal ; in other words, how many and what parts of a tri- 
 angle must be given in order to determine the size and shape of 
 the triangle. 
 
 1. If only one part, a side or an angle, is given, we can con- 
 struct as many different triangles as we please, all having this part. 
 Therefore, by one part the size and s-hape of a triangle is not 
 determined. 
 
 2. Likewise, if two parts are given, whether two sides, two angles, 
 a side and an adjacent angle, or a side and the opposite angle, it 
 is easy to show that an indefinite number of triangles may be 
 made, all having these two parts and yet differing in their other 
 parts (see Exercises 3-6 below) . Hence two parts are not suffi- 
 cient to determine the size and shape of a triangle. 
 
 3. If three parts are given, they may be, 
 (/.) The three angles ; 
 
 ('.) One side and two angles ; 
 
 (tit.) Two sides and the included angle ; 
 
 (/.) Two sides and an angle opposite to one of them ; 
 
 (v.) The three sides. 
 
 These cases require, each of them, a special investigation. 
 
 Exercises. 1. Draw three triangles having a common side. 
 
 2* Draw three triangles having a common angle. 
 
 3. Draw two triangles, both having for two of their sides 4O cm and 70. 
 
' 7'.] 
 
 CHAPTER IV. TRIANGLES. 
 
 83 
 
 4. Draw two triangles, both having for one side 50, and for an adjacent 
 angle 30. 
 
 5. Draw three triangles, all having for one side 45 cm , and for the opposite 
 angle 60. 
 
 Solution. Make an angle of 60. Then take 45 between the dividers, place 
 one point on one side of the angle, the other point on the other side. Join these 
 two points and the triangle is constructed. The other triangles are made in the 
 same way, taking care to use different points on the sides of the angle. 
 
 6. Draw two triangles, both having the angles 30 and 60. What kind 
 of triangles are they ? 
 
 71. Is a triangle determined if we know its three angles? 
 Compare the triangles ABC and DEF 
 (Fig. 66} ; as regards their angles, 
 A = D,B = E, C = F ( 58) ; that 
 is, the triangles have their angles equal 
 each to each. Now the triangles them- 
 selves are not equal ; they have the 
 same shape, but they differ in size. A B 
 
 In fact we may make as many tri- F - 66 
 
 angles as we please, all having the same 
 shape, but differing in size ; therefore, 
 
 By the three angles a triangle is not determined. 
 
 Two triangles which have their angles equal <each to each are 
 said to be mutually equiangular. 
 
 Exercises. 1. Draw two mutually equiangular triangles. 
 
 2, Can you draw two mutually equiangular triangles which differ in shape? 
 
 72. Problem. To construct a triangle, having given a side 
 and two angles. 
 
 When two angles are given, the third can always be found 
 (how?). We shall therefore suppose that the given angles are 
 both adjacent to the side. 
 
 Let a (Fig. 6?) be the given side, and 60 and 40 the given 
 
84 GEOMETRY FOR BEGINNERS. [ 7 2 
 
 angles. Draw a line A B a. At the points A and B make 
 angles equal respectively to 60 and 40. The sides of these 
 angles will meet at a point, C, and A B C is the triangle required. 
 Why cannot these sides be parallel, and hence never meet? 
 
 fif.67. a M 
 
 If we now construct another triangle, MN O, having the same 
 three parts, it will be equal to the triangle ABC. 
 
 Proof. Place MN O on A B C, so that MN falls wAB. M 
 will fall on A, and N on B t since MN = A B = a. The lines 
 MO and A C will coincide in direction, since the angles at J/and 
 A are equal (each 60), and the lines N O and B Cwill also coin- 
 cide in direction, since the angles at N and B are equal (each 40). 
 Therefore O must coincide with C, and the triangles will coincide 
 in all their parts. Point out the parts which are equal each to each. 
 
 The construction and proof would be the same for other values 
 of the given sides^and angles ; therefore, by a side and two angles, 
 a triangle is completely determined ; and also, 
 
 Theorem (I. Law of Equality). If in two triangles a side 
 and two angles are equal each to each, the triangles are equal. 
 
 Remark. The method used above to prove the two triangles 
 equal is called the Method of Superposition. When it can be em- 
 ployed to prove the equality of two magnitudes, it is a very easy 
 method, if we only take care to bear in mind what parts in the 
 two magnitudes are known to be equal. 
 
 Exercises. 1. Draw to reduced scale (on paper, I : 2000; on the black- 
 board, 1 : 200) a triangle, one side of which is 220'", and the adjacent angles 
 70 and 50. 
 
73-] CHAPTER IV. TRIANGLES. 85 
 
 2. Draw to reduced scale the outlines of a triangular field of which one 
 side is 6oo m , and the adjacent angles 80 and 47. 
 
 NOTE. Always write the scale employed by the side of the figure. 
 
 3o The same Exercise, one side of the field being 40, an adjacent angle 
 being 65, and the opposite angle being 37. 
 
 4. Draw a right triangle having given, 
 
 (2.) One leg = 8 dm , and the adjacent acute angle = 57. 
 
 (?'/.) One leg = 6o cm , and the opposite angle = 25. 
 
 (z'zY.) The hypotenuse = 90, and an adjacent angle = 42. 
 
 5. What limit is there to the sum of the two given angles in the problem 
 of this section in order that the solution may be possible ? 
 
 6. What are the given things or data 1 in the problem of this section ? 
 
 73. Problem. To construct a triangle having given two 
 sides and the included angle. 
 
 Let a and b (Fig. 68} be the given sides, and 50 the included angle. 
 
 Make an angle A = 50, take A B = a, A C = b, and join BC. 
 A B C is the triangle required. 
 Draw another triangle having 
 the three given parts, and then 
 prove by superposition that it 
 must be equal to A B C. 
 
 Hence, by two sides and the 
 included angle a triangle is com- 
 pletely determined ; and, 
 
 Theorem (II. Law of Equality). If in two triangles two 
 sides and the included angle are equal each to each, the triangles 
 are equal. 
 
 Exercises. 1. Draw (to reduced scale) a triangle with the sides 49 and 
 77 m , and the included angle 45. 
 
 2. Represent (to scale) on paper, and also on the blackboard, a triangle, 
 two sides of which are 140 and ioo m , and the included angle 55. Are the 
 two figures equal, equivalent, or similar? Are they mutually equiangular? 
 
 1 The singular of data is datum. The word is from the Latin, and means 
 given. 
 
86 GEOMETRY FOR BEGINNERS. [ 74. 
 
 3. In an isosceles triangle one of the equal sides is 32 cm , and the angL 
 at the vertex is 82; draw the triangle. 
 
 4. Draw a right triangle whose legs are i6 cm and 2O cm . What three 
 parts are here known? 
 
 5. Draw an isosceles right triangle whose legs are i6 cm each. What are 
 the three parts here given? 
 
 6. What is the greatest value which the given angle in the problem of 
 this section can have? 
 
 7 What are the data in the problem of this section? 
 
 74. Problem. To construct a triangle having given two 
 sides and an angle opposite to one of them. 
 
 There are two cases according as the given angle is opposite 
 (/'.) to the greater, or (it.) to the less side. 
 
 Fig. 69. 
 
 Case (/.). Let a and b (Fig. 69) be the given sides, and let 
 a > b ; also let the angle opposite to a be 70. Make an angle 
 CAB = 70, and lay off A C b \ this determines two corners, 
 A and C, of the triangle. The third corner must lie in the line 
 A B, and its distance from C must be equal to a ; therefore, it 
 must also lie in a circumference described from C as centre with 
 a radius equal to a. Hence it must be the intersection of this 
 circumference with the line AB. Now this circumference cuts 
 A B in two points, B and B' 9 so that we obtain two triangles, 
 A B C and A B' C. Of these, however, only A B C has the three 
 given parts; A B' C has, it is true, the given sides, but not the 
 given angle ; therefore, it is not the triangle required. 
 
8 74'] CHAPTER IV. TRIANGLES. 87 
 
 If we construct another triangle, M ' N O, having the same three 
 jarts, it will be equal to the triangle ABC. 
 
 Proof. Place A MN O on A A B C, making the equal sides 
 M O and A C coincide. MN will coincide in direction with A B 
 because OMN= CAB = 70 ; therefore N will fall in AB. But 
 N must also lie in a circumference having (now) C for centre, and 
 radius = a ; therefore N will coincide with B. And ON will 
 coincide with CB, and the triangles will coincide in all their parts. 
 Hence we see that by two sides and the angle opposite to the 
 greater side a triangle is completely determined ; and that, 
 
 Theorem (III. Law of Equality). If in two triangles 
 two sides and the angle opposite to the greater side are equal each 
 to each, the triangles are equal. 
 
 Case (//.). Let a and b (Fig. 70) be the given sides, and let 
 a < b ; also let the angle op- 
 posite to a = 42. 
 
 By proceeding as in Case 
 (/.) we obtain two triangles, 
 AB C and A B'C, both hav- 
 ing the three given parts, yet 
 
 differing in size and in shape. A '\ .-~''B 
 
 Therefore, by two sides and Fi - 7> 
 
 the angle opposite to the less side, a triangle is not determined. 
 
 There are, however, only two solutions, the triangles ABC and 
 AB'C. 
 
 . Exercises. 1. In the triangles A BC and AJB'C {Fig. 69) point out the 
 equal parts ; also the unequal parts. 
 
 2. What relation exists between the angles B AC and &A C (Fig. 69) ? 
 
 3. What kind of triangles are the triangles BC '' and N ON 1 (Fig. 69) ? 
 
 4. What equal parts have the triangles AC B and ACB 1 (Fig. 70) ? 
 
 5. Draw a triangle with the sides 6o cm and 9O cm , and the angle opposite 
 the greater side 76. 
 
 6. Construct a right triangle the hypotenuse of which is 8 dm , and one leg 
 is 6^. What are the three given parts ? 
 
GEOMETRY FOR BEGINNERS. 
 
 [75- 
 
 7. Construct a triangle with the sides 8 dm and 4 dm , and the angle opposite 
 to the greater side 80. 
 
 8. When are the data in Case ('.) of this section such that no triangle 
 is possible? 
 
 Ans. When the less side is shorter than the perpendicular from C to A B 
 (Fig. 70) ; also when the given angle is either right or obtuse. 
 
 9. When the less side is just equal to the perpendicular from C to A B, 
 what kind of a triangle is obtained? 
 
 75. Problem. To construct a triangle having given its 
 three sides. 
 
 Let a b c (fig. 71) be the given sides. Make A B a ; then A 
 and B are two corners of the triangle, and A B is one side. Since 
 the third corner must be at 
 the distance b from A, it must 
 lie in a circumference described 
 from A as centre with the ra- 
 dius b. For a like reason it 
 must also lie in a circumfer- 
 ence described from B as cen- 
 tre with the radius c. 1 
 
 Therefore it must be at the 
 intersection of these circum- 
 ferences. Now these circum- 
 ferences cut each other in two points, C and C', one above the 
 other below A B ; so that we obtain two triangles, ABC and 
 ABC', both having the three given sides. 
 
 These triangles, however, are equal. For, if we fold over ABC' 
 on A B, C' will be brought above A B, and, as it remains the inter- 
 section of circumferences having A and B for centres, and b and c 
 for radii, it will fall on C, and the triangles ABC and ABC 1 will 
 then coincide. 2 
 
 1 It is obvious that in solving this problem entire circumferences are not re- 
 quired ; short arcs near the points of intersection are sufficient. 
 
 2 See Exercise 5, page 92. 
 
 Fig. 71. 
 
76.] CHAPTER IV. TRIANGLES. 89 
 
 If we construct another triangle with the same three sides, it will 
 also be equal to ABC. For we can place it on A B C so that 
 one of its sides will coincide with AB, and then its third corner 
 must fall on C for the same reason that C' falls on C when ABC* 
 is folded over on A B. 
 
 From what precedes we conclude that, by the three sides a tri- 
 angle is completely determined ; also that, 
 
 Theorem (IV. Law of Equality). If in two triangles the 
 three sides are equal each to each, the triangles are equal. 
 
 Exercises. 1. The sides of a triangular garden measure 8o m , 65, and 
 45 m . Draw to scale a plan of the gai'den. 
 
 2. The sides of a triangular field are 2Oo m , 240, and 300. Draw to 
 scale a plan of the field. 
 
 3. Construct a triangle whose sides are 2O cm , 4O cm , and 6o cm . 
 
 4. Construct a triangle whose sides are 2O cm , 3O cm , and 6o cm . 
 
 5. When is the solution of the problem of this section impossible? (See 
 62.) 
 
 6. Construct an isosceles triangle whose base is 7O cm , and the sum of 
 whose other sides is 2 m . 
 
 7. Construct an equilateral triangle one side of which is i m . 
 
 8. Construct a triangle with the sides 3O cm , 3O cm , and 3O CU1 . What kind 
 of a triangle is it? 
 
 9. Construct a triangle with the sides 25 cm , 25 cm , and 40'". What kind 
 of a triangle is it? 
 
 10. Construct a triangle with the sides 2O cm , 3O cm , and 4.o cm . What kind 
 of a triangle is it? 
 
 11. Are two triangles equal if they have equal perimeters ? (Compare the 
 last three exercises.) 
 
 76. Problem. To make a triangle equal to a given triangle. 
 
 Take any three parts of the given triangle that completely de- 
 termine a triangle, and construct with them a new triangle ; it will 
 be equal to the given triangle. The best parts to take for this pur- 
 pose are the three sides. Show hew to construct the triangle with 
 them ( 75). 
 
90 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 77. 
 
 Exercises. 1. In what four ways can three parts be chosen with which 
 to construct a triangle? 
 
 2. Draw a triangle; then construct another equal to it. 
 
 Fig. 72. 
 
 77. Problem. At a given point D (Fig. 72) of a given 
 line D G, to make an angle equal to a given angle BAG. 
 
 Take on the sides of the given angle A M 
 = AN, and join MN. Then construct (by 
 the last problem) a triangle D E F equal to 
 the triangle AMN. The angle D, corres- 
 ponding to the angle A, will be equal to it. 
 
 Exercises. 1. What kind of a triangle is^ M N~! 
 Is it necessary for the solution of the problem to 
 make A MN a. triangle of this kind? 
 
 2. Make an angle, and then make another equal 
 to it. 
 
 3. Make an angle equal to a given angle with 
 j the protractor. 
 
 4 Make an angle equal to a given angle with the ruler and the square. 
 5. Make (with ruler and compasses) an angle of 60; an angle of 120. 
 
 78. Problem. Through a given point C (Fig. 73) to draw 
 a line parallel to a given line A B. 
 
 Through C draw any straight line CD, cutting AB in D. Then 
 
 draw (by the last problem) C P so 
 
 that the angle DCP = CDB. 
 
 CP is parallel to AB ( 57). 
 
 Exercises. 1. Draw a line, and then 
 \ through a point draw another line par- 
 
 \ allel to the first. 
 
 2. The same exercise, using ruler and 
 square instead of ruler and compasses. 
 3. Through the vertices of the angles of a triangle draw lines parallel to 
 the opposite sides, and prolong them until they meet. What kind of a figure 
 is thus formed? 
 
 / D 
 
 73- 
 
79-] CHAPTER IV. TRIANGLES. $1 
 
 79. A PROBLEM is a construction to be made according to 
 geometrical laws. 
 
 There is a general agreement that in the Science of Geometry 
 no instrument shall be employed in making a construction except 
 ruler and compasses ; if the problem cannot be solved by these 
 means it is not regarded as belonging to the science. In the Art 
 of Geometry that is, in the applications of Geometry to prac- 
 tical purposes, such as Drawing and Land-measuring various 
 other instruments which are found convenient are used, as divided 
 rules, protractors, theodolites, etc. 
 
 Thus, in practice, the problem of 77 is usually solved with the 
 protractor, or with the ruler and the square, and the problem of 
 78 with the ruler and the square. 
 
 In making a construction, all lines which are merely interme- 
 diate steps between the data and what is required should be drawn 
 dotted instead of full (see Figs. 72 and fj) . Such lines are called 
 auxiliary lines. 
 
 V. Some Consequences of the Equality of Triangles. 
 
 80. Let A B C (Fig. 74) be an isosceles triangle having A C 
 = B C, and let CD be a line 
 bisecting the angle C. This line 
 divides the triangle ABC into two 
 triangles, A C D and B C D. Com- 
 pare these two triangles ; they have 
 a common side CD, the side A C 
 = BC (why ?), and the angle c=d 
 (why?). Therefore the triangles 
 are equal (II. Law of Equality), 
 and among the equal parts we have a = b ; hence, 
 
 Theorem. /;/ an isosceles triangle the angles opposite to the 
 equal sides are equal. 
 
92 GEOMETRY FOR BEGINNERS. [ 80. 
 
 Corollary. In an equilateral triangle all the angles are equal ; 
 therefore each angle = 1 80 : 3 = 60. 
 
 Why does this corollary follow from the theorem? 
 
 Remark. A theorem consists of three parts : 
 
 (/.) The HYPOTHESIS (or hypotheses), or that which is given, or 
 assumed as true, at the start. 
 
 (ii.) The CONCLUSION (or conclusions), or that which is to be 
 proved. 
 
 (Hi.) The PROOF. 
 
 In the above theorem the hypothesis is : A C B C (Fig. 74) , 
 and the conclusion is : a b. The proof consists in showing that 
 the triangles A C D and BCD are equal, and hence inferring that 
 a = b. 
 
 Exercises. 1. In an isosceles triangle the bisector of the angle at the 
 vertex is perpendicular to the base, and bisects the base. 
 
 Hints. Use Fig. 74. The two hypotheses are: A C = B C, and CD bisects 
 A CB. The two conclusions are : C> A B, and CD bisects A B. Proof: show 
 that &ACD^ /\BCD\ hence ;w = = 90 (why ?), and AD BD. 
 
 2. The perpendicular let fall from the vertex of an isosceles triangle to 
 the base bisects the base, and also the angle at the vertex. 
 
 Hints. Here the hypotheses are : A C = B C, and CD _L A B. The conclu- 
 sions are : CD bisects A B, and CD bisects A C B. Prove by showing that A A CD 
 ^ A B CD. 
 
 3. The line joining the vertex of an isosceles triangle to the middle of the 
 base is perpendicular to the base, and bisects the angle at the vertex. 
 
 Hints. In this case what are the hypotheses ? the conclusions ? Prove by 
 showing (by IV. Law of Equality) that A A CD ^l\B CD. 
 
 4. What three conditions does the line CD {Fig. 74) fulfil? 
 
 5* Prove (with the aid of the theorem of this section) that the triangles 
 A B C and A B C' {Fig. 77, p. 88) are equal. 
 
 Hint. By drawing C C 1 we obtain two isosceles triangles, A C C' and B C C'. 
 
 6. Find the base angles of an isosceles triangle if the angle at the vertex 
 is (*.) 27 35', (it.} 75 18', (Hi.} 124 40'. 
 
 7. Find the angle at the vertex if one of the base angles is (i.) 15 I4 ; > 
 (ii.} 48 7 ', (Hi.} 83 4'. 
 
 8. In an isosceles right triangle find the base angles. 
 
3 1.] CHAPTER IV. - TRIANGLES. 93 
 
 9, Can you devise a way of finding the height of an object (tree, tower, 
 etc.) by means of an isosceles right triangle? 
 
 10 Show that in an isosceles triangle the exterior angles made by prolong- 
 ing the base are equal, and that the exterior angles at the vertex are also 
 equal. 
 
 11. Find the three angles of an isosceles triangle if the exterior angle at 
 the vertex is 146 19'. 
 
 12. Find the three angles of an isosceles triangle if the exterior angle at 
 the base is 120 50'. 
 
 13. Show that the exterior angles of an equilateral triangle are all equal, 
 and find the value of one of them. 
 
 14. Construct an isosceles triangle having given, 
 
 (a) The base and an adjacent angle ; 
 (<) The base and the opposite angle ; 
 (<r) The side and an angle at the base ; 
 (d} The side and the angle at the vertex. 
 
 81. Let us now transpose the hypothesis and the conclusion 
 of the theorem in 80 ; in other words, let us suppose that in a 
 triangle ABC (Fig. 75) the angles A and B are equal, and 
 inquire if the sides opposite to these angles must also be equal. 
 
 Let CD bisect the angle C. 
 Then the triangles A CD and 
 BCD are mutually equiangular 
 (why ?) . They also have the side 
 CD common, therefore they are 
 equal (I. Law of Equality), and 
 A C = B C. Hence, 
 
 Theorem. If in a triangle 
 two angles are equal, the opposite 
 sides are also equal, and the triangle is isosceles. 
 
 Corollary. If the three angles of a triangle are equal, the 
 triangle is equilateral. 
 
 Why does this corollary follow from the theorem ? 
 
 Remark. If we transpose the hypothesis and the conclusion 
 
94 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 82. 
 
 of a theorem, we obtain a new theorem called the CONVERSE of the 
 original theorem. The theorem of this section, for example, is the 
 converse of the theorem of the last section. 
 
 The converse of a true theorem is not necessarily true. Take, 
 for instance, the theorem, two vertical angles are always equal ; the 
 converse, two equal angles are always vertical angles, is not true. 
 
 Exercises. 1. In an isosceles triangle the bisectors of the base angles 
 form with the base another isosceles triangle. 
 
 2 Prove the converse theorem (namely, if in a triangle the bisectors of 
 the base angles form with the base an isosceles triangle, then the triangle is 
 also isosceles). 
 
 82. We will now compare in a triangle two unequal sides and 
 the opposite angles. 
 
 Let ABC {Fig. 76) be an isosceles triangle having A B = A C 
 and A B C = A C B. Prolong A C to any point D, and join B D. 
 
 Then A B D is a triangle having A D 
 >AB. Now compare the opposite 
 angles. Since ABD = ABC + 
 C BD, therefore ABD > ABC. 
 And since A DB = ACB-CBD 
 ( 66), therefore ADB < A CB, 
 andhence<^f^C. Therefore.^ Z> 
 > A D B. Here it is quite clear that 
 the two conditions, A D > A B, and 
 A B D > A D B, must always be ful- 
 filled together ; hence, 
 
 Theorem. I. Of two angles of a triangle the greater is oppo- 
 site to the greater side. 
 
 Theorem II. (Converse of Theorem I.). Of ftvo sides of a 
 triangle the greater is opposite to the greater angle. 
 
 Corollary. In a right triangle the hypothenuse is the greatest 
 side. 
 
 Fig. 76. 
 
CHAPTER IV. TRIANGLES. 
 
 95 
 
 Exercises, 1. From which theorem does the corollary follow? Why 
 does it follow? 
 
 2. What similar corollary follows for an obtuse triangle? 
 
 D F 
 
 Fig. 77, 
 
 83. Let CD JL AB (Fig. 77), and let C E, C F, C G, be 
 any other lines drawn from C to the 
 line A B. These lines are the hy- 
 pothenuses of the right triangles 
 CDF, CDF, CD G, respectively, 
 and therefore are, each of them, 
 greater than CD ( 82, Corollary) ; 
 therefore, 
 
 Theorem I. Of all lines which - 
 
 A 
 
 can be drawn from a point to a 
 straight line the perpendicular is the 
 shortest. 
 
 Corollary. Hence, the distance from a point to a straight line 
 is the length of the perpendicular let fall from the point to the line. 
 
 If DE = DF, then A CDE ^ A CDF (why?), therefore 
 C E = CF; that is, 
 
 Theorem II. Two oblique lines equally distant from the foot 
 of the perpendicular are equal. 
 
 Again, D G > D F. Likewise the angles CFD and CGD 
 are each acute (why?). And CFG, being the supplement of 
 CFD, must be obtuse. Therefore, in the triangle CFG we 
 have CFG > CGF; hence C G > CF ( 82, Theorem II.). 
 That is, 
 
 Theorem III. Of two oblique lines unequally distant from 
 the foot of the perpendicular, the more remote is the greater. 
 
 Exercises. 1. In each of these three theorems what is the hypothesis? 
 What the conclusion? 
 
 2. State and prove the converse of Theorem II. 
 
96 
 
 GEOMETRY FOR BEGINNERS. 
 
 [8 4 . 
 
 84. Let AB C and ABD (Fig. 78) be two isosceles trian- 
 
 gles having a common base AB. 
 Join C D. The triangles A CD and 
 B CD are equal (IV. Law of Equali- 
 ty). Let the triangle A CD be 
 folded over CD until it coincides 
 with the triangle C B D ; then A 
 will fall on B, and all the lines and 
 angles which fall upon each other 
 will be equal. Hence we have, (*'.) 
 # = and ; = </; (ii.)AE = BE; 
 (tit-) m = n, and therefore CD _L 
 AB. 
 
 Theorem. If two isosceles triangles have a common base, 
 the line drawn through their vertices (i.) bisects the angles at the 
 vertices ; (ii.) bisects the base ; (iii.) is perpendicular to the base. 
 
 Exercises. 1. Prove this theorem, when (as in 
 Fig. 79} the vertices of the two isosceles triangles 
 are both on the same side of the base. 
 
 2. If any number of isosceles triangles have a com- 
 mon base, the line passing through the vertices of two 
 of the triangles will pass through the vertices of all 
 the triangles. 
 
 3. If (Fig. 79) CAE= 64 and D A E = 36, find 
 all the other angles in the figure. 
 
 
 79' 
 
 85. Problem. To bisect a given angle B A C (Fig. So) . 
 
 Analysts. The theorem of 84 suggests a mode of solving 
 this problem. If we construct an isosceles triangle having the 
 given angle B A C for the angle at the vertex, and then upon its 
 base construct any other isosceles triangle, the line which joins 
 the vertices of the two triangles must, by 84 (i.) , be the bisector 
 required, 
 
86.] 
 
 CHAPTER IV. TRIANGLES. 
 
 97 
 
 Construction. With A as centre describe an arc cutting the 
 sides of the given angle in M and N. With A 
 
 M and N as centres, and equal radii, describe 
 arcs intersecting at a point D. Join A D : 
 it is the bisector required. 
 
 Exercises. 1. Prove, directly, that A AMD 
 ^A AND (Fig. 80), and hence that B A D = 
 CAD. 
 
 2. Make an acute angle, and then bisect it. 
 
 3. Make an obtuse angle, and then bisect it. 
 
 4. Draw a triangle, and then bisect the three 
 angles. In how many points do the bisectors 
 meet one another? 
 
 5* Divide an angle into four, and also into eight, equal parts. 
 
 6. Make, with ruler and compasses, an angle of 30; an angle of 45. 
 
 86. Problem. To bisect a given straight line A B (Fig. 81) . 
 
 Analysis. If we construct upon A B as a base any two isos- 
 celes triangles, the line joining their 
 vertices must, by 84 (ii.), bisect ^V^ 
 
 AB. 
 
 Construction. With A and B as 
 centres and equal radii, describe arcs 
 intersecting above A B in C, and be- 
 low A B in D. Join C D : it bisects 
 A B in E. 
 
 Fig. 81. 
 
 Exercises. 1. Bisect a line, first freehand, then with ruler and coin- 
 passes. 
 
 2. Bisect the three sides of a triangle, and then join the points of bisec- 
 tion to the vertices of the opposite angles. In how many points do the join- 
 ing lines (called the medians of the triangle) intersect? 
 
 3. Divide a line into four, and also into eight, equal parts. 
 
 4. Can you, by means of this problem, divide a line into six equal parts? 
 
 5. Bisect a line more than twice as long as the greatest opening of the, 
 Compasses ? 
 
GEOMETRY FOR BEGINNERS. 
 
 [87- 
 
 E 
 
 Fig. 82. 
 
 87. From 84 it appears that every point in the perpendicu- 
 lar erected at the middle of a line is equally 
 distant from the ends of the line. Moreover, 
 every point not in this perpendicular is un- 
 equally distant from the ends of the line. 
 
 For let D (Fig. 82) be such a point, C E 
 being the perpendicular. Since A D A C 
 + CD = B C+ C D, and since also B C + 
 C D> BD (why?), therefore AD> BD. 
 Hence, 
 
 Theorem. Every point equidistant from the ends of a straight 
 line is in the perpendicular which bisects the line. 
 
 88. Problem. To erect a perpendicular at a given point 
 C (Fig. 83) of a given line A B. 
 
 Analysis. If we find in A B two points, M and N, equally dis- 
 tant from C, and then find any other point D equidistant from M 
 
 and N, it is evident from 87 that 
 the line passing through C and D 
 will be the perpendicular required. 
 Construction. With the centre 
 C describe an arc cutting A Bin M 
 and N. With M and N as centres 
 
 M u ~~N and equal radii, describe arcs inter- 
 
 secting at a point D. Join CD. 
 
 / 
 
 83. 
 
 Note. In practice this construction is usually effected by means of the ruler 
 and the square (see 49). 
 
 Exercises. 1. In the above construction what is the least value which 
 the equal lines M D and A 7 D can have? 
 
 2. Erect a perpendicular at the end of a line without prolonging the line. 
 
 Construction. Let A C (Fig. 84} be the line. Construct on A C as base 
 an equilateral triangle ABC (how is this done?), and prolong A B to D, 
 making B D = A B. CD is the perpendicular required. 
 
 p roo f. Since A ABC is equilateral, ABC= 60; therefore, in A 
 BDC=fo (66). Now A B CD is isosceles (why?); 
 
89.] CHAPTER IV. TRIANGLES. 99 
 
 hence B CD = B D C- 30. Therefore A CD ~ A C B + B CD = 60 
 + 30 = 90. 
 
 3. Construct an equilateral triangle having the f 
 
 altitude 4O cm . / 
 
 Hints. Draw a line 40 long, erect at one end / 
 
 a perpendicular, at the other end make two angles of 30 / 
 
 each. 
 
 4-. Construct an isosceles triangle having given X 
 
 (<7) the base and the altitude ; (b) a side and the 
 altitude. 
 
 5. Bisect the three sides of a triangle, and then 
 erect perpendiculars at the points of bisection. In 
 how many points do they intersect? A Fig. 84. 
 
 89. Problem. To let fall a perpendicular from a 
 point C (Fig. 85} to a given line A B. 
 
 The analysis and the construction 
 are the same as in the last problem, 
 the only difference being that in 
 this case the point C is not on the 
 line A B. Give the analysis and the 
 construction in full. 
 
 Note. This construction is also usually ****.. 
 
 effected in practice by means of the ruler 
 
 and the square ( 49). Fig. 85. 
 
 Exercises. 1. Choose a point on each side of a line, and let fall per- 
 pendiculars from each point to the line. 
 
 2. Draw a triangle, and from the vertex of each angle let fall a perpen- 
 dicular to the opposite side. In how many points do the perpendiculars 
 intersect? 
 
 3. Through a given point draw a parallel to a given line by constructing 
 two perpendiculars. 
 
 90. Theorem. Two parallel lines are everywhere equally 
 distant from each other. 
 
 Proof. From any two points A and B (Fig. 86) of one of the 
 
100 GEOMETRY FOR BEGINNERS. [ 91. 
 
 lines let fall perpendiculars meeting the other line in C and D ; then 
 A _ B A C II B D ( 57, Corollary) . Join 
 
 / B C. The triangles ABC and 
 
 CD are equal ( 56 and I. Law of 
 Equality ) . Therefore A C = B D ; 
 that is, any two points of one line 
 
 c 
 
 Fig. 86. are equally distant from the other 
 
 line : hence the lines are every- 
 where equally distant from each other. 
 
 91. Let the problem be proposed : to find a point at a given 
 distance from a given point. 
 
 It is evident that any point in the circumference of a circle, de- 
 scribed with the given point as centre and the given distance as 
 radius, will satisfy the required condition, and will, therefore, be a 
 solution of the problem. It appears, then, that there are an inde- 
 finite number of solutions to the problem; hence the problem is 
 termed indeterminate. 
 
 The required point is limited in position to a certain line ; this 
 line is called the locus of the point. 
 
 Definition. The line (or lines) in which a point must be, in 
 order to satisfy a given condition, is called the Locus * of the point. 
 
 Again, let it be proposed : to find a point equidistant from two 
 given points. 
 
 It follows from 87 that the required point may be anywhere 
 in the perpendicular which bisects the straight line joining the 
 given points. The problem, therefore, is indeterminate, and the 
 locus of the point required is the above-mentioned perpendicular. 
 
 Generally speaking, when a point has to satisfy only one geomet- 
 rical condition, the problem is indeterminate, and the solution con- 
 sists in finding the locus of the required point. 
 
 1 Plural loci. 
 
92.] CHAPTER IV. TRIANGLES. 10 1 
 
 Exercises. 1, In the cases above considered, the points were supposed 
 to be confined to one plane : what would the loci be if this restriction were 
 removed ? 
 
 Ans. In the first case, the surface of a sphere with the given point as centre and 
 the given distance as radius ; in the second case, the plane perpendicular to the 
 line which joins the given points, and bisecting it. 
 
 2. Find the locus of a point at a given distance from a given straight line. 
 
 3. Find the locus of a point at a given distance from a given circumference. 
 
 4. Find the locus of a point at a given distance from a given plane surface. 
 
 5. Find the locus of a point equidistant from two given parallel lines. 
 
 6. Find the locus of a point equidistant from two given intersecting lines. 
 Hints. The locus consists of the bisectors of the angles made by the lines. 
 
 To prove this, let Pbe any point on either of the bisectors, P M and PN perpen- 
 diculars from P to the two lines, O the point where the lines meet; then show 
 that &POM^/\PON, and therefore that PMPN. Why is it that, although 
 there are four angles to be bisected, the locus will consist of only two straight lines ? 
 
 92. When a point has to be found which satisfies two condi- 
 tions, the problem (if possible at all) generally has one or two 
 solutions ; in other words, the problem is determinate. 
 
 Such problems are solved by constructing the loci of points 
 which satisfy each condition separately ; then the point, or points, 
 where the loci intersect, being common to both loci, will satisfy 
 both conditions, and will be the solution of the problem. 
 
 For example : find a point which shall be at given distances 
 from two given points. 
 
 The required point must be in a circumference having the first 
 given point for centre and the first given distance for radius ; and 
 it must also be in a circumference having the second given point 
 for centre and the second given distance for radius ; hence it must 
 be at the intersection of these two circumferences. 1 
 
 Since, in general, these circumferences intersect in two points, 
 there will be in general two solutions of the problem. When will 
 there be only one solution ? When will there be no solution at all ? 
 
 In the following exercises, after having given the solution, state 
 under what conditions (if any) the problem is impossible. 
 
 1 This reasoning has been already employed in \ 75. 
 
102 GEOMETRY FOR BEGINNERS. [ 93. 
 
 Exercises. 1. Find a point in a given straight line at a given distance 
 from a given straight line. 
 
 2. Find a point in a given straight line at equal distances from two other 
 straight lines. 
 
 3* Construct an isosceles triangle having a given base, and each of its 
 sides equal to three times the base. 
 
 4. Find a point at a given distance from a given point, and at the same 
 distance from a given straight line. 
 
 5. Construct a triangle having given the base, the sum of the other sides, 
 and one of the angles at the base. 
 
 6. Construct a triangle having given the base, the difference of the other 
 sides, and one of the angles at the base. 
 
 7. Find a point which shall be equidistant from three given points. 
 
 8. Find a point which shall be equidistant -from three given intersecting 
 straight lines (see 91, Exercise 6). Show that there are four solutions, 
 and construct the four points. How is the solution modified if two of the 
 given lines are parallel? 
 
 93. The complete solution of a problem consists of three 
 parts : 
 
 (*.) The ANALYSIS, in which we explain how a correct construc- 
 tion can be based upon known geometrical truths. 
 
 (it.) The CONSTRUCTION, in which, guided by the analysis, we 
 make the required lines, angles, etc., with the help of the ruler 
 and the compasses. 
 
 (in.) The DISCUSSION, in which we consider (a) whether there 
 may be more than one solution ; (b) whether the data may have 
 such values that no solution is possible ; (c) whether the solution 
 has any other peculiarities for particular values of the data. 
 
 The analysis of a problem that can be solved by the method of 
 loci has already been given ( 92) ; in all other cases a theorem 
 (or theorems) must first be found on which a correct construction 
 can be based. In this search nothing but experience and ingenu- 
 ity will ensure success ; but there is one general rule which will be 
 found very useful ; namely : 
 
93-] CHAPTER IV. TRIANGLES. 103 
 
 Rule. Suppose the solution effected, and draw a suitable figure ; 
 then trace the relations among the parts of the figure until some re- 
 lation is discovered which will give a clue to the right construction. 
 
 We will give an example to illustrate the application of this rule. 
 
 Problem. Through a given point to draw a line which shall 
 cut off equal lengths from the sides of a given angle. 
 
 Analysis. Let P (Fig. 87) be the given point, B A C the 
 given angle. Suppose that O P Q were the line required ; then we 
 know that A O Q must be an isos- 
 celes triangle (why ?) , and hence, that 
 if we bisected the angle A, O P Q 
 would be perpendicular to the bi- 
 sector (80, Exercise 1). This 
 suggests to us the correct construc- 
 tion. 
 
 Construction. Bisect the angle Fig. 
 
 A, let fall from P a perpendicular to 
 
 the bisector, and produce it until it meets the sides of the given 
 angle in O and Q : O P Q is the line required. 
 
 Discussion. Construct the figure for the case where P is not 
 between the sides of the given angle. For what position of P is 
 the solution impossible? 
 
 Exercises. 1. Draw a line which shall pass through a given point and 
 make equal angles with two given intersecting lines. 
 
 2. In a given straight line find a point which shall be equidistant from 
 two given points. 
 
 3. In one side of a triangle find a point which shall be equidistant from 
 the other two sides (see 91, Exercise 6). 
 
 4. In one side of an angle a point is given ; find in the same side another 
 point which shall be equidistant from the first point, and from the other side 
 of the angle. 
 
 5. From a given point without a given straight line draw a line making a 
 given angle with the given line. 
 
 6. Trisect a right angle (that is, divide it into three equal parts). 
 
104 GEOMETRY FOR BEGINNERS. [ 94, 
 
 VI. Applications. 
 
 94. If I wish. to find the distance from A to B (Fig. 88) it 
 is plain that the intervening pond will prevent me from using a 
 chain or other direct means of measurement. 
 
 In numerous instances direct measurement would be very 
 troublesome or quite impossible. On the ground obstacles, such 
 as houses, water, swamps, are often met with ; if we want to meas- 
 ure the distance from the earth to the moon, only one end of the 
 line to be measured is accessible ; while in the case of the distance 
 between the sun and a planet, the line is wholly inaccessible. 
 
 Here Geometry comes to our aid by giving us methods of indi- 
 rect measurement ; in which, for example, by measuring one line 
 we are able to learn the length of another. 
 
 How this can be done by the help of the laws of equal triangles, 
 we are now prepared to understand. 
 
 All cases of the indirect measurement of a line may be reduced 
 to four, 
 
 (/.) Both ends of the line are accessible ; 
 
 ('.) Only one end is accessible ; 
 
 (tit.) Both ends are inaccessible ; 
 
 (tv.) The line is wholly inaccessible. 
 
 95. Problem. To measure a line the ends of which only 
 are accessible. 
 
 Method I. Let A B (Fig. 88) be the line to be measured. 
 Choose a point C from which A and B are both visible. Measure 
 A C, B C, and the angle A C B. 
 
 Prolong A C to D, making C D = C A, and prolong B C to E, 
 making C E = C B. &DEC^&ABC (why?). There- 
 fore DE = A B, and its length (which can be measured directly) 
 is the distance required. 
 
96-] 
 
 CHAPTER IV. TRIANGLES. 
 
 105 
 
 Method II. Measure A C, B C, and A C B, as before. Then 
 construct on paper an angle 
 acb ACB; choose a suita- 
 ble scale of reduction, and lay 
 off on the paper the reduced 
 lengths a c and b c of the lines 
 A C and B C. Join a b : its 
 length gives, to the reduced scale, 
 the distance A B required. 
 
 For example : if A C = 8oo m , 
 
 Fig. 88. 
 
 D 
 
 B C iooo m , and the scale of 
 
 reduction is i : 5000, then a c = i6 cm , and b c = 2O cm . If now we 
 
 find that a b = 24 cm , then A B = 24 cm x 5000 = 1200 metres. 
 
 Remark. The first method is preferable, because any error 
 made in measuring a b is multiplied 5000 times in the result. 
 
 How can the length of the pond (Fig. 88) be found ? 
 
 96. Problem. To measure a line one end only of which is 
 accessible. 
 
 Method I. Let A B (Fig. 89) be the line to be measured, B 
 the end which is accessible. 
 At a point C, in A B pro- 
 longed, place a signal (a 
 pole or flag). Then place 
 a signal at a convenient 
 point D, and measure the 
 distances BD and CD. 
 Prolong CD to E, making 
 DE = DC, and prolong 
 B D to F, making DF = 
 DB. Place signals at E 
 and F. Then proceed in 
 the direction EF until a 
 
106 
 
 GEOMETRY FOR BEGINNERS. 
 
 [97- 
 
 point G is reached which falls in the line A D. Measure F G : 
 its length is equal to the distance A B required. 
 
 Proof: A D E F^D B C (II. Law of Equality). Therefore 
 angle DEF = D CB ; hence E G II A C ( 57, Remark). In 
 the triangles DAB and D F G, D B = D F, ADB = FD G 
 (why?), and AB D = DFG (why?) ; therefore A DAB^ 
 A DF G (I. Law of Equality), and F G = A B. 
 
 Method II. (By an isosceles triangle.) Choose a convenient 
 direction B H for running a straight line from B, measure the 
 angle A B H, and find in B H a point H at which the angle A HB 
 shall be equal to (180 ASH). Then B H = AB (why?). 
 What will be the value of A HB ifABH= 9 o? 
 
 Remark. In both methods three things have to be measured 
 directly : in Method I., three lines ; in Method II., one line and two 
 angles. But, if the means of measuring an angle are at hand, 
 Method II. is preferable, because it is easier to measure an angle 
 than a line. 
 
 97. Problem. To measure a line when both ends of it are 
 inaccessible. 
 
 Let A B (Fig. go) be the line, and let us suppose that a part 
 
 of the line between A and B is accessible, as shown in the Figure. 
 
 Method I. At a point C in A B erect a perpendicular C D 
 
 ( 88), and take DE = CD. 
 At E also erect a perpendic- 
 ular FG\ then F G will be 
 parallel to A B (why ?) 
 
 Find in FG the point F 
 which falls in the line B D, 
 and the point G which falls in 
 the line A D. Now &ACD 
 g* A ED G (why?), and A 
 Fi g . 90. B CD ^ A EDF (why?). 
 
CHAPTER IV. TRIANGLES. 
 
 107 
 
 Therefore FE = B C, and E G = A C ; and hence FE + E G 
 = BC + AC = AB,m the line to be measured. 
 
 Method II. (By isosceles triangles.) In C E find the points 
 H and K from which the directions of A and B respectively make 
 with C E the angle 45. Then C H -f C K = AS. Explain 
 fully why. 
 
 98. Problem. To measure a line which is wholly inacces- 
 sible. 
 
 Method I. Choose a convenient point C from which A and B 
 are both visible ; measure A C and B C as in 96 ; then measure 
 A'B'as in 95. 
 
 Method n. Choose C as before ; also a point F from which 
 A is visible, and a point 
 G from which B is visible. 
 Find E, the intersection 
 of A F and B C, and D, 
 the intersection of B G 
 and A C. Measure all 
 the angles at C, and also 
 the lines C F, C E, C G, 
 CD. Then represent on 
 paper the measured angles ^ 
 
 and lines (the latter to a 
 
 reduced scale) . By prolonging E F and D C till they meet, we 
 find the point on the paper which corresponds to A on the ground ; 
 and by prolonging in like manner D G and E C, we find the point 
 which corresponds to B. The length of A B on the paper gives 
 to the scale employed the required distance. 
 
108 GEOMETRY FOR BEGINNER^, 
 
 REVIEW OF CHAPTER IV. 
 QUESTIONS. 
 
 1, Define a plane figure; a polygon; its perimeter. 
 
 2. Define a triangle, and explain how it is named. 
 8. What six parts has every triangle? 
 
 4:. Distinguish between exterior and interior angles'* 
 
 5. The sum of any two sides of a triangle is greater than the third side. 
 
 6. Define equilateral, isosceles, and scalene triangles. 
 
 7. Define the terms base, vertex, and altitude. When will the altitude lie 
 outside the triangle? 
 
 8. The sum of the angles of a triangle is equal to 180. How can this 
 truth be illustrated? 
 
 9. Why can a triangle have only one right or one obtuse angle ? 
 
 10. Define acute, right, and obtuse triangles. 
 
 11. If two angles of a triangle are known how can the third be found? 
 
 12. In a right triangle what is the sum of the acute angles? 
 
 13. What is the complement of an angle? 
 
 14r. An exterior angle of a triangle = the sum of the opposite interior angles. 
 
 15. Define similar, equivalent, and equal magnitudes, and give examples of 
 each kind. 
 
 16. In two triangles what are corresponding side% (or angles) ? 
 
 17. What is the least number of parts which determine the size and shape 
 of a triangle? Of these how many at least must be sides? 
 
 18. In what cases do three parts fail to determine the triangle? 
 
 19. When are two triangles mutually equiangular? 
 
 20. Construct a triangle, having given, 
 
 (*'.) a side and two angles ; 
 
 (.) two sides and the included angle ; 
 
 (m.) two sides and the angle opposite to the greater side ; 
 
 (zz/.) the three sides. 
 
 21. State the four laws of equal triangles. 
 
 22. Make a triangle equal to a given triangle. 
 
 23. Make an angle equal to a given angle. 
 
 24. Draw a line parallel to a given line. 
 
 25. What is a problem ? What are auxiliary lines, and how distinguished 
 from others in the construction? 
 
CHAPTER IV. REVIEW, 109 
 
 26. In an isosceles triangle the angles opposite to the equal sides are equal. 
 Corollary. 
 
 27. If two angles of a triangle are equal the triangle is isosceles. Corollary. 
 
 28. What is the converse of a theorem? Give an example. 
 
 29. In a triangle the greater of two angles is opposite to the greater side, 
 and conversely. Corollary. 
 
 30. The perpendicular is the shortest line from a point to a straight line. 
 Corollary. 
 
 81. Oblique lines equally distant from the foot of the perpendicular are 
 equal. 
 
 32. Of two oblique lines unequally distant from the foot of the perpendicu- 
 lar, the more remote is the greater. 
 
 33. If two isosceles triangles have a common base the line joining their 
 vertices (z.) bisects the angle at the vertices, (zY.) bisects the base, (in.) 
 is perpendicular to the base. 
 
 34:. Bisect a given angle. 
 
 35. Bisect a given straight line. 
 
 36. Every point equidistant from the ends of a straight line is in the perpen- 
 dicular bisecting the line. 
 
 37. Erect a perpendicular at a given point of a given line. 
 
 38. Let fall a perpendicular from a given point to a given line. 
 
 39. Two parallel lines are everywhere equally distant from each other. 
 
 40. Give an example of an indeterminate problem. 
 
 41. Define the locus of a point, and illustrate the definition. 
 
 42. Give an example of a determinate problem. 
 
 43. How are problems solved by the method of loci? 
 
 44. What are the three parts of the complete solution of a problem? 
 
 45. What general rule is often useful in solving problems? 
 
 46. What four cases may occur in the indirect measurement of a line? 
 
 47. Measure a line the ends only of which are accessible. 
 
 48. Measure a line one end only of which is accessible. 
 
 49. Measui'e a line neither end of which is accessible. 
 
 50. Measure a line which is wholly inaccessible. 
 
 EXERCISES. 
 
 1. In an isosceles triangle the exterior angles at the base are equal. 
 
 2. In an isosceles triangle the bisector of an exterior angle at the vertex is 
 parallel to the base. 
 
110 GEOMETRY FOR BEGINNERS. 
 
 3. In an isosceles triangle the bisectors of the base angles (prolonged until 
 they meet the sides) are equal. 
 
 4. In what cases are we able, knowing one angle of a triangle, to find the 
 other two angles? 
 
 5. If in a right triangle the acute angles are 30 and 60, the hypotenuse is 
 equal to twice the smaller leg. 
 
 6. State and prove the converse of the preceding theorem. 
 
 7. Divide an isosceles triangle into two equal right triangles. 
 
 8. Make an angle equal to three times a given angle. 
 
 9. Make an angle of 15; also an angle of 150. 
 
 10. In what two ways can an angle of 22^ be constructed? 
 
 11. Name various angles between o and 180 which can be constructed by 
 means of the theorems and problems already given. 
 
 12. A straight railway passes within two miles of a town. A place is de- 
 scribed as four miles from the town and one mile from the railway. How 
 many places satisfy the conditions? 
 
 13. Place a line of given length between the sides of an angle so as to be 
 parallel to a given line. 
 
 14. Through a given point between two lines not parallel draw a line which 
 shall be bisected at that point. 
 
 15. Three lines meet in a point ; draw a line such that the parts of it inter- 
 cepted by the three lines shall be equal. 
 
 16. What axiom is implied in the last line of the proof of the theorem in 87? 
 
 17. The perpendiculars erected at the middle points of the sides of a triangle 
 meet in one point. , 
 
 Hints, Erect two of the perpendiculars ; show (by II. Law of Equality) that 
 their intersection is equidistant from the three corners of the triangle, then 
 make use of 80, Exercise 2. 
 
 18. The bisectors of the three angles of a triangle meet in one point. 
 Hints. Draw two bisectors; show by $ 91, Exercise 6, that their intersection 
 
 is equidistant from the sides of the triangle, and (by III. Law of Equality), 
 that the line joining the intersection with the third corner bisects the angle 
 at that corner. 
 
 19. Trisect a given straight line. 
 
 20. Illustrate by an example the application of the rule in 93. 
 
 21. Choose two stations on the ground, and find their distance by means of 
 (i.), 95; (.), 96; (iii^ 97; <>.), 98. 
 
 22. Can you find the height of an object (a tree, a church-spire, etc.) by 
 means of truths established in this chapter? How? 
 
 23 How many and what parts are sufficient to determine a right triangle? an 
 isosceles triangle ? an equilateral triangle ? an isosceles right triangle ? 
 
99-] CHAPTER V. QUADRILATERALS. Ill 
 
 CHAPTER V. 
 QUADRILATERALS. 
 
 CONTENTS. I. Sides and Angles of a Quadrilateral ( 99, 100). II. Different 
 Kinds of Quadrilaterals ( 101-105). HI- Construction of Quadrilaterals 
 ( 106-111). IV. Subdivision of a Lane ( 112-114). 
 
 J. Sides and Angles of a Quadrilateral. 
 
 99. A plane figure bounded by four straight lines is called a 
 QUADRILATERAL (A B CD, Fig. 92). 
 
 A straight line A C, which joins two 
 opposite corners of a quadrilateral, is 
 called a DIAGONAL. 
 
 Exercises. 1. How many sides and how 
 many angles has a quadrilateral? 
 
 2. How is a quadrilateral named or de- 
 noted? 
 
 3. What is the perimeter of a quadrilateral ( 59) ? 
 
 4. Into how many triangles is a quadrilateral divided by a diagonal? 
 
 5. How many diagonals can be drawn in a quadrilateral? 
 
 100. If we divide a quadrilateral AB CD (Fig. 92) into 
 two triangles by drawing a diagonal AC, it is evident that the sum 
 of the four angles of the quadrilateral is the same as the sum of 
 the six angles of the two triangles. Now the sum of the angles of 
 the two triangles is 360 (why?) ; hence, 
 
 Theorem. The sum of the angles of a quadrilateral is equal 
 to four right angles, or 360. 
 
 Exercises. 1. If the angles of a quadrilateral are all equal, what is the 
 value of each? 
 
 2. Three angles of a quadrilateral are 70, 40, and 120; find the fourth. 
 
112 GEOMETRY FOR BEGINNERS. [ IOI. 
 
 II. Different Kinds of Quadrilaterals. 
 
 101. With respect to the position of the sides there are three 
 kinds of quadrilaterals. 
 
 A quadrilateral which has no sides parallel is called a TRAPEZIUM 
 (Fig. 93,1.). 
 
 A quadrilateral which has only two sides parallel is called a 
 TRAPEZOID (Fig 93, II.). 
 
 A quadrilateral which has its opposite sides parallel is called a 
 PARALLELOGRAM (Fig. 93, III.). 
 
 Exercises. 1. Draw (free-hand) a trapezium, a trapezoid, and a paral- 
 lelogram. 
 
 2. Draw a trapezoid having two right angles. 
 
 102. Let AB CD (Fig. 94) be a parallelogram. Draw the 
 diagonal BD. The alternate angles m 
 and n are equal, and likewise the alter- 
 nate angles / and q ( 56). Therefore 
 A ABD ^ A CBD (I. Law of Equal- 
 
 D c ity), and AB = CD, and AD = BC. 
 
