UC-NRLF SB E7fl LIBRARY OF THE UNIVERSITY OF CALIFORNIA. GIFT OF 3.r.jX.r...^ Class \^M\^(^^^^ LESSONS IN GEOMETKY FOR THE USE OF BEGINNERS. BY G. A. HILL, A.M., AUTHOR OF A GEOMETRY FOR BEGINNERS. BOSTON: PUBLISHED BY GINN & COMPANY. 1888. Entered, according to Act of Congress, in the year 1887, by G. A. HILL, in the office of the Librarian of Congress, at Washington ELECTROTYPED BY J. 8. CUBBING & Co., BOSTON. BY GlNN & Co., BOSTON. PREFACE. THIS book has been prepared to meet the wishes of those who prefer a shorter and easier introductory course in Geometry than that given in the " Geometry for Beginners." The metric system of units is explained; but all exercises in metric units are confined to lessons placed at the ends of the chapters, and may be omitted if desired. The method of instruction is that best adapted to the mental condition of pupils between the ages of twelve and sixteen. The training in consecutive reasoning is introduced very gradually, and is confined mainly to the laws of equal triangles and a few of their simple applications. To one feature of the method the author desires to call spe- cial attention ; namely, the numerous exercises which involve the use of instruments and drawing to scale. It is assumed that every pupil is provided with ruler, divided scale, pencil compasses, triangle, and protractor. Any objection on the ground f of expense has been met by the publishers, who are prepared to supply, at a very low price, these instruments enclosed in a strong j wooden box. Experience shows that for the beginner of Geometry the care- ful execution of easy constructions is the most useful as well as the most interesting part of the daily lesson. This work calls into action the eye, the hand, and the judgment. It holds the attention. It is attended with the pleasing sense of the successful exercise of new-found power. Under these conditions progress in knowledge is sure to be rapid. The precept, " Do that you may know," finds here a pertinent application. 183667 IV PREFACE. No teacher, however, can expect to obtain the best results from this kind of work, unless he insists strenuously upon neatness and a reasonable degree of accuracy. Every teacher should lay down a standard of neatness and accuracy, remembering that his pupils are only beginners working with cheap instruments, and then should criticise unsparingly every drawing which, tried by his standard, is a slovenly or inaccurate piece of work. The contents of the book are so arranged that the course may be considerably abridged, if so desired. There are, in all, ninety- six lessons and fifteen drawing exercises. This makes abundant material for a course of three hours per week for a year, or, what is better for the pupil, a course of one hour per week for a year, and a course of two hours per week for the year follow- ing. The last two chapters, however, may be omitted, and like- wise the lessons in which metric units are used ; there will then be left sixty-five lessons and the drawing exercises, or a course of two hours per week for one year. Geometry, as here presented, should be studied before Algebra. If this is done, pupils, while learning the properties of figures and the measurement of areas and volumes, will see for themselves the great advantage of using letters to represent quantities. Thus the chief stumbling-block to every beginner of Algebra will be removed. The author takes this opportunity to express his warm thanks to the teachers who have had the kindness to read and correct the proof-sheets. A special acknowledgment of obligation for this assistance is due to Prof. G. A. WENTWORTH, of Exeter, N.H.; Prof. H. D. WOOD, of Trenton, Ga. ; Mr. J. E. CLARKE, of Chelsea, Mass. ; and Mr. E. H. NICHOLS, of Cambridge, Mass. G. A. HILL. CAMBRIDGE, Feb. 29, 1888. CONTENTS. CHAPTER I. INTRODUCTION. LESSON PAGE 1. Observation of a cube and a cylinder 1 2. Questions about prisms, pyramids, etc 4 3. Body, surface, line, point 5 4. Different kinds of lines, surfaces, and bodies 8 CHAPTER II. STRAIGHT LINES. 