 Also the angle A equals the angle C ; and, 
 
 since m n and p q, therefore m -\- n = p -\- q; that is, the 
 angles at B and D are equal. Hence. 
 
 Theorem I. Every parallelogram is divided by a diagonal 
 into two equal triangles. 
 
I03-] 
 
 CHAPTER V. QUADRILATERALS. 
 
 113 
 
 Theorem II. The opposite sides of .a parallelogram are equal. 
 Theorem III. The opposite angles of a parallelogram are 
 qual. 
 
 Corollaries. From Theorem II. it follows that, 
 
 1. Parallels between parallels are equal. 
 
 2. Perpendiculars between parallels are equal ( 57, Corol- 
 lary). 
 
 3. Parallels are everywhere equally distant (see 90) . 
 
 Exercises. 1. If two adjacent sides of a parallelogram are equal, what 
 follows as to the other sides? 
 
 2. What relation exists between two adjacent angles of a parallelogram? 
 Why? 
 
 3. If one angle of a parallelogram is a right angle, what must the others 
 be? Illustrate with a figure. 
 
 4 If one angle of a parallelogram is acute, what can be inferred of the 
 other angles? Illustrate with a figure. 
 
 5. One angle of a parallelogram is (*'.) 45, ('.) 54 1 8', (m.) 1 21 1 6' 
 44" ; find the other angles. 
 
 6. Prove the converse of Theorem II. 
 
 103. The sides of a parallelogram may be either equal or 
 -unequal, and the angles may be either right or oblique ; so there 
 are in all four kinds of parallelograms : the oblique unequal-sided 
 
 IV. 
 
 parallelogram or RHOMBOID (Fig. 95, I.), the oblique equal- sided 
 parallelogram or RHOMBUS (Fig. 95, II.), the right unequal-sided 
 
114 GEOMETRY FOR BEGINNERS. [ 104. 
 
 parallelogram or RECTANGLE (Fig. 95, III.), and the right equal- 
 sided parallelogram or Square (Fig. 95, IV.). 
 
 Exercises. 1. Give examples of parallelograms (surface of a table, roof 
 of a house, etc.), and state the kind in each case. 
 
 2. Draw (free-hand) a parallelogram of each kind. 
 
 3. One angle of a rhombus is 60; find the other angles. 
 
 4. One side of a rhombus is 24; find its perimeter. 
 
 5* The sides of a rectangular park are 400'" and 240. How many trees 
 must be set out in its perimeter if they are to stand io m apart? 
 
 104. In the parallelogram ABCD (Fig. 96), draw the 
 
 diagonals AD and BC ; how many 
 triangles are thus formed? A AOB 
 ^ C O D (why ?) ; therefore the 
 sides opposite to the equal angles 
 are equal; that is, AO = DO and 
 BO = CO : hence, 
 
 z?* Ji L ^ 
 
 Theorem. The two diagonals 
 of a parallelogram mutually bisect each other. 
 
 Exercises. 1. Find the angles which the diagonals of a square make 
 with the sides of the square. 
 
 2. In a square the diagonals are equal and perpendicular to each other. 
 8. In a rectangle the diagonals are equal and inclined to each other. 
 Hint, Draw the diagonals, and make use of $ 73. 
 
 4. In a rhombus the diagonals are perpendicular to each other. Are 
 they equal or unequal? 
 
 5. In a rhomboid the diagonals are inclined to each other. Are they 
 equal or unequal? 
 
 6. What kind of parallelograms are divided by the diagonals into isosce- 
 les triangles? Into right triangles? 
 
 105. Either side of a parallelogram may be regarded as the 
 base ; then a perpendiculai let fall to the base from any point in the 
 opposite side is the altitude of the parallelogram. 
 
io6.] 
 
 CHAPTER V. QUADRILATERALS. 
 
 115 
 
 In the parallelogram A B C D (Fig. 97) A B is taken as the 
 base, and either DE or FB or CG 
 is equal to the altitude. Why are 
 .DE, FB, and CD all equal? 
 
 In a rectangle whatever side be 
 taken as base the adjacent side is 
 the altitude. 
 
 In a square the base and altitude are equal. 
 
 In a trapezoid one of the parallel sides is taken as the base. 
 
 In a trapezium the terms base and altitude are not used. 
 
 III. Construction of Quadrilaterals. 
 
 106. Problem. To construct a square, having given one 
 side, a (Fig. 98). 
 
 Construction. At A make a right angle ; take AB = AD = a' t 
 
 then from B and D as centres, with a radius a 
 
 equal to a, describe arcs intersecting at a point 
 C. AB CD is the square required. 
 
 Proof 1 The four sides are each equal to a 
 by construction. Join B D, and make use of 
 102, Theorem I. 
 
 Construct another square with the same side 
 a then prove (by division into triangles, and 
 IV. Law of Equality) that the two squares are equal. 
 
 Hence, a square is completely determined if one side is known ; 
 and, 
 
 Theorem. Two squares are equal if they have a side equal. 
 
 Exercises. 1. Construct a square whose side shall be i dm . 
 
 2. Construct a square whose perimeter shall be i m . 
 
 3. Draw a rectangle ; then construct a square with the same perimeter as 
 that of the rectangle. 
 
 Fig. 98. 
 
 1 If the construction of a problem is given without a previous analysis, a proof 
 of the construction (unless it is quite obvious without it) ought to follow. 
 
116 GEOMETRY FOR BEGINNERS. [ 107. 
 
 4r Construct a square, having given its diagonal. 
 
 5. Construct a square, and upon each side an equilateral triangle. 
 
 6. Construct an equilateral triangle, and upon each of its sides a square. 
 
 107. Problem. To construct a rectangle, having given two 
 adjacent sides, a and b (Fig. 99) . 
 
 Construction and proof similar to those of the last section. 
 
 Construct another rectangle with 
 
 ~~~ b the same sides, and prove that the 
 
 two rectangles are equal. 
 
 Hence, a rectangle is determined if 
 we know two adjacent sides ; also, 
 Theorem. Two rectangles are 
 equal if they have tivo adjacent sides 
 equal each to each. 
 
 Exercises. 1. Construct a rectangle with the sides 4O cm and 6o cm . (If 
 on paper, use a reduced scale.) 
 
 2. Make a plan of a rectangular field whose sides measure 640 and 360. 
 
 3. How many rectangles can be constructed upon a given line as the 
 diagonal? 
 
 108. Problem. To construct a parallelogram, having given 
 two adjacent sides and the included angle. 
 
 Construction. Let a and b (Fig. 100) be the given sides, 70 
 
 _a the included angle. Make an angle 
 
 A = 70, then proceed as in the 
 last section. A B C D is the required 
 parallelogram. 
 
 Proof similar to that in 106. 
 Construct a second parallelogram 
 with the same given parts, and prove 
 that it is equal to the first. 
 
 Therefore, tivo sides and the in- 
 Fig. ioo. eluded angle completely determine a 
 
 parallelogram ; and, further, 
 
lOp.] CHAPTER V. QUADRILATERALS. 117 
 
 Theorem. Two parallelograms are equal if two sides and 
 the included angle of one are equal respectively to two sides and 
 tlce included angle of the other. 
 
 Exercises. 1. Construct a parallelogram, the two sides being 44 cm and 
 66 cm , and the included angle 1 10. 
 
 2. Construct a rectangle, having given 
 
 (a) A side and a diagonal ; 
 
 (b} A side and the opposite angle of the diagonals ; 
 
 (c) A diagonal and the angle between the diagonals. 
 
 3. Construct a rhombus, having given 
 
 (a) A side and an angle ; 
 
 (b) A side and a diagonal ; 
 (r) The diagonals ; 
 
 (^/) An angle and the diagonal through its vertex. 
 
 4. Construct a rhombus whose angles shall have the ratio 1 : 3. 
 
 109. Problem. To construct a trapezoid, having given 
 three sides (two being the parallel sides} and one angle. 
 
 Construction. Let a, b, c {Fig. a 
 
 JQI) be the given sides, 68 the 
 given angle. 
 
 Make an angle A = 68 ; take AB 
 = a, A D = b through D draw a 
 line parallel to A B, and on it take 
 D C = c. Join B C. A B C D is 
 the trapezoid required. 
 
 Exercises. 1. How many and what 
 
 parts determine a trapezoid? A r-,. B 
 
 rig. 101. 
 2. Construct a trapezoid with the sides 
 
 6o cm , 7O cm , 8o cm , and the angle between the first two sides equal to 75. 
 3* Construct a trapezoid, having given 
 
 (a) The two parallel sides and the altitude ; 
 (3) The two sides not parallel and the altitude ; 
 
 (c) The two parallel sides and one of the other sides ; 
 
 (</) One of the parallel sides and the two sides not parallel. 
 
118 GEOMETRY FOR BEGINNERS. [ IIO. 
 
 110. Problem. To construct a trapezium, having given 
 three sides and the two angles included by these sides. 
 
 Construction. If a, b, and c are the given sides, 60 and 75 
 the included angles, draw a line A B = b ; through A and B draw 
 lines making, with B B, the angles 60 and 75 respectively; lay 
 off on these lines the lengths A D a, and B C = c, and join the 
 points thus found. 
 
 Draw a figure to illustrate this construction. 
 
 Exercises. 1. Construct a trapezium with the sides i2O cm , 90, 70, 
 and the angles 67 and 83. 
 
 2. Construct a trapezium, having given 
 
 (a) Three angles and the two sides between the first and third 
 
 angles ; 
 
 () Four sides and one angle ; 
 (<:) Four sides and a diagonal. 
 
 111. Problem. To construct a quadrilateral equal to a 
 given quadrilateral A B CD (Fig. 102). 
 
 Analysis. If we draw the diagonal B D, we divide the given 
 quadrilateral into two triangles. If now we construct ( 76) two 
 new triangles equal respectively to these, and similarly placed, it is 
 
 evident that they will form, taken to- 
 gether, a new quadrilateral equal to 
 the given quadrilateral (Axiom II.). 
 
 Construction. Take E F = A B ; 
 from E and F as centres, with the 
 radii AD and B D respectively, de- 
 scribe arcs intersecting at a point H ' 
 
 Fig. IO2. 
 
 then from F and H as centres, with 
 
 B C and D C respectively as radii, describe arcs intersecting at a 
 point G. Lastly, join E H, H G, and G F. 
 
 Exercises. 1. Draw a trapezium, and then construct another equal to it. 
 2. When are two quadrilaterals equal to each other? 
 
112.] CHAPTER V. QUADRILATERALS. 119 
 
 IV. Subdivision of a Line. 
 
 112. Let a, b, c, d, e (Fig. 103) be parallel lines equidistant 
 from one another, and let them be cut by any other lines as in the 
 figure. 
 
 The perpendiculars A F, B G, C H, D K are equal (why?). 
 &ABF ^ A B C G ^ ., 
 
 4 M 
 
 A C DH a* A D E K 
 (why ?) ; therefore A B = 
 BC=CD = DE. 
 
 In like manner we can 
 prove that the parts cut 
 from any other line, M ' N 
 or P Q, are equal to each 
 other. Hence, 
 
 Theorem. Equidis- I J \ I Q 
 
 tant parallel lines cut equal pig. I0 j. 
 
 parts from any line that intersects them. 
 
 Exercise. In a right triangle the middle point of the hypotenuse is equi- 
 distant from the three corners of the triangle. 
 
 Hint. Through this middle point draw a line parallel to one of the legs. 
 
 113. Problem. To divide a straight line A B (Fig. 104) 
 into any number (say five} equal parts. 
 
 Construction. Through A draw any straight line A X, on which 
 lay off five equal parts of any 
 length. Join C, the last point 
 thus obtained, to B, and 
 through the other points of 
 division of A C draw lines 
 parallel to B C. They will 
 divide AB into five equal 
 parts. 
 
120 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 
 
 Proof. Make use of the theorem in 112. 
 
 Exercise. Divide a straight line into 3, 6, 7, 9, 10 equal parts. 
 
 114. Problem. To divide a straight line A B (Fig. 705) 
 into two parts which shall have a given ratio (say 3:4). 
 
 Construction. Draw through A any line A C, and lay off on 
 
 it 3 -f- 4 = 7 equal parts. Join the 
 last point C to B, and draw through 
 D, the third point of division from 
 A, a line DE parallel to B C ; it 
 intersects AB in E. Then, 
 
 E 
 Fig. 105. 
 
 B 
 
 Proof. Lines drawn through the 
 other points of division of A C, and parallel to B C, would (by 
 112) divide A E into three equal parts, and E B into four 
 equal parts. 
 
 Exercises. 1. Divide a line in the ratio 4 : 7. 
 
 2. Construct a rectangle, having given that the sum of two adjacent sides 
 is i6 m , and their ratio 2:3. 
 
 3. Construct a rhombus in which the sum of the diagonals shall be I2 m , 
 and one of them shall be twice the other. 
 
 4. The perimeter of a rectangular garden is 8o.4 m , and the longer side is 
 to the shorter as 5:7. Make a plan of the garden, and find the lengths of its 
 sides. 
 
 5. Divide a line into three parts which shall be to each other as 2 : 3 : 5. 
 Hint. Lay off on the auxiliary line 2 + 3 + 5 = 10 equal parts. 
 
 6. Construct a triangle whose sides shall be to each other as the numbers 
 3, 4, 5. Is this problem determinate or not? 
 
 7. Construct an isosceles triangle whose base shall be to one of the equal 
 sides as 3 : 4. 
 
 8. Divide a line into four parts which shall be to each other as the num- 
 bers i, 3, 4, 6. 
 
CHAPTER V. REVIEW. 121 
 
 REVIEW OF CHAPTER V. 
 SYNOPSIS. 
 
 1 A quadrilateral is a four-sided polygon. 
 
 2. The sum of its angles equals 360. 
 
 3. There are three classes of quadrilaterals : 
 
 (z.) The trapezium (no parallel sides); 
 (z'z.) The trapezoid (two parallel sides) ; 
 (z'z'z.) The parallelogram (both pairs of sides parallel). 
 4 There are four kinds of parallelograms : 
 
 (z.) The rhomboid (adjacent sides unequal, angles oblique); 
 
 (z'z.) The rhombus (sides all equal) ; 
 
 (z'z'z.) The rectangle (angles all right angles) ; 
 
 (zV.) The square (sides equal, and angles right angles). 
 
 5. A diagonal divides a parallelogram into two equal triangles.. 
 
 6. Hence, the opposite sides and also the opposite angles of a parallelo- 
 gram are equal. 
 
 7. The two diagonals of a parallelogram mutually bisect each other, and 
 are equal to each other in the rectangle and the square, perpendicular 
 to each other in the rhombus and the square. 
 
 8. Two squares are equal if they have a side equal ; two rectangles, if their 
 adjacent sides are respectively equal ; any two parallelograms, if they 
 have two sides and the included angle equal each to each. 
 
 9. Equidistant parallel lines cut equal parts from any line that intersects 
 them. 
 
 10. No. 9 enables us to divide a line into equal parts, or into parts which 
 shall be to each other in given ratios. 
 
 EXERCISES. 
 
 1. Divide a quadrilateral into a parallelogram and a triangle. 
 
 2. Divide a parallelogram into two equal triangles. 
 
 3* Divide a parallelogram into two equal parallelograms. 
 
 4. Divide a square into four equal isosceles triangles. 
 
 5. Divide a rhombus into four equal right triangles. 
 
 6. Divide a trapezoid in which the non-parallel sides are equal into a par- 
 allelogram and an isosceles triangk. 
 
122 GEOMETRY FOR BEGINNERS. 
 
 7. Divide a right triangle into two isosceles triangles. 
 
 8. The adjacent angles of a parallelogram are supplementary. 
 
 9. In a rhomboid one angle is 120; find the other angles. 
 
 10. If two opposite sides of a quadrilateral are equal and parallel, the figure 
 is a parallelogram. (Draw a diagonal. Use 56, 57, 73.) 
 
 11* The middle points of the four sides of a parallelogram are the corners 
 of a new parallelogram. If the first parallelogram is a rectangle the 
 second is a rhombus; if the first is a rhombus the second is a rectangle ; 
 if the first is a square the second is a square. 
 
 12. The bisectors of the angles of a parallelogram enclose a rectangle. (Ex- 
 ercise 8 and 64.) 
 
 13. Prove the converse of 102, Theorem III. 
 
 14. The three perpendiculars let fall from the corners of a triangle meet in 
 one point. 
 
 Hints. Through the corners of the triangle draw lines parallel to the oppo- 
 site sides, and prolong them till they intersect, forming a new triangle. 
 Then show (by means of \ 102, Theorem II., Axiom I., and 56, Corol- 
 lary) that the three altitudes of the given triangle are perpendicular to the 
 middle points of the new triangle, and make use of Exercise 17, page no. 
 
 15. Construct a rectangle, having given one side and the perimeter. 
 
 16. Construct a rhombus which shall have its obtuse angles twice as large as 
 those which are acute. 
 
 17. Construct a rhomboid, having given 
 
 (a) Two unequal sides and the included angle ; 
 () Two unequal sides and a diagonal ; 
 (c) One side and the two diagonals. 
 
 18. Place a line of given length between two given parallel lines so that it 
 shall pass through a given point. (Three different positions of the 
 point.) 
 
 19. Place a square within a given square so that one corner of it shall be at 
 a given point in a side of the given square. 
 
 20. In a square construct an equilateral triangle having one corner common 
 with a corner of the square, and the others in the sides of the square. 
 
 21. How many and what parts (sides or angles) are required to determine a 
 square ? a rectangle ? a rhombus ? a parallelogram ? a trapezoid ? a 
 trapezium ? 
 
 22. Divide a line into 3^ parts. 
 
 23. Construct a rectangle whose perimeter shall be i m , and the ratio of the 
 base and altitude 3 : 7. What are the lengths of its sides? 
 
CHAPTER VI. POLYGONS. 
 
 123 
 
 CHAPTER VI. 
 POLYGONS. 
 
 CONTENTS. I. Sides and Angles of a Polygon (115-117). II. Regular Poly- 
 gons ( 118-122). III. Symmetrical Figures ( 123). 
 
 J. Sides and Angles of a Polygon. 
 
 115. A POLYGON is a plane figure bounded by straight lines. 
 What are its sides and \\& perimeter ? ( 59). 
 
 Fig. 106. 
 
 As regards the number of sides, polygons begin with three 
 ( 60), and go on increasing without limit. As the number of 
 sides increases, the polygon approaches a circle in shape (see Fig. 
 106) in other words, the circle is the limit which the polygon 
 approaches (but never reaches) when the number of its sides is 
 increased more and more. 
 
124 GEOMETRY FOR BEGINNERS. [ > r6. 
 
 The triangle and the quadrilateral are polygons ; but, as they 
 have been already studied, we shall now apply the term chiefly to 
 those figures which have more than four sides. A polygon of five 
 sides is called a PENTAGON, of six sides a HEXAGON, of seven sides 
 a HEPTAGON, of eight sides an OCTAGON, of nine sides a NONAGON, 
 of ten sides a DECAGON, of twelve sides a DODECAGON, of twenty- 
 sides an ICOSAGON. 
 
 Which of these polygons are represented in Fig. 106 ? 
 
 Every polygon has as many angles as sides ; each side has two 
 adjacent angles ; each angle is formed by two intersecting sides. 
 
 A line joining two corners not in the same side is called a DIAG- 
 ONAL. 
 
 By drawing as many diagonals as possible from one corner of a 
 polygon we divide the polygon into triangles ; and Exercise i, below, 
 leads to the following general law : 
 
 The number of diagonals which can be drawn from one corner 
 of a polygon is THREE less than the number of sides ; and the num- 
 ber of triangles into which the polygon is divided is TWO less than 
 the number of sides. 
 
 Exercises. 1. Draw polygons of 3, 4, 5, 6, 7, 8 sides, and from a corner 
 of each all the diagonals possible ; then prepare three vertical columns, in 
 the first writing the number of sides of the several polygons, in the second 
 the number of diagonals, in the third the number of triangles. What general 
 conclusions follow from the results? 
 
 2. How many diagonals can be drawn from one corner of a polygon of 
 60 sides, and how many triangles will be formed? 
 
 3* Divide an octagon into three quadrilaterals and a triangle. 
 
 4. If you cut off a triangle from a hexagon with a diagonal, what kind of 
 polygon is. left? 
 
 5. How many diagonals in #//can be drawn in a quadrilateral? a penta- 
 gon? a hexagon? a decagon? 
 
 116. The angles of a polygon may be acute, right, obtuse, or 
 even convex. In the last case, they are usually termed reentrant 
 
1 1 6.] CHAPTER VI. POLYGONS. 125 
 
 angles. In Fig. 107, A is a polygon having one reentrant angle, 
 and B a polygon having two reentrant angles. 
 
 Fig. 207. 
 
 But whatever be the values of the angles, their sum in any poly- 
 gon always bears a simple relation to the number of sides. If a 
 polygon could be divided into as many triangles as it has sides, 
 then the sum of all its angles would evidently be equal to twice as 
 many right angles as there are sides ( 64). But since by the 
 last section it appears that the number of these triangles is two less 
 than the number of sides, it follows that the sum of all the angles 
 will be twice as many right angles as there are sides less two. 
 Hence, 
 
 Theorem. The sum of the angles of a polygon is equal to 
 twice as many right angles as there are sides less two. 
 
 Thus the hexagon (Fig. 108) consists of 6 2 = 4 triangles ; 
 and the sum of its angles = 4X2 = 8 right 
 angles = 720. 
 
 In general, if n be the number of sides (n 
 being any number), the following equation 
 or FORMULA enables us to find the sum of 
 the angles : 
 
 Sum of the angles (in degrees) = 180(n-2). [i.] ' p . 
 
 Exercises. 1. What polygon cannot have reentrant angles? 
 2. Find the sum of the angles of polygons of 3, 4, 5, 6, 8, 10, 12, 20 sides. 
 3* Find by division into triangles the sum of the angles of the polygon A 
 {Fig. 707). Also find the sum by means of Formula [i]. 
 
126 GEOMETRY FOR BEGINNERS. [ 117. 
 
 4. Find in the same two ways the sum of the angles of the polygon B 
 (Fig. 707). 
 
 5. Of how few sides can you make a polygon with one reentrant angle? 
 with two reentrant angles? 
 
 6. Make a hexagon with as many reentrant angles as possible. 
 
 7. Make a hexagon having only right angles or their multiples. 
 
 8. Draw a polygon such that one of its diagonals shall lie wholly outside 
 the polygon. 
 
 117. Problem. To make a polygon equal to a given poly- 
 gon ABCDEF (Fig. /op). 
 
 Analysis. Divide the given polygon into triangles by drawing 
 
 diagonals from one corner ; 
 then construct a series of 
 new triangles equal respec- 
 tively to those of the given 
 polygon, and similarly 
 
 placed. It is evident that 
 Fig. log. 
 
 these taken together will 
 form a new polygon equal to the given polygon (Axiom II.). 
 Draw the figures and give the construction in full. 
 
 Exercises. 1. Construct a pentagon equal to a given pentagon. 
 
 2. Construct an octagon equal to a given octagon. 
 
 3. When are two polygons equal to each other? 
 
 II. Regular Polygons. 
 
 118. Definitions. A polygon which has equal sides is 
 termed EQUILATERAL (Fig. no, A) ; a polygon which has equal 
 angles is termed EQUIANGULAR (Fig. no, B} ; and a polygon which 
 has equal sides and equal angles (Fig. no, C) is termed REGULAR. 
 
 Since the angles of a regular polygon are equal, each angle is 
 equal to the sum of all the angles divided by their number ; or, in 
 other words, divided by the number of sides of the polygon. 
 
 If we let n denote the number of sides, the sum of the angles 
 
1 1 9.] CHAPTER VI. POLYGONS. 127 
 
 (by 116) = 1 80 (n 2) = 180 w 360. Dividing this value 
 by n we obtain (Axiom V.) the formula, 
 
 RAA 
 
 Each angle of a regular polygon = 180 - . [ 2 .] 
 
 Fig. no. 
 
 Exercises. 1, What is the regular triangle called? the regular quadri- 
 lateral? 
 
 2. Can a triangle be equilateral without being equiangular? 
 
 3* What quadrilateral is equilateral but not equiangular? equiangular but 
 not equilateral? 
 
 4. Make three hexagons : the first equilateral but not equiangular ; the 
 second equiangular but not equilatei'al ; the third regular. 
 
 5. Find each angle of a regular polygon of 3, 4, 5, 6, 8, 10, 12, 20 sides, 
 and arrange the results in a vertical column. As the number of sides increases 
 how does the angle change? 
 
 6. Find the angle of a regular polygon of 360 sides. 
 
 7. Express Formula [2] in words. 
 
 8. One side of a regular dodecagon is i8 m ; find its perimeter. Give a 
 rule for finding the perimeter of a regular polygon when a side is known. 
 
 119. Regular polygons possess a very important property 
 stated in the following theorem : 
 
 Theorem. The bisectors of the angles of a regular polygon 
 meet in a point equally distant (i.) from all the corners, and also 
 (ii.) from all the sides. 
 
 Proof. Let the bisectors AO, BO (Fig. in or 112) of the 
 angles A, B, meet in <9, and join OC, OD, etc. The triangles 
 AOB, BOC are isosceles ( 81); therefore OA=OB. The 
 
128 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ "9- 
 
 same two triangles are also equal ( 73) ; therefore OC= OA. 
 Also, OCB= OAB \ one of the equal angles of the polygon; 
 therefore OC bisects the angle BCD. In like manner we can 
 prove that O>= OA, and bisects the angle CDE ; etc. 
 
 D 
 
 A G B 
 Fig. in. 
 
 A G 
 
 Fig. JI2. 
 
 (ii.) The distances from O to the sides of the polygon are the 
 perpendiculars let fall from O to the sides ( 83, Corollary) . Take 
 any two of these perpendiculars OG and OH; they are equal, 
 because A OGB ^ OHB (why?). In like manner we can 
 show that OH OJ = OK, etc. ; in other words, these perpen- 
 diculars are all equal. 
 
 Definitions. This remarkable point O in a regular polygon 
 is called its CENTRE ; the distance from the centre to a corner is 
 called the GREATER RADIUS ; the distance from the centre to a side 
 is called the LESS RADIUS ; the angle between two radii is called 
 the ANGLE AT THE CENTRE. 
 
 Corollaries. I. Every regular polygon can be divided into as 
 many equal isosceles triangles as it has sides (how ?) . 
 
 2. Every regular polygon can be divided into twice as many 
 equal right triangles as it has sides (how ?) . 
 
 3. The greater radii of a regular polygon are all equal, the less 
 radii are equal, and the angles at the centre are equal. 
 
 4. The angle at the centre is equal to 360 divided by the num- 
 ber of sides ; in other words, to the last term </ [2], 118. 
 
120.] 
 
 CHAPTER VI. POLYGONS. 
 
 129 
 
 
 Exercises. 1. Prove the preceding theorem for the case of a triangle. 
 Draw a figure and give the proof in full. (Compare this exercise with Exer- 
 cises 7 and 8, 92.) 
 
 2. Show that every regular polygon is composed of an even number of 
 equal right triangles. 
 
 3. Prove that in a regular polygon a radius, if prolonged, divides the poly- 
 gon into two equal parts. 
 
 4. Prove that the angle between two less radii is equal to the angle be- 
 tween two greater radii. What value has the angle between a greater radius 
 and a less radius? 
 
 6. Find the angle at the centre in an equilateral triangle? a square? a 
 pentagon? a hexagon? an octagon? a decagon? 
 
 6. If the angle at the centre is 10, how many sides has the polygon, and 
 what is the value of each angle? 
 
 7. If an angle at the centre is given, how can you find an angle of the 
 polygon? 
 
 8. If an angle of the polygon is given, how can you find an angle at the 
 centre? 
 
 NOTE. For the sake of reference, the chief data respecting the angles of 
 regular polygons are collected in the following table : 
 
 NUMBER OF SIDES. 
 
 SL'.M OF ANGLES. 
 
 EACH ANGLE. 
 
 ANGLE AT THE 
 CENTRE. 
 
 3 
 
 180 
 
 60 
 
 120' 
 
 4 
 
 360 
 
 9 
 
 9 
 
 5 
 
 54 
 
 108 
 
 72* 
 
 6 
 
 7 20 
 
 120 
 
 60 
 
 8 
 
 I080 
 
 135 
 
 45 
 
 10 
 
 I440 
 
 144 
 
 36 
 
 12 
 
 I800 
 
 150 
 
 30 
 
 20 
 
 3240 
 
 I62 ' 
 
 18 
 
 120. Problem. To construct a regular polygon having 
 given the number of sides. 
 
 Construction. Construct about a point as many angles as the 
 polygon has sides, each angle equal to the angle at the centre (how 
 is this angle found ?) ; lay off equal lengths on the sides of these 
 angles, and join the points thus determined. 
 
130 GEOMETRY FOR BEGINNERS. [ 121. 
 
 Exercises. 1. Is the above problem determinate or indeterminate ? 
 
 2. Construct (z.) an equilateral triangle ; (zY.) a square. 
 
 3. Construct a regular hexagon. Join the alternate corners by straight 
 lines. What figure do these lines enclose? Why? 
 
 4. Prove that the greater radius of a regular hexagon is equal to a side of 
 the hexagon. 
 
 5. Construct a regular octagon. Join the alternate corners by straight 
 lines. What figure do these lines enclose? Why? 
 
 6. Construct a regular nonagon. 
 
 7. Can you devise another way to construct a regular polygon? 
 
 121. Problem. To find the centre of a given regular poly- 
 gon. 
 
 Analysis. Two methods of solving this problem are suggested 
 by 118 ; what are they ? 
 
 A third method is based on the fact (see Figs, in and 112) 
 that, if the polygon has an even number of sides, a line through one 
 corner and the centre passes through the opposite corner ; while 
 if it has an odd number of sides, this line passes through the 
 middle of the opposite side. Hence, state rules for proceeding in 
 each case. 
 
 Exercises. 1. Find the centre of (*'.) an equilateral triangle; (M.) a 
 square ; (m.) a regular pentagon ; (iv.~) a regular hexagon. 
 
 2. Divide an equilateral triangle into 3 equal parts. 
 
 3 Divide a s x quare into 4 equal parts. Can you do this in more than 
 one way? 
 
 4. Divide a regular hexagon into 6 equal parts ; also into 12 equal parts. 
 
 5. Divide a regular octagon into 4, 8, 16 equal parts. 
 
 NOTE. The construction, etc., of regular polygons will be considered further 
 in the chapter on the Circle. 
 
 122. Applications. The stone pavements of large vesti- 
 bules, galleries, etc., and the patterns used in quilting, are illustra- 
 tions of the practical employment of regular polygons. 
 
 Not every regular polygon can be employed for these purposes ; 
 only those can be used whose sides touch at all points ; for, if this 
 
122.] 
 
 CHAPTER VI. POLYGONS. 
 
 131 
 
 were not the case, the surface would not be everywhere covered. 
 We meet most commonly with groups of squares, hexagons, or 
 octagons. Let us examine some cases. 
 
 1. Equilateral triangles (Fig. 113) can be used. For, since 
 each angle is equal to 60, it is obvious that the triangles may be 
 arranged in groups of six about a point, each group constituting 
 a regular hexagon. 
 
 2. The arrangement in squares (like a checker-board) is shown 
 in Fig. 114. 
 
 
 Fig. 113. 
 
 Fig. if 4. 
 
 Fig. 115. 
 
 3. Regular hexagons (Fig. 115} may be arranged in groups of 
 three about a point ; for the angle of the hexagon being 120, three 
 angles = 3 X 120 = 360. 
 
 4. Fig. 116 shows a pattern consisting of hexagons and equi- 
 lateral triangles combined. 
 
 Fig. 116. 
 
 Fig. 117. 
 
 5. Regular octagons alone are not sufficient; for the angle of 
 the octagon is 135. and twice 135 is only 270, and there remains 
 an angle of 90 to be filled. This can be done by the right angle 
 of a square (Fig. 
 
132 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 
 
 Fig. 118. 
 
 6. The angle of a regular decagon is 144. Two decagons 
 
 placed with one side common 
 would leave an angle of 360 2 
 X 144 = 72 to be filled. Since 
 no regular polygon has an angle 
 of 72, it follows that we cannot 
 pave or quilt with regular deca- 
 gons, either alone or combined 
 with other regular polygons. 
 
 7. The regular dodecagon may 
 be combined with the equilateral 
 triangle ; for the angle of the do- 
 decagon is 150, twice 150 is 300, and there remains 60, which 
 is just equal to the angle of the equilateral triangle (Fig. 118) . 
 
 8. Thus far we have spoken only of regu- 
 lar polygons. It is obvious that suitable com- 
 binations may also be made with polygons 
 which are not regular. Fig. 119 gives an 
 example. What is the name of the polygon? 
 
 Exercise. Show that a square, a hexagon, and a 
 dodecagon will fill up the space about a point ; and 
 make a pattern of these polygons. 
 
 III. Symmetrical Figures. 
 
 123. If the polygon ABCD is made to revolve about AD 
 through half a revolution, the figure AEFD, thereby obtained, is 
 
 said to be SYMMETRICAL with respect 
 to ABCD, and AD is called the 
 LINE OF SYMMETRY. 
 
 Two symmetrical plane figures are 
 always equal, but their equal parts 
 (sides and angles) follow in opposite 
 order with reference to the line of 
 symmetry. 
 
I 23.] CHAPTER VI. POLYGONS. 133 
 
 From what precedes it is obvious that in two symmetrical plane 
 figures, 
 
 I. Two symmetrically-situated sides are equal. 
 II. Two symmetrically-situated angles are equal. 
 
 III. The line joining two symmetrically-situated points is per- 
 pendicular to the line of symmetry and bisected by it (for example, 
 the lines BE, MN, CF, etc., Fig. 126}. 
 
 By means of III. we can easily construct upon one side of a 
 polygon, as a line of symmetry, a symmetrical polygon. Explain 
 how. 
 
 If we regard ABCDFE as one polygon, then AD is a line 
 of symmetry dividing it into two symmetrical polygons. 
 
 A line may divide a figure into parts which are equal and yet 
 not symmetrical. This is the case, for example, with the diagonal 
 of the parallelogram A BCD (Fig. 94). It divides the figure 
 into two equal triangles, but these triangles are not so placed with 
 respect to the diagonal, that they would coincide if folded over the 
 diagonal as an edge. 
 
 A figure may have more than one line of symmetry. Thus, in 
 Fig. 1 20, besides AD, the line MN is also a line of symmetry. 
 
 Exercises. 1. Point out in Fig. 120 symmetrically-situated lines and 
 angles. 
 
 2 Draw a parallelogram, and upon one side construct a symmetrical 
 parallelogram. 
 
 3* Construct upon one side of a hexagon a symmetrical hexagon. 
 
 4. Construct two symmetrical figures with reentrant angles. 
 
 5. Draw a line of symmetry in an isosceles triangle. 
 
 6. Draw all the lines of symmetry in an equilateral triangle. 
 7 Draw all the lines of symmetry in a square. 
 
 8. Draw all the lines of symmetry in a rectangle. 
 
 9. Make a trapezoid such that you can draw in it a line of symmetry. 
 
 10. Give instances of figures which, placed together with a side common, 
 will make a symmetrical figure. What is the line of symmetry? 
 
 11. Draw all the lines of symmetry in a regular hexagon. How many are 
 there ? 
 
134 GEOMETRY FOR BEGINNERS. 
 
 REVIEW OF CHAPTER VI. 
 SYNOPSIS. 
 
 I 
 
 1. A polygon is a plane figure bounded by straight lines. 
 
 2. Polygons receive different names according to the number of their sides. 
 
 3. The sum of the angles of a polygon = 180 X the number of sides less 
 two. 
 
 4. Two polygons are equal, if they can be divided into the same number 
 of triangles, similar each to each, and similarly placed. 
 
 5* A polygon is equilateral, if its sides are equal ; equiangular, if its angles 
 are equal ; regular, if both sides and angles are equal. 
 
 6, Each angle of a regular polygon may be found by dividing 360 by the 
 number of sides, and then subtracting the quotient from 180. 
 
 7. In every regular polygon there is a point called the centre, equally dis- 
 tant from its corners and also from its sides. 
 
 8 The greater radius of a regular polygon is the distance from the centre 
 to a corner ; the less radius the distance from the centre to a side ; the 
 angle at the centre the angle included between two greater radii. 
 9 Paving and quilting furnish examples of the practical use of regular 
 polygons. 
 
 10 Two plane figures are symmetrical, if either of them, when revolved 
 about a certain line as an axis through half a revolution, can be made 
 to fall upon and coincide with the other. The line is called the line of 
 symmetry. 
 
 11. There are figures in which several lines of symmetry may be drawn, each 
 dividing the figure into two symmetrical figures. 
 
 EXERCISES. 
 
 1. Make a hexagon having four right angles. 
 
 2. Make an octagon having only right angles or their multiples. 
 
 3. Make three polygons with reentrant angles. 
 
 4. Divide a hexagon into other figures in several different ways. 
 
 5. Find the sum of the angles of a polygon with 24 sides. 
 
 6. Find each angle of a regular icosagon. 
 
 7. Eight angles of a decagon are equal, and each of the other two is twice 
 one of the former ; find all the angles. 
 
CHAPTER VI. REVIEW. 135 
 
 8. In a dodecagon each angle taken in order is i 20' more than the 
 preceding ; find all the angles. 
 
 9. Find the angles of a pentagon, if they are to each other as the num- 
 bers I, 2, 3, 4, 5. 
 
 10. Draw a heptagon ; then construct another equal to it. 
 
 11. Construct a regular decagon. 
 
 12. Make a five-rayed star, and prove that the sum of the five acute angles is 
 equal to 1 80. 
 
 Hint. The star is made by prolonging the sides of a regular pentagon till 
 they meet, and then erasing the sides of the pentagon. 
 
 13. Make a seven-rayed star, and find the sum of the seven acute angles. 
 14:. Can you make a pattern of squares, pentagons, and icosagons? 
 
 15. Can you make a pattern of pentagons and decagons? 
 
 16. Construct a triangle, and upon its longest side a symmetrical triangle. 
 
 17. Construct a rhombus ; then construct upon one of its sides a symmetrical 
 rhombus. 
 
 18. Draw a polygon with two reentrant angles, and then construct a sym- 
 metrical polygon. 
 
 19. Make a right triangle in which a line of symmetry can be drawn. 
 
 20. How many lines of symmetry can be drawn in a regular polygon of any 
 number (n) of sides ? 
 
136 GEOMETRY FOR BEGINNERS. [ 124. 
 
 CHAPTER VII. 
 AREAS. 
 
 CONTENTS. I. Units of Area ( 124, 125). II. The Areas of Polygons ( 126- 
 135). III. Practical Exercises and Applications ( 136-141). IV. Theorem of 
 Pythagoras ( 142, 143). V. Transformation of Figures ( 144-150). VI. Par- 
 tition of Figures ( 151-155). 
 
 L Units of Area. 
 
 124. Surfaces, like lines, are measured by choosing a unit, 
 and then finding how often this unit is contained in the surface 
 which we wish to measure. 
 
 The unit chosen for this purpose must be a magnitude of the 
 same kind as that which is to be measured ; in other words, it 
 must be a surface. The most convenient units to employ are 
 squares whose sides are equal to the iinits of length. 
 
 Chief English units : the SQUARE INCH (sq. in.), the SQUARE FOOT 
 (sq. ft.), the SQUARE YARD (sq. yd.), and the SQUARE MILE. 
 30^ sq. yds. = i SQUARE ROD, and 160 sq. rods = i ACRE. 
 
 The metric units (with their abbreviations 1 and relative values) 
 are as follows : 
 
 SQUARE KILOMETER (qkm) = 1,000,000 square meters. 
 
 SQUARE HECTOMETER (qhm) = 10,000 
 
 SQUARE DEKAMETER (qdkm) = 100 
 
 SQUARE METER (qm) = i 
 
 SQUARE DECIMETER (qdm). = o.oi 
 
 SQUARE CENTIMETER (qcm) = o.oooi 
 
 SQUARE MILLIMETER (qmm) = o.oooooi 
 
 i Formed by prefixing q, the first letter of the Latin word quadra (a square) to 
 the corresponding linear abbreviations. 
 
I25-] 
 
 CHAPTER VII. - AREAS. 
 
 137 
 
 D 
 
 The square dekameter is usually called an AR (a), and the 
 square hectometer a HECTAR (ha). They are employed chiefly 
 in measuring land. 
 
 NOTE. i ba = 2.471 acres ; ii m = 10.764 sq. ft. ; ii dm = 15.5 sq. in., very nearly. 
 
 125. In the table of metric units above given, it will be 
 observed that each unit is 100 times greater than the unit next 
 following. The reason of this lies in the fact that, of the corre- 
 sponding linear units, one is 10 times greater than the other (see 
 page 41). 
 
 Suppose that A B (Fig. 121) repre- 
 sents a length of one meter. Con- 
 struct upon AB the square A B C D ; 
 then this square will represent one 
 square meter. Divide A B and A C 
 each into ten equal parts, and through 
 all the points of division draw lines 
 parallel to the adjacent sides. The 
 parallels through the points of division 
 
 of AB divide the square into 10 equal A Fia . I2I & 
 
 rectangulars, each io dm long and i dm 
 
 wide ; the other set of parallels subdivides each rectangle into 
 10 equal squares, each equal to one square decimeter ; there- 
 fore, one square meter must contain" exactly 10 X 10=100 square 
 decimeters. 
 
 The same reasoning will show that one square decimeter = 10 
 X 10 = 100 square centimeters, etc. ; and if the ratio of the two 
 linear units were any other number but 10, it may be shown in 
 the same way that the ratio of the corresponding units of surface 
 would be found by multiplying that number by itself; in other 
 words, by squaring it. 
 
 The ratio of two units of surface is always the square of the 
 ratio of the corresponding units of length. 
 
138 GEOMETRY FOR BEGINNERS. [ 126. 
 
 Exercises. 1. What is the ratio of a square foot to a square inch? a 
 square foot to a square yard? a square mile to a square foot? 
 
 2. How many square feet are there in one acre ? 
 
 3. How many acres are there in one square mile ? 
 
 4. What is the ratio of an ar to a square meter ? an ar to a hectar? a 
 square centimeter to a hectar? a square hectometer to a hectar? 
 
 5* Reduce to square meters the following: 4<i km ; I5 ha ; 9-87 a ; o.o84 ha ; 
 97,6751. 
 
 6. Reduce to hectars the following: 63<i km ; 39. 2 a ; 36ooi m ; $6,4-75 qAm > 
 4,ooo,oooi mm . 
 
 7. Subtract i a from i ha , and give the answer in square meters. 
 
 8. How many house-lots, each containing 2 a , are there in a field which 
 contains 8 ha ? 
 
 9. A man bought 3 ha of land at $200 per hectar, and sold it for $2.50 per 
 ar. Did he gain or lose, and how much? 
 
 10. Which is the greater farm, 200 hectars or 500 acres? 
 
 11. Divide 6 ha into 64 equal lots ; how many square meters in each lot? 
 
 12. What is the difference between 3 square meters and 3 meters square? 
 
 II. Areas of Polygons. 
 
 126. Definition. The number of times a unit of surface 
 is contained in any surface, followed by the name of the unif, is 
 called the AREA of the surface. 
 
 NOTE. Hence units of surface are often called units of area, as at the begin- 
 ning of this chapter. 
 
 In order to find the area of a surface, it might at first thought 
 seem necessary to apply a unit of area to the surface over and 
 over again as many times as possible, just as we measure a line 
 directly by using a yard-stick or a meter-rule. But this method 
 would be very tedious, and in many cases (for instance, a pond, 
 swamp, forest, etc.) utterly impracticable. 
 
 Fortunately, however, Geometry supplies us with indirect meth- 
 ods of measurement which are applicable to all surfaces. It teaches 
 that the value of areas depends upon the lengths of certain lines ; 
 whence it follows that areas can be found by performing certain 
 
127.] CHAPTER VII. AREAS. 139 
 
 simple operations upon the numbers which express the lengths of 
 these lines. 
 
 In explaining these indirect methods we begin with the square 
 and the rectangle, because they can be readily decomposed into 
 square meters, or into squares larger or smaller than the square 
 meter. 
 
 127. The SQUARE. If the side of a square contains a whole 
 number of meters, as six, it is easy to see that (by proceeding 
 exactly as in 125) we can first divide the square into six equal 
 rectangles, then each rectangle into six square meters; so that 
 the square will contain in all 6 X 6 = 36 square meters. 
 
 The same mode of reasoning may be used when the side con- 
 tains meters and subdivisions of a meter. Let, for instance, the 
 side = 6.35 ra ; then, in decomposing the square into smaller squares, 
 we may take as the unit of length one centimeter. In place of six 
 rectangles, there will be 635 ; and in place of 6 X 6 = 36 square 
 meters, there will be 635 X 635 = 403,225 square centimeters = 
 40.3225 square meters, a result obtained simply by multiplying 
 6.35 by itself. 
 
 In short, the number of units of area in a square is always found 
 by multiplying by itself the number of the corresponding units of 
 length in one of its sides. 
 
 Hence, when we multiply a number by itself, we are said to square 
 it ; and it is usual to express the preceding rule more briefly thus : 
 The area of a square is equal to the square of one of its sides. 
 
 Conversely, if the area is known, in order to find the value of 
 one side, we must find a number which, multiplied by itself, will 
 give the numerical value of the area ; in other words, we must 
 extract the square root of the given area. 
 
 For example : if the area of a square = 8.1225 square meters, 
 me side = V8.I225 = 2.85 meters. 
 
 Remark. For the sake of convenience we use the expression 
 
140 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 128. 
 
 "the square of AB" and we write AB*- These expressions do 
 not mean that the line AB is multiplied by itself, but that the 
 number of linear units in the line is multiplied by itself, the result 
 being the number of units of area in the square constructed upon 
 AB as a side. 
 
 128. The RECTANGLE. A rectangle can be decomposed into 
 units of area in the same way as a square, the only difference being 
 
 that the number of units of length in 
 its adjacent sides is not the same. 
 
 Thus, the rectangle ABCD (Fig. 
 122), the sides of which are A= y m , 
 AC=4 m , contains 7x4=28 square 
 meters. 
 
 The side AB is the base of the rect- 
 
 D 
 
 Fig. 122. 
 
 angle, and the side AC is the altitude ( 104) ; so that the number 
 of units of area in a rectangle is found by multiplying the number 
 of linear units in the base by the number of linear units in the 
 altitude. 
 
 Or, more briefly : The area of a rectangle is equal to the pro- 
 duct of its base by its altitude. 
 
 If the area and base are known, how can the altitude be 
 found? If the area and altitude are known, how can the base 
 be found? 
 
 Remark. In practice we must be careful to express both 
 base and altitude in terms of the same linear unit ; the area will 
 then be expressed in terms of the corresponding unit of area. 
 
 For example : the base of a rectangle = i2 m , the altitude = 8o cm ; 
 find the area. Here we must not multiply 12 by 80 ; but, if we 
 choose the meter as the linear unit, we must multiply 12 by 0.8, 
 because 8o cm = 0.8. Answer, 9.6 square meters. 
 
 If we take the centimeter as the linear unit, what will the 
 answer be? 
 

 
 CHAPTER VII. AREAS. 
 
 141 
 
 Fig. 123. 
 
 129. The PARALLELOGRAM. It is easy to transform a paral- 
 lelogram into an equivalent rectangle. 
 
 In the parallelogram ABCD (Fig. 123) erect perpendiculars 
 at the points A and B of the base ; 
 they cut, one of them the opposite 
 side, the other the opposite side pro- 
 longed, in the points F and E ; and 
 the figure ABEF is a rectangle having 
 the same base and altitude as the 
 parallelogram. 
 
 The right triangles AC E and B DP are equal (why?) ; hence, 
 whichever of these triangles we add to the quadrilateral ABCF, 
 we shall obtain the same area. If we add AAC, we obtain the 
 rectangle ABEF] and if we add ABFD, we obtain the paral- 
 lelogram ABCD: therefore, the parallelogram ABCD is equal 
 to the rectangle ABEF. 
 
 Theorem. A parallelogram is equivalent to a rectangle having 
 the same base and altitude. 
 
 NOTE. This transformation maybe easily shown to the eye by cutting from 
 stiff paper the trapezoid A CBF {Fig. 123), and the right triangle A E C, and then 
 placing the triangle in the two positions shown in the figure. 
 
 Corollaries. i. By combining this theorem with 128, we 
 see that the area of a parallelogram is equal to the product of its 
 base by its altitude. 
 
 2. All parallelograms having equal bases and 
 equal altitudes are equivalent. 
 
 Is the converse of Corollary 2 true ? Draw 
 a diagram to illustrate. 
 
 Construct several equivalent parallelograms 
 differing in shape. 
 
 Remark. In a rhombus the diagonals 
 bisect each other at right angles ( 104). 
 
 Construct about the rhombus ABCD (Fig. 
 124) a rectangle PQMN, by drawing through the corners of the 
 
 Fig. 124. 
 
142 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 130. 
 
 rhombus lines parallel to its diagonals. It is easy to show that this 
 rectangle is composed of eight equal -right triangles, of which the 
 rhombus contains four (how?). Now, since the base and altitude 
 of the rectangle are equal to the two diagonals of the rhombus, its 
 area is equal to the product of these diagonals. Hence, the area 
 of the rhombus is equal to half the product of its diagonals. 
 
 130. The TRIANGLE. A triangle ABC (Fig. 125) may be 
 regarded as half a parallelogram having an equal base and equal 
 altitude ( 101). We have only to draw through two corners, B 
 and C, of the triangle lines parallel to the op- 
 posite sides ; then, since &ABC=% ABDC, 
 and ABDC = AB X CE, therefore A ABC 
 = %AxC. That is, - 
 
 The area of a triangle is equal to half the 
 product of its base by its altitude. 
 
 Fi Corollary. All triangles having equal 
 
 bases and equal altitudes are equivalent. 
 
 NOTE. In finding the area of a triangle, we may either multiply the base by the 
 altitude, and then divide by 2, or we may multiply half the base by the altitude, or 
 we may multiply half the altitude by the base. For example : if the base = 14"', and 
 
 the altitude = 8 m , the area = 
 
 14X8 
 
 14 X - = 56 square meters. 
 
 131. The TRAPEZOID. A trapezoid ABCD (Fig. 126) is 
 divided by the diagonal BC into two triangles, ABC and 
 BCD, whose bases, AB and CD, are the 
 parallel sides of the trapezoid, and whose 
 common altitude CE is the altitude of the 
 trapezoid. 
 
 Since A ABC = AB X CE, 
 and A BCD = } CD X CE; 
 therefore, the trapezoid ABCD = %(AB+ CD} X CE. 
 That is, the area of a trapezoid is equal to half the product of 
 the sum of the parallel sides by the altitude. 
 

 
 CHAPTER VII. AREAS. 
 
 143 
 
 132. Any QUADRILATERAL. To find the area of any quadri- 
 lateral, divide it by a diagonal into two triangles, find the areas cf 
 these triangles, and add the results together. 
 
 133. A REGULAR POLYGON. We know ( 119, Corollary i) 
 that a regular polygon ABCDEF (Fig. 127) is composed of as 
 many equal isosceles triangles as there 
 are sides. Therefore its area may be 
 found by adding together the areas of 
 these triangles. The altitude of each 
 triangle is equal to the less radius of 
 the polygon, so that the area of each 
 triangle is half the product of its base 
 by the less radius. Therefore the area 
 of the polygon is half the product of the 
 sum of all the bases (that is to say, the 
 perimeter of the polygon) by the common altitude (that is to say, 
 the less radius of the polygon) . 
 
 The area of a regular polygon is half the product of its perimeter 
 by its less radius. 
 
 134. Any POLYGON. The area of any polygon may be found 
 by dividing it into triangles. Fig. 128 shows two ways of doing 
 this. In either case, if we 
 compute the area of each 
 triangle separately, and 
 then add the results, we 
 
 shall find the area of the \ \ / / \ / 
 
 polygon. 
 
 In each of these cases 
 
 Fig. I2J. 
 
 Fig. 128. 
 
 how many lines require to be measured directly in order to com- 
 pute the area? 
 
 But it is generally easier to find the area of an irregularly-shaped 
 
144 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 134- 
 
 polygon (as a field, farm, etc.) by division into right triangles and 
 
 trapezoids, as shown in Fig. 
 129. Join the two most dis- 
 tant corners of the polygon 
 by a straight line, and let fall 
 from the- other corners per- 
 pendiculars to this line taken 
 as a common base. The 
 polygon is thus divided into 
 right triangles and trapezoids, 
 the areas of which are then 
 computed separately, and 
 
 their sum is obviously equal to the area of the polygon. 
 
 For example : let (Fig. 129) Bb = 6.8 m ; Cc = io.6 m ; Dd = 
 io.i m ; Ff = 8.3 m ; Gg = 6.2 m ; Nh = 9.2 : also let Ab = 5.6; 
 th = 2 .6 m ; /^ = 4 .2 m ; ^= 4 .6 m ; gf = 3 m ; fd = 2.8 m ; dE 
 
 The work of computation may be arranged as follows : 
 
 
 FACTORS. 
 
 
 PARTS OF THE 
 POLYGON. 
 
 BASES, OR SUMS 
 
 
 PRODUCTS. 
 
 
 OF PARALLEL SIDES. 
 
 ALTITUDES. 
 
 
 &ABb 
 
 Bb= 6.8 
 
 Ab= 5.6 
 
 38.08 
 
 Trap. Bbc C 
 
 Bb + Cc= 17.4 
 
 bc= 6.8 
 
 118.32 
 
 Trap. CcDd 
 
 Cc -}- Dd= 20.7 
 
 c d= 10.4 
 
 215.28 
 
 l\DdE 
 
 Dd 10.1 
 
 <*= 5-8 
 
 58.58 
 
 &FfE 
 
 Ff= 8.3 
 
 /= 8.6 
 
 7I-38 
 
 Trap. Ffg G 
 
 *f+ Gg = 14.5 
 
 /^r= s 
 
 43-50 
 
 Trap. GghH 
 
 Gg+Hh~- 15.4 
 
 gh= 8.8 
 
 I35-5 2 
 
 kAhH 
 
 Ah- 8.2 
 
 ^>^= 9.2 
 
 75-44 
 
 
 2)756.10 
 
 Polygon ABCDEFGH = 
 
 378.o5qm 
 
135-3 
 
 CHAPTER VII. AREAS. 
 
 145 
 
 NOTE i. In the preceding table, in place of dividing each product by 2, we first 
 add the products, and then divide their sum by 2. 
 
 NOTE 2. In case the interior of the polygon is inaccessible, Fig. ijo illustrates 
 one way of meeting the difficulty. Construct a rectangle which shall completely 
 enclose the polygon, and, from all the corners of the polygon which are not in the 
 
 Fig. 130. 
 
 sides of the rectangle, let fall perpendiculars to the sides of the rectangle. In this 
 way a number of right triangles and trapezoids are formed, the areas of which can 
 be found in the usual way ; and it is evident that if the sum of their areas is sub- 
 tracted from the area of the rectangle, the remainder will be equal to the area of the 
 polygon. 
 
 135. It will now be clear in what way Geometry supplies us 
 with indirect methods of finding areas. In the first place, it teaches 
 that the area of a rectangle (the square is simply a rectangle with 
 equal sides) depends entirely on the lengths of its two sides, and 
 can therefore be found as soon as the lengths of these sides are 
 known ; it then shows that any parallelogram can be transformed 
 into an equivalent rectangle, that a triangle is half a parallelogram 
 of the same base and altitude, and that a trapezoid consists of two 
 triangles of equal altitude ; lastly, it shows that any polygon what- 
 soever may be decomposed into triangles and trapezoids. 
 
 Remark. In some cases the rule for finding an area may be 
 very briefly expressed by a formula in which letters are used to 
 represent the values of the various quantities. 
 
 Let 6" denote in all cases the area, a the side of a square, a and 
 b the adjacent sides of a rectangle, a and h the base and altitude 
 
146 GEOMETRY FOR BEGINNERS. [ 136. 
 
 of a parallelogram or triangle, a and c the parallel sides, h the 
 altitude of a trapezoid, p and r the perimeter and less radius of 
 a regular polygon. Then, 
 
 For the square, 
 For the rectangle, 
 For the parallelogram, 
 For the triangle, 
 For the trapezoid, 
 For the regular polygon, 
 
 S = a 2 , 
 S = a X b, 
 S = a x h, 
 S = \ a X h, 
 S = Ka + c) X h, 
 S=fpxr, 
 
 [3-1 
 [4-] 
 
 T5-] 
 [6.] 
 
 L7-] 
 [8.] 
 
 III. Practical Exercises and Applications. 
 
 NOTE. In solving most of the exercises, it is well to draw (free-hand) a figure 
 representing as nearly as possible the true shape of the polygon. 
 
 136. THE SQUARE (S = a*). 
 
 In a square, given, 
 
 1. The side = 23; find the area. 
 
 2. The side 35; find the area. 
 
 3. The side = 4| ft ; find the area. 
 
 4. The side = 5fy ds ; find the area. 
 6. The side = 6.25; find the area. 
 
 6. The side = 48.72; find the area. 
 
 7. The area =6259; find the side. 
 
 8. The area = 515. 29<i m ; find the side. 
 
 9. The area = 63.2025<3 m ; find the side. 
 
 10. The area = 54.76*; find the side in meters. 
 
 11. The area = i6.32i6 ha ; find the side in meters. 
 
 12. The area = 78.8544^; find the side in dekameters. 
 
 13. One side of a square garden measures 37.6; find its area. 
 
 14. A square park contains 64 acres. How many trees, 2oy ds apart, can 
 be set out around it ? 
 
 15. A square park contains 64 ha . How many trees, 2O m apart, can be set 
 out around it ? 
 
 16. The three dimensions of a room are each equal to 6.4. What will it 
 cost to paper its walls, if the price is $0.15 per square meter ? 
 
 17. Find the cost of looo panes of glass, each o.48 m square, if the glass 
 is worth $1.40 per square meter. 
 
1 3 7.] CHAPTER VII. AREAS. 147 
 
 18. Area of the State of New York = 47,156 square miles. Find the side 
 of a square having the same area. 
 
 19. Construct a square, and then find its area. 
 
 137. THE RECTANGLE (S = a X i>}. 
 
 In a rectangle, given, 
 
 20. Length = 6o m , breadth = 24; find the area. 
 
 21. Length = 2 m , breadth = 8o cm ; find the area. 
 
 22. Length = 8.5, breadth = 40; find the area. 
 
 23. Length - 92 dkm , breadth = i6.37 dkm ; nnc1 - tne area in hectars. 
 
 24. Length = 28f ft , breadth = i"jl ft ; find the area in sq. yds. 
 
 25. Length = 627?^, breadth = 435^'^; find the area in acres. 
 
 26. Area = 3601, length = 30; find the breadth. 
 
 27. Area = 5.6% length = i6 m ; find the breadth. 
 
 28. Area .= 50^, length = f km ; find the breadth. 
 
 29. Area = 6 acres, breadth = 44^ ; find the length. 
 
 30. Area = 767.24<i m , breadth = 24.5; find the length. 
 
 31. Area = 7.2 ha , breadth = o.25 kil ; find the length. 
 
 32. Find the area of the surface of a table 1.58 long, i.2 m wide. 
 
 33. How many ars in a field 87.5 by 45. 8 m ? 
 
 34. Find the area of a field 24O m long and T \ as wide. 
 
 35. The perimeter of a rectangle = 24, the length = 9.2; find the breadth 
 and the area. 
 
 36. If the area of a cellar 48 long is 15001, what is its breadth? 
 
 37. The perimeter of a rectangle = 86.2 m , the breadth = 12. 6 m ; find the 
 length and the area. 
 
 38. The area of a rectangle = I44 qm , the breadth = 6 m ; find the length and 
 the perimeter. 
 
 39. The area of a rectangle = S, the length a ; find the breadth and the 
 perimeter. 
 
 40. Construct a rectangle, and then find its area. 
 
 41. Find the area of the floor of the room. 
 
 42. The perimeter of a rectangle = 72, and the length is twice the breadth; 
 find length, breadth, and area. 
 
 43. If the perimeter = 96"*, and the length is to the breadth as 7:5, find 
 length, breadth, and area. 
 
 44. If the perimeter = 26, and the length is 2-5 m more than the breadth, 
 find length, breadth, and area. 
 
148 GEOMETRY FOR BEGINNERS. [ 137. 
 
 45. How much larger is a reotangle if the length is doubled ? if the breadth 
 is doubled ? if both are doubled ? 
 
 46. The sides of a rectangle are 42 and 6o m ; find the side of a square 
 equivalent to the rectangle. 
 
 47. If the side of a square I2 m , find the sides of an equivalent rectangle. 
 Is there any thing peculiar about this exercise ? 
 
 48. A square and a rectangle have the same perimeter, 6o m , and the rect- 
 angle is twice as long as it is wide. Find their areas. 
 
 49. Construct a rectangle i6 dm long, 4 dm wide ; then another, I5 dm , 5 dm 
 wide ; then another, I4 dm , 6 dm wide ; and so on, until you reach a square. 
 Then compare perimeters and areas. Which has the greatest area ? 
 
 50. A chain 8o m long is made to enclose a rectangle, one side being 25. 
 How much more area would it enclose if the figure were a square ? 
 
 51. A and B exchange fields of equally good soil. A's field is rectangular 
 in shape, 4 times as long as wide, and with a perimeter = iooo m ; B's field is 
 in the shape of a square, the perimeter being 8oo m . Which has the best of 
 the bargain ? 
 
 52. From a rectangular field 528 m by 240 a piece 6o m square is sold. 
 How much is left ? 
 
 Find the number of linear meters of carpeting required for floors of the 
 following dimensions : 
 
 53. Floor, 8 m by 6.6 m ; carpet, i m wide. 
 
 54. Floor, 9.5 by 7.7; carpet, 8o cm wide. 
 
 55. Floor, I2 m by 8.4 m ; carpet, i.2 m wide. 
 
 56. Floor, 24 ft by i6 ft 8 in ; carpet, 3O 5n wide. 
 
 Find the cost of carpeting rooms, their dimensions and the width and price 
 of carpeting being as follows : 
 
 57. Floor, io m by 6.5; carpet, go cm wide, at $1.25 per meter. 
 
 58. Floor, 11.4 by 5 m ; carpet, 1.4 wide, at $2 per meter. 
 
 59. Floor, 8 m by 7.37; carpet, 6o cm wide, at $0.75 per meter. 
 
 60. Floor, i8^ ft by 14^; carpet, |y d wide, at $1.60 per yard. 
 
 61. How much oil-cloth will it take to line the inside of a box 2.6 m by i.2 m 
 by o.8 m ? 
 
 62. How many tiles 8 in square will cover a floor 36 ft by 2O ft ? 
 
 63. What will it cost to paint the floor and walls of a room 8 m long, 6 m 
 wide, 3 m high, at one cent per square centimeter ? 
 
 64. Find the cost of glazing 12 windows, each 6 ft 6 in by 3 ft 4 in , at $0.50 
 per square foot ? 
 
 65. A room is io m long, 6 m wide, and 3-6 m high. There are four windows, 
 
138.] CHAPTER VII. AREAS. 149 
 
 each 2.2 by 1.2; two doors, each 3 by i.8 m ; and the base-board is 9O cm 
 high. Required (zV) the cost of plastering its walls and ceiling, at $0.40 per 
 square meter ; (zz.) the cost of papering its walls with paper 8o cm wide and 
 io m in a roll, at $1.75 a roll put on, and with a gilt border next the ceiling 
 that costs $0.20 per meter. 
 
 66. A man has a rectangular garden 68 m long and 42"* wide. He make:; 
 around it a path 3.4 wide. What area is left for other purposes ? 
 
 67. A field is 6oo m by 240. How much of its area will be taken by a 
 road 2O m wide passing through the middle and parallel to the longer side ? 
 
 68. Through the middle of a rectangular garden i8o m by 76, two paths 
 at right angles to each other and parallel to the sides are made, each i.4 m 
 wide. Find the area (z.) of the paths ; (z'z.) of what remains. 
 
 69. A mirror is 2.8 m by i.9 m , and the wooden frame is 4 cm wide. Find the 
 area of the glass and also of the frame. 
 
 70. A table 2.3 m by i.2 m is covered with cloth, which overlaps 3 cm on all 
 sides. How much cloth is needed ? 
 
 71. The bridge over the Elbe at Dresden is 400 long and io.3 m wide. 
 How many stones i8 cm by 15 are required to cover the roadway, the foot- 
 paths having together a width of 2.45? 
 
 72. How many shingles 8 in wide, laid 6 in to the weather, will cover the 
 roof of a house, the ridge being 52** long, the slant side of the roof 24 ft , and 
 the bottom course on each side being double ? 
 
 138. THE (oblique) PARALLELOGRAM (S = a X h) . 
 
 In a parallelogram, given, 
 
 73. Base = 27 cm , altitude = 22 cm ; find the area. 
 
 74. Base 91. 25 m , altitude = 40; find the area. 
 
 75. Base = 640 dkm , altitude = i8o dkm ; find the area in hectars. 
 
 76. Area = 84*, base = 105; find the altitude. 
 
 77. Area = i ha , altitude = I25 m ; find the base. 
 
 78. How far apart are the sides of a rhomboid containing 2lO^ m , if one of 
 these sides = I5 m ? 
 
 79. Perimeter of a rhombus 6 m , distance apart of two parallel sides = 
 75 cm ; find the area. 
 
 80. Find the area of a rhombus whose diagonals are 1 6 m and 1 2 m . 
 
 81. The area of a rhombus 24O ( i m , one diagonal = io m ; find the other. 
 
 82. Find the area of a rhombus if the sum of the diagonals = 2O m and their 
 ratio is 3 : 5. 
 
150 GEOMETRY FOR BEGINNERS. [ 139. 
 
 83. Construct a rhomboid, and then find its area. 
 
 84. Construct a rhombus, and then find its area. 
 
 85. Construct a rhombus and a square, both having the same perimeter. 
 Which has the greater area ? Prove your answer. 
 
 86. A rhombus and a rectangle have equal perimeters. A side of the 
 rhombus = 5 m , and a side of the rectangle = 3-5 m . Find the perimeter and 
 the two areas, if the altitude "of the rhombus 3.5. 
 
 139. THE TRIANGLE (S=^ax h). 
 
 In a triangle, given, 
 
 87. Base = 50 m , altitude = 40; find the area. 
 
 88. Base = 80.5, altitude = 6o m ; find the area. 
 
 89. Base = 2.4 km , altitude = i.875 km ; ^ n( ^ tne area * n nec tars. 
 
 90. Area = 45O ha , base = 5 km ; find the altitude. 
 
 91. Area = 4O ( i m , altitude = 12.5; find the base. 
 In a right triangle, given, 
 
 92. The legs 4 and 5.6; find the area. 
 
 93. The legs 34 and 25; find the area in ars. 
 
 94. The area = 15*, one leg = 50; find the other leg. 
 
 95. Sum of the legs = 2O m , their ratio = 2:3; find the area. 
 
 96. Sum of the legs = 70, and one is 6 times the other ; find the area. 
 
 97. Construct a triangle, and then find its area. 
 