5. Determination of straight lines. Use of the ruler 10 6. Vertical, horizontal, and inclined lines 12 7. The circle. Use of the compasses 15 8. Parallel lines, and their construction 18 9. Equal and unequal lines. Axioms of Geometry 21 10. Addition, subtraction, etc., of lines .24 11. Units of lengths 27 12. Measurement of straight lines 29 18. Drawing to scale 30 14. The metric units of length 32 15. Drawing and measuring with metric units 34 16. Review of Chapter II. begun 35 17. Review of Chapter II. concluded 36 CHAPTER III. ANGLES. 18. Definitions. The magnitude of an angle 37 19. Erecting perpendiculars 40 20. Dropping perpendiculars 42 21. Measurement of angles. Use of the protractor 43 22. Construction of angles. Bisecting an angle 46 23. Adjacent angles and vertical angles 48 24. Two parallel lines cut by a third line 50 VI CONTENTS. LESSON PAGE 25. Review of Chapter III. begun 52 20. Review of Chapter III. continued 53 27. Review of Chapter III. concluded (metric units) ..... 54 CHAPTER IV. TRIANGLES. 28. Definitions. Dimensions of a triangle 55 29. Construction of triangles. Practical applications .... 58 30. Construction of triangles. Practical applications .... 60 31. Sum of the angles of a triangle 62 32. Equivalence, similarity, and equality 64 33. Theorems. Method of equal triangles 67 34. Theorems 70 35. Theorems 72 36. Review of Chapter IV. begun 74 37. Review of Chapter IV. continued 75 38. Review of Chapter IV. continued 7(5 39. Review of Chapter IV. continued (metric units) 77 40. Review of Chapter IV. concluded (metric units) 78 CHAPTER V. POLYGONS. 41. Definitions. Sum of the angles of a polygon 79 42. Dimensions of quadrilaterals. Constructions 82 43. Parallelograms. Theorems and constructions 84 44. Parallelograms. Theorems and constructions 86 45. Polygons. Miscellaneous constructions 88 46. Regular polygons 90 47. Review of Chapter V. begun 92 48. Review of Chapter V. continued 93 49. Review of Chapter V. concluded (metric units) 94 CHAPTER VI. THE CIRCLE. 50. Definitions. Chords, arcs, and angles at the centre .... 95 51. Radius perpendicular to a chord. Constructions 98 52. Inscribed angles 100 53. Tangents 102 54. Two circles. Circumscribed and inscribed figures .... 104 CONTENTS. Vii LESSON PAGE 55. Centre of a regular polygon. Constructions 106 56. Circumference of a circle. Meaning of w 108 57. Review of Chapter VI. begun 110 58. Review of Chapter VI. continued Ill 59. Review of Chapter VI. concluded (metric units) 112 CHAPTER VII. AREAS. 60. Units of area. Area of a square 113 61. Area of a rectangle 116 62. Area of a parallelogram 118 63. Area of a triangle 120 64. Area of a polygon 122 65. Area of a circle 124 66. Theorem of Pythagoras - 126 67. Mean proportional. Transformation of figures 128 68. Transformation of figures 130 69. Review of Chapter VII. begun 132 70. Review of Chapter VII. continued 133 71. Review of Chapter VII. continued 134 72. Review of Chapter VII. continued 135 73. Review of Chapter VII. continued (metric units) .... 136 74. Review of Chapter VII. concluded (metric units) .... 138 CHAPTER VIII. RATIOS. 75. Definitions. Triangles and parallelograms compared . . . 139 76. Pivision of figures into parts 142 77. Proportions. Application to the circle 144 78. Similar triangles. Ratio of similitude 146 79. Applications. Areas of two similar triangles 148 80. Numerical relations in right triangles 150 81. Review of Chapter VIII. begun 152 82. Review of Chapter VIII. continued 153 83. Review of Chapter VIII. concluded (metric units) .... 154 CHAPTER IX. SOLIDS. 84. Planes and dihedral angles 155 85. The cube 158 80. The rectangular solid 160 Vlll CONTENTS, LESSON PAGE 87. The right prism 162 88. The right cylinder 164 89. The right pyramid . 100 90. The right cone . .' 108 91. The sphere 170 The sphere concluded 172 Review of Chapter IX. begun 174 92. 93. 94. Review of Chapter IX. continued 175 95. Review of Chapter IX. continued (metric units) 170 90. Review of Chapter IX. concluded (metric units) 178 Drawing Exercises 179 SIGNS AND ABBREVIATIONS. +, increased by. , diminished by. X, multiplied by. -h, divided by. =, is (or are) equal to. >, is (or are) greater than. <, is (or are) less than. II, parallel. _L, perpendicular. y' , the square root of. Z, angle. A, angles. A, triangle. A, triangles. Ax., axiom. T)ef., definition. Hyp., hypothesis. Con., conclusion. Const., construction. rt., right. LESSONS IN GEOMETRY CHAPTER I. INTRODUCTION. Lesson 1. 