 98. If the base of a triangle = 50, what must be its altitude in order that 
 its area may be the same as that of a rectangle whose sides are 40 and 25? 
 
 99. Compare (that is, find the ratio of) the areas of a square, and of a tri- 
 angle whose base is one side of the square and vertex is in the opposite side. 
 Does it matter where in the opposite side the vertex is ? Why not ? 
 
 100. Find the side of a square equivalent to a triangle with the base 90 
 and the altitude 20. 
 
 101. The base of a triangle = 6 m , altitude = 4. Find the base of a tri- 
 angle with the same altitude and twice as large ; 3 times as large ; 4 times 
 as large ; 10 times as large. 
 
 102. Two triangles have the same base, I2 m , and their altitudes are 8 m and 
 24. Find the ratio of their areas. 
 
 103. Two triangles are each I5 m high, and their bases are 20 and 30. 
 Find the base of a triangle equivalent to their sum. 
 
 104. From a triangle with base = 15, and altitude = 8 m , a triangle with 
 the base = 6 m is cut off by a line drawn from the vertex. What part of the 
 whole triangle is the triangle cut off ? 
 
14-] CHAPTER VII. AREAS. 151 
 
 105. Same exercise, making the altitude I2 m instead of 8 m . 
 
 106. Find the entire surface of a pyramid, with a square base (see page 4, 
 Fig. 2, IV.), if a side of the base = 2 m , and the altitude of one of the lateral 
 sides = 4 m . 
 
 107. The roof of a tower consists of 4 isosceles triangles. The base of 
 each triangle = 2. 2 m , the altitude = 5-4 m . How many square meters of tin 
 are required to cover the roof ? 
 
 140. THE TRAPEZOID (S 1 = [# + c] X /*), AND THE TRAPEZIUM. 
 
 108. Find the area of a trapezoid, if the parallel sides are 2O cm and 3O cm , 
 the altitude I2 cm . 
 
 109. Area of a trapezoid = 3691, the parallel sides are i6 m and 2^ m . Find 
 the altitude. 
 
 110. Area of a trapezoid = I24.8i m , altitude = 6.4 m , one of the parallel 
 sides = I2.8 m ; find the other. 
 
 111. A garden has the shape of a trapezoid. The parallel sides are 6o m 
 and 32, and their distance apart I24 m . Find the side of an equivalent square. 
 
 112. Construct a trapezoid, and then find its area. 
 
 113. A board is 8.5 long, o.45 m wide at one end, and o.36 m wide at the 
 other. Find its value at $0.50 per square meter. 
 
 114. A wall I2 m long is 3.15 high at one.end and 2."j m at the other. Find 
 its area. 
 
 115. The area of a field shaped like a trapezoid = I5 ha . The parallel sides 
 are i68.7 m and 144. How far apart are these sides ? 
 
 116. A court-yard has the shape of a trapezoid. The parallel sides are 
 20.4 and 18.2, and their distance apart is i6 m . How many stones, each 
 containing 5 < i dm , will be required to pave the yard ? 
 
 117. What will it cost to shingle a trapezoidal roof, the parallel sides 
 being 15.8 and n.6 m , their distance apart 6.2 m , and the price for shingling 
 being $1.20 per square meter ? 
 
 118. A diagonal in a trapezium = 32, and the altitudes of the triangles 
 made by the diagonal are i8 m and 2O m . Find the area of the trapezium. 
 
 119. Construct a trapezium, and then find its area. 
 
 120. The area of a four-sided field = 24 ha . The altitudes of the triangles 
 made by a diagonal are 140'" and 158. Find the length of the diagonal. 
 
 121. In a quadrilateral, a diagonal = 40, the altitude of one triangle = I2 m , 
 and the altitude of the other triangle is four times as great. Find the area of 
 the quadrilateral. 
 
152 GEOMETRY FOR BEGINNERS. [ 141. 
 
 141. POLYGONS (if regular, S=#xr). 
 
 The less radius r of a regular polygon has a fixed ratio (found by Trigo- 
 nometry) to the length of one side. 
 
 In a regular triangle, r = one side X 0.2887. 
 In a regular quadrilateral, r one side X 0.50x30. 
 
 In a regular pentagon, r= one side X 0.6882. 
 
 In a regular hexagon, r = one side X 0.8660. 
 
 In a regular octagon, r = one side X 1.2071. 
 
 In a regular decagon, r = one side X 1.5388. 
 
 In a regular dodecagon, r = one side X i .8660. 
 
 122. One side of a regular triangle 2o cm ; find the area. 
 
 123. One side of a square = 40; find the area. 
 
 124. One side of a regular pentagon = 5O cm ; find the area. 
 
 125. The perimeter of a regular hexagon = 3; find the area. 
 
 126. The perimeter of a regular octagon = i6 m ; find the area. 
 
 127. The perimeter of a regular decagon = 2OO m ; find the area. 
 
 128. One side of a regular dodecagon = 6o m ; find the area. 
 
 129. The area of a regular hexagon =&*, the less radius = 240; find one 
 side and the perimeter. 
 
 130. The area of a regular octagon i6o<i m , one side= 5 m ; find the less 
 radius. 
 
 131. Construct a regular pentagon, and then find its area. 
 
 132. Construct a regular hexagon, and then find its area. 
 
 133. Find the side of a square equivalent to a regular hexagon one side 
 of which = 4 m . 
 
 134. Find the entire surface of a hexagonal prism (see page 4, Fig. 2, III.), 
 if the altitude = i.2 m , and one side of the base = 4O cm . 
 
 135. Required the total area of the hexagons in Fig. //j, if a side of each 
 hexagon = 36"". 
 
 136. Compute the total area covered by the figures in Fig. 116, if a side of 
 each figure = 1 .4 m . 
 
 137. Compute the total area covered by the figures in Fig. 127, if a side 
 of each figure = 2 m . 
 
 138. The floor of a church is 30 by I5 m , and is paved with regular hex- 
 agonal stones, alternately white and black, a side of each stone being I2 cm . 
 How many stones of each kind were required, two per cent being added to 
 fill the corners ? 
 
CHAPTER VII. AREAS. 
 
 153 
 
 139. A garden in the form of an irregular hexagon can be divided into the 
 following triangles : 
 
 First triangle : base = 36.6 m , altitude = 6.6 m . 
 Second triangle : base 42.4, altitude = 2O m . 
 Third triangle : base = 42.4, altitude = 22 m . 
 Fourth triangle : base = 28.4 m , altitude 9.8 m . 
 How many ars does the garden contain? 
 
 140. Construct two equal heptagons ; then find the area of the first hepta- 
 gon by one of the methods illustrated in Fig. 128, and the area of the second 
 by the method illustrated in Fig. 129. IT the results do not agree, what is the 
 reason ? 
 
 IV. Theorem of Pythagoras. 
 
 142. If we construct a right triangle (Pig. 131), having for its 
 legs three and four units of length, we shall find that the hypote- 
 nuse will contain exactly five of these 
 units of length. If we then construct 
 squares upon the three sides of this 
 triangle, and divide these squares into 
 square units, as shown in the figure, 
 it will be evident that the square con- 
 structed upon the hypotenuse is just 
 equal to the sum of the squares con- 
 structed upon the two legs ; for, the 
 former square contains 25 square units, 
 and the latter squares 16 and 9 of 
 these units, respectively. 
 
 This is a simple example of a very useful theorem called, from 
 the name of the discoverer, the Theorem of Pythagoras. 1 
 
 Theorem. In every right triangle the square of the hypote- 
 nuse is equal to the sum of the squares of the two legs. 
 
 Several different proofs of this theorem have been devised, two 
 of which we shall give. 
 
 Fig. 131. 
 
 1 Pythagoras, born at Samos about 570 B.C., was one of the most renowned phi- 
 losophers and teachers of antiquity. 
 
154 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 
 
 Proof I. Upon the hypotenuse A C (Fig. 132} of a right 
 triangle ABC construct the square ACDE; from D and E 
 
 let fall upon AB prolonged the per- 
 pendiculars DF and EG ; also let fall 
 upon DFi\\Q perpendiculars C//and 
 EK. From this construction, it fol- 
 lows that the right triangles ABC, 
 AE^G, CDff, and DEK, which taken 
 in order we will call I., II., III., IV., 
 are equal (why?) ; also, that BCHF 
 is equal to the square upon BC, and 
 EGFK to the square upon AB (why ?). 
 Now the square ACDE the figure 
 ACHKE 4- the triangles III. and IV.; take from the square 
 these triangles, and place them in the positions of the triangles 
 
 I. and II., then it is clear that the 
 square will be transformed into the 
 figure GB CHKE, which is composed 
 of the squares BCHF and EGFK 
 of the two legs. Hence the square of 
 the hypotenuse ACDE must have 
 the same area as the sum of the 
 squares of the legs. 
 
 NOTE. This proof can be readily shown 
 in a material form by drawing Fig. 132 on 
 cardboard, cutting out the triangles I., II., III., 
 and IV., and then changing the places of two 
 of them as just explained. 
 
 Proof //. (Euclid's^ Proof ).- 
 Upon the sides of the right triangle ABC (Fig. ijj) construct 
 
 1 Euclid was a Greek geometer who lived about 300 B.C. He collected the scat- 
 tered geometrical knowledge then existing into a famous treatise known as Euclid's 
 Elements, a book still in use, but as unsuited for beginners as it is admirable for 
 rigor of logic, and for showing on how few axioms the science of Geometry can be 
 made to depend. 
 
 // 
 
1 43.] CHAPTER VII. AREAS. 155 
 
 squares, and draw BMA.AC. Join BD and C F. &ABD ^ 
 A CF (II. Law of Equality) . &ACF has the same base AF and 
 the same altitude AB as the square ABFG; therefore ( 130) 
 A A CF= %A 2 . And &ABD has the same base AD and the 
 same altitude DM as the rectangle ADMN; therefore &ABD 
 = rectangle ADMN. But if the halves, here the triangles 
 A CF and AB D, are equal, the wholes must also be equal (Axiom 
 II.) ; therefore AB* = rectangle ADMN. In the same way (by 
 joining A K and BE), it can be proved that BC* = rectangle 
 CE MN. Therefore AB* + B C~ = the sum of the rectangles 
 ADMN and CEMN = the square of the hypotenuse AC. 
 
 Corollaries. I. If from the square of the hypotenuse we 
 subtract the square of either leg, the remainder will be equal to the 
 square of the other leg. 
 
 2. If a and b denote the legs, c the hypotenuse, of a right triangle, 
 
 <2=a 2 + t 2 , a^c 2 -?, and P^^-a 2 . 
 
 3. A useful corollary follows from the second proof, namely : 
 If we construct squares upon the sides of a right triangle, and from 
 the vertex of the right angle draw a line perpendicular to the 
 hypotenuse, this line will divide the square of the hypotenuse into 
 two rectangles, each equal to the square of the adjacent leg. 
 
 143. PRACTICAL EXERCISES AND APPLICATIONS. 
 
 In a right triangle, given, 
 
 1. The legs = 24 and = 7; find the hypotenuse. 
 
 2. The legs = 35 and 12; find the hypotenuse. 
 
 3. The legs = 45 and 28; find the hypotenuse. 
 
 4:. The legs = 4.7 and = 1.104; find the hypotenuse. 
 
 5. The legs = 72 and = i8 m ; find the hypotenuse. 
 
 6. The legs = 2 dm and 15; find the hypotenuse. 
 
 7. The hypotenuse = 65, one leg= 33 m ; find the other leg. 
 
 8. The hypotenuse = 89, one leg 8o m ; find the other leg. 
 
 9. The hypotenuse = 65, one leg = 63 m ; find the other leg. 
 10. The hypotenuse = 42.6"% one leg = I94 m ; find the other leg. 
 
156 GEOMETRY FOR BEGINNERS. [ 143. 
 
 11. How long must a ladder be to reach the top of a wall 6 m high, if the 
 foot of the ladder is 3.2 from the bottom of the wall ? 
 
 12. A ladder 13 long leans against the top of a wall, its foot being 5 
 from the bottom of the wall. How high is the wall ? 
 
 13. The slant height of a roof = 14, its vertical height = 6.2 m . Find the 
 width of the building. 
 
 14. A triangle with the sides 3, 4, 5, is a right triangle. What kind of a 
 triangle is one with the sides 6, 8, 10 ? with the sides 9, 12, 15 ? Give other 
 instances. 
 
 15. If the perimeter of a right triangle = 96 m , and the sides are to each 
 as 3 : 4 : 5> find the sides (see page 50, Exercise 1 8). 
 
 16. If the legs of a right triangle are as 3:4, and the hypotenuse = 40, 
 find the legs 
 
 17. A rope 41 long is tied by one end to the top of a mast, and by the 
 other end to a ring on the deck 9 m from the foot of the mast. How high is 
 the mast ? 
 
 18. The longest side of a meadow in the shape of a right triangle cannot 
 be measured directly, by reason of a swamp. Compute its length, if the other 
 sides are 8o.6 m and i6o m . 
 
 19. How many meters of wall must be built to fenc'e in a garden in the 
 shape of a right triangle, the legs being 1 2O m and 35 ? 
 
 20. A garden has the shape of a right triangle. The hypotenuse i3O m , 
 one leg = 87.5. Find (z.) the area of the garden; (w.) the distance from 
 the right angle to the hypotenuse. 
 
 21. A cannon-ball, fired into the air at an angle of 45 with the horizon, 
 passes over 345 in one second. How high is it at the end of the second ? 
 
 22. Two travellers start together from the same place and travel, each 4 
 miles an hour, one due north, the other due east. When will they be 60 miles 
 apart ? How far apart will they be after each has travelled 80 miles ? 
 
 23. Find the area of an isosceles triangle, if one of the equal sides = io m 
 and the base = 8o m (see 80). 
 
 24. In an isosceles triangle, the base = i8.4 m , the perimeter 64. Find 
 the area. 
 
 25. Prove that in an isosceles right triangle the altitude is equal to half 
 of the base. 
 
 26. The area of an isosceles right triangle = 84<i m ; find the base. 
 
 27. The altitude of an isosceles right triangle = 2 m ; find the side of an 
 equivalent square. 
 
 28. In an ordinary roof the two sides are I2 m apart at the bottom, and the 
 vertical height is 4.5. Find the length of the rafters. 
 
143'] CHAPTER VII. AREAS. 1&7 
 
 29. The side of an equilateral triangle = 20 ; find the altitude and the 
 area (see 80) . 
 
 30. If the altitude of an equilateral triangle = 15, find one side. 
 Hint. The altitude, one side, and half the base, form a right triangle. 
 
 31. If a one side of an eqilateral triangle, prove that the altitude = \ ^'s/3 
 and the area = \ # 2 A/3- 
 
 32. A garden in the shape of an equilateral triangle contains 6cx3^ m ; find 
 the side, 
 
 33. Find the side of a square equivalent to an equilateral triangle whose 
 side = 86 m . 
 
 34. A square contains I2 ha . Find the side of the equivalent equilateral 
 triangle. 
 
 35. The side of a square 6 m . Find its diagonal. 
 
 36. What is the side of a square if the diagonal = 8 m ? 
 
 37. The diagonal of a rectangle = o.86 m , one side = 0.3. Find the other 
 side. 
 
 38. The diagonals of a rhombus are 4 and 6 m . Find its perimeter. 
 
 39. How far apart are the opposite corners of a floor I2 m by i8 m ? 
 
 40. What must be the least thickness of a log from which a rectangular 
 beam 24 cm by 40 can be obtained ? 
 
 41. How much is the diagonal of a square increased if the side is doubled? 
 trebled ? quadrupled ? 
 
 42. How much is the side of a square increased if the diagonal is doubled? 
 trebled? increased tenfold? 
 
 43. Find the less radius of a regular hexagon if the perimeter ico m . 
 
 44. Compute the side of a square equivalent to the sum of the squares whose 
 sides are 12 and i6 m . 
 
 45. Construct a square equivalent to the sum of two given squares. 
 
 46. The side of a square = 2O cm . Construct a square twice as large. Find 
 the length of its side, (z.) by measurement ; (zV.) by computation. 
 
 47. Construct a square equivalent to the sum of the squares whose sides 
 are 3O cm , 4O cm , 5o cm . 
 
 48. The side of a square i6 cm . Construct a square three times as large. 
 
 49. Construct a square equivalent to the difference of two given squares 
 (apply 74). 
 
 50. Construct a rectangle such that the square upon its diagonal shall be 
 five times the square upon one side. 
 
158 GEOMETRY FOR BEGINNERS. [ 144. 
 
 V. Transformation of Figures. 
 
 144. Definition. To TRANSFORM a figure is to change its 
 shape without changing its size ; in other words, to change it to an 
 equivalent figure. 
 
 What instance of this kind of change was given in 129 ? 
 
 There are two ways of effecting these changes : 
 
 1. By computation. Compute such parts of the new figure as 
 are necessary to determine it ; we can then construct it, if we wish 
 to do so. Examples of this method occur among the Exercises 
 in Parts III. and IV. of the present chapter (see Part III., Exer- 
 cises 1 8, 46, 47, 100, 133, and Part IV., Exercises 27, 33, 34). 
 
 2. By direct construction. This was the method employed in 
 129, and a few more useful examples will now be added. 
 
 145. Problem. To transform a triangle ABC (Fig. 134) 
 into an isosceles triangle. 
 
 Analysis. The vertices of all triangles 
 equivalent to ABC, and having AB as a 
 base, must lie in the line CD \\AB($ 130). 
 Of these triangles, that which has its vertex E 
 equidistant from A and B will be isosceles. 
 
 B Hence give the construction required. 
 Fig- 134- 
 
 Exercises. 1. Construct a triangle, and then 
 
 transform it into sen isosceles triangle. 
 
 2. Transform a given triangle into a right triangle. 
 
 3. Transform a given triangle into another having the angle 60. 
 4 Transform a given parallelogram into a rectangle. 
 
 5. Transform a given parallelogram into another having the angle 45. 
 
 146. Problem. To transform a triangle ABC (Fig. 135) 
 into another having a given base b. 
 
 Analysis. If on AB prolonged we take AD = b, and join 
 CD, A A CD will have the given base b, but will be too large by 
 
I47-] 
 
 CHAPTER VII. AREAS. 
 
 159 
 
 *35> 
 
 the &BCD. Therefore construct the &CED ^ &BCD ( 130, 
 
 Corollary) , and take it from A A CD, f_ 
 
 leaving A AE D, which is the trian- 
 gle required. 
 
 Give in full the construction re- 
 quired. 
 
 Exercises. 1. Solve this problem for 
 the case where the given base b is less than 
 the base of the given triangle. 
 
 2. Construct a triangle with the sides 4 dm , 6*, and 8 dm , and then trans- 
 form it, 
 
 (z.) Into an isosceles triangle with the base 5 dm . 
 (z'z.) Into a right triangle with a leg equal to 4 dm . 
 (z'z'z.) Into an equilateral triangle, 
 (zz/.) Into a triangle with the angle 50 and the base 3 dm . 
 
 3. Transform a given parallelogram into another having a given side. 
 
 147. Problem. To transform a triangle ABC (Fig. ij6) 
 into another having a given altitude h. 
 Analysis and construction to be 
 given by the learner (with the aid 
 of Fig. ij6). 
 
 Exercises. 1. Solve this problem for 
 a case in which the new altitude h is less 
 than that of the given triangle. 
 
 2. Construct a triangle with the sides 
 
 Fig. 136. 
 
 4 dm , 6 dm , and 8 dm , and transform it, 
 
 (z.) Into a triangle, the altitude 
 
 (z'z.) Into a triangle with the angle 
 
 70 and the altitude 4 dm . E 
 
 148. Problem. To transform 
 a triangle ABC (Fig. 137) into a A 
 rectangle. 
 
 Fig. 137. 
 
 Construction. At A and B erect perpendiculars to AB, and 
 
160 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 149' 
 
 through Z>, the middle of A C, draw EF II AB. Rectangle 
 ABEF = &ABC. 
 
 Proof. Show that each is half of the rectangle ABHG. 
 . Exercises. 1. Transform a triangle into a parallelogram with the an- 
 gle 60. 
 
 2. Construct a rectangle with the sides 6 dm and 8 dm , and then transform 
 it into (?'.) a triangle ; ('.) a right triangle ; (m.) an isosceles triangle ; 
 (z'z/.) an equilateral triangle. 
 
 149. Problem. To transform a rectangle ABCD (Fig. 
 138) into a square. 
 
 Construction. Make BE = BC, 
 and about O, the middle point of B E, 
 describe an arc cutting DA prolonged 
 in F. Join FB, and the square con- 
 structed upon FB will be equal in area 
 to the rectangle ABCD. 
 
 Proof. EBF is a right triangle. 
 For the triangles OEF and OBF are 
 isosceles (why?) ; hence EFO = FEO, 
 and OFB = OBF ( 80). By addi- 
 tion, EFO + OFB = EFB = FEO 
 -f OBF. But EFB + FEO + OBF 
 = 180 ( 64) ; therefore EFB = 90. 
 
 l&L. 
 
 Fig. 138. 
 
 Since EBFis a right triangle, it follows from 142, Corollary 3, 
 that ~FJ? = the area of the rectangle ABCD. 
 
 Exercises. 1. Construct a rectangle with the sides 9 dm and 4 dm , and 
 transform it into a square. Measure the side of the square. What is its 
 length ? What ought its length to be ? 
 
 2. Construct a triangle with the sides 4 dm , 6 dm , and 8 dm , and transform it 
 into a square. 
 
 3. Transform a given parallelogram into a square. 
 
 150. Problem. To transform any polygon ABCDE (Fig. 
 139} int a triangle. 
 
I5 1 -] CHAPTER VII. AREAS. 161 
 
 Construction. Draw a diagonal AD. Through E draw a line 
 parallel to AD and meeting 
 A B prolonged in F. Join 
 DF. Then is the quadri- 
 lateral BCDF equivalent 
 to the pentagon ABODE. 
 
 Repeat this construction 
 by drawing the diagonal 
 JSD, a parallel through C, 
 and then joining DG ; this 
 
 reduces the quadrilateral F &*39* 
 
 BCDE to the equivalent triangle FGD. 
 
 Proof. Left for the learner, with the aid of Fig. ijp and the 
 analysis of 146. 
 
 Corollary. Hence, any polygon may be transformed into a 
 square. By what three steps ? 
 
 This supplies another way of finding the area of a polygonal field. 
 Construct (to scale) the polygon on paper, transform it into a 
 square, and multiply one side of this square by itself. 
 
 Exercises. 1. Transform a hexagon into a triangle. 
 
 2. Transform a pentagon into a square. 
 
 3* Construct three equal octagons. Find the area of the first and second 
 by two of the methods given in 134, and the area of the third by transform- 
 ation into a square. 
 
 VI. Partition of Figures. 
 
 151. In buying and selling land it often becomes necessary 
 to run lines of division through estates, forests, fields, etc., dividing 
 them into smaller parts in some assigned way. 
 
 This is done either (/.) by computation (arithmetical partition) 
 or (".) by construction (geometrical partition) . 
 
 The first method, when practicable, is generally adopted. Let 
 us examine some cases. 
 
162 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 152. 
 
 152. TRIANGLES. Case 1. A line CD (Fig. 140) joining 
 the vertex of a triangle ABC to the middle point of the base AB 
 divides the triangle into two equal parts 
 ( 130, Corollary). 
 
 Again, it AE=. \AB, and EB = \ 
 AB, then &ACE = \kABC, and 
 &ECB = f &ABC (why?). And, 
 A ACE:& ECB = 1:2. In gen- 
 eral, 
 
 A line joining the vertex of a triangle to the base divides both 
 the triangle and its base into parts which have the same ratio. 
 
 What ratio have the triangles A EC and ABCt the triangles 
 ECBvrAABCt 
 
 Case 2. Line of division DE (Fig. 141) passes through two 
 sides of the triangle. 
 
 = I AC] then 
 i),and &ADE 
 
 = $ A ADC (Case i) ; therefore A ADE 
 = J X \&ABC= %t\ABC. 
 
 Again, if CE = CA, and CF=% CB, 
 then (by the same reasoning) A CEF \ 
 X %&ABC^ %&ABC; that is, A CEF 
 \&ABC=2\*i. In general, 
 The ratio of the triangle cut off to the entire triangle is equal to 
 the product of the ratios between the parts of the sides that meet 
 at the common vertex. 
 
 Case 3. Hence, if a definite part say, one-fourth of a tri- 
 angle is to be cut off by a line from side to side, this can be done 
 in various ways, because the fraction may be obtained as the 
 product of various pairs of factors; for example, and J, \ and |, 
 f and |, I- and ft, etc. 
 
 If, however, the line of division has to pass through a giver) 
 point in one side, then its position is determined. 
 
 B 
 
 F 
 Fig. 141. 
 
1 5 3-] CHAPTER VII. AREAS. 163 
 
 For example, if AD (Fig. 142} = \AB, and the AADE is 
 to be made equal to \AABC, then the line 
 of division will pass through a point in the 
 side AC, since A ADC would be too large 
 (how large is it?) ; and, since | -j- = |, the 
 line will pass through a point E such that 
 
 How must a line DF be drawn if A B DF 
 is to be made equal to \AABC1 
 
 153. PARALLELOGRAMS. Case 1. Line of division joins two 
 opposite corners. See 102. 
 
 Case 2. Line of division AE (Fig. 143) joins a corner to a 
 point in one side. 
 
 Let (Fig. 143} CE = J. CD, then 
 A ACE = \AACD ( 152, Case i). 
 But AACD = the parallelogram 
 
 ABCD ( 102);" therefore A ^ C 
 
 Figm I43 ^ ~~& = % the parallelogram ABCD. 
 
 CE must be what part of C.Z2, if 
 A ACE = f the parallelogram? 
 
 Case 3. Line of division joins two opposite sides and is parallel 
 to the other sides. This case is left for the learner to investigate. 
 How can a rectangle be divided into six equal parts ? 
 Case 4. Line of division EF (Fig. 144) joins two opposite 
 sides, and is not parallel to the other sides. 
 Let (Fig. 144) CE = CD, AF 
 %AB, then AACE = \AACD, and 
 A AEF = AADF (why?) = AABD ; 
 therefore A A CE -f- A A E F = the trap- 
 
 *^ P ^44'* 
 \ of half the parallelogram -f f of 
 
 half the parallelogram = J the parallelogram. 
 
164 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 154. 
 
 Case 5. Line of division FE (Fig. 145) joins two adjacent 
 sides. Let AE = AB, AF = f AC, 
 then ( 152, Case 2) &AEF = X f 
 X &ABC= AAC= T V the whole 
 parallelogram. 
 
 A B 
 
 Fig. i 45. 
 
 154. We will now give two or three 
 cases in which the partition is effected 
 by direct construction. The proofs of the constructions are left 
 to the ingenuity of the learner. 
 
 Problem I. To divide a triangle ABC (Fig. 146) from a 
 point D in one side into two equal parts. 
 
 Construction. }^ CD, bisect AB in E, draw EF II CD, 
 join DF, DFis the line of division required. 
 
 D E 
 
 Fig. 146. 
 
 Problem II. To divide a triangle ABC (Fig. 147) from a 
 point D in one side into three equal parts. 
 
 Construction. Join CD, trisect AB in E and F, draw EG 
 and FH both II CD, join DG and D H "; these last are the lines 
 of division required. 
 
 Problem III. To divide a trapezoid A B C D (Fig. 148) from 
 <f H D a point E in one of the parallel sides into 
 two equal parts. 
 
 Construction. Bisect the parallel sides 
 in F and G, make GH = EF and 
 join EH: it is the line of division re- 
 Fi/.i48. quired. 
 
1 55-] CHAPTER VII. AREAS. 165 
 
 Will this construction divide a parallelogram into two equal 
 parts ? See 153, Case 4. 
 
 How is this problem to be solved, if the line of division is to 
 start from a given point in one of the parallel sides ? 
 
 155. Practical Exercises. 
 
 1. Divide a triangle into five equal parts. 
 
 2. Divide a triangle into two parts having the ratio 4 : 7. 
 
 3. Divide a triangle into three parts which shall be to each other as the 
 numbers I, 3, 5. 
 
 4r. Through one corner of a triangle draw a line cutting off a part equal 
 to f the entire triangle. 
 
 5. How can a triangle containing i2Oi m be divided into parts containing 
 30<i m , 40i m , and -jo<i m ? 
 
 6. From a triangular field worth $1200 cut off a piece worth $400. 
 
 7. Through a triangular meadow, a ditch runs from one corner to a point 
 in the opposite side, 264^ from one end, 627y ds from the other. Find the 
 ratio of the two parts of the meadow to each other. 
 
 8. A piece of woodland in the shape of a triangle cost $6000, but is now 
 worth 50 per cent more than at first. It is required to run a line from one 
 corner to the opposite side, which is 1200 feet long, cutting off a portion 
 worth $1800. 
 
 9. A field has the shape of an isosceles right triangle, the equal sides 
 being each 400 long. Through the vertex of the right angle draw a line 
 cutting off a part which shall contain I hectar. 
 
 Through two sides of a triangle draw a line cutting off, 
 10. | of the triangle. 11. f of the triangle. 
 
 12. f of the triangle. 13. f of the triangle. 
 
 What ratio does the portion cut off bear to the whole triangle, if the line of 
 division cuts, 
 
 14. | from one side and \ from the other ? 
 
 15. f from one side and f from the other ? 
 
 16. f from one side and T 3 7 from the other ? 
 
 17. The legs of a right triangle are 24 and 40. Find the area of the 
 part cut off by a line which passes through points in the legs distant, respec- 
 tively, 6 cm and io cm from the vertex of the right angle. What is the ratio of 
 its area to that of the entire triangle ? 
 
166 GEOMETRY FOR BEGINNERS. [ 155. 
 
 18. Through the middle point in one side of a triangle draw a line dividing 
 it into parts which shall have the ratio 3:5. 
 
 19. Through the middle point of one side of a triangle draw lines dividing 
 the triangle into parts which shall be to each other as 1:2:3. 
 
 20. A, B, and C buy a triangular field ABC (Fig. 149). A pays $200; 
 
 B, $500; C, $300. They wish to divide it among 
 themselves, by lines starting at a point D, where 
 there is a valuable spring; A to take the upper part, 
 B the middle, and C the lower. Show how the 
 lines of division should be run if A D = 1 8o m , D C 
 300. 
 
 21. Divide two sides of a triangle, each into five 
 equal parts, and join in order the points of division. 
 Fig. i4Q. Then find the ratio of each of the smaller triangles 
 
 thus formed to the entire triangle. 
 
 22. What part of a parallelogram is the triangle cut off by a line drawn 
 from one corner to the middle of a side ? 
 
 23. Cut off f of a parallelogram by a line drawn through one corner. 
 
 24. From a parallelogram containing 6ooi m cut a triangle containing 2$o<i m . 
 
 25. Divide a rectangle whose sides are 36 and 2O m , by lines drawn from 
 one corner, into three parts which shall be to one another as 2:3:4. 
 
 26. Divide a parallelogram into seven equal parts. 
 
 27. Divide a parallelogram into two parts having the ratio 3 : 5. 
 
 28. What part of a parallelogram lies between J of one side and f of the 
 opposite side ? 
 
 29. A rectangular field is 500 by 300. A tree is in the longer side, 150 
 from one corner. Draw a line, beginning at the tree, which shall cut from 
 the field I hectar, adjoining the 150 part of the longer side. 
 
 30. A line joining two adjacent sides of a parallelogram cuts off a triangle 
 whose sides are ^ and f of the corresponding sides of the parallelogram. 
 What part of the latter is the triangle ? 
 
 31. Divide a trapezoid into five equal parts. 
 
 32. In a trapezoid A BCD the parallel sides are A B 25, CD = 3$ m ; 
 their distance apart = 2O m . Divide the trapezoid into two equal parts by a 
 line beginning at a point E in A B, such that AE= io m . 
 
 33. In a quadrilateral A BCD, the diagonal BC 40, the altitudes of the 
 triangles ABC and BCD are 2O m and 30. Divide the quadrilateral by a 
 line through B into two equal parts. 
 
CHAPTER VII. REVIEW. 167 
 
 REVIEW OF CHAPTER VII. 
 SYNOPSIS. 
 
 1. Surfaces are measured by choosing a unit and finding how often this unit 
 is contained in the surface to be measured. 
 
 2. The area of a surface is the number of times the unit is contained in the 
 surface followed by the name of the unit. 
 
 3. Units of area (except the acre) are squares whose sides are linear units. 
 
 4. The ratio between two units of area is the square of the ratio between 
 the corresponding linear units. 
 
 5. Geometry enables us to measure a surface indirectly by showing that its 
 area depends upon the length of certain lines. 
 
 6. Area of a square square of one side. 
 
 7. Area of a rectangle its base X its altitude. 
 
 8. A parallelogram = a rectangle of the same base and altitude ; its area = 
 its base X its altitude. 
 
 9. Parallelograms with equal bases and equal altitudes are equivalent. 
 
 10. Area of a rhombus = \ product of its diagonals. 
 
 11. A triangle = \ a parallelogram of same base and altitude; its area = its 
 base X half its altitude. 
 
 12. Triangles with equal bases and equal altitudes are equivalent. 
 
 13. Area of a trapezoid = \ sum of parallel sides X altitude. 
 
 14. Area of a regular polygon = \ perimeter X less radius. 
 
 15. Area of any polygon may be found by dividing it into triangles (two 
 ways) or into triangles and trapezoids. 
 
 16. In every right triangle the square of the hypotenuse is equal to the sum 
 of the squares of the two legs. (Two proofs.) 
 
 17. The perpendicular let fall from the vertex of the right angle upon the 
 hypotenuse divides the square of the hypotenuse into two rectangles, 
 equal respectively to the squares of the adjacent legs. 
 
 18. A figure may be transformed into another figure, either (?'.) by computing 
 such parts of the new figure as will enable us to construct it, or ('.) by 
 direct construction. 
 
 19. A square may be found equivalent to any polygon by first changing the 
 polygon to an equivalent triangle, then the triangle to an equivalent 
 rectangle, and then the rectangle to an equivalent square. 
 
 20. The division of figures into parts is effected, either (*'.) by arithmetical 
 partition t or (zV.) by geometrical construction. 
 
168 
 
 GEOMETRY FOR BEGINNERS; 
 
 21. A line through one corner of a triangle divides the triangle and the 
 opposite side into parts having the same ratio. 
 
 22. If a line is drawn between two sides of a triangle cutting off fractional 
 parts of the sides, the product of these fractions give the fractional part 
 which the triangle cut off is of the whole triangle. 
 
 23. Nos. 21 and 22 enable us to divide parallelograms and polygons in 
 various ways. 
 
 EXERCISES. 
 
 1. Compute the areas in square meters (as given below, in the last column 
 to the right) of the following regular polygons, one side of the polygon 
 in each case being supposed to be I meter : 
 
 2. 
 
 6. 
 
 Polygon 
 of 
 
 One Side. 
 
 Less Radius. 
 
 Area. 
 
 3 sides 
 
 
 0.2887 
 
 0-433 
 
 4 sides 
 
 
 0.5000 
 
 I.OOOO 
 
 5 sides 
 
 
 0.6882 
 
 1.7205 
 
 6 sides 
 
 
 0.8660 
 
 2.5980 
 
 8 sides 
 
 
 1.2071 
 
 4.8284 
 
 10 sides 
 
 
 I-S388 
 
 7.6942 
 
 12 sides 
 
 
 l.866o 
 
 11.1961 
 
 Same exercise, one side of each polygon being taken as 2 meters. 
 Hint. For the first case (3 sides), the less radius =2 X 0.2887, an d the area 
 _ 3 X 2 X 2 X 0.2887 _ z 732qm _ 4 times the area (0.433) found above for 
 
 the same polygon when the side is i m . 
 
 3. Same exercise, one side of each polygon being taken as 3 meters. 
 Hint. In the first case (3 sides), the area = 3 X 3 X 3 X 0.2887 _ g times 
 
 2 
 
 the area (0.433) of the same polygon if the side is jm. 
 
 4. From the above exercises derive a general rule for finding the area of a 
 regular polygon whose side contains any number (a) of linear units. 
 
 5. Two fields of equal value per unit of area have the shape, one of a square, 
 the other of a regular hexagon. If a side of each field = 136, and the 
 first field is worth $600.00, what is the second field worth ? What is the 
 value per square meter ? per hectar ? 
 
 Government lands are usually divided into tracts or townships 6 miles 
 square ; these are subdivided into 36 sections, and each section into half- 
 sections, quarter-sections, half quarter-sections, and quarter quarter-sec- 
 
CHAPTER VII. REVIEW. 169 
 
 twns. Find the area in acres of a township and of each of its sub- 
 divisions. 
 
 7. A quarter-section is fenced with a six-railed fence, the posts being I2 ft 
 apart. The rails cost $40.00 per thousand, the posts $60.00 per thou- 
 sand. Find the cost of the fence. 
 
 8. Of two kinds of paper, equal in quality, the first is 42 by 3O cm , and 
 costs $0.20 a quire ; the second is 6o cm by 4O cm , and costs $0.28 a quire. 
 Which kind is the cheaper ? 
 
 9. A man buys a rectangular garden 140 by 8o m . He surrounds the gar- 
 den by a wall o.8 m thick, and within the wall by a walk i.8 m wide ; and 
 through the middle of the garden he makes paths parallel to the sides, 
 each 1.5 wide. How much land remains for other uses ? 
 
 10. A piece of land is 160 square. Another piece of the same size has the 
 shape of a rectangle four times as long as it is wide. Find the difference 
 between their perimeters. 
 
 11. The perimeter of a right triangle = 6o m , the hypotenuse = 25. Find the 
 legs and the area. 
 
 12. The pejimeter of an isosceles triangle = 24, the base = 6 m . Find the 
 area. 
 
 In an isosceles right triangle, 
 
 13. The hypotenuse = 8 m ; find the area. 
 
 14. One leg = 32; find the area. 
 
 15. The area = 3Ooi m ; find the sides. 
 In an equilateral triangle, 
 
 16. One side = I2 to ; find the area. 
 
 17. The altitude = 6 m ; find the area. 
 
 18. The area = 1441; find the side and the altitude. 
 
 19. One side of a square = I4 m ; find the diagonal. 
 
 20. The diagonal of a square = i6 m ; find one side. 
 
 21. In a rhombus, a diagonal = one side = 5 m ; find the area of the rhombus. 
 
 22. Find the angles of the rhombus in the last exercise (see page no, Exer- 
 cise 5). 
 
 23. A square and a rhombus have the same side, 4. How much smaller is 
 the latter if its acute angle = 60 ? if its acute angle = 30? 
 
 24:* A rectangular field 2OO m long contains i ha . Find the distance apart of 
 two opposite corners. 
 
 25. The edge of a cube 4 m . Find the distance from a corner in the upper 
 face to the diagonally-opposite corner in the lower face. 
 
 26. What method of making a right angle with a cord is suggested by Fig. 
 
170 GEOMETRY FOR BEGINNERS. 
 
 Construct the following figures : 
 
 27. A triangle, if the area = 36oi cm , the altitude = 42 cm . 
 
 28. An equilateral triangle, if the area = 4OO9 cm . 
 
 29. A square, if the diagonal = 1.2. 
 
 30. A rectangle, if the area = 3001, and one side = 24 cm . 
 
 31. A rhombus, if the area = 24O<i cm , and one diagonal = 30. 
 Construct a square, 
 
 32. Three times as large as a given square. 
 
 33. Two and one-quarter times as large as a given square. 
 
 34. Equal to half the difference of two given squares. 
 
 35. Equivalent to a given equilateral triangle. 
 
 36. Equivalent to a given regular octagon. 
 
 Given a square with the side= 50; compute the side of 
 
 37. An equivalent equilateral triangle. 
 
 38. An equivalent regular pentagon. 
 
 39. An equivalent regular hexagon. 
 
 40. An equivalent regular decagon. 
 Transform a given triangle, 
 
 41. Into a right triangle. 
 
 42. Into an isosceles triangle. 
 
 43. Into an isosceles right triangle. 
 
 44. Into an equilateral triangle. 
 
 45. Into a rectangle with a given side. 
 
 46. Into a square. 
 Transform a square, 
 
 47. Into a parallelogram with a given angle. 
 
 48. Into a rectangle with a given side. 
 
 49. Into a rhombus with a given side. 
 
 50. Transform a given hexagon into a square. 
 
 51. What part of a triangle is cut off by a line which bisects two of the sides? 
 
 52. Divide a triangle into three equal parts by lines meeting at a given point 
 within the triangle, one line to pass through one corner. 
 
 53* Find the point within the triangle from which lines drawn to the corners 
 will divide the triangle into three equal parts. 
 
 54. Divide a parallelogram by lines through a corner into three equal parts. 
 
 55. Divide a parallelogram by lines through a point in one side into three 
 equal parts. 
 
 56. Divide a regular hexagon by lines through one corner into six equal parts. 
 
 57. Divide an irregular hexagon by a line through one corner into two equal 
 parts. 
 
$ 15^0 CHAPTER VIII. SIMILAR FIGURES. 171 
 
 CHAPTER VIII. 
 SIMILAR FIGURES. 
 
 CONTENTS. I. Proportional Lines and Figures ($ 156-158). II. Similar Tri- 
 angles ( 159-168). III. Problems and Applications ( 169-176). IV. Simi- 
 lar Polygons ($ 177-179). 
 
 J. Proportional Lines and Figures. 
 
 156. Review 37, 38, 112-114. 
 
 The lines a and b (Fig. 150) have the ratio 1:3; the lines c 
 and d have the ratio 2:6= 1:3, or the same ratio as the lines 
 a and b. By placing the sign of equality between the equal ratios 
 a : b and c : d, we obtain a proportion, a : b = c : d, which is read 
 a is to b as c is to d. 
 
 -2+ 4- 
 
 b 
 
 Fig. 150. 
 
 Definition. A PROPORTION is an equation between tivo equal 
 ratios. 
 
 The four terms of a proportion need not all be of the same kind ; 
 it is sufficient if the two terms of each ratio are alike in kind. 
 
 Two areas, for example, may have the same ratio as two lines, 
 and hence form a proportion with them. What instance of this 
 occurs in 152 ? 
 
 Again, two angles are to each other as the arcs described from 
 their vertices as centres and included between their sides. In this 
 proportion, the first ratio is that of two angles, the second that of 
 two lines. 
 
172 GEOMETRY FOR BEGINNERS. [ 157, 
 
 Let A, B, C, D be any four magnitudes (that is, things that can 
 be measured) , A and B however being alike in kind, and C and D 
 also being alike in kind ; and let a and b represent the numerical 
 values of A and B respectively in terms of a common unit, c and 
 </the numerical values of C and D respectively in terms of a com- 
 mon unit ; then, if a : b = c : d, we say that A> B, C, and D are in 
 proportion or are proportionals. Of the four terms a, b, c, and d 
 which express the proportion, a and d are called the extremes, b 
 and c the means. If b = c, b is called the mean proportional 
 between a and d. 
 
 157. A proportion, expressed, as just explained, by the num- 
 bers which are the numerical values of the magnitudes forming the 
 proportion, is subject to certain laws, among which the following 
 are useful for our purposes : 
 
 1 . If we write the proportion a : b = c : d in the form -=-,, 
 
 and then divide the equation i = i by the equation - = -, we 
 
 o a 
 
 obtain (Axiom V.) - = -, or b : a = d\ c. 
 a c 
 
 Law I. The truth of a proportion is not affected if the terms 
 of each ratio are transposed. 
 
 2. If we multiply both sides of the equation - = - by -, we 
 
 b d a 
 
 obtain (Axiom IV.), after cancelling common factors, - = -, or 
 
 d : b = c : a. If we multiply - = - by -, we obtain - = -, or 
 
 b d c c a . 
 
 a : c = b : d. 
 
 Law II. The truth of a proportion is not affected, if the ex- 
 tremes are transposed, or if the means are transposed. 
 
 3. If we multiply both sides of the equation - = - by the 
 
158.] CHAPTER VIII. - SIMILAR FIGURES. 173 
 
 product b x //, we obtain, after cancelling common factors, a X d 
 = b X c. 
 
 Law III. In every proportion the product of the extremes is 
 equal to the product of the means. 
 
 If in a proportion the two means and one extreme are given, 
 how can the other extreme be found ? 
 
 If in a proportion the two extremes and one mean are given, 
 how can the other mean be found ? 
 
 Exercises. 1. Verify the above laws with the lengths of the lines in 
 Fig. 150. 
 
 2. Verify the above laws in the case of the proportion 3:9 = 5:15. 
 
 8. In a proportion, the first, third, and fourth terms are 12, 25, 200; find 
 the second term. 
 
 4. Find a fourth proportional to the lengths 24, 50, and 8o m . 
 5. Find a length which shall be to 6o m as 5 : 6. 
 
 6. What relation exists between the side of a square and the sides of the 
 equivalent rectangle ? 
 
 7. Find the mean proportional between 9 and i6 m . 
 
 8. Show that the mean proportional between a and b is equal to "^Tab. 
 
 158. Let P and Q denote the areas of two squares, two par- 
 allelograms, or two triangles ; a and b the sides of the squares, or 
 the bases of the parallelograms or the triangles ; h and k the alti- 
 tudes of the parallelograms or of the triangles ; then, for 
 
 Two squares, P= a\ Q = P' } hence ^= -[ 
 
 Two parallelograms, P= aKh,Q = bKk] hence = a X h 
 
 Q b X k 
 
 Two triangles, P= 2**, Q = *^- } hence 1= **JL 
 
 2 2 Q b x k 
 
 Or, in words, 
 
 Two squares are to each other as the squares of their sides. 
 
 State in words the results for parallelograms and for triangles. 
 
174 GEOMETRY FOR BEGINNERS. [ 158. 
 
 In the cases of parallelograms and triangles, if (i) h = k, and 
 (2) a b, the preceding equations become, 
 
 (0 --- (2) - h 
 
 () Q *' Q-J 
 
 That is, 
 
 ( T ) Two parallelograms or two triangles, with equal altitudes, 
 are to each other as their bases. 
 
 (2) Two parallelograms or two triangles, with equal bases, are 
 to each other as their altitudes. 
 
 Exercises. 1. The legs of a right triangle are 24 and i8 m . In another 
 triangle, the base = 24 m , the altitude = i8 m . Compare the areas of the tri- 
 angles (that is, find the ratio of the areas). 
 
 2. The bases of two triangles of equal altitude are 25 and 3o m . Com- 
 pare the areas. 
 
 3. If the areas of two triangles of equal altitude are i6oi m and 2OO<i m , what 
 is the ratio of their bases? If the base of one 32, find that of the other. 
 
 4. If the bases of two triangles of equal altitude are 15"' and I9 m , find 
 the base of a triangle with the same altitude and equal in area to their sum. 
 
 5. The areas of two triangles with equal bases are 48o c i m and 3001. 
 Compare their altitudes. If the altitude of the larger = 30, find that of the 
 smaller, and also the base of either. 
 
 6. Two triangles have equal bases, and the altitude of one is ten times 
 that of the other. Compare their areas. 
 
 7. If two triangles have equal areas, compare their altitudes when the 
 base of one triangle is (z.) twice, (z'z.) four times, (z'z'z.) one-half, (zV.) three- 
 fifths, that of the other. 
 
 8. The altitudes of two equivalent triangles are 15 and 20. Compare 
 their bases. 
 
 9. Solve Exercises 6, 7, and 8, substituting " parallelograms " for " tri- 
 angles." 
 
 10. Compare a triangle with a parallelogram of the same base and altitude. 
 
 11. A triangle and a parallelogram have equal areas. Compare (z.) their 
 bases if they have equal altitudes; (z'z.) their altitudes if they have equal bases. 
 
 12. Compare the areas of a rhombus and the rectangle whose sides are 
 bisected by the corners of the rhombus (see Fig. 124}. 
 
 13 A square and a rhombus have equal perimeters. Compare their areas 
 if the altitude of the rhombus is two-thirds its side. 
 
1 5 9.] CHAPTER VIII. SIMILAR FIGURES. 175 
 
 14. A square and a rhombus have equal areas. Compare their perimeters 
 if the altitude of the rhombus is one-half its side. 
 
 15. Compare the areas of a square and an equilateral triangle having the 
 same side. 
 
 16. A square and an equilateral triangle have equal areas. Compare their 
 sides. 
 
 17. Compare the areas of a square, a regular pentagon, and a regular 
 hexagon, all having the same side. 
 
 II. Similar Triangles. 
 
 159. Two triangles which have the same shape and differ 
 only in size are called SIMILAR triangles. 
 
 The sign of similarity is ~. &ABC^>&DEFis read, the 
 triangle ABC is similar to the triangle D E F. 
 
 In order to discover the chief properties possessed by similar tri- 
 angles, take on one side AK (Fig. 151) of an angle KAL any 
 number, say five, equal parts, AB, BD, D F, FH, HK, and 
 through the points B, D, F, H, K, 
 draw parallel lines cutting A L in the 
 points C, E, G, /, L ; then ( 112) 
 AC=CE=EG= GI=IL. 
 
 The triangles ABC, ADE, etc., 
 differ in size, but have the same 
 shape ; hence are similar. 
 
 Now compare the angles and sides 
 of any two of these triangles, as A DE 
 and AKL. The angles are equal; K Fi L 
 
 for the angle A is common ; and of 
 
 the other angles, ADE = AKL, and AED^ALK ( 56). 
 As for the sides, it is evident from the construction that AD \AK 
 = 2:5, and likewise that A E : AL = 2 15. The same ratio also 
 exists between the sides D E and KL ; for, if we draw through 
 B, D, F, and PI lines parallel to AL, these lines will divide (by 
 112) DE into two equal parts, and KL into five equal parts, 
 
176 GEOMETRY FOR BEGINNERS. [ 159. 
 
 and the parts of DE will be equal to those of KL ( 102, Corol- 
 lary i) ; therefore, DE:XL = 2 15. Hence, A D : A K = A E 
 :AL = DE\ KL ; that is to say, the sides opposite equal angles 
 in the triangles are proportional. 
 
 The sides opposite equal angles are termed corresponding or 
 homologous sides. 
 
 We should obtain like results if the angle A had any other value, 
 or if the parallels BC, DE, etc., had any other direction, and also 
 whatever be the length or number of the equal parts AB, BD, etc. 
 
 Therefore, - 
 
 Theorem I. Two similar triangles are mutually equiangular, 
 and have their homologous sides proportional. 
 
 Corollary. By transposing ( 157, Law II.) the means of 
 the proportion AD:AK=AE:AL, we obtain AD:AE = 
 A K : A L. This shows that, in similar triangles the sides which 
 contain equal angles have the same ratio. 
 
 Two other theorems also follow from the preceding investiga- 
 tion : 
 
 Theorem II. If a line is drawn through two sides of a tri- 
 angle, parallel to the third side, a smaller triangle is formed which 
 is similar to the given triangle. 
 
 Theorem III. A line drawn through two sides of a triangle, 
 parallel to the third side, divides those two sides proportionally. 
 
 Exercises. 1 . The converse of Theorem III. is also true. State it, and 
 illustrate it by means of numbers. 
 
 2, The three sides of a triangle are 84, i2O m , and ioo m . Through a point 
 in the first side, 28 from the intersection of the first and second sides, a line 
 is drawn parallel to the third side. Find where it will cut the second side. 
 Find, also, the ratio between the triangle cut off and the entire triangle ( 152). 
 
 3* If the sides of a triangle are l6 m , 24, and 32, and a line is drawn 
 from the first side to the second side, connecting points distant 5 m and 7, 
 respectively, from the intersection of the first and second sides, will this line 
 be parallel to the third side? 
 
 4. Draw two similar triangles. Also, two equal triangles. Can two triangles 
 be equal without being similar? Can they be similar without being equal? 
 
l6o,] CHAPTER VIII. - SIMILAR FIGURES. 177 
 
 160. It appears from the last section that the similarity of 
 two triangles involves six conditions : the equality of three pairs 
 of angles and the equality of the ratios between three pairs of 
 sides. What six conditions does the equality of two triangles 
 involve ? 
 
 We have seen ( 72-75) that, in general, from the equality of 
 three parts in two triangles, we can conclude that the triangles are 
 equal in all respects. We shall now proceed to show that if two 
 triangles fulfil certain conditions out of the six above mentioned 
 as belonging to similar triangles, then these triangles must fulfil 
 the remaining conditions, and hence must be similar to each other, 
 
 161. In the triangles ABC and DEF (Fig. 152) let the 
 angle A == D, B = E, and therefore 
 C^F. Make CG ~ F>, and draw 
 GH\\AB. Then&GJ?C^ADF 
 (I. Law of Equality). But A ABC 
 ^ A GHC ( 159, Theorem II.). 
 Therefore A ABC ~ A DEF. 
 
 Hence >- Fig. 15*. 
 
 Theorem (I. Law of Similar- 
 
 ity) . Two triangles are similar if they are mutually equiangular. 
 
 Exercises. 1. Show that two triangles are similar if two angles of one' 
 are equal respectively to two angles of the other. 
 
 2. What condition suffices to make two isosceles triangles similar? two 
 right triangles? 
 
 8. Why are two equilateral triangles always similar? 
 
 4. Construct three similar triangles, each having the angles 60 and 80. 
 
 5. Construct upon a given line A B a triangle similar to a given triianglej 
 
 6. Two triangles are similar if their sides, taken two by two, are parallel. 
 
 7. Two triangles are s-imilar if their sides, taken two by two, are perpen- 
 dicular to each other. Point out the homologous sides. 
 
 Hints. Revolve one of the triangles about a corner through 90 ; then its sides 
 Become parallel respectively to the sides of the other triangle. See Exercise 6, 
 
178 GEOMETRY FOR BEGINNERS. [ 162. 
 
 162. In the triangles ABC and DBF (Fig. is 2), let AC: 
 DF=BC\EF, and let also the angle C = F. Make CG = 
 DF, and draw .#11 AB\ then AC:GC= BC\HC ( 159, 
 Theorem III.) Since the first three terms of this proportion are 
 the same as the first three of the proportion assumed at the start, 
 the fourth terms must also be equal; therefore CH = EF. Then 
 A GHC ^ &DEF (II. Law of Equality). But A ABC - 
 A GHC (why ?): therefore A ABC ~ A DEF\ and hence, - 
 
 Theorem (II. Law of Similarity). Two triangles arc 
 similar if they have an angle equal, and the sides which include 
 the angle proportional. 
 
 163. In the triangles ABC and DEF (Fig. 752), let AC 
 -.DF=BC:EF, AOBC, DF>EF, and B = E. Make 
 CG = DF, and draw GH\\AB; then AC-. GC = BC-.HC. 
 This proportion and the proportion assumed at first have the first 
 three terms equal, therefore their fourth terms must be equal, or 
 CH = EF. Whence, A GHC ^ DEF (III. Law of Equality) . 
 ButA^C^ A GHC, therefore kABC^&DEF. Hence, - 
 
 Theorem (III. Law of Similarity). Two triangles are 
 similar if two sides of one are proportional to two sides of the other, 
 and the angles opposite the greater sides are equal. 
 
 164. In the triangles ABC and DEF (Fig. 152], let 
 
 AC-.DF = BC-.EF, 
 and AC\DF= AB-.DE. 
 
 Make CG = DF, and draw GH II AB, then- 
 
 In .the first and third of these proportions, the first three terms 
 are equal, therefore the fourth terms must also be equal ; that is, 
 CH EF. For the same reason it follows from the second and 
 fourth proportions that GH =>E. Hence, A GHC ^ A DEF 
 
165.] CHAPTER VIII. SIMILAR FIGURES. 170 
 
 (IV. Law of Equality) ; but A ABC ~~ A GNC, therefore &ABC 
 ^DEF. That is, - 
 
 Theorem (IV. Law of Similarity). Two triangles are 
 similar if tJicir sides, taken in order, are proportionals. 
 
 Corollary. If each side of a triangle is two, three, four, etc., 
 times the homologous side of a similar triangle, then the sum of 
 the sides, that is, the perimeter, of one triangle is two, three, four, 
 etc., times the perimeter of the other triangle ; that is, 
 
 In similar triangles the perimeters are to each other as any two 
 homologous sides. 
 
 Exercise. Verify the truth of this Corollary by .two triangles, the sides of 
 one being 2 m , 4, 5, and those of the other 6 m , 12, 15. 
 
 165. Let the parallels AC, DF (Fig. 153), be cut by lines 
 OA, OB, OC, drawn through a common 
 point O. 
 
 &OAB ^ &ODE, 
 
 and 
 
 ( 56, and I. Law of Similarity). There- 
 
 fore,- / / _ V 
 
 ^ = :d^Land = <* B> 
 
 OE DE^ OE EF Fig.i 53 . 
 
 Whence 
 
 The same reasoning might be applied to any number of parallels 
 and lines intersecting them. Therefore, 
 
 Theorem, Lines drawn through a common point divide par- 
 allels into proportional parts. 
 
 Exercises. 1. If one of the parallels is divided into equal parts, how 
 will the others be divided ? 
 
 2. If a parallel to AC (Fig. 153) is drawn above O, cutting the lines 
 through O prolonged, will the theorem still hold true ? Can you prove ? 
 
180 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ i6 7 . 
 
 166. In the right triangle ABC, right angled at C (Fig. 
 
 draw CD _L A B ; it divides A B into two 
 parts or segments. The smaller right 
 triangles A CD and BCD are each 
 equiangular with respect to the triangle 
 ABC (why?); therefore they are equi- 
 angular with respect to each other. 
 Hence (I. Law of Similarity), 
 AACD- therefore AB \ AC = AC : AD. 
 therefore AB : BC ' = BC : BD. 
 
 154- 
 
 Theorems. If in a right triangle a perpendicular is drawn 
 from the vertex of the right angle to the hypotenuse : 
 
 1. Either leg is a mean proportional beeween the hypotenuse and 
 the adjacent segment. 
 
 II. The perpendicular is a mean proportional between the two 
 segments of the hypotenuse. 
 
 Exercises. In a right triangle let a and b denote the legs, c the hypote- 
 nuse, h the altitude with hypotenuse for base, / and q the segments of the 
 ihypotenuse. Find the other quantities, if, 
 
 .1. a = 4-5 m , h = 3.6. 3. a = 9.6, / o.784 m . 
 
 2. /& = o.8 m ,/ = i.5 m . 4. c= 
 
 167. Let (Fig. 155) kABC^&DEF, and draw the alti- 
 tudes CG and FH corresponding to 
 two homologous sides AB and DE 
 taken as bases. 
 
 By hypothesis, AB\DE = AC; 
 DF. The right triangles ACG and 
 DFH are similar (why?), there- 
 fore A C : DF = CG : FH. Hence 
 AB-.DE = CG'.FH. Thatis,- 
 Theorem. In two similar triangles, any two homologous sidet 
 taken as bases are to each other ax the corresponding altitudes* 
 
1 68.] CHAPTER VIII. SIMILAR FIGURES. 181 
 
 168. Let P and Q denote the areas, a and b the bases, h and 
 k the corresponding altitudes of two similar triangles. 
 
 Then (, 158), 
 But (| 167) 
 
 If for the fac 
 the equation becomes, 
 
 If for the factor - in the first equation we substitute its equal -, 
 k b 
 
 = y f = 
 
 <2 J S 2 ' 
 
 Since and / may be any two homologous sides taken as bases, 
 therefore, 
 
 Theorem. Similar triangles are to each other as the squares 
 of their homologous sides. 
 
 Exercises. 1. If the homologous sides of one triangle are each one-fifth 
 those of a similar triangle, compare the areas of the triangles. 
 
 2. The areas of two similar triangles are 4OO < i m and 1 296i m ; find the 
 ratio of two homologous sides. 
 
 8* Two triangular gardens are similar in shape. What is the easiest way 
 (z.) to compare their areas ? (zY.) to compute both their areas ? 
 
 4. A field has the shape of an equilateral triangle. From one corner I 
 wish to cut off a similar triangle equal in area to one-sixteenth of the whole ; 
 how must the line of division be run ? 
 
 Draw a triangle ; then construct a similar triangle equal in area to 
 
 5. 4 times that of the first triangle. 
 
 6. 2 times that of the first triangle. 
 7 \ that of the first triangle. 
 
 8. T 9 g- that of the first triangle. 
 
 9. Can you prove the theorem of this section by a method suggested by 
 152, Case 2 ? 
 
 Hints. Let a line cut i, , |, or any other fraction, from each of two sides of 
 a triangle; and use 159, Exercise i, and Theorem II. 
 
182 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ I6 9 . 
 
 IIL Problems and Applications. 
 
 169. Problem. To find the fourth proportional to three 
 given lines a, b, c {Fig. 156). 
 
 Construction. Make any angle O ; on its sides take 
 
 ^ 1 I 
 
 O DC 
 Fig. 156. 
 
 4. b > a, and c > b. 
 5. b > a, and c > a. 
 
 6. 
 
 7. 
 
 and draw BD II AC. OD 
 is the fourth proportional re- 
 quired. 
 
 Proof. Apply 159, 
 Theorem III. 
 
 Exercises. Give the con- 
 struction for the following cases : 
 
 1. b<a, and c> a. 
 
 2. b<a, and c = a. 
 
 3. b < c, and c =. b. 
 b> a, and c a. 
 
 7, b>a, and c = b. 
 
 170. Problem. 71? find a mean proportional to two given 
 lines a and b (Fig. 157} . 
 
 169 furnishes one solution (see Exercises 3 and 7). The 
 
 _2 more common solution is as fol- 
 
 _ lows : 
 
 Construction. Draw a straight 
 Xf> x line, and take A B = a, B C = b. At 
 
 \\ B erect a perpendicular. Bisect A C 
 
 \\ in O, and with a radius equal to O A 
 \\ describe an arc cutting the perpen- 
 ~~0 B ' c dicular in D. BD is the mean pro- 
 portional required. 
 
 Proof. Prove, as in 149, that ADC = 90; then apply 
 166. 
 
 ,*'' 
 
1 7 i.] 
 
 CHAPTER VIII. SIMILAR FIGURES. 
 
 183 
 
 171. Problem. To reduce or enlarge given lines a, b, c, 
 etc. (Fig. 158) , in a given ratio. 
 
 Construction. Let the ratio be 3 : 4 ; that is, let it be required 
 to reduce the lines each to 
 three fourths its present ~~~ 
 length. 
 
 Draw a straight line and ' 
 take on it four equal parts, 
 of which OA contains four, 
 OB three. At A and B 
 erect perpendiculars ; make 
 
 etc. 
 etc. 
 
 Join OC, OD, OE, 
 BF, BG, BH, etc., 
 
 O B A 
 
 Fig. 158. 
 are the reduced lengths required. 
 
 Proof. &OBF<^ AOAC, AO3G ^ AOAD, etc. (why?). 
 Then apply 159, Theorem I. 
 
 Exercises. 1. Draw four lines and reduce them in the ratio 3: 5. 
 2. Draw four lines and enlarge them in the ratio 5 : 2. 
 
 172. Problem. To divide a given line into equal parts. 
 
 One solution of this problem has been given ( 113) ; but, since 
 that solution requires the construction of several parallels, it is 
 tedious and liable to lead to errors. 
 
 The following construction is more 
 simple : 
 
 Construction. Let it be required 
 to divide MN (Fig. 159) into five 
 equal parts. Draw any line MO 
 through M-, then through a point 
 P of this line draw a line PQ II 
 MN, and take on this parallel any 
 five equal parts ending at a point Q. pig. 
 
 M 
 
 N 
 
184 
 
 GEOMETRY FOR BEGINNERS. 
 
 [Si7> 
 
 Through TV 7 arid Q draw a line cutting MO at O. Lastly, draw lines 
 through O and the points of division of PQ. These lines will 
 divide MN into five equal parts. 
 Proof. Apply 165. 
 
 Exercises. 1. Draw a line and divide it into eight equal parts. 
 2. Divide the line AB {Fig. 166} into ten equal parts. 
 Solution. Here the parts are so short that if obtained in the way above de- 
 scribed they would be indistinct. In this case pro- 
 
 C, , -D ceed thus : Erect at A and B perpendiculars, lay off 
 
 upon each ten equal parts, and through the points 
 of division draw lines parallel to A B. If now we 
 draw the diagonal AD, then the distances from the 
 points marked i, 2, 3, 4, etc., to the line BD will be 
 equal to fa, fa, T 3 o To> et c. of. the line A B. The 
 proof is left to the learner. 
 
 3. When (in making maps, plans, etc.) many 
 lines have to be very much reduced in length, it 
 becomes often convenient to employ scales of 
 equal parts, known among surveyors as PLAT- 
 TING SCALES. Explain how such a scale (shown 
 in Fig. 161}, containing one thousand parts, di- 
 vided decimally, is constructed and used. 
 
 Solution. Draw a straight line, and take on it 
 ten equal parts, A [B = B C= CD, etc., 1 each of these 
 lengths representing one hundred units of length 
 (meters). Divide AB into ten equal parts, each 
 Pip 160. part representing therefore ten units of length. Last- 
 
 ly, to represent single units of length, draw above 
 AB ten equidistant lines parallel to A B t erect perpendiculars at the points A, B, C, 
 
 9 8 7 6 
 
 c 
 
 Fig. 161. 
 
 1 Only a part of the scale can be shown in the figure. 
 
1 73-] CHAPTER VIII. SIMILAR FIGURES. 185 
 
 etc., and, after having subdivided FO into ten equal parts, join the points of division 
 of FO with those of A D by lines drawn obliquely, as shown in the figure. 
 
 The perpendicular OB, divided into ten equal parts by the parallels is num- 
 bered from L to 10, beginning at O\ then it follows (see Exercise 2) that the dis- 
 tance from i to the first oblique line measured on the parallel = i unit of length, 
 the distance from i to the second oblique line = II units, etc. ; and the distance 
 from 2 to the first oblique line = 2 units, the distance to the second oblique line = 
 12 units, etc.; and similarly for the other points of division of OB. 
 
 The point marked O is the zero of the scale ; hundreds are read off to the right, 
 tens to the left, and units on the vertical line. 
 
 From this explanation it is easy to see that a length of 348 meters measured on 
 the ground is represented upon the scale by the line mn ; also, that the length p q 
 corresponds to a distance of 216 meters, the length rs to 183 meters, etc. 
 
 4. Construct, with the aid of the scale in Fig. 161, a triangle whose sides 
 are 137, i6o m , and 225"*. 
 
 5. Draw a triangle, and determine the lengths of its sides by means of the 
 scale in Fig. 161. 
 
 6. Construct a scale of one hundred parts, divided decimally. 
 
 173. The measurement of the distance of an inaccessible 
 object is one of the most interesting applications of the laws of 
 Geometry. It gives us a good idea of the nature of those methods 
 which have enabled astronomers to measure the distance from the 
 earth to the moon, to the sun, and even to some of the fixed stars. 
 
 We have already seen ( 95-98) how distances may be meas- 
 ured indirectly by means of equal triangles. Now it is clear that, 
 if equal triangles are employed, the greater the distance to be 
 measured the larger the triangle which must be constructed ; so 
 that, when the distance is considerable, the construction of the 
 triangle becomes very inconvenient, and in many cases (for in- 
 stance, the distance from the earth to the moon) utterly impos- 
 sible. In such cases other methods must be found ; and these 
 methods are supplied by the laws of similar triangles. 
 
 In fact, the second method given for solving the problems of 
 95 and 98 is based upon the properties of similar triangles. 
 In 95, Method II., we construct on paper to a reduced scale a 
 fac-simile of the actual lines on the ground. What is \kasfac-simik 
 but a small triangle similar to the much larger one which contains 
 the distance sought? In other words, the triangles ABC and abc 
 
186 GEOMETRY FOR BEGINNERS. [ 174. 
 
 (Fig. 88) are similar; hence ac : ab =AC : A B, a proportion which, 
 by making (as on page 105) ac = i6 cnv , ab = 24 cm , AC = 8oo m , 
 
 becomes 16:24 = 800 : AB ; whence AB = 24 X 8oQ = i 2 oo ra . 
 
 16 
 
 Representing lines to scale on paper, all angles remaining un- 
 changed, always gives a figure similar to that formed by the unre- 
 duced lines, and is termed PLATTING TO SCALE. 
 
 Some examples will now be given of the indirect measurement 
 of distances by means of similar triangles. 
 
 174. Problem. To measure 
 the distance of two points A, X, sepa- 
 rated by a river. 
 
 At the point A (Fig. 162), sup- 
 posed accessible, erect (with a sur- 
 veyor's cross or other means) a line 
 _L AX, and measure its length to 
 any point B. Suppose, for exam- 
 ple, that AB i20 m ; take BC = 
 io m , and at C erect a second per- 
 pendicular CD, meeting the line B X at D. Measure CD ; let its 
 length be i8 m . Then it follows that, in all right triangles similar 
 to BCD that is to say, in all right triangles having an acute 
 angle equal to DBC the ratio of the legs must be 18 : IO. 1 
 
 Among these similar right triangles is the triangle ABX; there- 
 fore, AX-.AB = 18: 10; whence, AX=^2i6 m . 
 
 Exercises. 1. In Fig. 162, BC is made equal to io m solely for con- 
 venience of computation. If B C= 15, what would D C be found to be ? 
 
 2. If A is on the edge of the river, can you find A X by this method ? 
 What change in the position of the triangle B CD is necessary ? Give the 
 construction for this case. 
 
 3. How can AX be found by means of two equal right triangles ? 
 
 4. How can A X be found by means of an isosceles right triangle ? 
 
 1 In Trigonometry, 18 : 10, or {, is called the TANGENT of the angle DBC. 
 
I75-] 
 
 CHAPTER VIII. SIMILAR FIGURES. 
 
 187 
 
 175. Problem. To find the distance between two inac- 
 cessible points, X, Y. 
 
 Case 1 (Fig. i6j). Choose an accessible point, A, and meas- 
 ure the distances AX, AY, by 
 the preceding problem. Upon 
 AX measure a length Ax equal 
 to any fractional part (as T V, 5^, 
 etc.) of AX, and upon A Y meas- 
 ure Ay equal to the same frac- 
 
 tional part of A Y. Lastly, meas- 
 ure xy, which is parallel to X Y 
 (why ?) . Then the distance X Y can be found from the 
 
 Fig. 163. 
 
 If, for example, Ax = -fa AX, then XY = 10 xy. 
 
 Case 2. Where the line X Y prolonged is accessible (fig. 164) . 
 Erect at M. an accessible point, a line J_ MX, and mark the points 
 N where this line intersects the perpendicular let fall from Y, and 
 P, where it cuts XY prolonged. Measure MP and NP; let, for 
 example, MP = 360, NP = 200. 
 Take PO = io m ; at O erect OQ _L 
 PM, and measure the hypotenuse 
 PQ of the right triangle POQ. Sup- 
 pose, for example, that PQ = 24 ; 
 then, in all right triangles having an 
 acute angle equal to OPQ the ratio 
 of the leg corresponding to OP to the 
 hypotenuse is 10 : 24.* 
 
 PMX and PNY are two such tri- 
 angle? therefore, 
 
 N 
 Fig. 164. 
 
 M 
 
 10 : 24 = PM, or 360 
 
 10 : 24 = PN, or 200"= 
 
 and XY= PX - PY = 384* 
 
 PX, whence PX = 86 4 m ; 
 PY, whence PY = 48o m ; 
 
 1 In Trigonometry, the ratio 10: 2.4, or j 5 ^, is called the COSINE of the angle OPQ. 
 
iss 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 
 
 Exercises. 1. What is the advantage in Case 2 of making P = io m 
 rather than any other number ? 
 
 2. Explain how, with the aid of an instrument for measuring angles, the 
 distance X Fin Fig. 163 may be found by platting a similar triangle on paper. 
 By this method how many lines must be measured on the ground ? How 
 many by the other method ? 
 
 3. Given (Fig. 163} A X = 6oo m , A Y= 720, Ax = 5 AX, Ay = -fa AY, 
 xy-2i.iS m , XAY= 36. Find XY (z.) by a direct proportion; (.) by 
 platting a similar triangle to scale on paper. How nearly do the results 
 agree ? Why do they not exactly agree ? 
 
 4. Fig. zdj illustrates another method of measuring an inaccessible line 
 XY, a method by which only one line AB has to be measured on the 
 ground. Can you explain the method ? 
 
 Fig. 165. 
 
 176. Problem. To measure the vertical height of an ob- 
 ject (a tower, tree, church- spire, etc.) 
 
 1st Method. By means of a shadow. 
 
 Suppose that we wish to find the height A X of a tree (Fig. 
 166) which casts the shadow 'A B on the ground. Fix a staff ver- 
 tically in the ground near the tree, and measure its height mn. 
 Measure, also, the lengths of the shadows n o cast by the staff, and 
 AB cast by the tree. Now the ray of light XB is parallel to the 
 ray mo : whence it follows that ABX and mno are similar trian- 
 
1 76.] 
 
 CHAPTER VIll. SIMILAR FIGURES. 
 
 189 
 
 gles (why ?) . In other words, the shadows cast at the same 
 by different objects are proportional 
 to the heights of the objects. 
 
 Therefore no : AB = m n : A X. 
 If no = 0.9, AB = 24. 9 m , mn = 
 i.i7 m , then, 
 
 0.9 : 24.9 =1.17 : AX, 
 whence AX =32. 37. 
 
 NOTE. We may also reason thus : if 9O cm 
 of shadow correspond to H7 cin of height, then 
 l cm of shadow corresponds to -U_7 = 1.3^1 o f 
 height ; therefore 249o cm of shadow will cor- 
 respond to i.3 cm X 2490 = 3237 = 32.37 m . 
 
 2d Method. With the aid of an 
 instrument for measuring angles. 
 
 Let, for example, the height of 
 the tower (Fig. 167) be required. Place the instrument at a point 
 B, some distance from the foot of the tower, and measure the 
 angle abX, or angle of elevation, as it is termed. Measure AB, 
 the distance from the instrument to the foot of the tower. Then 
 construct on paper a right triangle similar to abX, and the side 
 of this triangle homologous to a X will give to the scale employed 
 the height aX, to which we must add A a, in order to obtain the 
 entire height. 
 
 pnnnnc 
 
 Fig. 166. 
 
 H 
 
 4- 
 
 " Fig. 167. Fig. 168. 
 
 3d Method. When the base of the object is inaccessible, choose 
 
190 GEOMETRY FOR BEGINNERS. [ 177. 
 
 a convenient place near the object, and measure a base line AB 
 (Fig. 1 68). Then measure the angles at the base, B and B A X, 
 and also the angle of elevation of the object XAO, and construct 
 on paper a triangle similar to XB A. This will give to scale the 
 length A X. Then, knowing A X and the acute angle X A O, con- 
 struct on paper the right triangle OA X, which will give to scale 
 the height OX required. 
 
 Here the unknown height OX depends on the construction of 
 two similar triangles, the first similar to the triangle ABX, the 
 second to the right triangle OAX. If the scale of reduction in 
 both cases is i mm to i m , the number of millimeters contained in 
 the line on paper which represents OX will be the same as the 
 number of meters in OX. 
 
 Exercises. 1. A monument casts a shadow of 44 at the same time that: 
 a rod 2 m high casts a shadow of i.33 m ; find the height of the monument. 
 
 2 In the 2d Method, why is the construction of a triangle on paper 
 avoided by placing the instrument so that the angle of elevation shall be 
 equal to 45 ? 
 
 3. What instruments are required in the 3d Method ? 
 
 4. Find the height of the building (Fig. 168} if A B = 250, angle B = 
 25 30', angle XA B = 104, and angle XA = 41 30'. 
 
 IV. Similar Polygons. 
 
 177. Two polygons are said to be SIMILAR if they have the 
 same shape. 
 
 In order to see what are the properties of similar polygons, 
 divide the polygon A B CD E (Fig. 169) into triangles by drawing 
 the diagonals AC, AD ; then begin at any point M of AB, and 
 make a new polygon by drawing MN II J3C, NO II CD, OP II DE. 
 The new polygon A MNO P is smaller than the original polygon, 
 but has the same shape, and is therefore similar to it. 
 
 By the same process, we can make a polygon AQRST larger 
 than the polygon ABCDE, and similar to it. 
 
I 77-] CHAPTER VIII. SIMILAR FIGURES. 191 
 
 Now the triangles AMN, ABC, AQR, are similar (why?); so 
 are the triangles AON, ADC,ARS\ 
 and lastly, the triangles A OP, ADE, 
 A S T. Whence it follows that, 
 
 Theorem I. Similar polygons 
 are composed of the same number of 
 triangles similar to each other and 
 similarly placed. 
 
 From this it follows, that similar 
 polygons are mutually equiangular. 
 
 Why, for example, is the angle A M B o 
 
 AMN equal to the angle ABC? Fig. 169. 
 
 the angle MNO equal to the angle BCD? 
 
 Compare now the sides in any two of the polygons, say the two 
 smaller ones. From the similar triangles AMN and ABC, 
 
 AM:AN = MN-.BC, 
 and also, AN: AC = MN: BC-, 
 
 and from the similar triangles A NO and AC D, 
 
 AN-. AC = NO: CD. 
 From the second and third proportions, it follows (Axiom I.) that 
 
 In this way, whatever be the number of sides, it may be shown 
 that the sides similarly placed are always proportional. 
 
 Theorem II. Similar polygons are mutually equiangular, 
 and have their homologous sides proportional. 
 
 Corollaries. I . Two regular polygons of the same number 
 of sides are similar. Why? 
 
 2. If each side of a polygon is 2, 3, 4, etc., times as great as 
 the homologous side of a similar polygon, then the sum of the sides, 
 that is, the perimeter of the first polygon, must be 2, 3, 4, etc., 
 times as great as the sum of the sides, or the perimeter, of the second 
 polygon. That is to say, the perimeters of similar polygons are to 
 each other as any two homologous sides. 
 
192 GEOMETRY FOR BEGINNERS. [ 178. 
 
 178. If two similar polygons, the sides of one of which are 
 double those of the other, are divided into triangles, by drawing 
 diagonals from the vertices of two equal angles, then each triangle 
 in the first polygon is (by 164) four times as great as the corre- 
 sponding triangle on the second polygon ; hence the sum of all 
 the triangles in the first polygon in other words, the area of the 
 first polygon is four times as great as the sum of all the triangles, 
 or the area, of the second polygon. Hence the areas of the two 
 polygons are to each other as 4 : i. But this is the ratio of the 
 squares of any two homologous sides. A like result would follow 
 if the sides had any other ratio than 2:1. Hence, - 
 
 Theorem. Similar polygons are to each other as the squares 
 of their homologous sides. 
 
 Therefore, if we have a polygonal figure on the ground, and 
 construct a similar figure on paper with its sides reduced to , ^, 
 |-, T V, etc., of their lengths, the area of the figure on the paper will 
 b e i> FJ TB> T<y<i> etc '> the area of the larger and similar figure on 
 the ground. 
 
 Exercises. 1. Compare the areas of two similar polygons with their 
 perimeters (see 177, Corollary 2). 
 
 2* Prove that the areas of regular polygons of the same number of sides 
 are to each other, (z.) as the squares of their sides ; (zY.) as the squares of 
 their perimeters. 
 
 3. If in two hexagonal parks a side of one is four times a side of the 
 other, how much larger is one park than the other ? Find, also, the ratio 
 of their perimeters. 
 
 179. Problem. To construct a polygon similar to a given 
 polygon. 
 
 There are various methods of solving this problem. 
 
 1st Method. When one side F G of the required polygon is 
 given. 
 
 Solution. By means of 177, Theorem I. Divide the given 
 polygon ABODE (Fig. 170) into triangles by diagonals; then 
 
1 79-] 
 
 CHAPTER VIII. SIMILAR FIGURES. 
 
 193 
 
 construct upon the given length FG, homologous to AB, a series 
 of triangles similar to those of the given polygon respectively, and 
 similarly placed. The poly- 
 gon FGHKL is the 
 polygon required. 
 
 Give the construction in 
 full. 
 
 2(1 Method. One side 
 F G being given, as before. 
 
 Solution. On one side 
 AB of the given polygon 
 ABCDE (Fig. 777) take 
 AAf=FG, construct the 
 polygon AMNOP^ ABCDE (how is this done?), and then 
 construct (by 117) upon the side FG the polygon FGHKL 
 & the polygon AM NO P. 
 
 Give the construction in full. 
 
 H 
 
 Fig. 170. 
 
 F 
 
 3d Method. When the homologous sides of the two polygons 
 are to be to each other in a given ratio. 
 
 Solution. Choose any point O, either within or without the 
 given polygon (see Figs. 172 and //j), and join it to all the cor- 
 ners by straight lines. Divide one of these lines at a point Af, so 
 that O M\ OA the given ratio, and then draw MN II AB, NP 
 II BC, etc. MNPQR is the required polygon.. 
 
194 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 179. 
 
 In Fig. 172, OM\ OA = i : 3 ; in Fig. 173. OM\ OA = 3 : 5. 
 Give the construction and proof for both cases. 
 
 Fig. 172. 
 
 Fig. 173. 
 
 Exercises. 1. Given a hexagon; construct a similar but larger hexagon. 
 
 2. Given an octagon; construct a similar but smaller octagon. 
 
 3. Upon the lengths 2O cm and 30 as homologous sides construct two 
 similar pentagons. 
 
 4. Draw a polygon, and then construct another whose sides shall be, 
 
 (z.) one-half those of the first polygon ; 
 (it.} two-thirds those of the first polygon ; 
 (tit.') twice those of the first polygon ; 
 (zz/.) three and one-half times those of the first polygon. 
 5* Draw a polygon, and then construct a similar polygon, 
 (z.) one-ninth as large as the first ( 174) ; 
 (z'z.) nine twenty-fifths as large as the first ; 
 (z'z'z.) four times as large as the first ; 
 (zV.) twice as large as the first. 
 
 6. Construct two similar pentagons which shall be to each other as 4: 9. 
 
 7. Construct two similar regular pentagons, one having three times the 
 perimeter of the other. What is the ratio of their areas? 
 
 8. Construct two regular hexagons about the same point as centre, making 
 the less radius of one = l\ times that of the other. 
 
 9. Construct two regular octagons, making the less radius of one = f that 
 of the other. What is the ratio of their perimeters ? of their areas ? 
 
CHAPTER VIII. REVIEW. 195 
 
 REVIEW OF CHAPTER VIII. 
 SYNOPSIS. 
 
 1. A proportion is an equation between two equal ratios. 
 
 2. The truth of a proportion is not affected if the terms of both ratios are 
 transposed. 
 
 3. The truth of a proportion is not affected if the extremes are transposed, 
 or if the means are transposed. 
 
 4. In every proportion the product of the extremes is equal to the product 
 of the means. 
 
 5. Two squares are to each other as the squares of their sides; two paral- 
 lelograms, or two triangles, as the products of their bases by their alti- 
 tudes. 
 
 6. If two parallelograms, or two triangles, have equal bases, they are to 
 each other as their altitudes ; if they have equal altitudes, they are to 
 each other as their bases. 
 
 7. Similar triangles are triangles that have the same shape. 
 
 8. Similar triangles are mutually equiangular, and have their homologous 
 sides proportional. 
 
 9. In two similar triangles, the sides that include equal angles have the 
 same ratio. 
 
 10. A line drawn through two sides of a triangle parallel to the third side 
 cuts off a smaller triangle similar to the entire triangle. 
 
 11. A line drawn through two sides of a triangle parallel to the third side 
 divides those sides proportionally. 
 
 12. I. Law of Similarity. Two triangles are similar if they are mutually 
 equiangular. 
 
 13. II. Law of Similarity. Two triangles are similar if they have an 
 angle equal, and the sides including this angle proportional. 
 
 14. III. Law of Similarity. Two triangles are similar if two sides of one 
 are proportional to two sides of the other, and the angles opposite the 
 greater sides are equal. 
 
 15. IV. Law of Similarity. Two triangles are similar if their sides, taken 
 in order, are proportional. 
 
 16. Lines through a common point divide parallels proportionally. 
 
 17. If in a right triangle a perpendicular is drawn from the vertex, of the right 
 angle to the hypotenuse, 
 
1.96 GEOMETRY FOR BEGINNERS. 
 
 (1) each leg is a mean proportional between the hypotenuse and the 
 adjacent segment ; 
 
 (2) the perpendicular is a mean proportional between the segments of 
 the hypotenuse. 
 
 18. The homologous sides of similar triangles are to each other (i) as the 
 corresponding altitudes, (2) as the perimeters. 
 
 19. Similar triangles are to each other as the squares of their homologous 
 sides. 
 
 20. Problem. To find the fourth proportional to three given lines. 
 
 21. Problem. To find the mean proportional between two given lines. 
 
 22. Problem. To reduce or enlarge given lines in a given ratio. 
 
 23. Problem. To divide a given line into equal parts. (Two ways.) 
 
 24. Problem. To measure the distance from an accessible to an inaccessi- 
 ble point. 
 
 25. Problem. To measure the distance between two inaccessible points. 
 (Two cases.) 
 
 26. Problem. To measure the height of an object. (Three methods.) 
 
 27. Similar polygons are polygons that have the same shape. 
 
 28. Similar polygons are composed of the same number of similar triangles 
 similarly placed. 
 
 29. Similar polygons are mutually equiangular, and have their homologous 
 sides proportional. 
 
 30. Regular polygons of the same number of sides are similar. 
 
 31. The perimeters of similar polygons are to each other as any two homo- 
 logous sides. 
 
 32. Similar polygons are to each other as the squares of their homologous 
 sides. 
 
 33. Problem. To construct a polygon similar to a given polygon. (Three 
 methods.) 
 
 EXERCISES. 
 
 1. A square and a rhombus have equal areas : find the ratio of their 
 perimeters, if the altitude of the rhombus is one-fourth that of the 
 square. 
 
 2. What single condition will make two right triangles similar ? also two 
 isosceles triangles ? 
 
 3. Given a triangle, two of whose angles are 55 and 80, construct a simi- 
 lar triangle one-fourth as large. 
 
CHAPTER VIII. - REVIEW. 197 
 
 4. Given a right triangle, an acute angle of which = 35; construct a 
 similar triangle four times as large. 
 
 5. Construct an isosceles triangle with the angle at the vertex = 30, and 
 one side = 6o cm . Then construct a similar triangle one-ninth as large. 
 
 6. Construct a triangle similar to a triangle whose sides are to each other as 
 the numbers 7, 8, II. 
 
 7. If a triangle is to be made three times as large as a given triangle, but 
 similar to it, what values must its angles have ? its sides? 
 
 8. Prove that parallels divide all lines that intersect them into proportional 
 parts. 
 
 Hints. This is an extension of 112. Compare also this theorem with the 
 theorem of 165. 
 
 9. The sides of a right triangle are 21, 28, and 35; find (z.) the seg- 
 ments of the hypotenuse made by a perpendicular let fall from the 
 vertex of the right angle, (z'z.) the length of this perpendicular. 
 
 10. Prove that the bisector of an angle of a triangle divides the opposite side 
 into parts that have the same ratio as the adjacent sides. 
 
 Hints. If ABC is the triangle, B D the bisector, prolong AB till it is met 
 
 by a parallel to the bisector throuh C. A A B D 
 A CBE\s, isosceles. 
 
 at E by a parallel to the bisector through C. A A B D ~ A ACE, and 
 
 11. The three medians of a triangle meet in one point. (A median is a line 
 drawn from a vertex and bisecting the opposite side.) 
 
 &ints. LetACbe the triangle. Draw two of the medians AD and BE 
 meeting at a point O, and then draw DF\\ AC meeting BE in F. Show 
 that A D F O ^ A A O E ; whence show that A O : O D =2:1. Draw 
 the third median CG, cutting AD in P, and show in like manner that 
 AP : PD II 2 : i ; therefore P must coincide with O. 
 
 The point O is called the centre of gravity of the triangle. 
 
 12. Construct a right triangle having given, 
 
 (.) The hypotenuse 70, and the ratio 3 : 2 of the legs. 
 (z'z.) The altitude 40, and the ratio 3 : 5 of the legs. 
 
 13. Construct on the diagonal of a rectangle an equal rectangle. 
 
 14. What is the ratio of the area ojf a field to that of its plan on paper, 
 
 (z.) if the scale of reduction is i cm to the metei ? 
 (zz.) if the scale of reduction is i mm to the meter ? 
 
 15. Wishing to find the distance from a point A on the mainland to an 
 island B, I erect at A a line A C JL A B, measure upon it a length A C = 
 900, and also measure the angle ACB, which I find to be 80. From 
 these data find the distance A B. 
 
 16. In Fig. s6j, let A X 542, A Y= 735, and the angle A = 57 30'; 
 find the distance XY. 
 
198 
 
 GEOMETRY FOR BEGINNERS. 
 
 17. A vertical rod 5 long throws upon a horizontal plane a shadow 5.7 
 
 long. At the same time a tower 
 throws a shadow 130 long. How 
 high is the tower ? 
 
 18. Explain how to find the height A B 
 (Fig. 174) by means of the right 
 triangles 4CandAD. 
 
 19. Explain how, without measuring 
 any angles, the distance X Y (Fig. 
 *75) can be found by means of 
 observations and measurements 
 performed on this side of the 
 water 
 
 20. The sides of a pentagon are 12, 20, n m , 15, and 22 m ; the perimeter 
 of a similar pentagon is i6 m ; find its sides. 
 
 Fig. ///. 
 
 21. Construct a polygon similar to a given polygon, and having to it the 
 ratio 4 : 9. 
 
 22. Construct a rectangle similar to one given rectangle and equal to another 
 given rectangle. 
 
l8o.] CHAPTER IX. THE CIRCLE. 199 
 
 CHAPTER IX. 
 THE CIRCLE. 
 
 CONTENTS. I. Sectors, Angles at the Centre, Chords, Segments ( 180-184). 
 II. Inscribed Angles (185-188). III. Secants and Tangents ($ 189-192). 
 IV. Two Circles ( 193, 194). V. Inscribed and Circumscribed Figures 
 ( 195-207). VI. Length of a Circumference ( 208-211). VII. Area of 
 a Circle ( 212-218). 
 
 I. Sectors, Angles at the Centre, Chords, Segments. 
 
 180. Review 26, 27, 28. 
 
 Some new definitions will now be given : 
 Definitions. I. The parts into which a circle (Fig. 176) 
 is divided by two radii are called SECTORS. 
 
 II. The angle between two radii is called an ANGLE AT THE 
 CENTRE. 
 
 III. A straight line joining two points of a circumference is 
 called a CHORD. 
 
 IV. A chord passing through the centre is called a DIAMETER. 
 
 V. A chord divides the circumference into two arcs ; if equal, 
 these arcs are called SEMI-CIRCUMFERENCES ; if unequal, the GREATER 
 <ind the LESS arcs, respectively. 
 
 A chord is often said to subtend the arcs into which it divides 
 the circumference. 
 
 VI. A chord divides the circle into two parts, called SEGMENTS ; 
 if equal, they are called SEMICIRCLES ; if unequal, the GREATER and 
 LESS segments, respectively. 
 
 In Fig. ij6 point out or name sectors, angles at the centre, 
 chords, a diameter, greater and less arcs, greater and less seg- 
 ments, semi-circumferences, and semicircles. 
 
200 GEOMETRY FOR BEGINNERS. [ 180. 
 
 Two radii form at the centre two angles ( 47, Note 2); but, 
 unless otherwise stated, the concave angle 
 (less than 180) is always to be under- 
 stood. To every such angle there corre- 
 sponds a definite arc, a definite chord, a 
 definite sector, and a definite segment. 
 Point out those which correspond to the 
 angle AOB (Fig. 176). 
 
 From what precedes several properties 
 Fig. 176. f tne circle follow as corollaries, or by 
 
 the method of superposition. 
 
 Corollaries. I. A circle has only one centre. 
 
 2. All radii of a circle are equal. 
 
 3. A diameter is twice a radius of the same circle. 
 
 4. All diameters of a circle are equal (why?) 
 
 5. An arc can be placed on an equal arc of the same circle (or 
 an equal circle) so as to coincide with it. 
 
 6. Circles are equal whose radii or whose diameters are equal. 
 
 Exercises. 1. Draw a figure of your own, and illustrate the definitions 
 of this section. 
 
 2. Of what is a circumference the locus ? 
 
 3 What determines (z.) the position, (zz.) the magnitude, of a circle ? 
 
 4. What three different positions can a point have as regards the circle ? 
 
 5* Given a circle, and the distance of a point from the centre ; how can 
 you tell whether the point will be inside, in, or outside the circumference ? 
 
 6. Draw a circle, having given its diameter. 
 
 7. In a given circle make a chord equal to a given line. 
 
 8. Make a few sectors differing in magnitude. 
 
 9. Make two sectors differing in magnitude, but having the same angle at 
 the centre. 
 
 10. What arc (greater or less) corresponds to a concave angle at the 
 centre ? to a straight angle ? to a convex angle ? What segment in each of 
 these cases ? 
 
 11. Make a few segments (greater and less) and name the corresponding 
 arcs and angles at the centre. 
 
l8l.] CHAPTER IX. THE CIRCLE. 201 
 
 12. If the angle at the centre is 200, what is the corresponding arc ? the 
 corresponding segment ? 
 
 13. If the angle at the centre is 180, what is the corresponding arc? 
 sector ? segment ? 
 
 14. Divide (free-hand) a sector into two equal parts. Are the parts also 
 sectors ? 
 
 15. Divide (free-hand) a segment into two equal parts. Are the parts also 
 segments ? 
 
 16. If you subtract a segment from the corresponding sector, what figure is 
 left? 
 
 17. Prove that a diameter is the longest chord in a circle. (Use 62.) 
 
 181. Theorem. In the same circle, or equal circles, fa 
 equal angles at the centre correspond equal arcs, chords, sectors, 
 and segments. 
 
 Draw a suitable figure, state hypothesis and conclusions (as 
 concisely as possible in terms of the letters marked on the figure), 
 and prove by superposition. 
 
 Remark. There are altogether four converse theorems. State 
 them, and prove them (by superposition) . 
 
 Hence it appears that the equality of any one pair of the five 
 elements mentioned in the theorem implies the equality of all the 
 other pairs. 
 
 Corollaries. I. If the angle at the centre is a straight angle 
 (180), the corresponding arc is a semi-circumference, and the 
 corresponding sector is a segment equal to a semicircle. 
 
 2. Since the sides of a straight angle form a diameter, every 
 diameter bisects a circle and also its circumference. 
 
 3. If the angle at the centre is a right angle, the corresponding 
 arc is half of a semi-circumference, and the corresponding sector 
 is half of a semicircle. 
 
 4. Two diameters at right angles to each other divide a circle 
 and also its circumference into four equal parts. 
 
 Definition. One-fourth of a circle is called a QUAPRANT. 
 The corresponding arc is the arc of a quadrant. 
 
202 GEOMETRY FOR BEGINNERS. [ 182. 
 
 Exercises. 1. If the angle at the centre is 1 80, what is the arc ? the 
 chord ? the sector ? the segment ? 
 
 2, How many degrees are there in an arc of a quadrant ? 
 
 8* What is the ratio of a quadrant and a half to the whole circle ? 
 
 4. In a given circle make (*'.) an arc equal to a given arc, (.) a sector 
 equal to a given sector, (in.) a segment equal to a given segment. 
 
 5* Make an arc of 225. 
 
 6. Make an arc equal to twice a given arc. 
 
 7. Make an arc equal to five times a given arc. 
 
 8. Make a sector equal to twice a given sector. 
 
 9. Can you make a segment equal to twice a given segment ? 
 
 182. If (Fig. 177) the angles AOM, MON, NOP, FOB, 
 COQ, QOR, ROD, are equal, then the 
 corresponding arcs and sectors are also 
 equal ( 181). And it is obvious (by 
 simple comparison, as in 37) that the 
 following ratios hold true : 
 
 Angles AOB : COD = 4:3; 
 
 Arcs AB-.CD =4:3; 
 
 Sectors A03: COD = 4 :$. 
 
 This reasoning may be applied to any 
 angles which have a common measure, and, in higher works on 
 Geometry, is extended to incommensurable arcs with the same 
 result. Hence, 
 
 Theorem. In the same circle, or equal circles, two arcs, or 
 two sectors, have the same ratio as the corresponding angles at the 
 centre. 
 
 Corollary. A sector has the same ratio to the entire circle 
 that the corresponding arc has to the entire circumference. 
 
 Exercises. 1. Make two arcs which shall be to each other in the ratio 
 
 2:5- 
 
 2. What ratios have the following sectors to the entire circle ? 
 ('.) A sector of 30. (m.) A sector of 90. 
 
 (?V.) A sector of 75. (iv.) A sector of 325. 
 
i8 3 .] 
 
 CHAPTER IX. THE CIRCLE. 
 
 203 
 
 Fig. 178. 
 
 183. Theorem I. The diameter perpendicular to a chord 
 bisects this chord and also the arcs subtended by it. 
 
 Proof. Draw radii to the ends of the chord, and apply 74. 
 To prove that the arcs are bisected, make use of 181 and 
 Axiom III. 
 
 What two other properties does this 
 diameter possess? 
 
 Theorem II. A perpendicular erected 
 at the middle of a chord passes through the 
 centre of the circle (or is a diameter). 
 
 Proof. This perpendicular passes 
 through every point which is equidistant 
 from the ends of the chord ( 87), and 
 the centre of the circle is such a point. 
 
 What other properties does this perpendicular have ? 
 
 Remark. We see (Fig. 178) that the middle of the chord, 
 the points of bisection of the major and minor arcs, and the centre 
 of the circle are in one straight line ; further, that this line is per- 
 pendicular to the chord ; and, finally, that it bisects the two angles 
 at the centre, their corresponding sectors and segments. This 
 makes, in all, eleven conditions which the line CD fulfils (or 
 eleven properties which it possesses). They are so related that, 
 if any two of them are known to be fulfilled, it follows that all the 
 others must also be fulfilled. 
 
 Problem. To find the centre of a given circle (or arc). 
 
 Analysis. Use the method of loci ( 92) in connection with 
 Theorem II. of the present section. 
 
 Corollaries. i. Through three points not in a straight line 
 a circle can always be drawn. (Compare Exercise 7, page 102.) 
 
 2. Two circumferences that have three points in common coin- 
 cide wholly. 
 
 Hence, a circle may be named by three letters at points on its 
 circumference. 
 
204 GEOMETRY FOR BEGINNERS. [ 184. 
 
 Exercises. 1. Bisect a given arc. 
 
 2. Bisect a given sector. 
 
 3. Bisect a given segment. 
 
 4. Find the centre of a given circle. 
 
 5. Find the centre of a given circle, using only one chord and one per- 
 pendicular. 
 
 6. Find the centre of a circle which shall pass through three given points. 
 
 7 . Given a curve, how can you ascertain whether it is the arc of a circle 
 or not ? 
 
 8. Find the locus of the centres of all the circles which pass through two 
 given points. 
 
 9. Draw a circle which shall pass through two given points, and whose 
 centre shall be in a given line. (In what case is there no solution ? In what 
 case is the problem indeterminate ?) 
 
 10. Draw a circle with a given radius and passing through two given points. 
 (Two solutions. In what case only one ? In what case no solution ?) 
 
 11. Find the locus of the middle points of a system of parallel chords. 
 
 12. Through a given point within a circle draw a chord which is bisected 
 at this point. 
 
 184. Let the radius OA (Fig. 179) remain fixed, and con- 
 ceive that a second radius turns about O 
 as a centre, starting from the position 
 OA, and coming successively into the po- 
 sitions OB, OC, OD, etc. The greater 
 the angle at the centre, AOB,AOC,AOD, 
 etc., the greater the corresponding chord 
 AB, AC, AD, etc. Likewise the greater 
 the chord the nearer it approaches the 
 p. centre O j and finally, when the moving 
 
 radius arrives at the position OE, the 
 
 chord passes through the centre, and becomes a diameter. In 
 
 this way we may arrive at the following conclusions : - 
 
 Theorems. I. The greater the angle at the centre, the greater 
 
 the corresponding chord (the angle being supposed less than 
 
 180). 
 
185.] CHAPTER IX. THE CIRCLE. 205 
 
 NOTE. The chord does not, however, increase in the same ratio as the angle 
 at the centre. The chord, for instance, of twice an angle is greater, but not twice 
 as great, as the chord of the angle itself. 
 
 II. The diameter is tJie longest chord of a circle. 
 
 III. Equal chords are equally distant from the centre. 
 
 IV. Of two unequal chords, the greater is nearer to the centre, 
 and conversely. 
 
 Exercise. Draw through a given point within a circle the longest chord ; 
 the shortest chord. 
 
 II. Inscribed Angles. 
 
 185. Definition. An angle B AC (Fig. r8d) whose vertex 
 is in the circumference of a circle, and whose sides are chords, is 
 called an INSCRIBED ANGLE. 
 
 To every inscribed angle corresponds a definite arc and a defi- 
 nite chord. The inscribed angle is often 
 said to " stand upon " its arc. 
 
 By an angle incribed in a segment 
 (DEFM) is understood an inscribed 
 angle (DEF) whose sides pass through 
 the ends of the arc of the segment. 
 
 Exercises. 1. Draw a few inscribed angles, 
 and point out the corresponding arcs and chords. 
 
 2. Inscribe several angles in the same seg- 
 ment. Fi S- *8- 
 
 186. Theorem. An inscribed angle is equal to one-half of 
 the angle at the centre which has the same arc. 
 
 Proof. The centre of the circle may lie (/.) in one of the sides 
 of the angle (Fig. 181), or (it.) between the sides of the angle 
 (Fig. 182), or (///.) without the sides of the angle (Fig. 183). So 
 there are three cases, but the second and third are closely con- 
 nected with the first. 
 
 Case (i.). Draw the radius OC, and use 66 and 80. 
 
206 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 186. 
 
 Case (ii.)- Draw the diameter AD, apply the proof for Case 
 /'.) twice, and add the results. 
 
 Fig. 181. 
 
 Fig. 182. 
 
 Fig. 183. 
 
 Case (m.)- Draw the diameter AD, apply the proof for Case 
 (/'.) twice, and subtract the results. 
 
 Corollaries. I. Inscribed angles which stand upon the same 
 arc are equal. 
 
 Proof. They are the halves of the same angle at the centre. 
 
 2. All angles inscribed in the same segment are equal. 
 
 Proof. They are inscribed angles standing upon the same arc ; 
 namely, the arc found by subtracting the arc of the segment from 
 tne whole circumference. 
 
 Upon what arc does the angle DEF (Fig. 180), inscribed in 
 the segment D EFM, stand? 
 
 3. An angle inscribed in a semicircle is a right angle. 
 
 Proof. The angle at the centre having the same arc is equal 
 to two right angles. 
 
 4. The sum of the angles inscribed, one in a greater, the other 
 in a less, segment, is equal to two right angles. Why ? 
 
 Illustrate by diagrams these four corollaries. 
 
 Exercises. 1. What is the value of an inscribed angle standing upon an 
 arc of 6o? 153? 320? 
 
 2. What value has an angle inscribed in a segment whose arc is equal to 
 90? 135 180? 270? 300? 
 
 3. Is the angle inscribed in a greater segment greater or less than 90? 
 What is its value inscribed in a less segment? 
 
187.] CHAPTER IX. - THE CIRCLE. 207 
 
 4. The arcs included by two parallel chords are equal. (Two methods of 
 proof.) 
 
 5. Parallel chords drawn from the ends of a diameter are equal. (Use 
 Corollary 3 of this section, and 72.) 
 
 G. Find the locus of the vertex of a right angle whose sides pass through 
 two given points. 
 
 Hints. See Corollary 3. Make the line which joins the given points the diame- 
 ter of a circle. 
 
 7. Construct a right triangle, having given its hypotenuse. Is the problem 
 determinate or not ? 
 
 8. Construct a right triangle, having given the hypotenuse, and the altitude 
 upon the hypotenuse as base. Is this problem determinate or not ? 
 
 9. Construct a right triangle, having given the middle point of the hypote- 
 nuse, the vertex of the right angle, and the length of one leg. 
 
 10. Find the locus of the vertex of a given angle whose sides pass through 
 two given points. 
 
 11. Find the locus of the middle points of all chords drawn through a given 
 point in the circumference of a given circle. (Use 183, Theorem II., and 
 Corollary 3 of the present section.) 
 
 187. If at any point D (Fig. 184) of the diameter AB of a 
 circle we erect a perpendicular DC, 
 and join AC and BC, the angle 
 ACB= 9 o( 186, Corollary 3); 
 whence it follows, from 166, that, 
 I. 
 
 II {C^tf X AD; 
 '\BC 2 =AB x BD. 
 
 Exercises. 1. State the above results in general terms as two theorems. 
 
 2. Verify the theorems for the case where D is at the centre of the circle. 
 Draw a diagram. 
 
 3. If (Fig. 184) 0=24 cm , and OD= io cm , find DC, AC, and BC. 
 
 188. Through a point P (Fig. 185) within a circle draw any 
 two chords APB and CPD. Join AD and C B. Show that 
 &APD ~ A CPB ( 54, 186, 161); whence PA:PD = 
 
208 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 
 
 Fig. 185. 
 
 PC:PB\ or, multiplying together the extremes and the means, 
 
 PA x PB = PC x PD. 
 Theorem. If chords are drawn 
 through any point within a circle, the 
 product of the two parts into which the 
 chord is divided by the point is the same 
 for all the chords. 
 
 Corollary. Among the chords which 
 may be drawn through P is the diameter 
 EPF; therefore, the constant value of 
 the above-mentioned product, for a given 
 position of P, is the product of the two parts of the diameter drawn 
 through P. 
 
 Exercises. 1. Prove 187, Theorem I., using this Corollary and 183. 
 
 2. If the radius of a circle = 24, find the product of the parts of any 
 chord drawn through a point io cm from the centre. 
 
 3. Prove that the angle APC (Fig. /#5), between two intersecting chords 
 AB and CD, is measured by half the arc A EC included between its sides + 
 half the arc DFB included between its sides produced. (Use 66 and 
 186.) 
 
 III. Secants and Tangents. 
 
 189. Definition I. A straight line of indefinite length 
 A B (Fig. 186} , which cuts the circumference of a circle in two 
 points is called a SECANT. 
 
 If we conceive the secant AB to turn 
 about the point A, towards the right, the 
 other point of intersection B will ap- 
 A proach nearer and nearer to A, and will 
 \ finally coincide with A. At this instant, 
 the secant takes the position CD, and 
 has only one point A in common with 
 the circumference. 
 
IQO-] CHAPTER IX. THE CIRCLE. 209 
 
 Definition II. A line which touches a circumference in one 
 point without cutting it is called a TANGENT. 
 
 Such a line is usually said to touch the circle ; and the point 
 where it touches the circle is called the point of contact. 
 
 Exercises. 1. In what relation does a secant stand to a chord ? 
 
 2. Can a straight line cut a circle in more than two points ? 
 
 3. What three different positions may a straight line have with respect to 
 a circle ? 
 
 190. Letj5C (Fig. 187) touch the circle at the point A. Join 
 A to the centre O, and also join any other 
 point of BC, as Z>, to O. Then O D> 
 OA (why?). That is, OA is shorter 
 than any other line which can be drawn 
 from O to the line BC; therefore ( 83), 
 OA i_ BC and BC _L OA. 
 
 Theorem. A tangent to a circle is 
 perpendicular to the radius drawn to the 
 point of contact. Fig. 187. 
 
 Exercises. 1 . Find the locus of the centres of circles which touch a given 
 line in a given point. 
 
 2. Find the locus of the centres of circles which have a given radius and 
 touch a given line. (Two lines. See 90.) 
 
 3. The radius perpendicular to a tangent bisects every chord parallel to 
 the tangent. 
 
 191. Problem. Through a given point to draw a tangent 
 to a given circle. 
 
 There are three cases ; for the given point may lie, (*.) within, 
 (ii.) in, (Hi.) without, the circumference. 
 
 Case (.) cannot be solved. Why ? 
 
 Case (ii.}. 190 supplies the means of solution. How? 
 Give the construction. 
 
210 GEOMETRY FOR BEGINNERS. [ 191. 
 
 Case (iii.). Construction. Let A (Fig. /<?) be the given 
 
 point, O the centre of the given 
 circle. Join A O, and upon A O 
 as a diameter describe a circle 
 cutting the given circle in B and 
 C. AB and A C are both tan- 
 gents to the given circle. 
 
 Proof. 186, Corollary 3, 
 and 190. 
 
 Corollary. The tangents 
 A B and A C (Fig. 188) are equal. 
 
 Proof. Show that the triangles ABO and A C O are equal. 
 
 Exercises. 1. Draw a tangent through a given point in the circumfer- 
 ence of a circle. 
 
 2. Draw tangents through a given point without the circumference of a 
 circle. 
 
 3. If tangents are drawn from a point without a circle, the line which 
 joins the point to the centre of the circle bisects the angle made by the tan- 
 gents. 
 
 4. If the angle between two tangents drawn through a point without a 
 circle is a right angle, find the angle at the centre formed by radii drawn to 
 the points of contact. 
 
 5* Draw a circle with a given radius that shall touch a given line in a given 
 point. (Two solutions.) 
 
 6. Draw a circle with a given radius that shall touch a given line and pass 
 through a given point (two, one, or no solution, according as the distance from 
 the given point to the given line is <, =, or > the diameter of the circle). 
 
 7. Describe a circle that passes through a given point and touches a given 
 line in a given point. (Where cannot the first point be ?) 
 
 8 To a given circle draw a tangent which, 
 
 (a) is parallel to a given line. 
 
 () is perpendicular to a given line. 
 
 (c) makes a given angle with a given line. 
 9. Describe a circle that touches two given intersecting lines, and, 
 
 (a) whose centre is in a given line. 
 
 () which touches one of the lines in a given point. 
 
 (f) which has a given radius. 
 

 
 CHAPTER IX. THE CIRCLE. 
 
 211 
 
 10. Describe a circle that touches two given parallel lines, one of them in 
 a given point. 
 
 11. Describe a circle that touches two given parallel lines, and passes 
 through a point between them. (Two solutions.) 
 
 12. Describe a circle that touches three given lines. (Four, two, or no 
 solution.) 
 
 A 
 Fig. 189. 
 
 192. Let (Fig. 189) AD be a tangent to a circle at A, and 
 AB any chord drawn from A. Draw 
 the diameter AC, and join B to the 
 centre O of the circle. CAD = 90 
 (why?) = arc ABC (measured in 
 degrees) . CAB=COB (why ?) 
 = arc C B (measured in degrees). 
 The angle between the tangent and 
 the chord BAD = CAD CAB. 
 Therefore, BAD = arc ABC - % 
 arc C B = arc BD. 
 
 Theorem. The angle between a tangent and a chord drawn 
 through the point of contact, is measured in degrees by half the arc 
 included between its sides. 
 
 Exercises. 1. Prove that the obtuse angle BAR (Fig. 189) made by 
 the tangent and the chord is measured by half the arc included between its 
 sides. 
 
 2. The angle made by two secants that meet without the circle is measured 
 by half the difference of the arcs included between its sides. 
 
 Hint. Join two alternate points in which the secants meet the circle. Use 
 66 and 130. 
 
 3. The angle made by two tangents drawn through a point without a circle 
 is measured by half the difference of the arcs included between its sides ( 66 
 and the Theorem above). 
 
 4:. The less arc between the points of contact of two tangents drawn 
 through a point without a circle = 120. Find the angle between the tan- 
 gents. 
 
 5. The angle made by a tangent and a secant that meet without the circle 
 is measured by half the difference of the arcs included between its sides. 
 
212 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 193. 
 
 IV. ~ Two Circles. 
 
 193. The relative position of two circles depends upon the 
 positions of their centres and the lengths 
 of their radii. 
 
 Definitions. I. Two circles that 
 have the same centre (Fig. 190) are said 
 to be CONCENTRIC. 
 
 II. The plane figure lying between their 
 circumferences is called a RING. 
 
 III. Two circles that have different cen- 
 tres are said to be ECCENTRIC. 
 
 IV. The line joining their centres is called the LINE OF CENTRES. 
 
 194. Let (Fig. 191} O and P be the centres of two eccentric 
 circles, with the radii r and r 1 respectively, and let us proceed to 
 examine the different positions which these circles will have, as the 
 distance of their centres OP = c is supposed continually to in- 
 crease. There are five different cases. 
 
 Case (i.) (Fig. 192} . c < r r'. If the distance of the centres 
 is less than the difference of the radii, the smaller circle lies wholly 
 within the larger. 
 
 Fig. 191. 
 
 Fig. 192. 
 
 193- 
 
 Case (*.) (Fig. 193) . c=r r 1 . If the distance of the cen- 
 tres is equal to the difference of the radii, the circles touch, or are 
 tangent, internally. 
 
1 94.] 
 
 CHAPTER IX. THE CIRCLE. 
 
 213 
 
 Case (Hi.'} (Fig. 193) . c > r r 1 and < r + /. If the distance 
 of the centres is greater than the difference, but less than the sum 
 of the radii, the circles intersect in two points. 
 
 Case (if.) (Fig. 194) . c = r + /. If the distance of the cen- 
 tres is equal to the sum of the radii, the circles touch, ^ or are 
 tangent, externally. 
 
 Fig. 194. 
 
 Fig. 195. 
 
 Case (v.) (Fig. 195}. c>r+r'. If the distance of the cen- 
 tres is greater than the sum of the radii, the circles lie wholly 
 outside, each other. 
 
 Exercises. 1. Can you give a reason why two circles cannot meet in 
 more than two points ? 
 
 2. What different positions can two circles have relatively to each other ? 
 
 3. Let r and r 1 be the radii of two circles, c the distance of their centres. 
 What relative positions have the circles for the following values of r, r' , and c ? 
 
 ()r=s, ^=3, c = 8. 
 0) rT, r'-^, c = 2. 
 (c} r = 6, r r =2, <r=io. 
 (</)r=8, r'=3, <:= 6. 
 0) ^=9, r'=5, r= 4. 
 
 4. Describe, with the radii 35 cm and 25 c 
 each other (z.) internally, (zV.) externally. 
 
 5. In a given circle make two circles such that they shall both touch the 
 given circle internally,' and also touch each other externally. 
 
 6. With three given radii, m, n, and /, describe three circles which shall 
 mutually touch one another externally. 
 
 Hint. First construct a triangle having the sides m + n, m + p, and n -\- p. 
 
 (/) 
 
 r = 
 
 6, 
 
 r /_ 
 
 6, 
 
 ^= 8. 
 
 (d 
 
 ?* ^^ 
 
 10, 
 
 r'= 
 
 3, 
 
 ir= 7. 
 
 (A) 
 
 r = 
 
 5 
 
 r /_ 
 
 5 
 
 f = 10. 
 
 (0 
 
 r 
 
 12, 
 
 r'= 
 
 7. 
 
 r= 9. 
 
 (/) 
 
 r = 
 
 9, 
 
 r /_ 
 
 2, 
 
 c = 5- 
 
 two circles which shall touch 
 
214 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 
 
 V. Inscribed and Circumscribed Figures. 
 
 195. Definitions. I. A circle is said to ^INSCRIBED in a 
 polygon when its circumference touches all the sides of the polygon ; 
 and the' polygon is said to be CIRCUMSCRIBED about the circle (Fig. 
 196). 
 
 II. A circle is said to be CIRCUMSCRIBED about a polygon when 
 its circumference passes through all the corners of the polygon ; 
 and the polygon is said to be INSCRIBED in the circle (Fig. 197) . 
 
 In Fig. 196, the sides of 
 the polygon are tangents 
 to the circle ; in Fig. 197, 
 they are chords of the cir- 
 cle. 
 
 Fig. 196. 
 
 Fig. 197. 
 
 NOTE. In treating this sub- 
 ject we shall first suppose the poly- 
 gon to be given, and the circle 
 inscribed or circumscribed ($ 
 
 196-201) ; and then suppose the circle to be given and the polygon to be 
 
 inscribed or circumscribed ( 202-207). 
 
 196. Problem. To inscribe a circle in a given triangle 
 ABC (Fig. 198}. 
 
 Construction. Bisect the angles A and B, and from the inter- 
 section O of the bisectors draw 
 OD1.AB. O is the centre of 
 the required circle, OD the ra- 
 dius. 
 
 Proof. Draw O E J_ A C, 
 O F B C, and j oin O C. Show 
 that A A OD ^ A A O E, and 
 A B O D ^ A B O F, whence 
 
 A D * OD = OE = OF; therefore, 
 
 the circle with O as centre and 
 OD as radius touches the sides of the triangle in D, E, and F. 
 
19 7] CHAPTER IX. THE CIRCLE. 215 
 
 Compare this problem with 92, Exercise 8. 
 
 Corollary. A circle can always be inscribed in a triangle. 
 
 197. Problem. To circumscribe a circle about a given tri- 
 angle. 
 
 Analysis, by means of 87 and 92. 
 
 Draw a figure, and give the construction. 
 
 Compare this problem with 92, Exercise 7. 
 
 Corollary. A circle can always be circumscribed about a 
 triangle. 
 
 Exercise. Prove that if a triangle is equilateral, the radius of the circum- 
 scribed circle=: twice the radius of the inscribed circle (see 112, Exercise). 
 
 198. Theorem. If a quadrilateral A B C D (Fig. 199) is 
 circumscribed about a circle, the sum of two opposite sides is equal 
 to the sum of the other two sides. 
 
 Proof. Apply 191, Corol- 
 lary ; then add the equal parts 
 of the sides together, obtaining, 
 for the result, AB + DC= BC 
 -f AD. 
 
 Corollary. If in a quadri- 
 lateral the sum of two opposite 
 sides is equal to the sum of the 
 other two sides, a circle can be in- " F - 
 
 scribed in the quadrilateral. 
 
 Exercises. 1. How is the centre of the inscribed circle found ? 
 
 2. In what two kinds of quadrilaterals can a circle always be inscribed ? 
 
 3. Draw a rhombus, and inscribe in it a circle. 
 
 199. Theorem. If a quadrilateral is inscribed in a circle, 
 the sum of the opposite angles is equal to 180. 
 
 Proof. Draw a figure of an inscribed quadrilateral, then draw 
 a diagonal, and apply 186, Corollary 4. 
 
216 GEOMETRY FOR BEGINNERS. [ 2OO. 
 
 Corollary. If in a quadrilateral the sum of two opposite 
 angles = 180, a circle can be circumscribed about the quadri- 
 lateral. 
 
 Exercises. 1. How is the centre of the circumscribed circle found ? 
 
 2. About what two kinds of quadrilaterals can a circle always be circum- 
 scribed ? 
 
 3. Draw a rectangle, and circumscribe about it a circle. 
 
 200. Theorem. A circle can always be inscribed in a 
 regular polygon. 
 
 Proof. Use 119. What is the radius of the circle, and what 
 point is its centre ? 
 
 Exercises. 1. Show how to find the centre of the inscribed circle. 
 
 2. Inscribe a circle in a regular hexagon. 
 
 3. Within a square describe four circles touching each other, and also the 
 sides of the square. What is the common radius of the circles ? 
 
 201. Theorem. A circle can always be circumscribed about 
 a regular polygon. 
 
 Proof. Use 119. What is the radius of the circle, and what 
 point is its centre ? 
 
 Exercises. 1. Show how to find the centre of the circumscribed circle. 
 
 2. Circumscribe a circle about a regular hexagon. 
 
 3. Circumscribe a circle about a regular octagon. 
 
 202. In order to inscribe regular polygons in circles, or to 
 circumscribe them about circles, it is necessary to divide the cir- 
 cumference of the circle into as many equal parts as there are 
 sides in the polygon. 
 
 Problem. To divide with ruler and compasses the circumfer- 
 ence of a circle into equal parts. 
 
 This problem can be solved only in the following cases : 
 
 Case 1. Two, four, eight, sixteen, thirty- two, etc., equal parts. 
 
202.] 
 
 CHAPTER IX. THE CIRCLE. 
 
 217 
 
 Solution. A diameter bisects the circumference ( 181, Corol- 
 lary 2); two diameters perpendicular to each other divide it into 
 four equal parts ( 181, Corollary 4); repeated bisection of the 
 arcs will divide it into eight, sixteen, thirty-two, etc., equal parts. 
 
 Draw a circle, and divide the circumference into two equal parts ; 
 four equal parts ; eight equal parts. 
 
 Case 2. Three, six, twelve, twenty-four, forty-eight, etc., equal 
 parts. 
 
 Solution. If we construct in a 
 circle (Fig. 200) an equilateral tri- 
 angle OAB, having each side equal 
 to the radius, it is clear that, since 
 AOB = 60, the arc AB = the 
 circumference. Therefore, the ra- 
 dius applied as a chord six times 
 to the circumference will divide it 
 into six equal parts. 
 
 The alternate points of division, 
 as A y C, E, divide the circumference 
 into three equal parts (Axiom II.) . 
 
 How can the circumference be divided into twelve equal parts ? 
 into twenty-four? into forty-eight? 
 
 Divide a circumference into six equal parts ; twelve equal parts. 
 
 Case 3. Five, ten, twenty, forty, 
 eighty, etc., equal parts. 
 
 Solution (Fig. 201) . Bisect A O 
 in C, and with C as centre, and a 
 radius equal to CD, describe an arc 
 cutting OB in E ; OE applied as a 
 chord ten times to the circumference 
 will divide it into ten equal parts, and 
 DE applied as a chord five times 
 will divide it into five equal parts. 
 
 Fig. 200. 
 
218 GEOMETRY FOR BEGINNERS. [ 203. 
 