1. Why is this block of wood (Fig. 1) called a body ? Why is it also called a cube? The block is called a body because it occupies space, or, in other words, "takes up room." Everything which occupies a portion of space is called a body. The block is called a cube on account of its shape. Every bod\' having the shape of this block is called a cube. 2. Point out and name its dimensions. The block can be measured in three main directions : (1) From left to right ; (2) From front to back ; (3) From top to bottom. These three measurements are called the dimensions of the block ; one is FlG - called the length, another the breadth, the other the thickness. The dimension which is measured straight upwards or downwards is also called the height of the block. 2 LESSONS IN GEOMETRY. 3. Are the dimensions of a cube equal or unequal ? The three dimensions of this cube (Fig. 2) seem to be equal. If we measure them with a foot rule, we shall find that they are equal. The dimensions of a cube are equal. 4. Describe the surface of the cube. The cube is limited or bounded by its surface. When we look at the cube, it is its surface only which we see. When we handle the cube, it is its surface only which we touch. The surface of the cube con- sists of six distinct parts, called faces: The front and the back faces ; The right and the left faces ; FIG. 2. The upper and the lower faces. The faces have no thickness ; each face has only two dimensions, length and breadth, or length and height. The faces of the cube are flat or plane surfaces. 5. Describe the edges of the cube. Each face of the cube is bounded by four edges. Each edge is the place where two of the faces meet. There are in all twelve edges. The edges of the cube are lines; they have neither breadth nor thickness ; they have only one dimension, length. The edges of the cube are straight lines. 6. Describe the corners of the cube. Each edge of the cube is limited by two corners. Each corner is the place of meeting of three edges. There are in all eight corners. The corners of the cube are points; they have no dimen- sions, neither length, breadth, nor thickness. They have only place or position. INTRODUCTION. 3 FIG. 3. 7. The body represented in Figure 3 is called a cylinder. Point out and name its dimensions. The cylinder has three dimensions. The di- mension measured straight upwards or down- wards is called the length or height. The other dimensions are called the breadth and thickness ; the}' are measured, one of them from left to right, the other from front to back. The breadth is not everywhere the same ; neither is the thickness. But in such cases we always choose the greatest breadth and the greatest thickness. The greatest breadth and the greatest thickness are equal, as we can see by measuring them with a foot rule. 8. Describe the surface of the cylinder. The cylinder is bounded on all sides by its surface. The surface has no thickness ; it has only two dimensions, length and breadth. The surface consists of three parts : two flat or plane sur- faces, called the bases ; and a surface extending all around the cylinder between the bases, called the convex surface. The convex surface is not flat ; it is a curved surface. 9. Describe the edges of the cylinder. The cylinder has two edges. These edges are the bounda- ries of the bases. They are also the places where the lateral surface meets the bases. The edges are lines, and have one dimension, length. These edges are curved lines. The cylinder has no corners. 10. Measure with a foot rule the dimensions of the cylin- der w T hich is given you. 4 LESSONS IN GEOMETRY. Lesson 2. 1. The body represented in Figure 4 is called a prism. Point out and name its dimensions. Are they equal or unequal? Which is the longest? Which the shortest? Describe carefully the surface, the edges, and the corners, as those of the cube were described in Lesson 1. 2. Examine in like manner the four-sided prism (Fig. 5). 