 The proof that this solution is correct cannot be given here. 
 
 How will the division into ten equal parts enable us, without the 
 use of DjE, to make the division into five equal parts? 
 
 How can a circumference be divided into twenty equal parts ? 
 
 Divide a circumference into five, and into ten equal parts. 
 
 Remark. With the aid of a protractor (see./7^. 46) a cir- 
 cumference may be divided with sufficient exactness for all practical 
 purposes into any number of equal parts.. Suppose, for instance, that 
 the number of equal parts is 25 ; then the angle at the centre cor- 
 responding to each part = ^ = 14 24'. Construct with the pro- 
 tractor, placed so that its centre shall coincide with the centre of 
 the circle, an angle of 14 24' ; the arc of the given circle con- 
 tained between the sides of this angle will be ^V the entire cir- 
 cumference. 
 
 NOTE. An angle of i (among others) cannot be constructed with the ruler 
 and compasses. The question may here occur : how is a protractor itself gradu- 
 ated into degrees ? One way is as follows : divide with ruler and compasses the 
 entire circumference into six equal parts, and then into five equal parts. The dif- 
 ference between two of these parts (i ^) = -J^, the circumference = J^ the semi- 
 circumference. Bisect twice in succession each of these fifteen parts ; this divides 
 the semi-circumference into sixty equal parts ; finally, each of these sixty parts is 
 trisected by repeated trials, opening or closing a little the compasses till the right 
 opening is obtained. But the most perfect method of graduating a circle into 
 degrees and subdivisions of a degree is by means of nicely-adjusted screw-motion 
 in machines made for this express purpose. 
 
 Exercises. 1. Find all the angles between o and 360 which can be 
 constructed by Case I. 
 
 2. Find all the angles which can be constructed by Case 2. 
 
 3. Find all the angles which can be constructed by Case 3. 
 
 4. What angle can be trisected with ruler and compasses ? Trisect it. 
 
 5. Divide with the protractor a circumference into seven, nine, fifteen 
 equal parts. 
 
 203. Problem. To inscribe in a given circle a regular 
 polygon {Fig. 202). 
 
 Construction. Divide the circumference into as many equal 
 
20 4 .] 
 
 CHAPTER IX. THE CIRCLE. 
 
 parts as the polygon has sides, and join in order the points of, 
 division by straight lines. 
 
 Proof. The sides AB,BC, etc., 
 are equal, and the angles ABC, 
 BCD, etc., are equal ( 181, Re- 
 mark) . Therefore, the inscribed fig- 
 ure is a regular polygon ( 118). 
 
 Remark. By means of the fol- 
 lowing construction, the side of any 
 inscribed regular polygon (except the 
 triangle and the square) can be found 
 to a close degree of approximation. Fi s- 202 - 
 
 Divide the diameter AB (Fig. 203) of the given circle into as many 
 equal parts as there are sides in 
 the polygon to be inscribed. 
 Draw CD .AB. Prolong A B 
 to E and CD to F, making A E 
 = DF= one of the equal parts. 
 Join EF y and finally join the 
 intersection G of EF with the 
 circumference to the third point 
 
 of division H of the diameter ; 
 
 GH is the side of the polygon required. 
 
 Fig. 203. 
 
 Exercises. Inscribe in a given circle, 
 
 1. A square. 4, A regular polygon of seven sides. 
 
 A regular pentagon. 5. A regular polygon of nine sides. 
 
 3. A regular octagon. 6. A regular polygon of eleven sides. 
 
 204. Problem. To circumscribe about a given circle a reg- 
 ular polygon (Fig. 204). 
 
 Construction. Divide the circumference into as many equal 
 parts as the polygon has sides, and draw tangents through all the 
 points of division. 
 
220 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 205 
 
 Proof. Show that the quadrilat- 
 erals ABCO, CDEO, etc., are equal 
 (by the method of superposition) . 
 
 Exercises. Circumscribe about a given 
 circle, 
 
 1. An equilateral triangle. 
 
 2. A regular hexagon. 
 
 3. A regular heptagon. 
 
 4. A regular nonagon. 
 
 5. A regular decagon. 
 
 6. A regular dodecagon. 
 
 205. Regular polygons are most easily constructed by employ- 
 ing circles. 
 
 Problem. To construct a regular polygon, having given the 
 
 number of sides, and the length AB 
 {Fig. 2Oj) of one side. 
 
 Solution. Find the centre O of 
 the polygon ( 121), and about O, 
 with a radius = OA, describe a cir- 
 cle then apply AB to the circum- 
 ference as many times as there are 
 sides in the polygon. 
 
 Remark. Every regular poly- 
 gon is composed of a series of equal 
 right triangles ( 119, Corollary 2). 
 The sides of one of these triangles, 
 
 AOD (Fig. 205), are the greater radius, the less radius, and 
 the half-side of the polygon. It follows, from 142, that these 
 sides cannot have any values which we may choose to give them, 
 but are so related that ~AO* = ~AI? + ~OL? . The following table 
 gives the values (near enough for most purposes) of the less 
 radius and the side, the greater radius in each case being taken 
 equal to 100. 
 
 Fig. 205. 
 
20 5 .] 
 
 CHAPTER IX. THE CIRCLE. 
 
 221 
 
 Polygon 
 of 
 
 Greater 
 Radius. 
 
 Less Radius. 
 
 One Side. 
 
 3 sides 
 
 100 
 
 5 
 
 173 
 
 4 sides 
 
 loo 
 
 71 
 
 141 
 
 5 sides 
 
 100 
 
 8l 
 
 Il8 
 
 6 sides 
 
 100 
 
 87 
 
 IOO 
 
 7 sides 
 
 100 
 
 90 
 
 87 
 
 8 sides 
 
 100 
 
 92 
 
 77 
 
 9 sides 
 
 IOO 
 
 94 
 
 68 
 
 10 sides 
 
 TOO 
 
 95 
 
 62 
 
 12 sides 
 
 IOO 
 
 96 
 
 52 
 
 Exercises. Upon a line 2O cm long construct, 
 
 1. A regular pentagon. 3. A regular octagon. 
 
 2. A regular hexagon. 4. A regular decagon. 
 Construct a regular polygon, having given, 
 
 6. Perimeter = i.2 m , number of sides = 8. 
 
 6. Greater radius = 6o cm , less radius = 3O cm . 
 
 7. Greater radius = 5O cm , one side = 26 cm . 
 
 8. Less radius = 27 cm , one side = 39 cm . 
 
 9. Greater radius = one side. 
 
 10. One side = 36, one angle = 135. 
 
 11. Greater radius = 45 cm , angle at the centre = 30. 
 
 12. Less radius = 32 cm , angle at the centre = 40. 
 
 13. Since a regular polygon is composed of a series of equal right triangles, 
 any two parts which determine one of these triangles will determine the entire 
 polygon. What are the six parts of one of these triangles ? What two parts 
 can always be found if the number of sides is known ? How ? 
 
 14. Give all the cases of two parts that determine a regular polygon. 
 
 15. The radius of a circle = io cm . Find the side of the inscribed equilate- 
 ral triangle ; also a side of the inscribed regular decagon. 
 
 16. The perimeter of a square = 4O cm . Find the radii of the inscribed 
 and of the circumscribed circles. 
 
 17. In a regular hexagon the line which joins the middle points of two 
 opposite sides is 2 m long. Required the perimeter of the hexagon. 
 
 18. A square, whose perimeter = 40, is changed into a regular octagon 
 by cutting isosceles right triangles from its corners. Find the sides of one of 
 Ihese triangles. 
 
222 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 206. 
 
 19. One side of a marble slab in the shape of a regular octagon = 42. 
 Find the radius of the inscribed circle. 
 
 20. In a circle whose radius = I5 cm , a regular hexagon and a regular 
 dodecagon are inscribed. Find the difference of their perimeters. 
 
 206. Problem. Having given the side A B (Fig. 206} of an 
 inscribed regular polygon, to construct the side of an inscribed regu- 
 lar polygon having twice as many 
 sides, and to compute its length. 
 
 Construction. Bisect the arc 
 AB in D, and join AD. AD 
 is the side of the polygon re- 
 quired. 
 
 Proof. Apply 181. 
 Computation. If AB and 
 AO are known, then in the right 
 triangle AOC we can compute 
 the value of CO (how?). Sub- 
 tract CO from DO, and we ob- 
 tain CD; lastly, in the right tri- 
 angle A CD, in which AC and CD are known, compute AD 
 (how?). 
 
 Exercises. 1. In a circle whose radius = i m a regular hexagon is in- 
 scribed. Find the side and the perimeter of the regular inscribed dodecagon. 
 
 Solution.], n this case, A B {Fig. 206} =A O = i m ; A C= 0.5 ; CO - 
 
 Vi 2 o.S 2 = 0.86602541"; CD = DO CO=i 0.8660254 = o. 1339746 ; A D 
 
 = Vo.5 2 + o. 1339746* = 0.51763818^. The perimeter of the do- 
 decagon = 12 X 0.51763818 = 6.2ii658 m . 
 
 2. Find the area of the triangle A DB (Fig. 206}. 
 
 3. If the radius of the circle = i m , compute the side and the perimeter of 
 the inscribed regular polygon of twenty-four sides. 
 
 The radius of a circle = io m . Find, 
 
 4. The perimeter and the area of the inscribed square. 
 
 5. The perimeter and the area of the inscribed regular octagon. 
 
CHAPTER IX. THE CIRCLE. 
 
 223 
 
 207. Problem. Having given the side AB (Fig. 207) of 
 an inscribed regular polygon, to construct the side of the circum- 
 scribed regular polygon having 
 the same number of sides, and 
 to compute its length. 
 
 Construction. Bisect the arc 
 AB in D ; through D draw a 
 tangent, and prolong it until it 
 meets OA and OB prolonged 
 in E and in F. EF is the side 
 of the polygon required. 
 
 Proof Conceive the same 
 construction carried out for all 
 the equal arcs subtended by the 
 sides of the inscribed polygon ; then prove that the triangles OEF, 
 OFG, etc., are equal (by showing that the right triangles which 
 are their halves are all equal). 
 
 Computation. AB and EF are parallel, both being perpen- 
 dicular to OC; therefore, &AOB ^ &EOF, and OC ': OD = 
 AB : EF. In this proportion, OD and A B are given, and OC 
 can be found from the right triangle AOC (how ?) . E F can then 
 be computed (how?). 
 
 Exercises. 1 In a circle whose radius = i m a regular hexagon is in- 
 scribed. Find the side and the perimeter of the regular circumscribed 
 hexagon. 
 
 Solution. If (Fig. 207} OD=i, then AI?=i,zr\d OC= V^?* AC 1 
 V* 2 ~" -5~ = 0.8660254 ; hence 0.8660254 : i = I : E F. E F 1. 1547005 = one 
 side of the circumscribed polygon. And the perimeter = 6 X 1.1547005 = 6.928203. 
 
 2. In a circle whose radius = i m the side of the inscribed regular dodeca- 
 gon has been found ( 206) equal to 0.5176318. Find the side and the 
 perimeter of the circumscribed regular dodecagon. 
 
 The radius of a circle = io m . Find, 
 
 3. The perimeter and the area of the circumscribed square. 
 
 4. The perimeter and the area of the circumscribed regular octagon. 
 
224 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 208. 
 
 VI. Length of a Circumference. 
 
 208. In order to find the length of a circumference, we make 
 use of two general truths, which are sufficiently obvious from what 
 precedes. 
 
 I. The perimeter of an inscribed regular polygon is always less, 
 and that of a circumscribed regular polygon is always greater, than 
 the length of the circumference. 
 
 II. The greater the number of sides, the nearer the perimeter in 
 either case approaches to the length of the circumference. 
 
 If the radius of the circle = i, the perimeter of the inscribed 
 regular hexagon = 6 ( 202, Case 2), and that of the circum- 
 scribed regular hexagon = 6.928203 ( 207) ; therefore, the length 
 of the circumference is more than 6, but less than 6.928203, times 
 the length of the radius. 
 
 If now we inscribe and circumscribe regular polygons of twelve 
 sides, twenty-four sides, forty-eight sides, etc., and compute to the 
 sixth decimal place the values of their perimeters, we shall obtain 
 the following results : 
 
 Number of 
 
 PERIMETER OF THE POLYGON. 
 
 Sides in 
 the Polygon. 
 
 Inscribed. 
 
 Circumscribed. 
 
 6 
 
 6.000000 
 
 6.928203 
 
 12 
 
 6.211658 
 
 6.430782 
 
 24 
 
 6.265257 
 
 6.319320 
 
 4 8 
 
 6.278700 
 
 6.292172 
 
 9 6 
 
 6.282066 
 
 6.285430 
 
 192 
 
 6.282905 
 
 6.283746 
 
 384 
 
 6.283115 
 
 6.283325 
 
 7 68 
 
 6.283168 
 
 6.283220 
 
 1536 
 
 6.283181 
 
 6.283194 
 
 3072 
 
 6.283183 
 
 6.283187 
 
208.] CHAPTER IX. THE CIRCLE. 225 
 
 Since the length of the circumference always lies between the 
 perimeters of the inscribed and circumscribed polygons having the 
 same number of sides, and the perimeters of these polygons having 
 3072 sides agree in value in the first five decimal places, therefore 
 the number 6.28318 expresses the length of the circumference 
 correctly to five decimal places ; that is, the length of the circum- 
 ference is 6.28318 times the length of the radius, or 3.14159 times 
 the length of the diameter. 
 
 The number which expresses the ratio of the circumference of 
 a circle to its diameter, is often denoted, for the sake of conve- 
 nience, by the Greek letter TT l ; that is, TT = 3.14159, very nearly. 
 
 For most practical purposes we may take T: = 3^ = ^ 2 -, or, deci- 
 mally, - = 3.14. When greater accuracy is required, more decimal 
 places must be used. If more than five decimals are desired, 
 they may be obtained by continuing the above process, or much 
 more easily by means of other methods known to mathematicians. 
 
 But whatever method be used, the exact value of the ratio of the 
 circumference of a circle to the diameter cannot be found. The 
 farther the computation is carried, the nearer we approximate to 
 the true value, but the series of decimals which we obtain never 
 comes to an end. This amounts to saying that the circumference 
 of a circle and its diameter are two incommensurable lines ( 38). 
 
 NOTE. The problem of finding the ratio between the circumference of a circle 
 and its diameter exercised the minds of geometers in very early times. In an ancient 
 
 Brahmin work, called Ayeen Akbery, the value is assigned as 2i?Z = 3.1416. Archi- 
 
 1250 
 
 medes, the greatest geometer of antiquity (died 212 B.C.), using the method of 
 inscribed and circumscribed polygons, up to polygons of 96 sides, found that 
 the value lay between 3^ and 3^. Ludolph of Cologne, with prodigious labor, 
 carried the computation to polygons of 32,212,254,720 sides, which gives the value 
 correct to 35 decimal places. Metius found it to be 355, which is correct to six 
 
 decimal places. And Professor Richter of Elbing, Germany, has computed the 
 value of TT to 500 decimal places ! 
 
 But there is no practical advantage in obtaining more than the first five or six 
 places ; the value TT = 3.14159 is so near the truth that the error made in computing 
 
 l TT (pronounced like /) is the first letter of a Greek word meaning circumference. 
 
226 GEOMETRY FOR BEGINNERS. [ 2 09. 
 
 the circumference of a circle with a diameter of one mile is considerably less than 
 one inch. Accordingly, this problem, which engaged the attention of geometers 
 so deeply when their methods of approximation were less perfect, has now sunk 
 to the rank of those useless questions over which only persons ignorant of geo- 
 metrical science are willing to waste their time. 
 
 209. Let r, d, and c denote the numbers which express, in 
 terms of the same unit, the lengths of the radius, the diameter, and 
 the circumference, respectively, of a circle ; then, from what pre- 
 cedes, 
 
 C = ir(l= 2irr. [lO.] 
 
 Read this formula in words. If r is given, how can c be found ? 
 If c is given, how can r and d be found? 
 
 For example, if r=4 m , then d= 8 m , and c = 8 X 3.14 = 25.12. 
 
 Conversely, if <= 20 m , then d - - = 6.36, and r= 3.i8 m . 
 
 3- J 4 
 
 It is obvious, from Formula [10], that if either r or </be doubled, 
 trebled, quadrupled, etc., c will be doubled, trebled, quadrupled; 
 etc. ; in other words, 
 
 The circumferences of two circles are to each other as their radii, 
 or as their diameters. 
 
 210. An arc may be measured in degrees, minutes, and sec- 
 onds (angular measure), or its length may be expressed (like that 
 of any line) in meters, feet, or other linear units. 
 
 It follows, from 181, that, - 
 
 Length of an arc : whole circumference = corresponding angle at 
 the centre : j6o. 
 
 By means of this proportion such questions as the following may 
 be solved : 
 
 i. In a circle whose radius = 5, find the length of an arc 
 
 of 45. 
 
 The circumference = iox 3. 14 = 31. 4. Length of arc of 45 : 
 
 31.4 = 45 : 360. Length of the arc = 3I ' 445 = 3-92 m . 
 
21 lo] CHAPTER IX. THE CIRCLE. 227 
 
 2. In the same circle, how many degrees, etc., are there in an 
 arc 6 m long? 
 
 The above proportion in this case becomes, 6 : 31.4 = No. 
 of degrees in the arc : 360. No. of degrees in the arc = 
 
 31-4 
 
 =68 
 
 211. PRACTICAL EXERCISES. 
 
 ( When not otherwise specified, take TT = 37-) 
 
 Find the circumference of a circle, if the radius is equal to, 
 
 1. 7 m . 4. 6 T 3 T ft . 7. 3- 2 7 m ( 7r = 
 
 2. 236. 5. #". 8. 75n(7r = 3-14159). 
 
 3. 4.38 km . 6. ^ milc . 9. 8425* (^ = 3.1415926). 
 Find the radius and the diameter of a circle, if the circumference is equal 
 
 to, 
 
 10. I7 m . 13. I5i ft . 16. 9.74 (TT= 3.14). 
 
 11. 162.4. 14. ii in . 17. 64 cm (TT= 3.14159). 
 
 12. 2.16*". 15. i mile . 18. 40^0=3.1415926). 
 
 19. How many revolutions does a car-wheel, the diameter of which = 1.4, 
 make in going a distance of I32 km ? 
 
 Ans. Circumference of wheel = = 4.4 m ; No. of revolutions = ! 
 
 7 4-4 
 
 = 30,000. 
 
 20. If the driving-wheels of a locomotive have a radius of 3 ft , how many 
 revolutions will they make in going from New York to Boston, a distance of 
 236 miles? 
 
 21. The diameter of a wheel = 75 cm , and in going a certain distance it was 
 observed to make 3000 revolutions. Required the distance. 
 
 22. What is the diameter of a circular reservoir, if a man in walking around 
 it makes 840 paces, and each pace = 0.625? 
 
 23. What must be the diameter of a round dining table for eight persons, 
 if 77 cm is allowed to each person ? 
 
 24. How many nails are required to fasten the cloth covering about the 
 edge of a circular table 1.4 in diameter, the distance between two nails to 
 be 4 cm ? 
 
 25. A toothed wheel 3-756 m in diameter has 360 teeth. Find the distance 
 between the centres of two teeth. 
 
228 GEOMETRY FOR BEGINNERS. [ 211. 
 
 26 On the circumference of a wheel are 36 teeth, and the distance between 
 the centres of two teeth = i8 mm . Find the diameter of the wheel. 
 
 27. The radii of three concentric circles are i, 2, and 3 m . Find their 
 circumferences. 
 
 28. The diameters of two concentric circles are I2 cm and 2O cm . Find the 
 diameter of a circle whose circumference lies half-way between the other 
 two. 
 
 29. The diameters of two concentric circumferences are 8 cm and io cra . 
 Find the circumference lying midway between them. 
 
 30. If the diameters of the ends of a smooth log are 3 dm and 5 dm , what is 
 the diameter of a section midway between the ends ? 
 
 31. Find the exterior and interior circumferences of a circular tank, the 
 exterior diameter being i.o8 m and the thickness of the side 3 cm . 
 
 32. A circular reservoir is dug g.6 m in diameter, j.2 m deep. It is then 
 lined with stones 28 cm long, I2 cm wide, 8 cm high. Find the number of stones 
 required, allowing i cm , outside measurement, between the stones for mortar. 
 
 33. An iron wheel 1.26 in diameter has teeth upon its circumference, 
 distant 6 cm from centre to centre ; (z.) how many teeth are there, and ('.) 
 what, is the length of the circumference in which the highest points of the 
 teeth lie, the height of the teeth being 7 cm ? 
 
 34. If the length of the minute-hand of a watch =i8 mm , what distance 
 does its end pass over in one day ? in one year ? 
 
 35. What distance is traversed by a point in the circumference of a water- 
 wheel 2 m in diameter, if the wheel makes 1400 revolutions in one hour ? 
 
 36. The earth turns on its axis once a day. If its diameter = 8000 miles, 
 find the velocity per second of a point on the Equator. 
 
 37. What ought to be the diameter of a millstone which is to make 
 100 revolutions per minute, if experience teaches that to secure good grind- 
 ing a point on the circumference should have a velocity of about 8 m per 
 second ? 
 
 38. Draw two concentric circles with the radii 25 cm and 4O cm ; then de- 
 scribe a concentric circle whose circumference equals the sum of the two other 
 circumferences. Find, also, its length. 
 
 39. What is the radius of a circle whose circumference is equal to the 
 difference between the circumferences of circles 24 and 45 cm in diameter ? 
 
 40. Find the diameter of a circle whose circumference is equal to five 
 times that of a circle 3O cm in diameter. 
 
 41. Describe two circles, making the circumference of one equal to one- 
 quarter that of the other. 
 
211.] CHAPTER IX. THE CIRCLE. 229 
 
 42. The circumferences of two concentric circles are 50 and 8o m . How 
 far apart are they ? 
 
 43. Of two toothed wheels which work together, the first has fifty, the 
 second eighty teeth. How many revolutions will the first make while the 
 second makes sixty ? What diameter has the first, if that of the second = 
 1.32? 
 
 Hint. The distance between the centres of the teeth on both wheels must be 
 the same. 
 
 44. The diameter of a toothed wheel having 100 teeth = 6o cm . How 
 many teeth must a wheel have, which, when connected with the first wheel, 
 makes ten revolutions while the first makes one ? And what must be its 
 diameter ? 
 
 45. Of two belted wheels in a machine, the first makes twelve revolutions 
 while the second makes five. If the diameter of the first wheel g6 cm , find 
 that of the second. 
 
 46. Invent and describe an arrangement of toothed wheels, such that, 
 if the first wheel revolves once a day, the last wheel will revolve once a 
 second. 
 
 47. The hind wheels of a carriage are i.5 m in diameter. How many 
 revolutions will they make while the fore wheels, i m in diameter, make 2500 
 revolutions ? 
 
 48. In a circle 2 m in diameter find the length of the arc subtended by 
 one side of, 
 
 (z.) The inscribed square. 
 (.) The inscribed regular pentagon, 
 (m.) The inscribed regular octagon. 
 (zV.) The inscribed regular decagon. 
 
 49. Diameter of a circle = 3-3 m . How long is an arc of 140 ? 
 
 50. What is the diameter of the circle if an arc of 40 has a length 
 of 8.3 m ? 
 
 51. What is the radius of a circle in which each degree of the circumfer- 
 ence has a length of i cm ? 
 
 52. \Vhat is the length of i on a protractor, the diameter of which = 48 cm ? 
 
 53. Find the ratio between the lengths of degrees on two concentric cir- 
 cles, the radii of which are i dm and 8 dm . 
 
 54. How many degrees, etc., are there in an arc equal in length to the 
 radius of the circle ? 
 
 55. Find the ratio between an arc of a quadrant and the corresponding 
 chord. 
 
230 GEOMETRY FOR BEGINNERS. [ 212. 
 
 VII. Area, of the Circle. 
 
 212. The area of a circle is always less than that of a circum- 
 scribed regular polygon, and greater than that of an inscribed 
 regular polygon ; moreover, the areas of these polygons approach 
 nearer to each other, and to the area of the circle, as the number 
 of their sides is increased. We might, therefore, find the area of 
 a circle by a process similar to that employed for finding the 
 length of the circumference. The following method, however, is 
 much better : 
 
 Circumscribe a regular polygon about a circle (as in Fig. 204], 
 and let r = its less radius = radius of the circle. Now it is clear, 
 that by increasing the number of sides in the polygon, we can make 
 its area approach the area of the circle as nearly as we please, but 
 we cannot make it absolutely equal to the area of the circle ; in 
 
 other words, 
 
 rthat value which the area of the polygon 
 Area of the circle = -< approaches, but never absolutely reaches, as 
 
 v. the number of its sides is increased. 
 
 This value of the area of the polygon is termed its limiting value, 
 or simply its LIMIT ; so that we may write the above equation more 
 briefly, thus : 
 
 Area of the circle = limit of area of the polygon. 
 
 By 133, area of the polygon = J r x its perimeter. This equa- 
 tion is true, no matter how great or how small the number of sides. 
 Now, by increasing the number of sides, the perimeter can be made 
 to approach the circumference of the circle as nearly as we please, 
 but can never quite reach it. Therefore, 
 
 Limit of area of the polygon = %r X circumference of the circle. 
 Therefore (Axiom I.), 
 
 Area of circle = J r X its circumference. 
 
214.] CHAPTER IX. - THE CIRCLE. 231 
 
 213. If we substitute for the circumference the values 
 2 TT r, we obtain the very useful formula, 
 
 Area of a circle = 1 ird 2 = irr 2 . [n.] 
 
 For example : if the radius /-= 8 m , the area 6"= 3.14 X 64 = 
 200.96**. 
 
 From Formula [n] we find that the values of d and of r in 
 terms of and - are, 
 
 Read these formulas in words. 
 
 F 20 / - ' 
 
 For example : if S = 2O qm , then r = A - = V6.37 = 2.52. 
 
 \3-i4 
 
 In Formula [n] assume any value for r (or d), then double it, 
 treble it, quadruple it, etc., and work out for each case the cor- 
 responding value of the area of the circle. We find that the areas 
 are to each other as the numbers i, 4, 9, 16, etc. That is, 
 
 The areas of circles are to each other as the squares of their radii 
 (or their diameters} . 
 
 214. Problem. To transform a given circle into a square. 
 
 By computation. Find the area of the circle, extract its square 
 root, and the result will be one side of the square required. 
 
 Let a = the radius of the given circle ; then its area = ~ a 2 , and 
 \he side of the equivalent square = V^^ 2 = a V~ = a Vs. 14159 
 6=3x1.77245. 
 
 By construction. Draw a straight line equal to half the circum- 
 ference, and then find, by 170, a mean proportional between 
 this line and the radius ; this will be a side of the required square. 
 
 Remark. A square which shall be exactly equivalent to a 
 given circle cannot be found, because the value of TT cannot be 
 exactly found. In other words, the problem of " squaring the 
 circle " admits only of an approximate solution. But the value 
 
232 
 
 GEOMETRY FOR BEGINNERS. 
 
 [215- 
 
 ^ = 3.14159 gives a result far more accurate than could be actually 
 constructed with ruler and compasses. 
 
 Exercises. 1. Draw a circle, and then find the side of an equivalent 
 square, (z.) by computation, (z'z.) by construction. How do the results 
 agree ? 
 
 2* Construct a square, and then make a circle equivalent to the square. 
 3. Find the ratio of the areas of a square and a circle, if, 
 
 (z.) A side of the square is equal to the radius of the circle. 
 (zY.) A side of the square is equal to the diameter of the circle. 
 
 215. Upon the sides of the right triangle ABC (Fig. 208), 
 as diameters, describe circles. By 
 142, 
 
 If we multiply the terms of this 
 equation by TT, the equality is not 
 destroyed (Axiom IV.), and we ob- 
 tain, 
 
 TT x TC 2 = i-xZ# 2 4- J* X ~BC\ 
 Now these three quantities, taken 
 in order, are the areas of the circles 
 described upon the hypotenuse and 
 the two legs as diameters. 
 
 Theorem. The area of the circle described upon the hypote- 
 nuse of a right triangle as diameter is equal to the sum of the areas 
 of the circles described upon the two legs as diameters. 
 
 NOTE. This is a special case of a very general theorem, which may be called 
 the Generalized Theorem of Pythagoras, namely : if upon the sides of a right trian- 
 gle we construct any three similar figures, the area of the figure constructed upon the 
 hypotenuse will be equal to the sum of the areas of the figures con 
 
 Fig. 208. 
 
 legs. 
 
 constructed upon the 
 
 Exercises. 1. Make a circle equal to the sum of two given circles. 
 
 2. Make a circle equal to twice a given circle. 
 
 3. Make a circle equal to the difference of two given circles. 
 
 4. Make a circle equal to half of a given circle. 
 
2 1 6.] 
 
 CHAPTER IX. THE CIRCLE. 
 
 233 
 
 A 
 
 216. Problem. To find the area of a circular sector AOB 
 (Fig. 209], having given the radius of the circle and the angle of 
 the sector. 
 
 Solution. From 182, Corollary, and 
 212. it follows that the area of a sector 
 is found by multiplying its arc by half the 
 radius of the circle. 
 
 Example. Let the radius = 14, and 
 the angle of the sector = 40. Then, 
 
 Circumference of circle = 2 x 3! X 14. 
 
 Arc of sector = ^ Ty X circumference. 
 
 Area of sector = its arc x 7 = 68.44 qm . 
 
 Compute the sector for the following 
 values of the angle : 90, 180, 270. 
 
 217. Problem. 71? find the area of a circular segment 
 C ED (Fig. 209), having given the radius of the circle -, the angle 
 at the centre COD, and the height E F of the segment. 
 
 Solution. Compute the areas of the sector COD and the tri- 
 angle COD. Then, segment C ED = sector COD triangle 
 COD. 
 
 Example. Let the radius = i, the angle COD = 40, and 
 
 the height EF= 6 cm . Then ( 216), the sector COD = ^L 
 
 360 
 
 = - = o.349 qm . In the triangle COD, the altitude OF = OE 
 
 9 
 
 EF= 0.94 ; by 187, I., half the base CF = \lEF(2 - EF) 
 = 0.342 ; the base CD = 0.684 ; and the area of the triangle 
 = '94 X 0.684 = . 32I48qm . Therefore, the area of the seg- 
 ment = 0.349 0.32148 = o.o2752 qm : an answer correct enough 
 for all practical purposes. 
 
 NOTE. The height E F (Fig. 209} of a segment, and the angle COD at the 
 centre, are not independent of each other, but are connected by a relation which 
 it is the business of Trigonometry to investigate. 
 
234 GEOMETRY FOR BEGINNERS. [ 2 1 8. 
 
 218. PRACTICAL EXERCISES. 
 
 ( Take TT = 3^, unless otherwise stated.) 
 
 Find the area of a circle, having given, 
 
 1. The diameter = I i m . 4. The radius = 
 
 2. The diameter = 4.37. 5. The radius = 
 
 3. The diameter = 9. 0. The radius = 4| milcs . 
 Find the area of a circle (TT = 3.14159), having given, 
 
 7. The diameter = 8.346. 9. The radius = 27.19. 
 
 8. The diameter = i6.37 km . 10. The radius = 5 i ft . 
 Find the radius of a circle, having given, 
 
 11. The area = 100^. 14. The area = 2ooo s i- <*. 
 
 12. The area = 946<i m . 15. The area = 682.4i- in . 
 
 13. The area = 23.37^. 16. The area = 72^- miles . 
 
 17. The radius of a circle = 7. Find the area, ('.) taking TT= 3}, (.) 
 7r =3- I 4 I 59- What is the difference (in square centimeters) of the results? 
 Which result is more correct? 
 
 18. The diameter of a circular park is i km . Find the difference in its 
 areas computed, (j.) taking TT 3!; (.) taking TT = 3.14159. 
 
 19. A man buys a circular field, diameter = i6oo ft , for four cents per square 
 foot. What difference will there be in the price paid for the field according 
 as TT is taken equal to 3^ or equal to 3.14159? 
 
 20. Describe a circle, and then find its area. 
 
 21. Find the area of a circle if its circumference = 8o m . 
 
 22. The circumference of a tree measures 2.6 m . Find the area of a cross- 
 section. 
 
 23. If the length of a circular race-course is to be i mile , how many acres 
 of land will it cover? 
 
 24. If the circumference of a garden-bed = 48.25 m , what is its area? 
 
 25. Around the banks of a circular pond are planted 256 poplars, 6 m apart. 
 Find the area of the pond. 
 
 26. What will it cost to cover a round table the radius of which = 75, 
 W Sie cloth is 6o cm wide and costs $2.00 per meter? 
 
 Find the radius, the circumference, and the area of a circle equal to, 
 
 27. The sum of two circles whose diameters are 5 and 6 m . 
 
 28. The sum of two circles whose areas are 6oi m and 8o f i m . 
 
 29. The difference of two circles whose radii are 3 m and 4. 
 
 *JO The difference of two circles whose circumferences are 33 and 44. 
 
2l8.] CHAPTER IX. THE CIRCLE. 235 
 
 Find the radius of a circle, 
 
 31. Three times as large as a circle whose radius = 7 m . 
 
 32. Five and one-half times as large as a circle whose diameter = I2.6 m . 
 
 33. One-ninth as large as a circle whose circumference = 45.23. 
 
 34:. If a horse, tied by a rope 4 m long to a stake driven in the ground, 
 eats all the grass which he can reach in three hours, how long must the rope 
 be in order that the grass may last twelve hours? 
 
 35. If the diameters of the holes in the side of a tank are as 3 : I, how 
 much more water will flow in an hour out of one than out of the other? 
 
 36. If 80 flowers will grow on a circular garden-bed, and I wish another 
 bed on which 800 of the same flowers will grow, what diameter must it have 
 compared with that of the first bed? 
 
 37. Find the area of a circular ring (Fig. 209}, the radii of the circles 
 being 6 and 7. 
 
 38. Find the area of a ring, the two circumferences being 2O m and 30. 
 
 39. Describe two concentric circles; then find the area of the circular ring. 
 
 40. In the middle of a circular pond 400 in diameter stands a circular 
 island 90 in diameter. Find the area of the surface covered by the water. 
 
 41. Find the area of a ring if the outer circumference = 85 cm and the 
 breadth of the ring = io cm . 
 
 42. What will it cost to have a walk i m wide about a circular reservoir 
 64 in diameter, the price of paving being $2.50 per square meter? 
 
 43. A target consists of a black circle o.75 m in diameter, surrounded by a 
 white ring 0.28 wide. Find the total area covered by the target. 
 
 44. The outer and inner circumferences of a round tower are 17.2 and 
 I2.8 m . Find the ground area covered by the wall. 
 
 45. How wide is a ring if the areas of the two circles are 2OOi m and 641? 
 
 46. If a ring is to be 4 m wide, and is to contain Scfl, what must be the 
 radii of the two circles? 
 
 47. To the shorter sides of a table, 2 m by 1.25'", semicircular leaves are 
 added. Find the total area of the table. 
 
 48. Find the total area of the figure formed by adding semicircles to the 
 four sides of a square, one side of which = i.2 m . 
 
 49. A square, a rectangle, and a circle have the same perimeter, 8 m . Find 
 the three areas, the rectangle being two-thirds as wide as it is long. 
 
 50. The circumference of a cifcle is equal to the perimeter of a square 
 whose side is io m . Find the difference in their areas. 
 
 NOTE. Of all plane figures with equal perimeters the circle has the greatest 
 area. 
 
236 GEOMETRY FOR BEGINNERS. [ 2l8. 
 
 51. A circle and a square have the same area, i2Cfl m . Find the difference 
 between the circumference of the circle and the perimeter of the square. 
 
 52. If a circle, a square, a regular hexagon, and a regular octagon each 
 contain io.4 a , find the differences between the perimeters of the polygons and 
 the circumference of the circle. 
 
 53. A circular piece of lead is recast in the shape of a square, the thick- 
 ness remaining the same. If the radius of the circle = 45 cm , find a side of the 
 square. 
 
 54. Find the diameter of a circle equivalent to a regular octagon whose 
 side = 2 m . 
 
 55. The radius of a circle = 8 m . Find the difference between its area and 
 the area of the inscribed regular hexagon. 
 
 56. In a picture-frame 6o cm square is a circular picture, the radius of which 
 = 2O cm . Find the area of the part not covered by the picture. 
 
 57. In a circle whose radius = 3O cm , find the area of a sector of 80. 
 
 58. If the radius of a circle = 76, find the area of a sector whose arc is 
 2O cm long. 
 
 59. How large is a sector if its arc contains 120 and is 40 long? 
 
 60. What part of the entire circle is the sector whose arc equals the radius 
 of the circle? 
 
 61. If the area of a sector of 45 = 2 i m , find the radius of the circle. 
 
 62. The radius of a circle = 3, the height of a segment = i.968 m , and the 
 angle of the segment = 98. Find the area of the segment. 
 
 In a circle whose radius 4 m , find the area of the segment cut off by the 
 side of, 
 
 63. The inscribed equilateral triangle. 
 G4. The inscribed square. 
 
 65. The inscribed regular hexagon. 
 
 66. The inscribed regular octagon. 
 
CHAPTER IX. REVIEW. 237 
 
 REVIEW OF CHAPTER IX. 
 
 SYNOPSIS. 1 
 
 I. Sectors, Angles at the Centre, Chords, Segments. 
 
 180. Review of 26-28. Definitions of a sector, an angle at the centre, a 
 chord, a diameter, a semi-circumference, greater and less arcs, a seg- 
 ment, greater and less segments. Six corollaries. 
 
 181. Equal angles at the centre, their arcs, chords, sectors, and segments. 
 Five theorems. Four corollaries. Definition of a quadrant, 
 
 182. Unequal angles at the centre, their arcs and sectors. One theorem. 
 One corollary. 
 
 183. Properties of a diameter perpendicular to a chord. Two theorems. 
 One problem. Two corollaries. 
 
 184. Equal and unequal chords, their angles at the centre, and their dis- 
 tances from the centre. Four theorems. 
 
 II. Inscribed Angles. 
 185. Definition of an inscribed angle. 
 
 186. Measure of an inscribed angle. One theorem. Four corollaries. 
 187. Application of 166 to the circle. Two theorems, 
 188. Product of the parts of a chord passing through a point within a 
 circle. One theorem. One corollary. 
 
 III. Secants and Tangents. 
 
 189. Definitions of a secant and a tangent. 
 
 190. Relation between a tangent and a radius drawn to the point of con- 
 tact. One theorem. 
 
 191. Tangent to a circle through a given point. One problem. One corol- 
 lary. 
 
 192. Measure of the angle between a tangent and chord. One theorem. 
 
 IV. Two Circles. 
 
 193. Definition of concentric and eccentric circles, a ring, the line of centres. 
 194. Relative positions of two eccentric circles. Five cases. 
 
 1 Here, and in the remaining chapters, the synopsis is presented in a condensed 
 form which may be made the basis of a more or less extended review according 
 to the discretion of the teacher. 
 
238 GEOMETRY FOR BEGINNERS. 
 
 V. Inscribed and Circumscribed Figures. 
 
 195. Definitions of inscribed and circumscribed figures. 
 
 196. Circle inscribed in a triangle. One corollary. 
 
 197. Circle circumscribed about a triangle. One corollary. 
 
 198. Relation between the sides of a circumscribed quadrilateral. One 
 theorem. One corollary. 
 
 199. Relation between the angles of an inscribed quadrilateral. One theo- 
 rem. One corollary. 
 
 200. Circle inscribed in a regular polygon. One theorem. 
 
 201. Circle circumscribed about a regular polygon. One theorem. 
 
 202. Division of a circumference into equal parts. Three cases. 
 
 203. Regular polygon inscribed in a circle. 
 
 204. Regular polygon circumscribed about a circle. 
 
 205. Construction of regular polygons. 
 
 206. An inscribed regular polygon, and the inscribed regular polygon with 
 twice as many sides. 
 
 207. An inscribed regular polygon, and the circumscribed regular polygon, 
 with the same number of sides. 
 
 VI. Length of a Circumference. 
 
 208. The general truths which enable us to find the length of a circumfer- 
 ence. Method of applying these truths. Value of TT. 
 
 209. Formula for the length of a circumference. Relation between two 
 circumferences, their radii, and their diameters. 
 
 210. Two ways of estimating the length of an arc. 
 
 211. Practical exercises. 
 
 VII. Area of a Circle. 
 212. Method of finding the area of a circle. 
 213. Formula for the area of a circle. Relation between the areas of circles, 
 
 their radii, and their diameters. 
 214. A circle and the equivalent square. 
 215. Relation between the areas of circles described upon the sides of a 
 
 right triangle as diameters. 
 216. Area of a sector. 
 217. Area of a segment. 
 2ia Practical exercises. 
 
CHAPTER IX. REVIEW. 239 
 
 EXERCISES. 
 
 1. In a circle, as the angle at the centre increases, the corresponding arc, 
 chord, sector, and segment also increase. Which increase at the same 
 rate as the angle, and which do not ? 
 
 2. Upon a given straight line as a chord describe a segment which shall 
 contain a given angle. 
 
 Hints. If AB is the given line, draw A C, making B A C equal the given 
 angle. Then find the centre of the circle which has AB for a chord, and 
 which A C touches at A. Then apply 189 and 192. 
 
 3. Construct a right triangle, having given the hypotenuse and a line in 
 which the vertex of the right angle must lie. 
 
 4. In a given circle draw a chord parallel to a given straight line and having 
 a given length. 
 
 5. Find the locus of all points from which tangents drawn to a given circle 
 shall have a given length (73, 190). 
 
 6. Given two circles ; find the point from which tangents to the circles 
 shall have the same length. 
 
 7. Draw a circle which shall touch two given parallel lines and pass through 
 a given point lying between the lines. (Two solutions.) 
 
 8. About a given point describe a circle that shall touch a given circle. 
 
 (The point may be either without or within the given circle.) 
 
 9. About each of two given points describe a circle so that the two circles 
 shall touch each other, and one of them shall touch a given straight line. 
 
 Hint. First construct the circle which touches the line, and then use the 
 preceding exercise. 
 
 10. With a given radius describe a circle which shall touch a given circle and 
 pass through a given point. (Discuss the problem.) 
 
 11. Construct a triangle, having given two sides and the radius of the cir- 
 cumscribed circle. 
 
 12. Construct a triangle, having given two angles and the radius of the in- 
 scribed circle. 
 
 13. Divide a given circle into four equal parts by describing concentric circles. 
 
 14. If the perimeter of a sector is equal to the circumference of the circle, 
 find the angle at the centre. 
 
 15. If the area of a sector is equal to the square of the radius of the circle, 
 find the angle at the centre. 
 
 16. If the perimeter of a sector is equal to twice the diameter of the circle, 
 find the ratio of the sector to the circle. 
 
240 GEOMETRY FOR BEGINNERS. [ 219. 
 
 CHAPTER X. 
 THE ELLIPSE. 
 
 CONTENTS. I. The Properties of the Ellipse ( 219-221). II. Construction oi 
 Ellipses ( 222-224). 
 
 I. The Properties of the Ellipse. 
 
 219. Among curved lines the ellipse ranks in importance next 
 to the circle. 
 
 Fig. 210 shows how an ellipse may be described. Fasten a 
 
 piece of paper upon a smooth sur- 
 face, stick in two pins at any dis- 
 tance apart, and tie to the pins the 
 ends of a string longer than the 
 distance apart of the pins. Then 
 describe the curve with a pencil, as 
 shown in the figure, taking care to 
 keep the string constantly stretched 
 to its full length. 
 
 As the point of the pencil moves 
 Fig. 210. around the pins, its distance from 
 
 either one of the pins is constantly 
 
 changing ; but it is clear that the sum of its distances from the two 
 pins must always remain the same. 
 
 Thus, in Fig. 210, AP + BP = AQ + BQ = AR -f B R, etc. 
 Definition. The ELLIPSE is a curve such that the sum of the 
 distances of every point in it from two fixed points is the same. 
 The two fixed points are called the foci. 
 The point O, half-way between the foci, is called the centre. 
 Define the ellipse regarded as the locus of a moving point. 
 
220.] 
 
 CHAPTER X. THE ELLIPSE. 
 
 241 
 
 The distance OA or OB from the centre to either of the foci, is 
 called the eccentricity of the ellipse. 
 
 The less the eccentricity, that is to say, the nearer the foci are 
 to each other, the wider the ellipse becomes in proportion to its 
 length, and the more nearly it approaches to the form of the circle. 
 
 Describe several ellipses with their foci at different distances from 
 .each other. 
 
 What does the ellipse become if the foci coincide, or what comes 
 to the same thing, if only one pin is used, to which both ends of 
 the thread are fastened? 
 
 NOTE i. The ellipse is a more beautiful curve than the circle. Ellipses, or 
 curves closely resembling ellipses, are much used in the arts; in cabinet work 
 (tables, mirrors, etc.), in various designs and patterns, in the arches of bridges, and 
 in architecture. It is to the skilful use of the ellipse and still more complex curves, that 
 the superiority of Greek art over that of other nations is largely due. Astronomy 
 presents us with examples of ellipses on the grandest scale. Our earth and all 
 the planets of the Solar System revolve around the sun in elliptical paths in obe- 
 dience to the law of gravitation, the sun being at one of the foci. 
 
 NOTE 2. The ellipse is connected with the circle in a remarkable way. A 
 section of a cylinder (page 4, Fig. 3) made by a plane parallel to the bases is (like 
 the bases) a circle ; but if the section is inclined to the bases, the boundary of the 
 section is an ellipse; and the more inclined the section, the longer the ellipse in 
 proportion to its breadth. All this is easily shown by making sections through a 
 wooden cylinder. 
 
 If we look at a circle on paper with one eye placed directly in front of the 
 centre of the circle, and then gradually turn the paper till its edge comes in front of 
 the eye, we shall find that, as the paper is turned around, the circle will present the 
 appearance of an ellipse which becomes narrower and narrower, and tends to pass 
 finally into a straight line. In this way we pass, by imperceptible gradations, from 
 the circle to the straight line. 
 
 220. In an ellipse CDEF (Fig. 211), the line CD, which 
 passes through the foci, and is lim- 
 ited by the ellipse, is called the 
 Major Axis ; and the line E F, 
 which passes through the centre O 
 perpendicular to the major axis, 
 and is limited by the ellipse, is 
 called the Minor Axis. 
 
 I. From the definition of the 
 ellipse, it follows that AC + BC = A D + B D, or 2 AC -f 
 
242 GEOMETRY FOR BEGINNERS. [ 221. 
 
 AB = 2 B D -f A B j whence (Axiom III.) 2 A C = 2 23 D, and 
 AC= BD; that is, - 
 
 The vertices of an ellipse are at the same distance from the foci. 
 
 II. Since OA=OB, and AC = BD; therefore (Axiom II.), 
 OA + AC= OB + BD, or OC = OD; that is, - 
 
 The centre of an ellipse bisects the major axis. 
 
 III. Because (Fig. 210), AP+ BP = AC + BC, and ^C = 
 ^Z>; therefore, ^/> + BP=BD + BC = CD-, that is, 
 
 77/<? sum of the distances of any point of the ellipse from the foci 
 is equal to the major axis. 
 
 IV. Since A A O E ^ A B O E, therefore A E = B E ; and since 
 AE -f BE = CD, therefore 2AE^ CD, and AE = CZ> = 
 BE ; that is, 
 
 The ends of the minor axis are equidistant from the foci, the 
 distance being equal to half the major axis. 
 
 V. Since A A OE ^/^AOF, therefore OE = OF ; that is, 
 The centre of an ellipse bisects the minor axis. 
 
 Exercises. 1. To what are the three sides of the right triangle AOE 
 {Fig. 2ii\ respectively equal ? 
 
 2. What kind of plane figure \&AEBF (Fig. 211} ? Why ? 
 
 3. Given the axes of an ellipse, find the positions of the. foci. 
 
 4. Given the major axis and the positions of the foci, find the minor axis. 
 
 5. Semi-major axis = 18; semi-minor axis= I2 m . Find the eccentricity. 
 
 6. Semi-major axis = 1. 3 m ; eccentricity = 0.34. Find the semi-minor axis. 
 1. Semi-minor axis= 2.9; eccentricity = 0.9. Find the semi-major axis. 
 8. Major axis = 58 dm ; minor axis = 45 dm . How far apart are the foci ? 
 9 The path of the earth about the sun (at one focus) is an ellipse whose 
 
 axes are 20,657,70x3 and 20,655,100 geographical miles. How can the least 
 and the greatest distances of the earth from the sun be computed ? 
 
 221. AREA OF AN ELLIPSE. If we describe circles about the 
 axes of an ellipse as diameters (see Fig. 213), it is obvious that 
 the area of the ellipse is less than that of the circle with the semi- 
 major axis as radius, and greater than that of the circle with the 
 
221.] CHAPTER X. THE ELLIPSE. 243 
 
 semi-minor axis as radius. Now it can be proved that the area of 
 the ellipse is exactly equal to that of the circle whose radius is a 
 mean proportional between the semi-axes of the ellipse. 
 
 If a and b are the semi-axes of an ellipse, r the radius of the 
 circle equal in area to the ellipse, then a : r r : b, or r 2 a b. 
 The area of this circle = -r 2 ( 213) ; therefore the area of the 
 ellipse = TC r 2 ; or, since r 2 = a b, 
 
 Area of the ellipse = irab. [12.] 
 
 That is to say, the area of an ellipse is equal to the continued 
 product of its semi-axes and the number TT. 
 
 Exercises. Find the area of an ellipse, if, 
 
 1. The semi-axes are I2 dm and 8 dm . Find the area. 
 
 2. The axes are 8.5 and 4.5. Find the area. 
 
 3. The major axis = 9, the eccentricity = 1.5. Find the area. 
 
 4. The minor axis = 2 m , the eccentricity = i m . Find the area. 
 
 5. The major axis = 5, the focal distance = 3. Find the area. 
 
 6. The area= 681, the major axis = I2 m . Find the minor axis and the 
 eccentricity. 
 
 7. How much cloth will cover an elliptical table 2.4 long, i.3 m wide ? 
 
 8. The famous amphitheatre at Verona, built by the Emperor Domitian, 
 and which seated 24,000 spectators, has for its base an ellipse whose axes are 
 133 and 105. Find the area of this ellipse. 
 
 9. The section of an arch has the shape of a semi-ellipse i6 m long, 4.8 m 
 high. Find the area of the section. 
 
 10. The diameter of a circle = 7. Upon this diameter as major axis an 
 ellipse is constructed half as large as the circle. Find its minor axis. 
 
 11. A pond in the shape of an ellipse, with the axes 6o m and 44, is changed 
 to a circle with the radius 6o m . How much larger is it than before ? 
 
 12. Find the radius of a circle equal in area to an ellipse whose axes are 
 I3 m and 9. 
 
 13. Find the side of a square equal in area to an ellipse whose axes are 
 I7 m and io m . 
 
 14. Find the area of the largest ellipse that can be made out of a board 
 2 m long and o.6 m wide. 
 
 15. The axes of an ellipse are 4 m and 3. Upon these axes as diameters 
 circles are described. Find the areas of all three figures. 
 
244 GEOMETRY FOR BEGINNERS. [ 222. 
 
 II. Construction of Ellipses. 
 
 222. Problem. To construct an ellipse, having given the 
 major axis and the eccentricity. 
 
 Let {Fig. 212} CD be the given major axis, O its middle point. 
 3^3 Make OA = OB the given ec- 
 
 centricity ; then A and B are the 
 foci. Find the points E and F, 
 } D such . that the distance from each 
 of them to A and to B is equal 
 to the semi-major axis ; E and 
 F are the ends of the minor 
 axis. 
 
 In order to find more points of the ellipse, assume any points 
 m, 11, p, etc., lying in the major axis between A and O. With a 
 radius equal to Cm, describe arcs from each focus as centre, both 
 above and below CD\ and likewise do the same with a radius 
 equal to Dm ; the four points marked i, where these arcs intersect, 
 are four points of the ellipse. For, by this construction, the sum 
 of the lines drawn from either of the points marked i to the two 
 foci is equal to the major axis. (See 220, III.) 
 
 In like manner, by the use of n, the points marked 2, and by the 
 use of/, those marked 3, are found. 
 
 By drawing (free-hand) through all these points a curved line, 
 making the curvature as nearly uniform as possible, we obtain a 
 curve which approaches the nearer to the exact ellipse required, 
 the greater the number of points that are previously determined. 
 
 Exercises. 1. How can an ellipse be described by continuous motion 
 ( 219). (Gardeners sometimes use this method to trace the outline of an 
 elliptical flower-bed.) 
 
 2 Construct the ellipse whose major axis = 8o cm and eccentricity = 3O cm , 
 by finding sixteen points in the ellipse. 
 
 i 1 * Construct the ellipse whose major axis = 6o cm and eccentricity = i6 cm , 
 fourteen points. 
 
22 3 .] 
 
 CHAPTER X. THE ELLIPSE. 
 
 245 
 
 223. Problem. To construct an ellipse, having given the 
 axes, by the aid of circles described about the axes as diameters. 
 
 Let AB and CD {Fig. 213) be the given axes intersecting at O. 
 Describe circles about A B and CD 
 as diameters. Through O draw any 
 radii O E, OF, etc. Through the 
 points where these radii cut the 
 inner circle draw lines II AB, and 
 through the points where the radii 
 cut the outer circle draw lines II 
 CD. The intersection of each 
 pair of these lines is a point in the 
 required ellipse ; so that nothing 
 remains but to draw (free-hand) 
 through all the points of intersection 
 a smooth curve. 
 
 Fig. 213. 
 
 Exercises. 1. Construct an ellipse having for its axes 72 and 4O cm , and 
 find the positions of the foci. 
 
 2. Construct an ellipse whose major axis is to the minor axis as 5 : 4, and 
 find the positions of the foci. 
 
 224. Problem. To construct with arcs of circles a curve 
 that resembles an ellipse. 
 
 Draw a line A D (Fig. 214) , and divide it into three equal parts, 
 AB, BC, CD. With a radius equal to 
 one of these parts describe about B and 
 C as centres two circles, cutting each 
 other in E and F. Through these A\ 
 points and the two centres draw the 
 lines EG, EH, FI, FK. Then de- 
 scribe about E and F as centres, with 
 a radius equal to EG, the arcs 
 
 Fig. 214. 
 
 IK. The curve AGPID IK is the required curve. 
 
246 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 22 4 . 
 
 The elliptical curves employed in the arts are usually constructed 
 in this or a similar way from arcs of circles. By compounding 
 arcs of circles, in different ways, a great variety of curves may be 
 constructed. Some examples are added below, in Figs. 215222. 
 Explain how each figure is constructed. 
 
 Fig. 215. 
 
 Fig. 216. 
 
 Fig. 217. 
 
 Fig. 218. 
 
 Fig. 219. 
 
 Fig. 220. 
 
22 4 /! 
 
 CHAPTER X. THE ELLIPSE. 
 
 247 
 
 Fig. 22T. 
 
 Fig. 222. 
 
 Exercises. 1. Construct an elliptical qirve with arcs of circles. 
 
 2. Construct the curve shown in Fig. 215. 
 
 3. Construct the curve shown in Fig. 216. 
 
 4. Construct the curve shown in Fig. 217. 
 
 5. Construct the curve shown in Fig. 218. 
 
 6. Construct the curve shown in Fig. 219. 
 
 7. Construct the curve shown in Fig. 220. 
 
 8. Construct an aval like that in Fig. 221. 
 
 9. Construct a spiral like that in Fig. 222. 
 
 10. Find the area of an oval like that of Fig. 221, if the diameter AB = 4. 
 Explain each step of the solution, stating the truth on which it depends. 
 
 SYNOPSIS OF CHAPTER X. 
 
 219. Definition of the ellipse, its foci, its centre, its eccentricity. Relation 
 
 of the ellipse to the circle. 
 220. Definition of the axes of an ellipse. Relations between the ends of 
 
 the axes, the foci, and the centre. Five general relations. 
 221. Area of an ellipse. Formula for the area. 
 
 222. Construction of an ellipse, given the minor axis and the eccentricity. 
 223. Construction of an ellipse with the aid of circles, given the axes of the 
 
 ellipse. 
 224. Construction of a curve resembling an ellipse by means of arcs of 
 
 circles. Other instances of the use of circular arcs in constructing 
 
 curves. 
 
248 GEOMETRY FOR BEGINNERS. [ 225. 
 
 CHAPTER XI. 
 PLANES. 
 
 CONTENTS. I. Straight Lines in Space ( 225-227). II. A Plane ( 228-230). 
 III. A Plane and a Straight Line ($$ 231-235). IV. Two Planes ($$ 236-239). 
 V. Three Planes ($ 240). VI. Solid Angles ( 241). 
 
 I. Straight Lines in Space. 
 
 225. Thus far the figures studied (triangles, polygons, circles, 
 etc.) are such as can be represented in their true shape on paper 
 or the blackboard ; they are figures which lie wholly in one plane. 
 
 We shall now proceed to consider some of the properties of 
 figures which are not confined to one plane. Every actual body 
 which we can see or handle is an example of such a figure. 
 
 We shall begin by examining the relations of lines and planes in 
 space to one another. 
 
 NOTE i. In this part of Geometry i\\e first and cAief difficulty encountered by 
 beginners lies in conceiving and holding firmly in the mind the true relative posi- 
 tions of the lines and planes under consideration. Without the power to do this 
 the proofs of theorems amount to mere words ; with it the proofs are easily mastered ; 
 in fact many of the theorems will seem self-evident. Accordingly, in the brief space 
 here devoted to the subject, rigorous proofs will not always be given, but care will 
 always be taken to set forth the truth so that it shall be completely realized in 
 thought. The geometric imagination must be exercised and strengthened before 
 the mind is prepared to study the subject with all the rigor of logic. 
 
 NOTE 2. An additional difficulty arises from the fact that in representing 
 magnitudes in different planes upon a single plane surface (that of paper or the 
 blackboard) these magnitudes must, in general, be changed in size or shape or 
 both. Lines and angles must be altered in size, plane figures in both size and 
 shape. The rules for doing this correctly are based on the principles of Geometry, 
 and constitute the Art of Perspective, of Isometric Drawing, etc. There is one 
 general rule of Isometric Drawing which it is useful to bear in mind : a square, not 
 in the plane of the drawing, is represented in that plane by a rhombus, a rectangle 
 by a parallelogram, and a circle by an ellipse. Beginners should make a habit of 
 comparing the figures in the book, and those which they draw themselves, with actual 
 lines and planes placed in the proper relative positions, until they are able to recog- 
 nize at once the true relative positions of the parts of a drawing, and to make similar 
 drawings themselves. For lines, wires and stretched threads may be employed; 
 for planes, paper, cardboard, a thin piece of board, glass, the floor and sides of the 
 
226.] CHAPTER XI. PLANES. 249 
 
 room ; for bodies, suitable models made of wood or cardboard (see figs. 1-6) or 
 constructed simply of stiff wire for the edges. 
 
 NOTE 3. Some of the relations of lines and planes have been noticed in Chap- 
 ter I. and in 24, 25. The learner should refresh his memory, if necessary, by 
 reviewing them. 
 
 226. Two straight lines in space may have either of three 
 relative positions, as follows : 
 
 (/.) They may be parallel to each other. 
 
 (//.) They may intersect each other. 
 
 (///.) They may be neither parallel nor intersecting. 
 
 In the first two cases the lines must lie in the same plane ( 29). 
 Give examples of both cases. If the lines intersect, they may be 
 either perpendicular to each other or inclined to each other ( 48). 
 
 In the third case the two lines do not lie in the same plane. A 
 line from north to south on the table, and a line from east to west 
 on the floor, is an example of two such lines. 
 
 Let AB and CD (Fig. 22j) represent two lines which are not 
 parallel and do not intersect, and let E 
 be the point of A B which is nearest to 
 CD, and F the point of CD which is 
 nearest to AB. Then the line E F is 
 the shortest distance between the lines ABA E . B 
 
 and CD, and is a common perpendicular Fig, 223. 
 
 to both lines. 
 
 NOTE. In Fig. 223, A B is to be regarded as in the plane of the paper, CD as 
 nearly perpendicular to this plane. 
 
 227. Let ABC and DEF (Fig. 224) be two angles in space, 
 and let AB II DF, and BC II F F. 
 Conceive the angle ABC to move 
 so that its vertex B keeps in the line 
 BE, and AB and CB to remain 
 parallel to their first positions. When 
 B reaches F, it is evident that A B 
 will coincide with DE, and CB with 
 
250 GEOMETRY FOR BEGINNERS. [ 228. 
 
 FE, and therefore the angle ABC will coincide with the angle 
 DEF. 
 
 Therefore, the theorem of Plane Geometry ( 58) that angles, 
 whose sides are respectively parallel and directed the same way 
 from the vertex, are equal, also holds true of any two angles .in 
 space. 
 
 II. A Plane. 
 
 228. The surface of still water, the surface of a table, the floor, 
 or the blackboard, are examples of plane surfaces, or planes. Give 
 another example. How can a plane be generated by motion? 
 (See 28.) 
 
 How was a plane denned in 28 ? And what is the test of a 
 plane surface? 
 
 The following is the more abstract but more precise definition 
 commonly given by mathematicians : 
 
 Definition. A PLANE is a stirface such that the straight line 
 which joins any two points in the surface lies wholly in the surface. 
 
 A plane (like a straight line) may be conceived to extend indefi- 
 nitely. The surface of the table may be extended in thought as 
 far as we please, and in such a way that the surface which we 
 imagine will be everywhere a part of the same plane as the actual 
 surface which we see. 
 
 Exercise. Give an instance of a vertical plane, a horizontal plane, an 
 inclined plane. 
 
 229. When we conceive a plane as containing a given point 
 or straight line, we are said to pass the plane through the point or 
 the line. 
 
 Through one point an endless number of planes in all conceiva- 
 ble positions may be passed. 
 
 Through two points, also, we may pass as many different planes 
 
230.] CHAPTER XI. PLANES. 251 
 
 as we please ; for if we pass a plane through the line joining the 
 points A and B (Fig. 22 5) , and then rotate the plane about this 
 line as an axis, it will assume different positions, in all of which it 
 contains the line, and therefore the points A and B. 
 
 But if a third point C be given, through which the plane must 
 also pass, it is evident that as we rotate the plane there is only one 
 position in which it will contain all three points. Through three 
 points not in the same straight line only one plane can be passed ; 
 and the same is obviously true of a straight line and a point not in 
 the line. 
 
 It is likewise true of two intersecting lines and of two parallel 
 lines ; for if we pass a plane through one of the lines, and then turn 
 the plane about this line, there is only one position in which it can 
 contain the other line. 
 
 Therefore, a plane is determined, 
 
 1 . By three points not in the same straight line. 
 
 2. By a straight line and a point not in the line. 
 
 3. By two intersecting lines. 
 
 4. By two parallel lines. 
 
 230. Any two planes which are passed through a straight line 
 AB (Fig. 2 25} have this line for their 
 intersection ; and it is quite clear that 
 through a curved line only one plane 
 at most can be passed. 
 
 The intersection of two planes is 
 always a straight line. i ^^^ 
 
 Exercises. 1. How many planes can Fig. 225. 
 
 intersect in the same straight line ? 
 
 2. How many planes can be passed through a curve drawn on the black- 
 board ? 
 
 3. Make (with a wire) a curve such that no plane can be passed through it 
 
252 GEOMETRY FOR BEGINNERS. [ 231. 
 
 III. A Plane and a Straight Line. 
 
 231. A straight line may be either, 
 
 (/.) Parallel to a plane. 
 
 ('.) Perpendicular to a plane. 
 
 (Hi.) Inclined to a plane. 
 
 An edge of the ceiling is parallel to the floor ; a plumb-line is 
 perpendicular to the surface of still water ; a rafter of a common 
 roof is inclined to the level ground. Give other examples. 
 
 In Fig. 226, the line AB is parallel to the plane M. 1 The line 
 PO is perpendicular to M t and the lines PC and PD are inclined 
 to M. 
 
 Exercises. 1. Draw a figure like Fig. 226, to illustrate the different 
 positions which a straight line may have relative to a plane. 
 
 2. Hold a pencil, (z.) parallel, (.) perpendicular, (iii) inclined, to the 
 table. 
 
 232. Definition. A straight line and a plane are parallel, 
 if they cannot meet however far extended. 
 
 This condition will be fulfilled if the straight line A B (Fig. 226) 
 
 is parallel to a straight line C E in 
 the plane ; for, since A B can never 
 meet CE, and can never leave the 
 plane determined by AB and CJl 
 ( 229,4), it can never meet the plane 
 M in which C E lies. Hence, 
 
 I. A straight line is parallel to a 
 p- 22 fr plane if it is parallel to a straight 
 
 line drawn in the plane. 
 
 II. If two lines are parallel, every plane passed through one of 
 the lines will be parallel to the other line. 
 
 What exception is there to this last statement ? 
 
 1 In naming a plane, a single letter, as M or N, is usually sufficient. 
 
233-] 
 
 CHAPTER XL PLANES. 
 
 253 
 
 Exercises. 1. How many lines parallel to a plane can be drawn through 
 a point outside the plane ? 
 
 2. If two lines are parallel to a plane, are they parallel to each other ? 
 Give illustrations. 
 
 233. A straight line, perpendicular or inclined to a plane, will 
 (prolonged, if necessary) meet the plane in a point. This point 
 is called the foot of the line. 
 
 Definition. A straight line is perpendicular to a plane, if it 
 is perpendicular to every straight line that can be drawn through 
 its foot in the plane. 
 
 It is also said to be normal to the plane. 
 
 A line having this property can always be drawn from a point 
 P (Fig. 227) to a plane M. For, 
 among all the lines that can be 
 drawn from P to meet the plane, 
 there will be one, PO, shorter than 
 any other. If we draw through its 
 foot O any straight lines AB, CD, 
 etc., PO will be the shortest line 
 from P to each of these lines ; there- 
 fore ( 83), perpendicular to each 
 and all of them. 
 
 Hence, a perpendicular measures the distance from a point to a 
 plane. 
 
 Through the intersection O (Fig. 228) of two lines, AB and CD, 
 draw any line O E _L CD, and revolve OE about CD, keeping 
 OE _L CD ; there is one, and only 
 one, position in which the line will 
 also be J_ AB. Let OP be this 
 position ; then O P, being the only 
 line through O perpendicular to both 
 A B and CD, must be perpendicular 
 to the plane passed through A B and 
 CD. Therefore, 
 
 227. 
 
 Fig. 228. 
 
254 GEOMETRY FOR BEGINNERS. [ 234. 
 
 If a straight line is perpendicular to two straight lines drawn 
 through a point of the line, it is perpendicular to the plane contain- 
 ing those two lines. 
 
 This proposition enables us to solve the problem : 
 To make a plane perpendicular to a straight line. 
 We have only to draw through any point of the line 
 two perpendiculars, and then to pas:; a plane through 
 these perpendiculars. Explain how a carpenter would 
 proceed in order to saw squarely in two the beam of 
 wood shown in Fig. 229. 
 
 Exercises. 1. Give examples of a line perpendicular to 
 
 a P lane ' 
 
 2. When a right triangle revolves about one of its legs, 
 
 what is generated by the other leg ? 
 
 3. At a point of a straight line in space, how many perpendiculars can be 
 erected ? From a point not in the line, how many perpendiculars can be let 
 fall to the line ? 
 
 4. Find the distance from a point P to a plane, if its distance from a 
 point A in the plane = a, and the distance of this point from the foot of the 
 perpendicular let fall from the first point to the plane = b. Solve for the 
 case where a = 17, b = 8. 
 
 5. Prove that lines drawn from a point meeting a plane at equal distances 
 from the foot of the perpendicular let fall from the point are equal ; and that 
 of two lines drawn to unequal distances from the foot of the perpendicular, 
 the more remote is the greater. 
 
 6. Prove the converse of the first part of the last exercise. 
 
 7. Find the locus of points in a plane equidistant from points not in the 
 plane. 
 
 8. Find the locus of points equidistant from two given points. 
 
 9. Find the locus of points equidistant from three given points. 
 
 234. Let A B and CD (Fig. 236) be normal to the plane J/, 
 and suppose AB to move along AC, keeping parallel 1 ) its first 
 position. Then its position relative to the plane will not change, 
 that is, it will remain normal to the plane ; and' when A coincides 
 
'350 
 
 CHAPTER XI. PLANES. 
 
 255 
 
 with C, AB will coincide with CD. Therefore, AB II CD. 
 That is, 
 
 Two lines that are normal to a plane are parallel to each other. 
 
 The converse is also true. State it. 
 
 Of 
 
 Fig. 2jo. 
 
 Fig. 231. 
 
 235. Lines neither parallel nor perpendicular to a plane are 
 inclined to the plane. 
 
 Let A O (Fig. 2ji) be inclined to the plane M. Draw AB _L M, 
 and join OB ; then OB is called the PROJECTION of the line AO 
 upon the plane. When is the projection of a line equal in length 
 to the line ? less than the line ? When is it a point ? 
 
 The acute angle A OB, which A O makes with its projection BO, 
 is the smallest angle which A O makes with any line drawn through 
 its foot on the plane M. To see this more clearly, revolve BO 
 about O in the plane M. The angle which the revolving line 
 makes with AO will continually increase, becoming equal to a 
 right angle AOC after one-fourth of a revolution, and attaining 
 its greatest obtuse value AOD after half a revolution. During the 
 remaining half of the revolution what changes does the angle 
 undergo ? 
 
 The angle which a line inclined to a plane makes with its pro- 
 jection in the plane, is called the inclination of the line to the 
 plane. 
 
 NOTE. What is said above may be illustrated by drawing lines through a 
 point on a sheet of paper, and placing the end of the pencil held in an inclined 
 position against this point. If in this position we let the pencil drop, the angle 
 which it describes in falling is the inclination of the pencil to the table. 
 
256 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 236- 
 
 IV. Two Planes. 
 
 236. Two planes may be either, 
 
 (/.) Parallel to each other/ 
 
 ('.) Perpendicular to each other. 
 
 (V.) Inclined to each other. 
 
 The floor and ceiling of the room are parallel planes ; the floor 
 and a side of the room are perpendicular to each other ; the floor 
 and the roof of the house are inclined to each other. Give other 
 examples. 
 
 Exercises. 1. In Fig. 232, which planes are parallel? perpendicular? 
 inclined ? 
 
 2 Draw two planes of each kind. 
 
 3. Hold two planes (books, thin sticks of wood, etc.) parallel ; perpen- 
 dular ; inclined. 
 
 Fig. 232. 
 
 Fig- 233. 
 
 237. Definition. Two planes are parallel if they cannot 
 meet however far extended. 
 
 This condition is fulfilled if the planes M and N (Fig. 2jj) are 
 both perpendicular to the same line AB. For if we pass a third 
 plane through AB, it will intersect M in a line CD \-AB (why?), 
 and N in a line EF ^LAB. Therefore, CD and EF t being in 
 the same plane, are parallel ( 57), and can never meet. The 
 same reasoning may be applied to all planes that can be passed 
 through AB ; therefore, the planes J/and TV can never meet. 
 
238.] CHAPTER XI. PLANES. 257 
 
 Therefore, 
 
 I. Planes that have a common normal are parallel. 
 The converse is also true. State it. 
 
 It is also evident from what precedes, that, 
 
 II. The intersections of two parallel planes by a third plane are 
 parallel lines. 
 
 If, also, CE II DF (Fig. 233), then CDFE is a parallelogram 
 ( 101), and AC = BD ( 102) ; that is, - 
 
 III. Parallel lines between parallel planes are equal. 
 
 Among these equal parallel lines are the common normals 
 ( 234), which measure the distance apart of the planes; there- 
 fore, 
 
 IV. Parallel planes are everywhere equally distant. 
 
 238. Definition. Two planes are perpendicular to each 
 other if a normal to either plane, erected at any point of their 
 intersection, lies in the other plane. 
 
 The angle made by the normals to the two planes, erected at the 
 same point of the intersection, is a right angle (why ?) . 
 
 Let the planes M and N (Fig. 234} intersect in the line AB, 
 and let CD be normal to M and lie in N. 
 
 Draw D E normal to N\ then C D E is a right angle (why?) ; 
 therefore DE must lie in M, the 
 plane containing all lines perpen- 
 dicular to CD at D. 
 
 Further, let G be any other point 
 of the intersection, and draw GF nor- 
 mal to My and G H normal to N. Then 
 GF II DC ( 234) ; and since DC 
 lies in the plane N, and G also lies in 
 N, therefore GF must lie in N. For 
 like reasons, GH lies in M. 
 
 Comparing these results with the definition, we see that, 
 
258 GEOMETRY FOR BEGINNERS. [ 239. 
 
 1. Two planes must be perpendicular, if a single normal to one of 
 the planes erected at a point of the intersection lies in tJie other plane. 
 
 II. Every plane passed through a normal to a plane is perpen- 
 dicular to that plane. 
 
 Exercises. 1, How can a plane perpendicular to a given plane be passed 
 through a given point ? 
 
 2. How can a plane perpendicular to a given plane be passed through a 
 straight line, (z.) when the line lies in the given plane ? (zY.) when it is 
 inclined to the given plane ? (zYY.) when it is parallel to the given plane ? 
 
 239. Two inclined planes M and N (Fig. 235} intersect each 
 other and form what is called a DIHEDRAL ANGLED This angle 
 
 may be conceived as generated by 
 the revolution of one of the planes M 
 about the line A B until it comes into 
 the position of the other plane N. 
 The planes M and N are called the 
 faces of the angle ; the intersection 
 
 rAB is called the edge of the angle. 
 A dihedral angle is always measured 
 by the angle of two lines. If OC is J_ 
 
 A~ ~~~ AB, during the revolution of M about 
 
 AB, OC describes the angle COD, 
 
 which has the same value wherever the point O is situated in the 
 line AB ( 227). This angle may therefore be used to measure 
 the dihedral angle made by the planes. That is, 
 
 A dihedral angle is measured by the angle made by two lines 
 drawn through any point of the edge, perpendicular to the edge, one 
 line in one plane, the other in the other plane. 
 
 Exercises. 1. Illustrate, by the leaves and edges of a book, dihedral 
 angles, and how they are measured. 
 
 2. If the angle of the lines which measure a dihedral angle is 90, what 
 position do the planes have ? 
 
 1 From Greek words meaning two-faced. 
 
240.] 
 
 CHAPTER XI. PLANES. 
 
 259 
 
 V. Three Planes. 
 
 240. Three planes may have five different relative positions, 
 shown in Figs. 236-238. 
 
 (/.) In Fig. 236, the planes are parallel to one another. 
 
 
 
 7 
 
 V 
 
 Fig. 236. 
 
 Fig. 237. 
 
 (.) In Fig. 237, the two planes J/and N are parallel, and the 
 third plane P (or Q) intersects M and N in parallel lines. (See 
 237.) 
 
 (///.) In Fig. 237, the planes N, P, and Q have a common line 
 of intersection. 
 
 (iv.) In /3g-. 2J7, J/, P, and <? intersect in three lines, which 
 are parallel to one another. 
 
 Fig. 238. 
 
 Fig. 239. 
 
 .) In Fig. 238 there are also three lines of intersection. These 
 lines meet in a common point O j so that O is a point common to 
 all the planes. 
 
260 GEOMETRY FOR BEGINNERS. [ 241. 
 
 If two of the planes J/and N (Fig. 239) are both perpendicu- 
 lar to the third plane R, it follows, from 238, that their inter- 
 section must be a normal to the plane R ; for the normal to R, 
 erected at the point O, must lie in both M and N, 
 
 Exercise. Give an example of three parallel planes ; or' two planes, both 
 perpendicular to a third plane. 
 
 VI. Solid Angles. 
 
 241. When three or more planes meet in one point O (Fig. 
 241) they are said to form at that point a SOLID ANGLE. The 
 planes are called the faces, their lines of in- 
 tersection the edges, and the angles at (9, 
 made by the two edges of each plane, the 
 plane angles of the solid angle. 
 
 If there are only three planes, the angle 
 is usually called a trihedral angle. 
 
 In order to form a trihedral angle from 
 three angles in one plane, AO B = a, AOC 
 = b, BOD c (Fig. 241), of which we 
 will suppose a to be the greatest, turn the 
 plane AOC about A O, and the plane AOD 
 
 about BO, till they meet, if possible, and the lines CO and DO 
 coincide. 
 
 If b -f- c = a, CO and DO will just coincide when turned through 
 half a revolution, so that both again lie in the plane of the paper ; 
 if b + f<a, CO and DO cannot be made to coincide. In both 
 cases no trihedral angle will be formed. But if b -f- c > a, CO and 
 DO will coincide in a line OF, either in front of the plane of the 
 paper (Fig. 242), or behind it (Fig. 243), according as the motion 
 is forward or backward '; and a trihedral angle is formed. And, 
 since a is supposed the greatest of the three angles, it follows that 
 a + b > c, and that a -f- c > b. Hence, in every trihedral ang -he 
 sum of any two of the plane angles is greater than the third. 
 

 
 CHAPTER XL PLANES. 
 
 261 
 
 If the motion is forwards, the trihedral angle shown in Fig. 242 
 is formed ; if backwards, the trihedral angle shown in Fig. 243. 
 These two angles are composed of equal plane angles, taken in 
 
 Fig. 241. 
 
 Fig. 242. 
 
 Fig. 243. 
 
 order ; yet they cannot be made to coincide, because their equal 
 parts follow in reverse order, in one, from left to right; in the 
 other, from right to left. 
 
 They stand to each other in the same relation as an object to its 
 image in a mirror, or as the right hand to the left. Two such solid 
 angles are said to be symmetrical. 
 
 Two solid angles are equal, if they are composed of equal plane 
 angles taken in the same order. We may conceive two equal 
 solid angles placed one upon the other so as to coincide in all 
 their parts. 
 
 Whatever be the number of plane angles which compose a solid 
 angle, their sum, estimated in degrees, must be less than j6o; if 
 it were equal to 360, the plane angles would necessarily lie in one 
 plane, and would not form a solid angle. 
 
 Exercises. 1. Give examples of solid angles. 
 
 2. In a solid angle, how is the angle which two faces make with each 
 other measured? 
 
 3. Draw a solid angle having three faces ; four faces ; six faces. 
 
262 GEOMETRY FOR BEGINNERS. 
 
 REVIEW OF CHAPTER XL 
 
 I. Straight Lines in Space. 
 
 225. Nature of Solid Geometry. 
 226. Two straight lines in space. 
 227. Two angles in space. 
 
 II. A Plane. 
 
 228. Definition of a plane. 
 
 229. Determination of a plane. 
 
 230. Intersection of two planes. 
 
 III. A Straight Line and a Plane. 
 
 231. Positions of a straight line with respect to a plane. 
 
 232. The straight line parallel to a plane. 
 
 233. The straight line perpendicular to a plane. 
 
 234. Two normals to a plane. 
 
 235. The straight line inclined to a plane. 
 
 IV. Two Planes. 
 
 236. Relative positions of two planes. 
 
 237. Two parallel planes. 
 
 238. Two perpendicular planes. 
 
 239. Two inclined planes. 
 
 V. Three Planes. 
 240. Relative positions of three planes. 
 
 VI. Solid Angles. 
 
 241. Solid angles. Symmetrical solid angles. Equal solid angles. 
 
242.'] CHAPTER XII. GEOMETRICAL BODIES. 263 
 
 CHAPTER XII. 
 GEOMETRICAL BODIES. 
 
 CONTENTS. I. Polyhedrons ( 242,243). II. The Prism, Cylinder, Pyramid, 
 and Cone ( 244-250). III. The Sphere ($ 251, 252). 
 
 J. Polyhedrons. 
 
 242. A limited portion of space regarded as possessing -solely 
 form and magnitude is called a GEOMETRICAL BODY, or a SOLID 
 (see 7). 
 
 In this sense a room is a body as much as a stone or piece of 
 wood. Speaking exactly, it is the space occupied by the air, etc., 
 in the room, and limited by the floor, sides, and ceiling of the room. 
 
 A geometrical body bounded on all sides by planes is called a 
 POLYHEDRON. 
 
 Give examples of polyhedrons. 
 
 The bounding planes are called the faces of the polyhedron, 
 their intersections are called the edges, and the intersections of 
 the edges are called the corners. 
 
 The faces that meet at each corner form a solid angle, composed 
 of as many plane angles as there are faces. 
 
 A polyhedron must have at least four faces ; for it takes at least 
 three planes to form a solid angle, and a fourth plane is required 
 to close the opening between the three planes. 
 
 A polyhedron with four faces is called a TETRAHEDRON ; with five 
 faces, a PENTAHEDRON ; with six faces, an HEXAHEDRON ; with 
 eight faces, an OCTAHEDRON ; with ten faces, a DECAHEDRON ; with 
 twelve faces, a DODECAHEDRON ; with twenty faces, an ICOSAHEDRON. 
 
 A very simple relation exists between the faces, corners, and 
 edges of a polyhedron. 
 
264 GEOMETRY FOR BEGINNERS. [ 243. 
 
 If we examine the polyhedrons in Fig. 2, page 4, we find that, 
 No. I. has 5 faces, 6 corners, and 9 edges. 
 II. has 6 " 8 " " 12 " 
 
 III. has 8 " 12 " " 1 8 " 
 
 IV. has 5 " 5 "8 " 
 
 In each of these cases, the number of faces 4- number of cor- 
 ners = number of edges + 2 - 
 
 This relation is perfectly general, and is called Euler's Theorem. 1 
 Let F denote the number of faces, C the number of corners, E the 
 number of edges ; then we may state Euler's Theorem as follows : 
 
 In every polyhedron, F + C = E + 2. [13.] 
 
 243. A polyhedron whose faces are equal regular polygons, 
 and whose solid angles are all equal, is called REGULAR. 
 
 It is easy to show that only five regular polyhedrons are possible. 
 
 We know already that at least three faces are necessary to form 
 a solid angle ; also, that the sum of the plane angles at each corner 
 must be less than four right angles ( 241) . 
 
 The angle of the equilateral triangle = 60, and 3 X 60 180 ; 
 therefore, three such triangles may form a solid angle. Likewise, 
 four such triangles may be used, since 4 X 60 = 240 ; also five 
 such triangles, since 5 X 60 = 300. But 6 X 60 = 360, so that 
 six, or more than six, equilateral triangles cannot form a solid angle. 
 
 Therefore, only three regular polyhedrons are possible whose 
 faces are equilateral triangles. 
 
 The angle of a square = 90. Since 3 X 90 = 270, while 
 4 x 90 = 360, three is the only number of squares that can 
 form a solid angle. 
 
 The angle of a regular pentagon = 108. Since 3 x 108 324, 
 while 4 x 1 08 = 432, only three regular pentagons can be used 
 to form a solid angle. 
 
 1 Euler was a famous German mathematician, born 1707, died 1783. 
 
2430 
 
 CHAPTER XII. GEOMETRICAL BODIES. 
 
 265 
 
 The angle of a regular hexagon = 1 20. Since 3 X 120 = 360, 
 three hexagons placed together with a corner common would lie 
 in one plane and could not form a solid angle. Therefore, no 
 solid angle can be formed with regular hexagons ; still less with 
 regular polygons having more than six sides. 
 
 Fig- 247- 
 
 Therefore, only five regular polyhedrons are possible. They 
 are, 
 
 (1) The regular Tetrahedron (Fig. 244), bounded by four 
 equilateral triangles. 
 
 (2) The regular Octahedron (Fig. 245}, bounded by eight 
 equilateral triangles. 
 
 (3) The regular Icosahedron (Fig. 246), bounded by twenty 
 equilateral triangles. 
 
 (4) The regular Hexahedron (Fig. 247}, bounded by six 
 squares. 
 
 (5) The regular Dodecahedron (Fig. 248), bounded by twelve 
 regular pentagons. 
 
266 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 243- 
 
 To DEVELOP 1 the surface of a solid is to draw in one plane the 
 various parts of the surface, so that when folded up they shall form 
 the solid. 
 
 The surface of the solid, when drawn in one plane, is called the 
 DEVELOPMENT of the surface. 
 
 The surfaces of many solids cannot be developed ; but when 
 this can be done, it is easy to construct by this means models of 
 the solids. 
 
 The surfaces of the regular polyhedrons are all developable, and 
 their plane models are shown in Figs. 249-253. 
 
 Fig. 249. 
 
 Fig. 250. 
 
 Fig. 251. 
 
 Fig. 252. 
 
 253- 
 
 To make models of these solids, draw on cardboard the diagrams 
 shown in these figures, cut them out of the cardboard, and then 
 cut half through the edges of the adjacent polygons. The figures 
 will then readily fold up into the shape of the solids, and can be 
 retained in shape by means of glue or paste. 
 
 Literally, -unroll. 
 
2 44-] 
 
 CHAPTER XII. GEOMETRICAL BODIES. 
 
 267 
 
 Exercises. 1. Give examples of polyhedrons. 
 
 2. How many edges have each of the regular polyhedrons? 
 
 3. How many corners have each of the regular polyhedrons? 
 
 4. Verify Euler's theorem for each of the regular polyhedrons? 
 
 5. Construct models of the regular polyhedrons. 
 
 NOTE. In making the model of the regular dodecahedron, the diagonals, shown 
 dotted in Fig. 253, may be used to advantage. 
 
 II. The Prism, Cylinder, Pyramid, and Cone. 
 
 244. If the polygon ABODE (Fig. 254) move along the 
 line AF t keeping parallel 
 to its first position, its sides 
 will describe parallelograms, 
 and the polygon will gener- 
 ate a solid called a Prism. 
 
 Definition. A PRISM 
 is a polyhedron bounded by 
 parallelograms and by two 
 
 Fig. 254. 
 
 equal and parallel polygons. 
 
 The parallelograms are 
 called the lateral faces, their lines of intersection the lateral edges, 
 and the two polygons the bases of the prism. 
 
 The height of the prism is the distance (HL) between the bases. 
 
 A prism is called triangular, quadrangular, etc., according as 
 the base is a triangle, quadrilateral, etc. 
 
 A prism is oblique if the lateral edges are inclined to the bases 
 (Fig. 255); right, if they are normal to the bases (Fig. 256); and 
 regular, if it is a right prism with a regular polygon for its base. 
 
 A parallelepiped is a prism whose bases are parallelograms. It 
 is, therefore, a solid bounded by six parallelograms. 
 
 Fig. 255. 
 
 Fig. 256. Fig. 257. Fig. 258. 
 
 Fig. 259. 
 
268 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ MS- 
 
 What is an oblique parallelepiped (Fig. 257) ? A right parallele- 
 piped (Fig. 258} ? 
 
 A rectangular parallelepiped is a right parallelepiped, all of 
 whose faces are rectangles. 
 
 A cube is a right parallelepiped, all of whose faces are squares 
 (Fig. 259). 
 
 Exercises. 1. Can you give an example of a prism? What kind of 
 a prism is it ? 
 
 2. In Fig. 234, point out lateral faces, lateral edges, the bases, the height. 
 
 3. Name a prism with a base of five sides, six sides, eight sides. 
 
 4. What are the lateral faces of a right prism? Why is a lateral edge 
 equal to the height of the prism? 
 
 5. How many faces, edges, corners, in a prism with a base of three sides, 
 four sides, six sides, eight sides, twenty sides? Apply Euler's formula to each 
 one of these cases. 
 
 6. Draw the following : 
 
 (*'.) Oblique triangular prism. (z>.) 
 
 (z'z.) Right pentagonal prism. (&*) 
 
 (z'z'z.) Right hexagonal prism. (z/z'z.) 
 
 (zV.) Regular hexagonal prism. (viii.} 
 
 Oblique parallelepiped. 
 Right parallelepiped. 
 Rectangular parallelepiped. 
 Cube. 
 
 245. If a circle (Fig. 260) move in the direction normal to 
 its own plane, keeping always parallel to 
 its first position, it generates a solid called 
 the right circular Cylinder. The same 
 solid is also generated by a rectangle 
 ABCD (Fig. 257), which revolves about 
 one side CD. 
 
 Definition. A right circular CYL- 
 INDER is the solid produced by the revolu- 
 tion of a rectangle about one of its sides. 
 
 NOTE. In future, by the word "cylinder" the right circular cylinder will be 
 
 Fig. 260. 
 
 meant. 
 
 A cylinder is bounded by a curved surface called the convex 
 surface, and by two equal and parallel circles called the bases. 
 The line joining the centres of the bases is called the axis ; its 
 length is equal to the height of the cylinder. 
 
246.] 
 
 CHAPTER XII. GEOMETRICAL BODIES. 
 
 269 
 
 246. If the polygon ABODE (Fig. 261} move along the 
 line AO, keeping parallel 
 to its first position, but 
 diminishing in size at a 
 uniform rate, without al- 
 tering in shape, until it is 
 reduced to a point O, its 
 sides will describe trian- 
 gles, and the polygon will 
 generate a solid called a 
 
 Pyramid. 
 
 Fig. 261. 
 
 Definition. A PYRAMID is a polyhedron bounded by triangles 
 that have a common vertex, and by a polygon whose sides are the 
 bases of the triangles. 
 
 The triangles are called the lateral faces ; their lines of intersec- 
 tion, the lateral edges ; their common vertex, the vertex of the 
 pyramid ; and the polygon, the base of the pyramid. 
 
 The heightQi the pyramid is the distance (OF) from the vertex 
 to the base. 
 
 A pyramid is triangular, quadrangular, etc., ac- 
 cording as its base is a triangle, quadrilateral, etc. 
 
 A regular pyramid is a pyramid which has a 
 regular polygon for its base, and in which the 
 perpendicular from the vertex passes through the 
 centre of the base (Fig. 262). The distance OM 
 from the vertex to a side of the base is called 
 the slant height. 
 
 Fig. 262. 
 
 Exercises. 1. In Fig. 261. point out lateral faces, lateral edges, the 
 base, the vertex, the height. 
 
 2. Name a pyramid with a base of five sides ; 6 sides ; 8 sides. 
 
 3. What kind of a pyramid is the tetrahedron (Fig. 244) ? 
 
 4. What kind of triangles are the faces of a regular pyramid ? 
 
 5. Draw (*'.) a triangular pyramid ; (it.} a regular hexagonal pyramid. 
 
270 GEOMETRY FOR BEGINNERS. [ 247. 
 
 247. If a circle (Fig. 263) move in a direction normal to its 
 
 own plane, keeping parallel to its 
 first position, but diminishing in 
 size at a uniform rate till it is 
 reduced to a point O, it will gener- 
 ate a solid called a right circular 
 Cone. 
 
 The same solid is also generated 
 Fig 263 by a right triangle AOB, which 
 
 revolves about the leg OB. 
 
 Definition. A right circular CONE is the solid produced by 
 the revolution of a right triangle about one of its legs. 
 
 NOTE. In future, by the word " cone " a right circular cone will be meant. 
 
 A cone is bounded by a curved surface called its convex surface, 
 and by a circle called its base. The point O is the vertex of the 
 cone, and the line OB joining the vertex to the centre of the base 
 is the axis ; its length is equal to the height of the cone. 
 
 The length of the line AO (Fig. 263} is called the slant height. 
 
 248. From the mode of generation of the four bodies now 
 described, Prism, Cylinder, Pyramid, Cone, it follows that a 
 section made by a plane parallel to the base is, 
 
 \ ': 
 
 Fig. 264. Fig. 265. ~ Fig. 266. 
 
 In a prism, a polygon equal to the base. 
 
 In a cylinder, a circle equal to the base {Fig. 264). 
 
 In a pyramid, a polygon similar to the base (Fig. 271} . 
 
 In a cone, a circle smaller than the base (Fig. 272). 
 
 The section of a cylinder made by a plane inclined to the base 
 
249-] CHAPTER XII. GEOMETRICAL BODIES. 271 
 
 is an ellipse (Fig- 265} . Such sections are made when two cylin- 
 ders intersect each other, as in the case of two pipes meeting at 
 right angles (Fig. 266) . 
 
 In a cone, a section inclined to the base cuts the surface in 
 an ellipse (Fig. 268} , if it does not meet the base ; in a parabola 
 (Fig. 269), if it meets the base and is parallel to the side of the 
 cone ; in a hyperbola (Fig. 270) , if it meets the base and is not 
 parallel to the side of the cone. 
 
 Fig. 267. Fig. 268, 
 
 The four sections shown in Figs. 267-270 namely, circle, 
 ellipse, parabola, hyperbola are sometimes called the conic 
 sections. 
 
 NOTE. These sections may be shown to the eye by holding a cone partly 
 under water, at first with the axis vertical, then with the axis more and more 
 inclined. If the water is colored red, the effect is better seen. 
 
 Exercises. What is the section made by a plane passing through, ^ 
 
 1. Two lateral edges of a prism ? 3. The axis of a cylinder ? 
 
 2. Two lateral edges of a pyramid ? 4. The axis of a cone ? 
 
 249. A plane parallel to the base of a pyramid, or of a cone, 
 divides the body into a smaller pyramid or cone, and a body called 
 the frustum of the pyramid or cone (Figs. 271 and 272) . The 
 base and the section parallel to the base are called the lower and 
 upper bases of the frustum ; their distance from each other is the 
 height of the frustum. 
 
 If we know the height PQ (Fig. 271} of a frustum of a pyramid, 
 and two homologous sides AB and EF of the bases, we can find 
 the height OP of the entire pyramid, as follows : 
 
 Through OP and the edge OA pass a plane intersecting the 
 
272 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 250. 
 
 bases of the frustum in the parallel lines APand EQ (why paral- 
 lel?), and draw E M II OB. By 159, 
 
 = _ d OE = 
 ' ' 
 
 PQ EA' EA AM 
 
 Therefore, 
 
 OQ EF 
 = 
 
 And 
 
 PQx EF PQxEF 
 = AUT 
 OP=OQ+QP. 
 
 Fig. 271. 
 
 Fig. 272. 
 
 Exercises. 1. What are the faces of the frustum of a pyramid ? 
 
 2. If (Fig. 271} PQ = S m , A=4 m , EF= 2 m , find OQ and OP. 
 
 3. Given the height PQ (Fig. 272} of the frustum of a cone, and the radii 
 of the bases ; find the height of the cone cut off, and also the height of the 
 entire cone. 
 
 Hint. Draw BC \\ OP, and use the similar triangles OBQ andAC. 
 
 4. If (Fig. 272} PQ=4 m , AP = 3 m , BQ_ = i m , find OQ and OP. 
 
 250. The surfaces of the prism, the pyramid, the cylinder, and 
 the cone are all developable. 
 
 Fig. 273 shows the development of a regular hexagonal prism ; 
 Fig. 2J4 that of a regular pyramid whose base is a square. 
 
 The auxiliary lines in the two figures serve to illustrate how they 
 are to be constructed. Fig. 273 is the development of a regular 
 hexagonal pyramid. How is it constructed ? 
 
250.] 
 
 CHAPTER XII. GEOMETRICAL BODIES. 
 
 273 
 
 The development of the convex surface of a cylinder is a rect- 
 angle ; for we can roll up a rectangular sheet of paper, for exam- 
 ple, into the shape of a cylindrical surface ; and conversely, a cylin- 
 drical surface can be unrolled into a rectangle (Fig. 276). 
 
 Fig. 274. 
 
 275- 
 
 Fig. 276. 
 
 Fig. 277. 
 
 Fig. 278. 
 
 The development of the convex surface of a cone is a circular 
 sector whose arc is the circumference of the base of the cone, and 
 whose radius is the slant height of the cone (Fig. 278). 
 
 Fig. 276 shows the development of the entire surface of a cylin- 
 der ; Fig. 277, that of the entire surface of a cone ; Fig. 277, that 
 of the entire surface of the frustum of a cone. 
 
 Exercises. Make models of the following solids : 
 
 1. A regular triangular prism. 
 
 2. A regular hexagonal prism. 
 
 3. A right pi-ism, not regular. 
 4:, A rectangular parallelepiped. 
 5* A regular hexagonal pyramid. 
 
 6. A cube. 
 
 7 A cylinder. 
 
 8. A cone. 
 
 9. A frustum of a pyramid. 
 10. A frustum of a cone. 
 
274 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 25L 
 
 Fig. 279. 
 
 III. The Sphere. 
 
 251. Definition. A SPHERE is the solid produced by the 
 revolution of a semicircle (or circle) about the diameter (Fig. 279). 
 
 NOTE. Spin a piece of money rapidly on the table, and it will present the 
 appearance of a sphere. 
 
 The semi-circumference PMQ describes the surface of the 
 sphere ; and it is clear that this surface 
 must be everywhere equally distant 
 from O, the centre of the semicircle 
 This point O is called the centre of th< 
 sphere. 
 
 Therefore, a sphere may also be de- 
 fined as a solid bounded by a surface, 
 which is everywhere equally distanl 
 from a point called the centre. 
 
 What is a radius of a sphere ? a diameter ? 
 An orange may be taken as a good example of a sphere. If we 
 cut an orange into a series of parallel slices, 
 the planes of division between the slices 
 are circles differing in size, the largest being 
 that which passes through the centre of the 
 orange. 
 
 Every section of a sphere made by a plane 
 is a circle. 
 
 If the section passes through the centre 
 of the sphere, it is called a great circle ; 
 in all other cases, it is called a small circle. 
 The equal parts into which a great circle divides the sphere are 
 called Hemispheres. 
 
 In Fig. 2 So, the sections PAQB and AGB H are great circles ; 
 the sections CEDF2cs\& MS NT me small circles. 
 
 All great circles of a sphere are equal; they have the same centre, 
 the same radius, the same diameter, as the sphere itself. 
 
251.] CHAPTER XII. GEOMETRICAL BODIES. 275 
 
 All great circles divide each other into two equal parts ; every 
 great circle divides the sphere and its surface into two equal parts. 
 
 Small circles are the smaller the further they are from the centre 
 of the sphere. 
 
 Every small circle divides the sphere and its surface into two 
 unequal parts. 
 
 The centre of any circle of the sphere must lie in the diameter 
 of the sphere which is normal to the plane of the circle ; the two 
 points where this diameter meets the surface of the sphere are called 
 the Poles of the circle. In Fig. 280, the points P and Q are the 
 poles of what circles ? 
 
 On the earth (or terrestrial sphere) , parallels of 'latitude are small 
 circles, the equator is a great circle, and the north and south poles 
 are the poles of all these circles. 
 
 All points in the circumference of a circle (great or small) are 
 equidistant from either one of its poles, whether this distance be 
 measured in a straight line or along the arc of a great circle pass- 
 ing through the point and the pole. Since the latter way of meas- 
 uring distances is much more convenient on a sphere, it is always 
 employed. Thus, in Fig. 280, the distance from the point C to 
 the pole Pis the length of the arc PC. 
 
 Of all lines that can be drawn on the surface of a sphere from 
 one point to another, the arc of the great circle passing through 
 the two points is the shortest. In this respect, the arc of a great 
 circle is to the surface of a sphere what the straight line is to the 
 plane. 
 
 By means of arcs of great circles (never small circles) , spherical 
 triangles and spherical polygons may be drawn on the surface of a 
 sphere. 
 
 The portion of the surface of a sphere comprised between the 
 circumference of two parallel circles is called a Zone. 1 The dis- 
 
 1 From a Greek word, meaning belt or girdle. 
 
276 GEOMETRY FOR BEGINNERS. [ 252. 
 
 tance between the planes of the circles is the height of the zone, 
 and the circumferences of the circles are its bases. If one of 
 the planes touches the sphere without cutting it (or is a tangent 
 plane), the zone is called a zone of one base. Point out zones in 
 Fig. 280. 
 
 Exercises. 1. Give examples of spheres (globes, balls). 
 
 2. Draw a figure of a sphere with lines in it to illustrate the definitions of 
 this section. 
 
 3. How will a sphere be divided by three great circles mutually perpen- 
 dicular to each other ? 
 
 4. In what zones is the surface of the earth divided ? Which are zones 
 of one base ? 
 
 252. An orange, after the peel is removed, is easily divided 
 into wedge-shaped slices, separated by natural planes of division. 
 These planes are planes of great circles intersecting in a common 
 diameter. In the fruit, the number of the planes is limited ; but, 
 in any sphere, we may imagine as many such planes as we please, 
 all passed through a common diameter, and each cutting the sur- 
 face of the sphere in the circumference of a great circle. 
 
 Fig. 281 shows several such planes passing through the diameter 
 PQ. They intersect the surface in the 
 circumferences, the front parts of which 
 are the semi-circumferences PA Q, PBQ, 
 etc. 
 
 These semi-circumferences are called 
 Meridians, and the portion of the sur- 
 face of the sphere comprised between two 
 meridians is called a Lune. Point out 
 lunes in Fig. 281. 
 
 Terrestrial Meridians are the circum- 
 ferences of great circles on the earth's surface made by planes 
 passing through the earth's axis. 
 
 The surface of a sphere is not developable, but a close approxi- 
 
CHAPTER XII. GEOMETRICAL BODIES. 
 
 277 
 
 mation to it can be constructed by drawing a series of equal lunes 
 in one plane, as follows : 
 
 Divide a line AB {Fig. 282), which must be made equal to 3^ 
 times the diameter of the proposed sphere, into twelve equal parts, 
 and add nine of these parts to each end of the line. Then, with a 
 
 U 
 
 A 
 
 V 
 
 A 
 
 A 
 
 \l 
 
 i ii in 
 
 t-M- 
 
 Fig. 282. 
 
 radius equal to ten of these parts, describe arcs from the points 
 marked i, 2, 3, etc. ; and, likewise, arcs from the points marked 
 I., II., III., etc., as shown in the figure. These arcs will inclose 
 lunes which, cut out and properly folded up, will very nearly form 
 a complete spherical surface. 
 
 It is by a process of this sort that the surfaces of balloons, base- 
 balls, globes, etc., are made. 
 
 Exercises. 1. Describe how a hemisphere would be generated by 
 motion. 
 
 2. Make a model of a sphere. 
 
278 GEOMETRY FOR BEGINNERS. 
 
 REVIEW OF CHAPTER XII. 
 
 I. Polyhedrons. 
 
 242. Definition of the Polyhedron. Specific names of polyhedrons. Euler's 
 
 theorem. 
 243. The five regular polyhedrons. Development of their surfaces. How 
 
 models of them are made. 
 
 II. The Prism, Cylinder, Pyramid, and Cone. 
 
 244. Definition of the Prism. Different kinds of prisms. The parallele- 
 piped. 
 
 245. Definition of the Cylinder. 
 
 246. Definition of the Pyramid. Different kinds of pyramids. 
 247. Definition of the Cone. 
 
 248. Sections of the prism, cylinder, pyramid, and cone. 
 249. Frustum of a pyramid or a cone. 
 
 250. Development of the surfaces of these four solids. How models of 
 them are made. 
 
 III. The Sphere. 
 
 251. The Sphere defined. Section of a sphere made by a plane. Great 
 
 and small circles. Spherical triangles and polygons. Zones. 
 252. Meridian lines. Lunes. How a model of a sphere may be made. 
 
253-] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 279 
 
 CHAPTER XIII. 
 SURFACES AND VOLUMES OP BODIES. 
 
 CONTENTS. I. Surfaces of Bodies ( 253-258). II. Volumes of Bodies ($$259- 
 274). III. Exercises and Applications ( 275-279). 
 
 J. Surfaces of Bodies. 
 
 253. THE PRISM. The lateral surface consists of parallelo- 
 grams ; the bases are polygons. Therefore, 
 
 I. Lateral surface = sum of the areas of the parallelograms of 
 which it is composed. 
 
 II. Total surface = lateral surface -\- areas of the bases. 
 
 Exercises. Find the total surface of a cube if, 
 
 1. The edge = 8 m . 3. The edge = 4 ft . 
 
 2. The edge = 3.5 m . 4. The edge = 7f in . 
 
 5. If the edge of a cube contain a unit of length, how many units of area 
 are there in the entire surface ? Ans. 6 a 2 . 
 
 6. The total surface of a cube 6ooi m . Find the length of the edge. 
 
 7 . If the total surface of a cube contains S units of area, how many units 
 of length are there in one edge ? Ans. -\J S -+ 6. 
 
 8. Find the total surface of a rectangular parallelepiped if the length = 
 15, the breadth = 9 m , the height 6 m . 
 
 9. Find the total surface of a rectangular parallelepiped if the length = a, 
 breadth = b, height c. Ans. 2 ab + 2 ac -f 2 be. 
 
 10. How many bricks, 8 in by 3!, are required to line the bottom and sides 
 of a reservoir 60^ long, 24** wide, 12^- deep ? 
 
 Find the total surface of the following regular prisms : 
 
 11. Triangular prism, height = 76 cm , side of base = 40. 
 
 12. Quadrangular prism, height = 6 ft , side of base = 27 in . 
 
 13. Hexagonal prism, height = 2.46 m , side of base 84 cm . 
 
 Hints. Find less radius of base by use of $ 202, Case 2, and $ 142, and area 
 of base by $ 133. 
 
280 GEOMETRY FOR BEGINNERS. [ 254. 
 
 14, How would Exercise 13 be solved if the less (or the greater) radius 
 of the base were given instead of a side of the base ? 
 
 15, The total surface of a regular hexagonal prism = 822<i m , a side of the 
 base io m . Find the height (see Table, p. 221). 
 
 16, In a regular hexagonal prism, the height = /, a side of the base = a. 
 Find the total surface. Ans. 6a/i+ 3 2 V3- 
 
 254. THE CYLINDER. The development of the convex sur- 
 face is a rectangle {Fig. 276} whose base is equal to the circum- 
 ference of the base of the cylinder, and whose altitude is equal to 
 the height of the cylinder. The bases of the cylinder are equal 
 circles. Therefore ( 128),- 
 
 I. Convex surface = circumference of base X height of cylinder. 
 
 II. Total surface convex surface + twice the area of either 
 base. 
 
 Exercises. Find the convex surface, the areas of the bases, and the total 
 surface of the following cylinders : 
 
 1. Height = 3 m , radius of base = i m . 
 
 2. Height = 96 cm , diameter of base = 6o cm . 
 
 3. Height 4.5 m , circumference of base = 3.64 m . 
 
 4. Height = 34 m , area of base = i6.74<i m . 
 
 5. Height = 9^ ft , radius of base 6 ft . 
 
 6. Height = h, radius of base = r. 
 
 7. Height = h, diameter of base d. 
 
 8. How many sq. yds. of sheet tin are required to cover the convex surface 
 of a circular tower 40** high, the diameter of the base being 8 ft ? 
 
 9. The length of a hollow tube = 3 m , the outer diameter = 36, the thick- 
 ness of the metal = 2 cm . Find the area of the inner convex surface. 
 
 255. THE PYRAMID. The lateral surface consists of triangles ; 
 the base is a polygon. Therefore, 
 
 I. Lateral surface = sum of the areas of the triangles of which 
 it is composed. 
 
 II. Total surface = lateral surface -f area of the base. 
 Exercises. Find the total surface of the following regular pyramids : 
 1, Triangular pyramid, slant height = 68 cm , side of base 42. 
 
256.] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 281 
 
 2. Square pyramid, slant height = 1 5, side of base = 3 m . 
 
 3. Octagonal pyramid, slant height = 7.o6 in , side of base = 1. 44. 
 
 4. How can the above exercises be solved if the less (or greater) radius 
 of the base is given in place of a side ? (See Table, p. 221.) 
 
 5. The height of a regular triangular pyramid = 8 m , and the side of the 
 base = 2 m . Find the total surface. 
 
 Hints. Find less radius of base by means of the Table, p. 221, and then the 
 slant height by means of 142. 
 
 6. Height of a regular hexagonal pyramid = i8 m , side of the base = i-5 m . 
 Find the entire surface. 
 
 7. The vertex of a pyramid with a rectangular base is vertically above the 
 middle point of the base. Find the total surface if the height = 6 m , and the 
 sides of the base are 2.4'" and i.6 m . 
 
 8. The height of the frustum of a regular pyramid with a square base = 
 l6 m , and the sides of the bases are 4 and 2 m . Find the entire surface. 
 
 Hints. 'The. lateral surface consists of four equal trapezoids. Find their com- 
 mon altitude by means of 142. 
 
 256. THE CONE. The developed convex surface is a circular 
 sector {Fig. 277} whose arc is equal to the circumference of the 
 cone, and whose radius is equal to the slant height. The base of 
 the cone is a circle. Therefore ( 216), 
 
 I. Convex surface = circumference of base X half the slant 
 height. 
 
 II. Total surface = convex surface -f- area of the base. 
 
 Exercises. Find the convex surface, the area of the base, and the entire 
 surface of the following cones : 
 
 1. Slant height = 5, radius of base = 2 m . 
 
 2. Slant height = 60^, diameter of base = I2 ft . 
 
 3. Slant height = I2.62 m , circumference of base 2.27 m . 
 
 4. Slant height k, radius of base = r. 
 
 5. Slant height = k, diameter of base = d. 
 
 6. Height = I2 m , radius of base = 5 (see 142). 
 
 7. Height = /i, radius of base = r. 
 
 8. How many square yards of sail-cloth are required to make a conical 
 tent 6o ft high, if the diameter of the base is 40**. 
 
 9. Two cones have equal bases, but one is three times as high as the other. 
 Compare their convex surfaces. 
 
282 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 257- 
 
 257. FRUSTUM OF A CONE. The development of this surface 
 is a part of a circular ring (Fig. 283) whose arcs A B and CD are 
 
 the circumferences of the 
 
 //Yx bases of the frustum, and 
 
 / / \ \ whose distance apart is 
 
 equal to the slant height 
 of the frustum. We shall 
 first find the area by the 
 method of reasoning used 
 in 212 (a method which 
 we shall have occasion to 
 use again hereafter); and 
 shall then transform the 
 value of the area thus obtained into two other useful forms. 
 
 Construct a series of trapezoids, such as the three shown in Fig. 
 283, by drawing radii and parallel pairs of tangents (why are they 
 parallel ?) . The common altitude of these trapezoids = distance 
 between the arcs slant height of frustum. By increasing the 
 number of such trapezoids we can plainly make the sum of their 
 areas approach the area of the frustum as nearly as we please, 
 although it will never quite reach it. That is to say, 
 
 Surface of the frustum = limit of the sum of the trapezoids. 
 By 131 and by addition, the sum of the trapezoids = sum 
 of their parallel sides X slant height. As the number of trapezoids 
 is increased, the sum of their parallel sides approaches, but never 
 quite reaches, the sum of the arcs AB and CD. Therefore, 
 
 {(AB + CD) X slant height 
 Limit of sum of the trapezoids = -H 
 
 (. of frustum. 
 
 Therefore (Axiom I.), 
 
 I. Convex surface of frustum = (AJ3 X CD) X slant height 
 = (sum of circumferences of bases) X slant height. 
 
 The result can be put into another form, as follows : Let a frus- 
 tum be generated by the revolution of the trapezoid AX BY (Fig. 
 
257-] CHAPTER Xlil. SURFACES AND VOLUMES OF BODIES. 283 
 
 284) about the side XY. Through M, the middle point of AS, 
 draw MN\\ AX. MN is the radius of a circu- 
 lar section of the frustum midway between the 
 bases. Draw BP II XY, and intersecting MN 
 
 (why?). Put BY= a, MQ = b ; then MN = 
 a+b,AX=a + 2b. By 209,- 
 
 Flg. 284. 
 
 Circumference of upper base = 2 TT a. 
 " lower " 
 
 11 middle section = 2ira + 2-n-b. 
 Whence, (sum of circumferences of bases) = 4 
 
 = 2Tra -f- 2Trb = circumference of middle section. 
 
 Therefore, 
 
 II. Convex surface of frustum = circumference of middle section 
 X slant height. 
 
 Draw MR\.AB\ AMNR ~ AAP (why?); therefore, 
 MN: MR = BPor its equalXY: AB; whence, 
 
 MN x AB = MR x XY. 
 
 If we multiply both sides of this equation by 2 TT, we have 
 AB= 2-rrMR x XY. 
 
 The first side of this equation is the value of the convex surface 
 given in II. ; the second side is the product of the circumference 
 whose radius is MR and the height of the frustum. Therefore, 
 
 III. Convex surface of frustum = circumference whose radius 
 is the length of the perpendicular erected at the middle point of the 
 side of the frustum and terminated by the axis X height of frustum. 
 
 Exercises. 1. Prove that Theorems II. and III. hold true of the convex 
 surface of an entire cone. 
 
 Hint. In the proof of II., make a ~ o, because there is no upper base. 
 
 2. Prove that Theorem III. holds true of the convex surface of a cylinder. 
 
 3. How is the total surface of a frustum found ? 
 
284 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 258. 
 
 Find the convex surface, and also the total surface, of the following frus- 
 tums of cones : 
 
 4. Slant height = I i m , radii of bases 5 and 3 m . 
 
 5. Slant height = 4 ft , diameter of bases i8 in and io in . 
 
 6. Slant height = k, radii of bases, a and b. 
 
 7. Slant height = k, radius of middle section r. 
 
 8. The height of a frustum = I2 m , the radii of the bases are y m and 2*. 
 Find the convex surface. 
 
 Hint. It is evident, from Fig. 284 and from 142, that slant height = 
 
 9, The height of a frustum h, the radii of the bases are a and b. Find 
 the convex surface. 
 
 258. THE SPHERE. Inscribe in the semicircle X BY (Fig. 
 with a radius OX r, half of a regular polygon with an 
 even number of sides (here four). Draw the 
 less radii OM, ON, OP, OQ, and from the 
 vertices A, B, C, let fall the perpendiculars 
 AD, BO, CE. 
 
 If the semi-circumference and the sides of 
 the polygon revolve about X Y, the former will 
 describe the surface of a sphere, and the sides 
 of the polygon will describe the convex surfaces 
 of cones or frustums of cones. Which two 
 sides will describe the former? 
 
 By 257, III. (including Exercise i), 
 
 Surface described by side XA = z-n-OM x DX. 
 
 X OD. 
 X EO. 
 " " " CY= 2irOQ x YE. 
 
 Adding these equations, and remembering that OM=ON 
 OP=OQ,we have, 
 
 Total surface described by sides of polygon = 2irOM X XY. 
 As the number of sides is increased, the total surface described 
 
 Fig. 285. 
 
258.] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 285 
 
 by the sides approaches the surface of the sphere as its limit. 
 That is, - 
 
 ( limit of surface described by sides of the 
 Surface of sphere = < 
 
 ( polygon. 
 
 As the number of sides is increased, the less radius O M ap- 
 proaches r as its limit. Therefore, 
 
 Limit of surface described by } vv 
 
 r 2 7T r X -A X . 
 
 the sides of the polygon J 
 
 Therefore (Axiom I.), 
 
 Surface of the sphere = 2tcr X XY= 2-n-rX 2r = ^irr 2 . 
 
 Or (209,213),- 
 
 I. Surface of a sphere = circumference of a great circle X the 
 diameter ^=four times the area of a great circle. 
 
 The above method applied to a part of the semicircle, as, for 
 instance, the arc ABC, which, by revolving, describes a zone, 
 gives as the result, 
 
 II. Surface of a zone = circumference of a great circle X height 
 of zone. 
 
 Exercises. Find the surfaces of the following spheres : 
 
 1. Radius = 5. 4, Diameter 44. 
 
 2. Radius = 3| in . 5. Diameter = I2 ft . 
 
 3. Radius = r. 6. Diameter = d. 
 
 7. Find the surface of the earth, taking the diameter as 8000 miles. 
 
 8. If the surface of a sphere = 4001, find the radius. 
 
 9. Find the total surface of a hemisphere if the radius = 6 m . 
 
 10. Find the total surface of a hemisphere if the radius = r. 
 
 11. How much leather will it take to cover a base-ball if the diameter = 
 4 p? 
 
 12. If the radius of a sphere = 8 m , find the area of a zone the height of 
 which = 2 m . 
 
 13. Height of a cylinder = diameter of its base. Compare its total sur- 
 face with that of the largest sphere that can be inscribed in it. 
 
 Ans. Surface of the sphere = f surface of the cylinder. 
 NOTE. This curious fact was discovered by Archimedes more than 200 B.C. 
 
286 GEOMETRY FOR BEGINNERS. [ 259. 
 
 II. Volumes of Bodies. 
 
 259. In order to compare the sizes or magnitudes of bodies, 
 we select a body of known size as the common term of comparison 
 or unit, and then find how many times this unit is contained in 
 other bodies. 
 
 Definition. The number of times the unit is contained in a 
 body, followed by the name of the unit, is called the VOLUME of the 
 body. 
 
 If, for instance, the cubic inch is the unit, and one body will 
 contain it four times while another body will contain it twelve 
 times, the volumes of the bodies are four cubic inches and twelve 
 cubic inches ; and one of the bodies is exactly three times as large 
 as the other. 
 
 The most convenient units of volume are cubes whose edges are 
 equal to the units of length. 
 
 The CUBIC INCH (cub. in.), the CUBIC FOOT (cub. ft.), and the 
 CUBIC YARD (cub. yd.), are such units. The BUSHEL and the 
 GALLON are not such units. The bushel contains 2150.42 cub. in., 
 the (United States) liquid gallon contains 231 cub. in., and the 
 (United States) dry gallon contains 268.8 cub. in. 
 
 In the metric system, the units of volume are always cubes 
 whose edges are units of length. The three most used are the 
 CUBIC METER (cbm.), the CUBIC DECIMETER (cdm.), and the 
 CUBIC CENTIMETER (ccm.). 
 
 The cubic decimeter is often called a LITER (1.). A HECTO- 
 LITER (hi.) = 100 liters. 
 
 The cubic meter, when used for measuring wood, is called a 
 STERE. 
 
 NOTE. i cubic foot = 0.0283 cubic meter. I cubic meter 35.3156 cubic feet. 
 I bushel = 35.24 liters. i liter = 0.02838 bushel, 
 
 i liquid gallon = 3.785 liters. I liter 0.264 gallon. 
 
 Exercises. 1. The bushel contains 4 pecks. How many cubic inches in 
 one peck ? 
 
260.] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 287 
 
 2. The gallon of either kind is divided into four quarts, the quart into two 
 pints, the pint into four gills. How many cubic inches in the quart, pint, and 
 gill, both liquid and dry ? 
 
 3. Reduce 6000 cubic feet to cubic meters. 6000 cubic meters to cubic feet. 
 
 4. If I pay $1.20 per bushel for wheat, what must I ask per hectoliter, in 
 order to make a profit of 25 per cent ? 
 
 5* A hectoliter of wine will fill how many pint bottles ? 
 
 260. To measure the volume of a body is to find how many 
 times it will contain the unit of volume. 
 
 The volume of a liquid body may be found directly by repeat- 
 edly filling a vessel which is known to hold exactly a unit ; as, for 
 example, a quart pot or a gallon jug. But this method could not 
 be applied to bodies of fixed shape, such as a stick of timber or a 
 cannon-ball. 
 
 Geometry teaches how volumes are measured indirectly, by 
 showing that they depend upon the lengths of certain lines (for 
 instance, the length, breadth, and height of a parallelepiped), and 
 can, therefore, be found by performing simple operations upon 
 the "numbers which express the lengths of these lines (compare 
 126). 
 
 We shall deduce a general rule for computing the volume of any 
 prism or cylinder by treating as special cases, 
 
 (z.) The rectangular parallelepiped ; 
 
 (.) The right parallelepiped ; 
 
 (/.) The oblique parallelepiped ; 
 
 (iv.) The triangular prism ; 
 
 (v.) Any prism ; 
 
 (vi.) The cylinder ; 
 
 and another general rule for computing the volume of any pyramid 
 or cone by treating as special cases, 
 
 (/.) The triangular pyramid ; 
 
 (.) Any pyramid ; 
 
 (in-) The cone. 
 
288 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 26l. 
 
 Fig. 286. 
 
 261. THE RECTANGULAR PARALLELOPIPED. In the rectangu- 
 lar parallelepiped AP (Fig. 286) let the edges be AB = 5, AD 
 = 3 m , AM= 7 m . Then the base A BCD con- 
 tains i5 qm ( 128); and upon each square me- 
 ter of the base we can construct a pile of 7 
 cubic meters, such as DQ, so that the solid 
 must contain exactly 7 X 15 = 105 cubic me- 
 ters. 
 
 A like result would be obtained if the edges 
 had any other values. If there were fractions 
 of a meter, we would proceed as shown in find- 
 ing the area of a square ( 127). Therefore, 
 the number of units of volume in a rectangular parallelepiped is 
 found by multiplying the number of units of area in the base by 
 the number of linear units in the height. Or, more briefly, 
 Volume of a rectangular parallelepiped = base X height. 
 If a, b, and c denote the length, the breadth, and the height 
 of the solid, then the base = a X b t and therefore the volume = 
 a X b X c. 
 
 In the case of the cube, a = b c ; therefore, 
 Volume of a cube = a X a X a = a 3 . 
 
 Here a? is a short way of expressing that a is taken as a factor 
 three times, or is raised to the third power. Hence the third power 
 of a number is often called the cube of the number. 
 
 Remark. Since the edge of a cubic meter = 10 decimeters, 
 therefore, i cubic meter =10X10X10 1000 cubic decime- 
 ters (liters). For a like reason, i cubic decimeter = 1000 cubic 
 centimeters. In general, 
 
 The ratio between two units of volume is the cube of the ratio 
 between the corresponding units of length. 
 
 Exercises. Find the volumes of the following cubes : 
 
 1. Edge = 8 m . 3. Edge = 72 cm . 5. Area of base = 36 f i m . 
 
 2. Edge = 3.17. 4. Edge = 9| in . 6. Area of base = S. 
 
262.] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 289 
 
 Find the volumes of the following rectangular parallelepipeds : 
 
 7. Length = 4.5, breadth = 7, height = 3.4. 
 
 8. Length = /, breadth = b, height = h. 
 
 9. Length = breadth = /, height = h. 
 
 10. Area of base = 8ii m , height = 4. 
 
 11. Area of base = S, height = h. 
 
 12. How many cubes, each with an edge - I* 1 , can be put into a box 
 I2 dm by gdm by 6 dm , inside measurement ? 
 
 13. Find the number of cubic decimeters of wood in a board 5 long, 
 o.6 m wide, 0.03 thick. What kind of a solid is the board ? 
 
 14. How high must a box 5 dm long and 2 dm wide be to hold 30 liters? 
 
 15. Volume of a cube = 74,o88 ccm . Find the edge. 
 
 16. Volume of a cubical tank = 474.552 ccm . Find its depth. 
 
 17. Volume of a cube = V. Find the edge. 
 
 IS. Two dimensions of a rectangular parallelepiped are 9.37 and 5 m . 
 Find the third dimension. 
 
 19. A room is 8 long, 6 m wide, 2.5"' high. How many cubic meters does 
 it contain ? How much more space would it inclose if the floor had the same 
 perimeter, but were in the shape of a square ? 
 
 20. How many stones 3 dm long, i-5 dm wide, O.75 dm thick, are required to 
 build a wall 40 long, 3'" high, and o.5 m thick ? 
 
 21. How many liters of water have fallen in a garden 42 long and 32 m 
 wide, if a vessel standing in it is filled to the height of I2 cm ? 
 
 22. How many hogsheads will a tank lo* 1 by I2 ft by 9 ft hold ? (i hogs- 
 head = 63 gallons.) 
 
 23. Why does one cubic foot contain 1728 cubic inches. 
 
 262. THE RIGHT PARALLELOPIPED. If from the right paral- 
 lelepiped A G (Fig. 287) we take away the 
 
 prism B G, by passing through the edge n^\ A/ 
 
 BF a plane BFMN perpendicular to the - fi pf -j -p^{ 
 
 base, and then place this prism on the left 
 in the position A H, the right parallelepiped 
 
 will be transformed into an equivalent rec- | /-;;- : '-" 
 
 4-n r** m 1 1 or* r^if nll^l^-TiTT^ri K a win rr tn f* en VY\ P> n Pi O"ni" ^* " 
 
 tangular parallelepiped having the same height A ^ 2? 
 
 and an equivalent base ( 129). There- Fig. 287. 
 fore, by 261, - 
 
 Volume of a right parallelepiped = base X height. 
 
290 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 263. 
 
 Since area ABCD AB X B M, therefore, the volume of the 
 right paraUelopipedAG = AB X B MX BF= AB X BF X BM 
 = area AB EFx BM = a rectangular face taken as base x the 
 corresponding height. 
 
 Exercises. 1. Find the volume of a right parallelepiped if the height = 
 Ii m , and the base is a parallelogram having the length I2 m and the altitude 5. 
 
 2. A wall 2O m long, 3 high, is o.8 1M thick at the bottom and 0.4 thick at 
 the top. Find how many cubic meters it contains. 
 
 Hint. Consider the wall a right parallelepiped having for base a trapezoid 
 whose parallel sides are o.8 m and.o.4" 1 . 
 
 263. THE OBLIQUE PARALLELOPIPED. Let AG (Fig. 288) be 
 an oblique parallelepiped. Through the corner E pass a plane 
 perpendicular to the edge EF, cutting from the parallelepiped the 
 
 solid AMDNEHO, and place 
 this solid on the right in the 
 position BPCQFGR. By 
 this operation, the oblique par- 
 allelepiped is transformed into 
 an equivalent right parallele- 
 piped, having the same height, 
 and a rectangular base EFRO 
 equivalent to the base EFGH 
 of the oblique parallelepiped 
 ( 129). Now the volume of 
 the right parallelepiped = its base EFR O x the height ( 262) ; 
 therefore, 
 
 Volume of the oblique parallelepiped = base EFGH X height. 
 
 264. THE TRIANGULAR PRISM. If through the edges AB and 
 GH of the oblique parallelepiped AG (Fig. 288) we pass a plane, 
 it will divide the solid into two oblique triangular prisms. The 
 same plane divides the equivalent right parallelepiped into two right 
 triangular prisms. These right prisms are equivalent, each to its 
 
265.] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 291 
 
 corresponding oblique prism, because each is formed by cutting 
 off (by the plane through E _L the edge EP) a portion of the 
 oblique prism, and transferring it to the right of the figure. Now 
 these right prisms are also equivalent to each other ; for, if one of 
 them is inverted, it is then clear that it has precisely the same 
 shape as the other, and that its faces are equal and parallel to those 
 of the other. Therefore, the two oblique prisms are equivalent, 
 and hence each is half of the parallelepiped. Take ADEN as the 
 base of the parallelepiped ; then the distance between the planes 
 A D EH and B CFG is its height, the two prisms have the same 
 height, and their bases are the triangles A EH and. ADH, each of 
 which is half the parallelogram ADEN. So that the volume of 
 either prism (half the volume of the parallelepiped) is equal to its 
 base (half the base of the parallelepiped) X its height. In general, 
 
 Volume of a triangular prism = its base X its height. 
 
 Exercises. Find the volume of a regular triangular prism, having given, 
 
 1. Height X 75 cm , area of base = i4O ccm . 
 
 2. Height h, area of base = S. 
 
 3. Height = 8.5 m , one side of the base = 3. 
 
 4. Height = h, one side of the base = a. 
 
 5. The bases of a triangular prism 8o cm high are isosceles right triangles, 
 a leg of which = I2 cm . Find the volume of the prism. 
 
 6. If the volume of a triangular prism with isosceles right triangles for 
 bases = 6oo cbm , and the height = 14, find (z.) area of the base, (zY.) perime- 
 ter of the base. 
 
 265. ANY PRISM. Any prism (Pig. 289) can be divided into 
 triangular prisms by passing planes, as shown in the 
 figure, through one of the lateral edges. The trian- 
 gular prisms have the same height as the entire 
 prisms, and the sum of their faces is equal to the 
 base of the entire prism. Therefore, by 264 and 
 by addition, it follows that the 
 
 Volume of any prism = its base X its height. 
 
 Hence, any two prisms having equivalent bases Fig. 289. 
 and equal heights are equivalent. 
 
 A 
 
292 GEOMETRY FOR BEGINNERS. [ 266. 
 
 Exercises. Find the volumes of the following regular prisms : 
 
 1. Hexagonal prism, height = I7 m , side of base = 2 m . 
 
 2. Hexagonal prism, height =: h, side of base = a. 
 
 3. Octagonal prism, height = 85 cm , side of base io cm 
 
 4. Octagonal prism, height //, side of base = a. 
 
 266. THE CYLINDER. If we inscribe in the bases of a cylin- 
 der regular polygons of the same number of 
 sides, so that their sides, taken pair by pair, are 
 parallel, and then pass planes through each pair 
 of sides, these planes with the polygons will form 
 a prism inscribed in the cylinder (Fig. 290}. 
 Now the greater the number of lateral faces in 
 the inscribed prism, the nearer does its volume 
 approach that of the cylinder ; that is to say, - 
 
 Volume of cylinder = limit of volume of in- 
 Fig. 290. scribed prism. 
 
 The volume of the prism = its base X its height ; 
 and its base has the base of the cylinder for its limit. Therefore, 
 limit of the volume of the inscribed prism base of cylinder x its 
 height. 
 
 Hence (Axiom I.), 
 
 Volume of the cylinder = its base X its height. 
 
 Exercises. Find the volumes of the following cylinders : 
 
 1. -Height = 6 m , area of base = 421. 
 
 2. Height = 8 m , radius of base = 2 m . 
 
 3. Height = 64 cm , diameter of base = i6 cm . 
 
 4. Height = 7^, circumference of base = 44 in . 
 
 5. Height = h, radius of base = r. 
 
 6. Height = h, radius of base = d. 
 
 7. The volume of a cylinder = 2OO cbm , the height = 13"'. Find the radius 
 of the base. 
 
 8. The volume of a cylinder = V, the height = h. Find the radius of the 
 base. 
 
 {). A cylindrical pail holding exactly one liter is i8 cm high. Find the 
 diameter of the base. 
 
267.] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 
 
 293 
 
 10. A hollow iron tube is 6 m long, the outer diameter = 92, the inner 
 diameter = 68 cm . How many cubic decimeters of iron are there ? 
 
 11. How long is an iron tube, if the inner diameter = 3.5 cm , the outer cir- 
 cumference = I5 cm , the volume = i2O ccm ? 
 
 12. A vessel in the shape of a cylinder is to be made from sheet-iron 2 cm 
 thick. The vessel is to be i' n high, and to hold 100 liters. What must be its 
 outer diameter ? 
 
 267. THE TRIANGULAR PYRAMID. Let S-ABC and S 
 -MNO {Fig. 291) be two triangular pyramids standing on th: 
 same horizontal plane, and having equivalent bases ABC and 
 MNO, and equal heights. Divide the common height into any 
 number of equal parts, and through the points of division pass 
 planes parallel to the bases of the pyramids. Since the bases are 
 equivalent, it follows, from the mode of generation of a pyramid 
 ( 246), that any two sections, as DEF and PQR, situated at 
 the same distance from the bases, will also be equivalent. 
 
 In the two pyramids construct triangular prisms, such as DEFGH1 
 and PQRTUV, having for upper base one of the sections; for 
 lateral edges, lines parallel to SA or SM-, and for a common 
 height, the distance apart of two sections. It follows, from 265, 
 that these prisms will be equivalent, taken pair by pair, in the 
 two pyramids ; therefore, 
 the sums of their volumes 
 will be equal, no matter 
 how many or how few of 
 them there may be. But, 
 the greater their number 
 the nearer the sums of 
 their volumes approach the 
 volumes of the two pyra- 
 mids ; that is to say, 
 
 Volume of each pyramid =: limit of sum of volumes of the in- 
 scribed prisms. 
 
 N 
 
 Fig. 291. 
 
294 
 
 GEOMETRY FOR BEGINNERS. 
 
 [ 267. 
 
 Now the sums of these volumes remain always equal ; therefore 
 their limits, or the volumes of the pyramids, must also be equal. 
 
 Hence, two triangular pyramids having equivalent bases and 
 eqital heights are equivalent. 
 
 Now let S-ABC (Fig. 292) be any triangular pyramid. If 
 we pass a plane through BC II SA, and a plane through S II ABC, 
 
 we form a triangular prism 
 A B CSP Q. This prism is 
 divided by the plane SBC, 
 and a plane passed through 
 C and SP, into three trian- 
 gular pyramids, S-ABC, 
 P- SB C, and C - SPQ. Of 
 
 Fig. 292. these, S-AB C and C-SPQ 
 
 are equivalent because they 
 
 have equivalent bases and equal heights. Also, P-SBC and 
 C-SPQ are equivalent, because we may take *S for the common 
 vertex, and then they have equal heights (the distance from S 
 to the plane B C P Q) and equivalent bases (the equal triangles 
 PBC and PC Q). Therefore, each pyramid = the prism. Now 
 the volume of the prism = base x height ; therefore, the volume 
 of the pyramid S-ABC (which has the same base and the same 
 height as the prism) = % base X height. Therefore, 
 Volume of a triangular pyramid '= % its base X its height. 
 
 Exercises. 1. Height of a regular triangular pyramid I5 m , one side 
 of the base = 2 m . Find the volume. 
 
 2. He.ight of a regular triangular pyramid = h, one side of the base - a. 
 Find the volume. 
 
 3. A triangular pyramid 9O cm high has an isosceles right triangle for base, 
 ote leg of which 4O cm . Find the volume of the prism. 
 
 4. A triangular pyramid with an isosceles right triangle for base is 3.4 
 high. Its volume = 8o cm . Find (?.) the area of the base ; (?V.) the perime- 
 ter of the base. 
 
 5. The frustum of a regular triangular pyramid is 4 m high, and two homolo- 
 gous sides of the bases are 2 rn and i.25 m . Find its volume. (See 249.) 
 
268.] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 
 
 295 
 
 268. ANY PYRAMID. Any pyramid whatever can be divided 
 into triangular pyramids by passing planes through 
 one of the lateral edges, as shown in Fig. 293. 
 These triangular pyramids all have the same height 
 as the entire pyramid, and the sum of their bases 
 = the base of the entire pyramid. 
 
 Therefore, . 
 
 Volume of any pyramid = -J base X height. Fig. 293. 
 
 Exercises. 1. Find the volume of a regular hexagonal pyramid, if the 
 height = 36 cm and a side of the base = 8 cm . 
 
 2. Find the volume of a regular octagonal pyramid, if the height = 44"* 
 and a side of the base = 4 m . 
 
 3. Solve Exercise I, if the height = h and a side of the base = a. 
 
 4. Solve Exercise 2, if the height = h and a side of the base = a. 
 
 5. The frustum of a regular hexagonal pyramid is 2.7 high, and two 
 homologous sides of the bases are i.2 m and 7O cm . Find the volume. 
 
 6. A vessel has the shape of the frustum of a regular four-sided pyramid. 
 It is i8 cm high, and the outer edges of its bases are 24 cm and i6 cm . How 
 many liters of water will it hold if the material of which it is made is 2 cm thick ? 
 
 269. THE CONE. Inscribe in the base of a cone (Fig. 
 a regular polygon, and pass planes through the 
 sides of this polygon and the vertex of the cone. 
 These planes with the polygon form an in- 
 scribed regular pyramid whose volume ap- 
 proaches the nearer to that of the cone, the 
 greater the number of sides in the regular poly- 
 gon ; that is, 
 
 Volume of cone = limit of volume of inscribed 
 pyramid. 
 
 But volume of any pyramid = its base X its 
 height. And in this case the base has the base 
 of the cone for its limit. Therefore, limit of volume of inscribed 
 pyramid = base of cone X height. Therefore (Axiom I.), 
 
 Volume of a cone = % its base X its height. 
 
 Fig. 294. 
 
296 GEOMETRY FOR BEGINNERS. [ 270. 
 
 Exercises. Find the volumes of the following cones : 
 
 1. Height = 15, area of base = 6o c i m . 
 
 2. Height = I2 rn , radius of base ~ 3 m . 
 
 3. Height = h, radius of base = r. 
 
 4. Height 74 cm , diameter of base = 4.6. 
 
 5. Height h, diameter of base = d. 
 
 6. Height 6 ft , circumference of base = ijjft. 
 
 7. Slant height = 5, radius of base = 3. 
 
 8. Slant height = k, radius of base r. 
 
 9. Find the volume of a cone whose slant height = diameter of the base 
 
 - 2. 
 
 10. Volume of a cone 8oo cbm , radius of base i.8 m . Find the height. 
 
 11. Find the volume of the frustum of a cone, if the radii of the bases are 
 7-5 cm and 5 cm , and the height 25 cm . (See 249.) 
 
 12. A vessel has the shape of the frustum of a cone i m high, and the inner 
 circumferences of the bases are 5 and 4 m . How many cubic meters of water 
 will it hold ? 
 
 13. What part of a cone remains if a cone two-fifths as high as the entire 
 cone is cut off ? 
 
 14. How many gallons of water will a vessel hold which has the shape 
 of the frustum of a cone, the height being 3 ft , the diameters of the bases i8 in 
 and I2 in ? 
 
 15. A Dutch windmill in the shape of the frustum of a cone is I2 m high. 
 The outer diameter at the bottom and the top are i6 m and I2 m ; the inner 
 diameters I2 m and io m . How many cubic meters of stone were required to 
 build it ? 
 
 16. The chimney of a factory is 32 high, and the outer circumferences of 
 its bottom and top are 20 and I4 m , while the bore is, throughout, 2 m wide. 
 How many cubic meters of brick does it contain ? 
 
 270. The results obtained in H 261-266 may be summed 
 up in one general rule, as follows : 
 
 Volume of a prism or a cylinder = base X height. 
 
 And the results obtained in $ 267-269 may be summed up in 
 one general rule, as follows : 
 
 Volume of a pyramid or a cone = J base x height. 
 
271.] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 297 
 
 271. THE SPHERE. In order to find the volume of a sphere, 
 conceive a series of pyramids formed, having for a common vertex 
 the centre of the sphere, and for bases the faces of a polyhedron 
 circumscribed about the sphere (that is, a polyhedron whose faces 
 touch the surface of the sphere, each in a single point). By in- 
 creasing the number of these pyramids, we can make their sum 
 approach the volume of the sphere as nearly as we please ; or, in 
 other words, 
 
 Volume of sphere = limit of sum of the pyramids. 
 
 Now each pyramid = height X base, and their common height 
 = radius of the sphere ; therefore, sum of the pyramids = radius 
 of sphere x sum of the faces. The limit of the sum of the faces 
 (when their number is increased more and more) = the surface of 
 the sphere. Therefore, limit of the sum of the pyramids = radius 
 X surface of the sphere. Therefore (Axiom I.), 
 
 Volume of a sphere == radius X surface of sphere. 
 
 Exercises. 1. Find the volume of a sphere if the radius = 6 m . 
 
 2. Find the volume of a sphere if the radius = r. 
 
 3. Find the volume of a sphere if the diameter 14. 
 
 4. Find the volume of a sphere if the diameter = d. 
 
 5. Find the volume of a sphere if the circumference of a great circle = io ft . 
 
 6. Required the volume of the largest sphere that can be turned from a 
 cubical block of wood whose edge = i dm . How much of the wood is lost ? 
 
 7. The volume of a sphere = i8o cbm . Find the radius. 
 
 8. The volume of a sphere = V. Find the radius. 
 
 9. Find the volume of the largest sphere that can be turned from a cylin- 
 der of wood whose diameter = its height I2 cm . 
 
 10. The diameter of a cylinder = its height = h. Compare the volume of 
 this cylinder with that of the largest sphere that can be inclosed in it. 
 
 Ans. Volume of the sphere = f volume of the cylinder. 
 NOTE. This curious fact was discovered by Archimedes about 230 B.C. 
 
 11. Find the difference between the volumes of two concentric spheres 
 whose radii are 4 m and 5. 
 
 12. In two concentric spheres, the volume of the larger = twice that of the 
 smaller. If the radius of the smaller = i m , find the radius of the larger. 
 
298 GEOMETRY FOR BEGINNERS. [ 272. 
 
 272. The volume of any body, whatever be its shape, may be 
 found by weighing it, and then dividing the weight by the weight 
 of one unit of volume. The quotient will be the number of units 
 of volume in the body. 
 
 In the metric system, the most common units of weight are, 
 
 The GRAM (g.)= the weight of i ccm of pure water (at 4 C.). 
 The KILOGRAM (kg.)= weight of i cdm of pure water (at 4 C.). 
 
 Since i cdm = iooo ccm , i kilogram = 1000 grams. 
 
 The ratio of the weight of any substance to the weight of the 
 same volume of pure water is called the SPECIFIC GRAVITY of that 
 substance. 
 
 SPECIFIC GRAVITIES OF SOME COMMON SUBSTANCES. 
 
 Platinum . 
 
 . 22.00 
 
 Brass (sheet) . 
 
 8.40 
 
 Boxwood 
 
 . 1.20 
 
 Gold . 
 
 IQ 35 
 
 Steel .... 
 
 780 
 
 Ice ... 
 
 O Q2 
 
 Mercury 
 
 . 13.60 
 
 Iron (cast) . . 
 
 / *** 
 
 7-25 
 
 Oak . . . 
 
 . 0.90 
 
 Lead . . 
 
 "-35 
 
 Zinc .... 
 
 7.19 
 
 Alcohol . . 
 
 . 0.80 
 
 Silver . . 
 
 . 10.51 
 
 Marble . . . 
 
 2.70 
 
 Pine . . . 
 
 o-57 
 
 Copper . . 
 
 . 8.80 
 
 Glass .... 
 
 2.60 
 
 Cork . . '. 
 
 . 0.24 
 
 The gram and the kilogram being by definition the weights of 
 unit volumes of water, it follows that the above numbers express 
 the weights of i ccra of the substance in grams, or of i cdm (i liter) of 
 the substance in kilograms. 
 
 Whence it is clear that in the metric system, 
 
 T7 / * t j 
 
 Volume of a body = _ 
 
 its specific gravity 
 
 the volume being expressed (/.) in cubic centimeters, or (it.) in 
 cubic decimeters (liters), according as the weight is expressed 
 (*'.) in grams or ('.) in kilograms. 
 
 The most important English units of weight are the POUND (Ib.) 
 avoirdupois, the OUNCE (oz.) = jV 1 *, and the TON = 2Ooo lb . 
 
272.] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 299 
 
 They are connected (nearly enough for practical purposes) with 
 the weight of unit volume of water by the relation, 
 
 Weight of i cubic foot of 'water = 62^ lbs = iooo oz . 
 
 The weight in pounds of a cubic foot of any other substance is 
 found by multiplying its specific gravity by 62^, and the weight in 
 ounces by multiplying its specific gravity by 1000. 
 
 NOTE. 283g T.OZ (nearly); i k s = 2.2^3 (nearly). 
 
 Exercises. 1. Find the volume of a piece of granite which weighs 4 k s. 
 
 2. A piece of lead weighs 24 k s. Find its volume. 
 
 3. What is the volume of 848 of silver ? 
 
 4. Required the edge of a cubical vessel which will hold 2OOO lbs (i ton) 
 of mercury. 
 
 5. Find the edge of a marble cube weighing 184^. 
 
 6. An iron cylinder 2.$ m long weighs 68o k s. Find its diameter. 
 
 7. Find the edge of a cube of cork which weighs the same as an iron ball 
 22 cm in diameter. 
 
 8. How many liters will a vessel hold if the vessel weighs when empty 
 I.5 k S, and when full of water i-5 k s ? 
 
 9 An oak cone 40 high weighs 2 k s. Find its diameter. 
 
 10. What is the diameter of a cannon ball which weighs 64 lbs ? 
 
 11. How can the weight of a body be found if its volume and its specific 
 gravity are known ? 
 
 12. Find the weight of I M of water, i hl of alcohol, and i cub ft of mercury. 
 
 13. Find the weight of a rectangular parallelepiped of copper, o.34 m long, 
 o.n m wide, o.O3 m thick. 
 
 14. Find the weight of an oak cone, if the radius of the base 32 cm and 
 the height = 9O cm . 
 
 15. Find the weight of a hollow brass cube, if the inner diameter = ^ 
 and the thickness of the brass = i cm . 
 
 16. If a cube, whose edge = 8 cm , weighs 508, find the weight of the 
 largest cylinder that can be turned from it. Also the weight of the largest 
 cone. Also the weight of the largest sphere. 
 
 17. How can the specific gravity of a body be found, if its weight and its 
 volume are known ? 
 
 18. A cubical block of pine wood, an edge of which I2 cm , weighs i k . 
 Find the specific gravity of the wood. 
 
 19. If a mass of ice containing 27O cbm is known to weigh 229,000^, find 
 the specific gravity of ice. 
 
 20. Find the specific gravity of cast iron, if 2^ cutt weigh 953 lbs . 
 
300 GEOMETRY FOR BEGINNERS, [ 273. 
 
 273. Equivalent solids have the same volume (size) ; similar 
 solids, the same shape ; equal solids, the same volume and the 
 same shape. 
 
 Equal polyhedrons must fulfil three conditions : 
 
 (/.) They must have the same number of faces, equal each to 
 each. 
 
 (i/.) Their homologous edges must be equal. 
 
 (m.) Their homologous solid angles must be equal. 
 
 Similar polyhedrons must also fulfil three conditions : 
 
 (/.) They must have the same number of faces similar each to 
 each. 
 
 ('.) Their homologous edges must be proportional. 
 
 (iii.) Their homologous solid angles must be equal. 
 
 In both cases, there are the same number of edges and also of 
 corners. 
 
 Conversely, if all three of the above conditions hold true of two 
 polyhedrons, then these polyhedrons are equal or similar, as the 
 case may be. 
 
 There are four important classes of similar solids : 
 (/.) All cubes and regular polyhedrons of the same name are 
 similar. 
 
 (//.) In order that two regular prisms or two regular pyramids 
 may be similar, they must have the same number of faces, and their 
 homologous lateral edges must be proportional to the homologous 
 sides of their bases. But the homologous lateral edges in the 
 prisms are obviously proportional to the heights, and in the pyra- 
 mids they are easily proved to be proportional to the heights, by 
 reasoning as in 249 ; and the homologous sides of the bases are 
 (by 167) proportional to the less radii of the bases. Therefore, 
 two regular prisms or pyramids with the same number of sides are 
 similar, if their heights are proportional to the less radii of their 
 bases. 
 
274-] CHAPTER xm. SURFACES AND VOLUMES OF BODIES. 301 
 
 (in.) By regarding the cylinder or the cone as the limit of the 
 inscribed regular prism or regular pyramid, when the number of 
 its sides is increased more and more ( 266, 269), it becomes 
 evident that two cylinders or two cones are similar, if their heights 
 ire proportional to the radii of their bases. 
 
 (iv.) All spheres are similar. 
 
 274. If the edge of a cube = a, the (total) surface = 6 a 2 , and 
 the volume a 3 . 
 
 Now, if we have other cubes whose edges are 2 a, 3 a, 4 a, etc., 
 their surfaces will be 4, 9, 16, etc., times that of the first cube, since 
 (2tf) 2 = 4# 2 , (3<?) 2 = 9# 2 , (4#) 2 = i6a 2 , etc.; and their volumes 
 will be 8, 27, 64, etc., times that of the first cube, since (2 a) s = 80 3 , 
 (3^) 3 = 27# 3 , (4) s = 6 4 # 3 , etc. 
 
 Bearing in mind that 4, 9, 16, etc., are the sqttares, and 8, 27, 
 64, etc., the cubes, of 2, 3, 4, etc., we see that in cubes the surface 
 is proportional to the square, and the volume proportional to the 
 cube, of the edge. 
 
 What has here been shown to be true of the surface and volume 
 of the cube in respect to its edges, can be shown to be true of 
 the surface and volume of all similar solids in respect to any 
 homologous lines (edges, heights, radii, diameters, etc.). 
 
 That is to say, 
 
 1. The surfaces of two similar solids are proportional to the 
 SQUARES of any two homologotis lines. 
 
 II. The volumes of two similar solids are proportional to the 
 CUBES of any two homologous lines. 
 
 Exercises. 1. Compare the surface and the volume of a cube whose 
 edge = 2 m with those of a cube whose edge = i m . 
 
 2. The edge of a cube = i m . Find the edge of a cube, (V.) having twice 
 the surface, (') having twice the volume of the first cube. 
 
 3. Required the depth of a cubical tank which will hold 64 times as much 
 as a similar tank whose depth = 3 m . 
 
302 GEOMETRY FOR BEGINNERS. [ 274. 
 
 . 4. Compare (i.) the surfaces, (zY.) the volumes, of the rectangular 
 parallelepipeds whose edges are a, b, c, and 3 a, 3 b, 3 . 
 
 5. Given two similar cylinders whose diameters are d and 4^, and whose 
 heights are h and 4/2. Prove that Laws I. and II., stated above in italics, hold 
 true of them. 
 
 6. Same exercise applied to two similar cones. 
 
 7 . Given two spheres with radii r and 5 r. Prove that the above laws hold 
 true of their surfaces and their volumes. 
 
 8. The edge of a polyhedron = 8 m , and the homologous edge of a similar 
 polyhedron = 5 m . How many times will the first polyhedron contain the 
 second ? 
 
 9. A box is 9O cm long, 58 cm wide, and 43 cm deep. Find the dimensions of 
 a similar box twice as large. 
 
 10. Find the volume of the box in the last exercise if its three dimensions 
 are doubled. 
 
 11. Find^its volume if one dimension is doubled. If two dimensions are 
 doubled. 
 
 12. I wish to make an iron cylinder, similar to a cylinder whose diameter 
 = 26 cm and whose height = 64 cm , but three times as large. What must be 
 its dimensions ? How many kilograms of iron are required ? 
 
 13. How far from the vertex of a cone 5o m high must a plane parallel to 
 the base be passed in order that the cone cut off may be exactly y^Vo f tne 
 entire cone ? 
 
 14. Divide into two equivalent parts, by a plane parallel to the base, a 
 cone 65 cm high, and having a base whose diameter = 30. 
 
 15. If the radius of a sphere = 37 cm , what will be the radius of a sphere 
 ten times as large ? 
 
 16. I have a cylindrical boiler 6o ft long and 8 ft in diameter, and I wish 
 another boiler similar in shape but f as large. What must be its dimensions ? 
 
 17. If the edge of a regular tetahedron = i m , it can be proved that the 
 surface = 1.732^, and the volume = o.i I78 cbm ; find the surface and the volume 
 if the edge = 7. 
 
 18. If the edge of a regular octahedron - i m , it can be proved that the 
 surface = 3.4641 l m , and the volume = 0.471 4 cbm ; find the surface and the 
 volume if the edge = 4 cm . 
 
275-] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 303 
 
 III. Exercises and Applications. 
 275. PRISMS AND CYLINDERS. 
 
 1. How many steres of wood in a cubical pile one edge of which = I2 m ? 
 
 2. Total surface of a cube = 8o*i cm ; find its volume. 
 
 3. Total surface of a cube = 5; find its volume. 
 
 4. Volume of a cube 2OO cbm ; find its surface. 
 
 5. Volume of a cube = V; find its surface. 
 
 6. A diagonal on one face of a cube = I2 cm ; find the volume of the cube. 
 
 7. Edge of a cube = 4 cm ; find the diagonal through the cube. 
 
 8. Edge of a cube = a ; find the diagonal through the cube. 
 
 9. A diagonal through a cube = 3O cm ; find the edge. 
 
 10. Find the edge of a cork cube which will weigh the same as l ccm of 
 iron. (See Table, page 298.) 
 
 11. Volumes of two cubes are as 8 : 27; what is the ratio of their edges ? 
 
 12. Find the edge of a cube half as large as a cube whose edge = i6 cm . 
 
 13. Edges of two cubes are 6 cm and I2 cm ; find the edge of a cube equal 
 to their sum. 
 
 14-. Edges of two cubes are I2 cm and i6 cm ; find the edge of a cube equal 
 to their difference. 
 
 15. Find the edge of a cubical tank which holds looo 111 of water. 
 
 16. A water tank has a rectangular base 3.2 m by 2.5. How many liters 
 of water does it contain when the height of the water i.24 m ? 
 
 17. A tank has a square base one edge of which = 3 m . How deep is the 
 water when the tank contains iSoo 1 ? 
 
 18. A cubical reservoir whose edge 24 contains how many hectoliters 
 of water when the water is 4 deep ? 
 
 19.- If water enters the above reservoir, through a tube, at the rate of I6O 1 
 per minute, in what time will it be full ? 
 
 20. If it is found that in 24 hours the reservoir loses by evaporation a 
 layer 2 cm in depth, how many liters does it lose ? 
 
 21. What time will be required to fill the reservoir, taking account of 
 evaporation ? 
 
 22. How large a field can be irrigated by the reservoir full of water if 
 8oo cbm of water are required for every hectare ? 
 
 23. If the dimensions of a rectangular parallelepiped are a, If, and c, find 
 the length of the diagonal from an upper corner to the lower opposite corner ? 
 
304 GEOMETRY FOR BEGINNERS. [ 275. 
 
 24. What is the cost of digging a cellar n.5 m long, 8.8 m wide, and 2.62 m 
 deep, at the rate of $0.80 per cubic meter ? 
 
 25. What is the weight of a marble block 2.75 m long, o.35 m wide, and 
 2.25 thick ? 
 
 26. A body weighs less in water than in air by exactly the weight of its 
 pwn volume of water. If a body i.2 m long, o.62 m wide, and 0.4 thick weighs 
 4Oo k in air, what will it weigh in water ? 
 
 27. Find the specific gravity of the body in the last exercise. 
 
 28. How much will a block of marble 2 m long, i.6 m wide, and o.8 m thick 
 weigh under water ? 
 
 29. Which will weigh the most in water, an iron cube whose edge = I2 cm 
 or a rectangular parallelepiped of lead I5 cm long, io cm wide, and 8 cm thick ? 
 
 30. In a room 6.2 m long, 4.75 wide, and 2.88 m high are 6 persons. In 
 what time will the air in the room become unfit for respiration if each person 
 breathes 20 times a minute, and at each breath vitiates 4OO ccm of air ? 
 
 31 What is the weight of the air in the above-mentioned room ? (Specific 
 gravity of the air = 0.0013.) 
 
 32. What must be the height of a box 2.5 m long and 2 m wide, that it may 
 hold 22.5 cbm ? 
 
 33. A rectangular vessel 48 cm long and 36 cm wide contains 56 liters of 
 water. What is the depth of the water ? 
 
 34. Into a corn-bin 8 m long and 5.2 wide, 56 hl of corn are put; find the 
 height. 
 
 35. The total surface of a cube 54 km ; find the sum of its edges. 
 
 36. Ten leaden cubes, each with an edge 2O cm , are melted into one 
 cube; find its edge. 
 
 37. If the dimensions of a trunk are i m , I5 m , and 2 m , find the dimensions 
 of a similar trunk that will hold eight times as much. 
 
 38. Find the dimensions of a similar trunk holding twice as much. 
 
 39. A wooden trough is to be made, whose inner dimensions are to be, 
 length 3.35, breadth i.6 m , depth i m ; the ends are to be io cm thick, the sides 
 8 cm , and the bottom I2 cm . (?'.) How many cubic metres of wood are required ? 
 (zY.) How many liters will the trough hold? If the trough is made of oak 
 wood, what will it weigh? (iii.') when empty? (iv?) when full of water? 
 
 40. What will it cost to dig a ditch around a field in the shape of a square, 
 containing 4 hectares, if the ditch is to be 2 m deep, i m wide at the bottom, 
 i.4 m at the top, and the price is $0.25 per cubic metre of earth thrown out? 
 
 41. How many cubic metres of manure are required to cover with a layer 
 2 cm thick a triangular field whose base = 840'" and altitude = 320"*? 
 
275-] CHAPTER XIII. - SURFACES AND VOLUMES OF BODIES. 305 
 
 42. If 78oo hl of gravel are spread upon a rectangular meadow 560 by 
 280, how much gravel is required to make a layer of the same thickness 
 upon a trapezoidal field whose parallel sides are 920 and 74O m , and alti- 
 tude 250? 
 
 43. Through an orifice in the side of a vessel 6 cm square, water flows at 
 the rate of i6 m per second. How many hectoliters will flow out in one 
 minute ? 
 
 44. Through a rectangular orifice I2 cm by 8 cm water flows at the rate of 
 2o m per second. How much water will flow out in one hour? 
 
 45. If one bushel of wheat makes 48 Ibs. of flour, and one barrel of flour 
 contains 196 Ibs., how many barrels of flour can be made from the contents of 
 a bin 2O ft long, io ft wide, and 6 ft deep, filled with wheat ? 
 
 46. How many tons of ice can be packed in a building 60** long, 40** wide, 
 and 30*"' high ? 
 
 47. How many tons of coal will a bin 2O ft long, i6 ft wide, and I2 ft deep 
 contain, allowing 4O cub ft to a ton ? 
 
 48. How many gallons of water will a cistern 8 ft deep, and 6 ft square at 
 the bottom, hold ? 
 
 49. A merchant imports 4O hl of wine. What must be the dimensions, in 
 feet, of a cubical tank which will just hold it ? 
 
 50. After a shower, I find a dish, placed in the open air, filled with water 
 to the depth of half an inch. How many tons of water have fallen upon one 
 square mile, the fall being supposed everywhere the same ? 
 
 51. How many bricks 8 in X 4 in X 2 in will be required to build a wall 50** 
 long, 3o ft high, and i6| in thick, allowing in in each dimension for the mortar? 
 
 52. How many cords of wood in a pile 60** long, 8 ft wide, and 7 ft high ? 
 NOTE. One cord of wood is a pile 8 ft long, 4 ft wide, and 4 ft high. 
 
 53. A stick of timber is I4 in X 18^ at the end. What length of it will 
 contain 30 cubic feet ? 
 
 54. Find the contents, in board feet, of a board i8 ft long, I5 in wide, and 
 2 in thick ? 
 
 NOTE. One board foot is i ft long, i ft wide, and i in thick. 
 
 55. Find the cost of 20 planks, each i6 ft long, i8 in wide, and 3 in thick, at 
 $2.00 per hundred feet, board measure. 
 
 56. How many feet of lumber, board measure, will be required to make a 
 box 6 ft X 4 ft X 3 ft , outside measurement, the thickness of the lumber to be 
 
 57. Find the volume of a 3-sided regular prism if a lateral edge = a side 
 of the base = 24. 
 
306 GEOMETRY FOR BEGINNERS. [ 275. 
 
 58. In a regular hexagonal prism the height = 3 times the less radius of 
 the base = 6o cm . Find the height of a similar prism, made of glass, which 
 weighs 6 k s. 
 
 59. How high must a regular hexagonal prism be to hold one hectoliter if 
 one side of the base = 64 ? 
 
 60. A cylindrical vessel, the area of whose base = 51, contains 75 hl of 
 water when full ; find the height of the vessel. 
 
 61. Give the dimensions of a cylindrical vessel which will hold 1000 liters. 
 Is this question determinate or indeterminate? Why? 
 
 62. Required the dimensions of a similar cylinder that will hold ^ as much? 
 
 63. If a vessel holding one hectoliter is a cylinder whose height is equal 
 to the diameter of the base, find the height. 
 
 64. A cylindrical cistern 345 1 " deep and i.i m wide is to be lined with 
 stones, each of which, mortar included, measures I2 cm by 6 cm . How many 
 stones are required? 
 
 65. A cylindrical boiler is 58 m long and has a diameter of 0.9. Find the 
 area of the convex surface exposed to the fire, if this surface intersects the 
 circular end of the boiler in an arc of 130. 
 
 66. If a glass cylinder o.35 m in diameter and i m in height is filled with 
 water, and the water is then poured into another cylinder o.8 in in diameter, 
 how high will the water rise in the latter vessel? 
 
 67 . Out of a cylindrical reservoir 8.8 m in diameter water flows at the rate 
 of 2 1 per second. Through what distance will the surface of the water fall in 
 one hour? 
 
 68. About the convex surface of a cylinder o.8 m in diameter and i.2 m high, 
 a cord 2 mm in diameter is wound until the surface is completely covered. 
 How many meters of cord are required? 
 
 69. A well is to be I5 m deep and 2. 3 in diameter. The price for digging 
 it is to be $0.40 per cubic meter for the first three meters of depth, $0.25 more 
 for the next three meters, and so on, $0.25 being added to the price for each 
 successive three meters of depth. Find the cost of digging the well. 
 
 70. A cast-iron cylinder 3 long and 2O cm in diameter is reduced in a 
 turning lathe to a diameter of i8 cm . Find the loss in weight. 
 
 71. The horizontal section of a bath-tub i m high is an ellipse whose axes 
 are i.8 m and o.64 m . How many liters of water will it hold? (See 221.) 
 
 72. How large a cylinder can be made by rolling up a rectangular sheet 
 of zinc 8o cm by 6o cm , if the height of the cylinder is 8o (;m ? How large will 
 the cylinder be if the height is 6o cm ? 
 
 73. An iron cylinder 8o cm long and weighing 56 k s is bored through the 
 centre until of the iron is removed; find the thickness of what is left. 
 
276.] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 307 
 
 74. A cylindrical vessel 4 high and i.6 m in diameter is half-full of water. 
 Through an orifice at the bottom 4 cm in diameter the water is running out at 
 the rate of 6 m a second, and through a pipe 2 cm in diameter it is running into 
 the vessel at the rate of I2 m a second; find the level of the water after one- 
 quarter of an hour. 
 
 75. What must be the diameter of a supply-pipe, for the vessel in the last 
 exercise, which will supply water as fast as it runs out? 
 
 76. What must be its diameter in order that the vessel may be filled in 
 ten minutes? 
 
 77. What must be its diameter in order 
 that the vessel may be emptied in one hour? 
 
 78. The piston of a pump is 36 in diam- 
 eter, and moves through a space of 5O cm . 
 How many liters of water are thrown out by 
 100 strokes? 
 
 79. When a body is placed under water, 
 in a cylinder 6o cm in diameter, the level of the 
 water is observed to rise 30 {Fig. 295) ; find 
 the volume of the body. 
 
 80. How much will a brass cylinder weigh 
 under water if the height = 64 cm and the diam- 
 eter of the base = 40? (See Exercise 26.) 
 
 Fig. 295. 
 
 276. PYRAMIDS AND CONES. 
 
 81. In a regular hexagonal pyramid a side of the base = i m , and the height 
 of the pyramid^ i.8 m ; find (z.) the total surface; (z'z.) the volume. 
 
 82. The roof of a summer-house has the shape of a regular octagonal 
 pyramid ; its height = 15**, and a side of its base = 9 ft . Find the cost of the 
 boards, i in thick, necessary to make it, at 2 cents a board foot. 
 
 83. A vat has the shape of an inverted frustum of a square pyramid; a 
 side of the upper base = 4, a side of the lower = 2 m . and the depth of the 
 vat = 5. How much water will it hold? 
 
 84. The pyramid at Ghiza, near Memphis in Egypt, is about 144 high; 
 the base is a square whose side = 187, and the top also a square whose side 
 = 3.7. If the pyramid were solid, what would be its volume? 
 
 85. Divide a pyramid 24'" high into two equal parts by a plane parallel 
 to the base. 
 
 86. Find (z.) the total surface, (zY.) the volume, of a regular tetrahedron 
 whose edge = 8o cm . 
 
308 GEOMETRY FOR BEGINNERS. [ 276. 
 
 87. The base of a pyramid is a square, and its sides are equilateral tri- 
 angles ; find its volume if a side of the square = 4O cm . 
 
 88. The edge of a regular octahedron 2 m ; find (z.) its surface; (zz.) its 
 volume; (zYz.) the ratio of its volume to that of a cube having the same edge. 
 
 89. A hill, having the form of a cone, is to be removed, and the. base 
 divided into building-lots. The slant height of the hill= 120.8, and the cir- 
 cumference of Jhe base = 750. (z.) How many cubic meters of earth must 
 be removed? (zz.) If the base is divided into six equal lots, how large will 
 each be? 
 
 90. The height of a cone = 86 cm , and the slant height is twice as much; 
 find (z.) the area of the base; (z'z.) the volume of the cone. 
 
 91. A tower has a conical roof 5.8 m high and 12.6 in diameter; find its 
 area. 
 
 92. Three leaden cones, each 4O cm high, and having for diameters I2 cm , 
 24 cm , and 36, respectively, are melted together and cast into one having the 
 same height; find its diameter. 
 
 93. If the three cones of the last exercise had been cast in the shape of a 
 cube, what would be the edge of the cube. 
 
 94. How far from the base must a cone of sugar, 54 cm high and i8 cm in 
 diameter, be cut in two, parallel to the base, in order that the parts may be 
 equal? 
 
 95. Show how the cone of the last exercise may be divided by planes par- 
 allel to the base into three equal parts. 
 
 96. Out of a circular piece of sheet-tin 86 cm in diameter a sector with the 
 angle 150 is cut, and this is then rolled into the shape of a cone; find the 
 volume of this cone. 
 
 97. Explain how you would find the height of a cone having given the 
 circumference of the base and the slant height ? 
 
 98. I wish to find the value of a straight pine tree that tapers to a point. 
 the circumference of the base being 4 m . The tree casts a shadow 30 Ion ; 
 at the same time that the shadow of a vertical rod i.2 m long is i.G" 
 What is the tree worth at $8 per cubic meter (branches, etc., being regarded 
 as worth the price of felling and trimming) ? 
 
 99. A granite column has the shape of the frustum of a cone; the base is. 
 2.i2 m in circumference, the top i.5 m , and the height = 5 m . Find its weight. 
 
 100. A vessel holding I2 1 has the form of the frustum of a cone; the 
 lower base is 28 cm in diameter, and the upper base 24'"; find its height. 
 
 101. How much sheet tin is required to make a speaking-tube i.42 m long, 
 and 43 cm and 36 cm in diameter at the ends? 
 
277-] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. ' 309 
 
 102. How much metal is required to make a tin can, the diameter of the 
 bottom being 36 cm , that of the open top i6 cm , and the height being 48 cm ? 
 
 103. Assuming the cask {Fig. 296} to consist of tw^o frustums of cones 
 placed with their greatest bases together at the 
 
 middle, find its volume, having given that the di- 
 ameter of the middle section = 76, the diameter 
 of each end = 66 cm , and length = i m . Will the j 
 answer obtained be too large or too small? 
 
 104. A more accurate value for the volume of 
 
 a cask may be found by use of the formula, pjg % 2 g6. 
 
 Volume = i TT h (r 1 + 2 ^), 
 
 where h the length, r radius of one end, R radius of the greatest 
 section. Find the volume of the preceding cask by means of this formula. 
 
 105. How many bottles of wine, each holding 0.6^, can be filled from 
 a cask in which the greatest and the least diameters are 86 cm and 62 cm , respect- 
 ively, and the length i.5 m ? 
 
 106. How many gallons will a cask hold whose length is 3^, and greatest 
 and least circumferences I2 ft and 9 ft , respectively? 
 
 277. THE SPHERE. 
 
 107. How many bullets 15 in diameter can be cast from io k s of lead? 
 
 108. The English national debt is about $4,500,000,000. What would be 
 the diameter of a sphere of gold having the same value, assuming i k s of gold 
 as worth $720? 
 
 109. The diameter of the sun is about 113 times that of the earth. Find 
 the volume of the sun, assuming the diameter of the earth to be 8000 miles. 
 
 110. Eighty bullets, equal in size, are thrown into a cylindrical vessel 
 68 cm in diameter containing water. If the level of the water rises 2 cm , find 
 the diameter of a bullet. 
 
 111. A body sinks in water till the weight of the water displaced weight 
 of the body. How heavy is a sphere o.2 m in diameter which floats half under 
 water? 
 
 112. Find the weight of a wooden sphere 0.36 in diameter if one-sixth 
 of its bulk is below the water-level when it floats on water. 
 
 113. If a wooden sphere 2O cm in diameter is reduced to one-half its 
 original volume in a lathe, what will then be its diameter? 
 
310 GEOMETRY FOR BEGINNERS. [ 278. 
 
 114. Find the diameter of a sphere made by melting together three 
 spheres whose diameters are i.2 m , o.8 m , and o.4 m respectively. 
 
 115. A sphere o. m in diameter is reduced in size by turning till its 
 diameter is one-twelfth less than at first. How much is the volume reduced? 
 
 116. The outer circumference of a hollow sphere = i.4 m , the thickness 
 =r 24 mm . Find the interior volume. 
 
 117. An arch in the form of a hemisphere has an inner diameter of 
 4.8, and its thickness 0.7. Find how many cubic meters of stone it 
 contains. 
 
 118. What is the value of a hemispherical copper kettle i m inside diame- 
 ter, if the thickness of the copper = i cm , and copper is worth $0.75 per 
 kilogram ? 
 
 119. The diameter of a balloon = 30; the material of which it is made 
 weighs o.5 k s per square meter; and the car attached below weighs 40^. Find 
 its ascensional force if air weighs 1.38 per liter, and the balloon is filled with 
 coal gas which is half as heavy as air. 
 
 Hint. The ascensional force = weight of the air displaced by the balloon 
 weight of the balloon, the car, and the gas in the balloon. 
 
 120. What must be the outer diameter of a hollow sphere 25 mm thick, in 
 order that it may hold just 60 liters? 
 
 121. How many bodies as large as the moon could be made from the 
 earth, taking the diameter of the latter body as four times that of the former? 
 
 122. If a drop of soap and water 4 cm in diameter is blown into a bubble 
 I5 cm in diameter, find the thickness of the bubble. 
 
 123. The two frigid zones contain 0.082 of the earth's surface. What is 
 the height of each zone? What is the diameter .of the Arctic circle? 
 
 278. EQUIVALENT SURFACES AND SOLIDS. 
 
 124. A cube and a sphere have the same volume; which has the greater 
 surface? 
 
 125. A cube and a sphere have the same surface; which has the greater 
 volume ? 
 
 The radius of the base of a cone = io cm , the height of the cone 2O cm . 
 Find, 
 
 126. The edge of an equivalent cube. 
 
 127. The height of an equivalent cylinder, etc., having the same base. 
 
 128. The radius of an equivalent sphere. 
 
278.] CHAPTER XIII. SURFACES AND VOLUMES OF feODIES. 31 1 
 
 129. How high must a right prism be, whose base is a rectangle i6 cra 
 X I2 cm , that it may be equivalent to a cube whose edge = io cm ? 
 
 130. A cube whose edge = io cm is cut into two equal parts by a plane 
 parallel to the base, and the parts are then placed with the half sides together 
 so as to form a prism. Compare this prism and the cube as regards (z.) vol- 
 umes ; (zY.) surface ; {in.} bases ; (z'?y.) heights. 
 
 Note i. Equivalent prisms have not always equivalent surfaces. 
 
 Note 2. In equivalent prisms (also cylinders, pyramids, and cones), the bases 
 are inversely proportional to the heights (that is, the less the base fas greater the 
 height, and conversely, so that the product of the two factors, base and height, 
 shall be always equal). 
 
 131. In two equivalent cylinders the ratio of the bases is 3: 5. What is 
 the ratio of the heights ? 
 
 132. The dimensions of a cylinder are : diameter of the base = 7 cm , height 
 _ i2 cm . Find the height of an equivalent cylinder, the diameter of its base 
 being 6 cm . 
 
 133. Compare the surfaces of the two bodies in Exercise 129. Which is 
 the greater, and by how much ? 
 
 Note. Among all equivalent four-sided right prisms the cube has the least 
 surface. 
 
 134. A cube with an edge of 8 cm has the same surface as a right prism 
 whose base is a rectangle 6 cm X 4 cm ; compare their volumes. Which is the 
 greater, and by how much? 
 
 Note. Among all four-sided right prisms with equivalent surfaces the cube 
 has the greatest volume. 
 
 135. The base of a square pyramid is 6 cm X 6 cm , and the height = I2 cm ; 
 find the height of an equivalent prism with an equal base. 
 
 136. Transform a cylinder io cm in diameter and I2 cm high into an equiva- 
 lent square pyramid with a base 8 cm X 8 cm . 
 
 137. Transform a cube with an edge of 8 cm into an equivalent cylinder 
 the diameter of whose base shall be io cm ; then compare the surfaces of the 
 two bodies. 
 
 138. Transform the above cube into a cylinder with an equivalent surface, 
 the diameter of its base to be io cm ; then compare the volumes of the two 
 bodies. 
 
 139. Transform a cylinder, in which the diameter of the base and the 
 height are each equal to io cm , into an equivalent cylinder having a height of 
 i6 cm ; then compare the surfaces of the two cylinders. 
 
 Note. Among equivalent cylinders that has the least surface in which the diam- 
 eter of the base and the height are equal. 
 
312 GEOMETRY FOR BEGINNERS. [ 279. 
 
 140. Slant height of a cone = 8 cm , and diameter of its base = 6 cm ; find the 
 height of a cylinder 4 cm in diameter, having an equivalent convex surface. 
 
 141. Transform a sphere 6 cm in diameter into an equivalent cylinder, the 
 diameter of its base to be the same as that of the sphere ; then compare their 
 surfaces. 
 
 142. A leaden sphere 9 cm in diameter is recast in the shape of a cube; 
 find the edge of the cube. 
 
 143. Another sphere of the same size is recast in the form of a cylinder 
 5 cm in diameter; find the height of the cylinder. 
 
 144. What must be the height of a cylinder whose base is 4 cm in diame- 
 ter if it has the same surface as that of a sphere 4 em in diameter? 
 
 145. A sphere is io cm in diameter, and the base of an equivalent cylinder 
 has the same diameter ; compare the surfaces of the two bodies. 
 
 Note. Among all equivalent bodies the sphere has the least surface. 
 
 146. A sphere and a cylinder have equivalent surfaces; the sphere is 
 lo cm in diameter, and so likewise is the base of the cylinder ; compare the 
 volumes of the two bodies. 
 
 Note. Among all bodies having equivalent surfaces the sphere has the greatest 
 volume. 
 
 279. RATIOS OF SURFACES AND OF SOLIDS. 
 
 147. Two prisms have bases each containing 36 ( i m ; what is the ratio of 
 their volumes if their heights are io cm and I5 cm ? 
 
 148. Compare in general the volumes of two prisms if they have equiva- 
 lent bases. 
 
 149. Compare in like manner the volumes of two cylinders, two pyramids, 
 and two cones, with equivalent bases. 
 
 150. The volumes of two prisms with equivalent bases are 36o cbm and 
 56o cbra ; what is the ratio of their heights? If the height of one is ay ' 11 , find 
 that of the other. 
 
 151. A cone 8 cm high is reduced in size till the height is 3.5 cm , the base 
 remaining the same. What fractional part of the whole is left? 
 
 152. Into the ends of a cylinder I2 cm long two cone-shaped boles 3"" 
 deep are bored. If the cylinder weighed at first 6ooS, what will it weigh after 
 the holes are made? 
 
 153. If a cylinder and a cone have equal bases, what must be the ratio of 
 their heights in order that their volumes may be equal? 
 
 154. The heights of two prisms are each 2O cm ; compare their volumes if 
 the base of one is four times that of the other. 
 
279-] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 313 
 
 155. Compare in general the volumes of two prisms, two cylinders, two 
 pyramids, or two cones, having equal heights. 
 
 156. A cone 6 cm iu diameter is turned in a lathe until the diameter is re- 
 duced to 5 cm , the height remaining unchanged; what part of the whole cone 
 is left? 
 
 157. If the above is turned till it is reduced to one-fourth its original vol- 
 ume, what will be its diameter? 
 
 The edge of a cube = io cm ; find, 
 
 158. Volume of the largest cylinder that can be made from it. 
 
 159. Volume of the largest sphere. 
 
 160. Volume of the largest cone. 
 
 161. Compare the volumes of the above cube, cylinder, sphere, and cone. 
 The diameter of a cylinder = its height = io cm ; find, 
 
 162. Volume of the largest sphere that can be turned from it; 
 
 163. Volume of the largest cone. 
 
 164. Compare the volumes of the above cylinder, sphere, and cone. 
 
 165. What part of a hemisphere is the largest cone that can be made 
 from it? 
 
 166. If the diameter and the height of a cylinder are equal, compare the 
 convex surface with the areas of the bases. 
 
 167. If the diameter and the height of a cone are equal, compare the 
 convex surface with the area of the base. 
 
 168. If the convex surface of a cylinder sum of areas of the bases, com- 
 pare the height with the radius of the base. 
 
 169. I want a cylinder whose convex surface shall be six times the area of 
 the base. Find the ratio of the height to the radius of the base. 
 
 170. What relation must exist between the height of a cone and its diam- 
 eter in order that the convex surface may be equal in area to the base? 
 
 171. Compare the convex surfaces of a cylinder and a cone having equal 
 bases and heights. Compare, also, their total surfaces. 
 
 172. Compare the surfaces of two spheres whose diameters are 6 cm and 
 7 cm . What, in general, is the relation between the surfaces of spheres differ- 
 ing in diameter? 
 
 173. A cylinder and a hemisphere have equal bases and heights. Com- 
 pare (z.) their convex surfaces ; (z'z.) their total surfaces ; (m.) their volumes. 
 
 174:. The diameter of a cylinder = its height = io cm ; find the volume of 
 the largest rectangular parallelopiped that can be made from it. 
 
 175. Compare the volumes of a sphere and the largest cube that can be 
 made from it. 
 
3 14 GEOMETRY FOR BEGINNERS. 
 
 REVIEW OF CHAPTER XIII. 
 
 I. Surfaces of Bodies. 
 
 253. Surface of a prism. 
 
 254. Surface of a cylinder. 
 
 255. Surface of a pyramid. 
 
 256. Surface of a cone, 
 
 257. Surface of the frustum of a cone. 
 
 258. Surface of a sphere. 
 
 II. Volumes of Bodies. 
 
 259. Definition of Volume. Units of volume. 
 
 260. Measurement of volumes. Analysis of the different cases to be 
 
 considered. 
 
 261. Volume of the rectangular parallelepiped. 
 
 262. Volume of the right parallelepiped. 
 
 263. Volume of the oblique parallopiped. 
 
 264. Volume of the triangular prism. 
 
 265. Volume of any prism. 
 
 266. Volume of the cylinder. 
 
 267. Volume of the triangular pyramid. 
 
 268. Volume of any pyramid. 
 
 269. Volume of the cone. 
 
 270. Results of 261-269 summed up in two rules. 
 
 271. Volume of a sphere. 
 
 272. Relation between volume and weight. 
 
 273. Equivalent solids, similar solids, equal solids. 
 
 274. Two laws applicable to similar solids. 
 
 III. Exercises and Applications. 
 
 275. Prisms and cylinders. 
 
 276. Pyramids and cones. 
 
 277. The sphere. 
 
 278. Equivalent surfaces and solids. 
 
 279. Ratios of surfaces and solids. 
 
THIS BOOK IS DUE ON THE LAST DATE 
 STAMPED BELOW 
 
 AN INITIAL FINE OF 25 CENTS 
 
 WILL BE ASSESSED FOR FAILURE TO RETURN 
 THIS BOOK ON THE DATE DUE. THE PENALTY 
 WILL INCREASE TO SO CENTS ON THE FOURTH 
 DAY AND TO $1.OO ON THE SEVENTH DAY 
 
 OVERDUE.. 
 
 MAR 24 1934 
 
 
 ft PR p 1934 
 
 
 
 
 ... 
 
 
 *^* I3df 
 
 
 MAY 1 3 1941 M 
 
 
 
 
 $ 
 
 
 .... 
 
 
 - VWov'5'D? 
 
 
 ^ 
 
 
 Aoftk \ 
 
 
 ^& ' 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 LD 21-100>?i-7,'33 
 
YB 17293 
 
 800542 
 
 
 UNIVERSITY OF CALIFORNIA LIBRARY