3. Examine in like manner the six-sided prism (Fig. 6). 4. Examine in like manner the four-sided pyramid (Fig. 7) . 5. Examine in like manner the cone (Fig. 8). 6. Examine in like manner the frustum of a cone (Fig. 9). 7. Examine in like manner the sphere (Fig. 10). FIG. 8. FIG. 9. INTRODUCTION. 5 - Lesson 3. 1, What is a bndy. as understood in Geometry ? A body, in common language, is a portion of space filled with something which we can see, touch, handle, etc. ; for example : a pencil, a book, a stone, the sun. But in Geome- try we pa} 7 no attention to the matter of which a bod}' is composed ; we study simply its size and its shape. In Geometry, a body means simply a limited portion of space. Geometrical bodies are sometimes called solids, 2, How many dimensions has a body? How are they named? Every body has three dimensions. The words used for naming dimensions are length, breadth (or ividth) , thickness, height, and depth. Usually the longest dimension is called the length, and the shortest the thickness. The terms height and depth are used only for dimensions measured straight upwards or downwards. 3. Define the terms surface, line, and point. A surface is a space magnitude having only two dimen- sions. A line is a space magnitude having only one dimension. A point is a position in space without magnitude. The boundaries of a body are surfaces. The boundaries of a surface are lines. The limits of a line are points. 4. Define the intersection of two surfaces. When two surfaces meet or cut each other, the place where they meet is called their intersection. The intersection of two surfaces is ahvays a line. For example : the faces of the cube intersect in the edges. 6 LESSONS IN GEOMETRY. 5, Define the intersection of two tines. The place where two lines meet or cut each other is also called their intersection. The intersection of two lines is always a point. For example : the edges of the cube intersect at the cor- ners ; two lines on paper which meet always meet in a point. 6, What are the parts of bodies, surfaces, and tines ? If we divide a body into parts, each part, however small, will also be a body having length, breadth, and thickness. In this way we never could obtain a part so small that it was a surface, a line, or a point. The parts of a body are bodies; the parts of a surface are surfaces; and the parts of a line are lines. Therefore the surface of a body is not a part of the body. If any number of flat surfaces were placed one upon another, they would have no thickness, and would form but a single surface. The common boundary which separates water from oil resting on it is neither water nor oil nor any other sub- stance ; it is not a body, but a surface. Likewise a line is not a part of a surface, and a point is not a part of a line. 7, How are points and tines represented to the eye ? Strictly speaking, a point can- not be seen, because it has no magnitude. On paper we repre- sent a point by a small dot, and name it by a letter, as A. The dot is really a very small body, but we do not regard its dimen- sions, and think of it as having position only. INTRODUCTION. 7 A line is represented by a narrow mark, and named generally by two letters, one placed at each end of the line. Surfaces are represented by their bounding lines, and solids by their bounding surfaces. 8. What is the path of a point ? When we move the point of a pencil on paper, it leaves be- hind a trace which we call a line ; imagine the point of the pencil to be a point as understood in Geometry, and then the trace or path of the pencil would be a geometrical line. Who has not observed that if the red-hot point of an iron rod be moved rapidly in the dark, the path of the point appears as a luminous line ? The path of a moving point is a line. 9. Can you think of a body shaped like a cube ? And one shaped like a cylinder? And one shaped like a sphere? 10. Is the space within the room a body? Why? 11. Name the dimensions of this room ; the dimensions of the floor ; the dimensions of one of the walls. 12. Name the dimensions of a pencil ; a cistern ; a brick wall. 13. Choose among the bodies before you one which has two of its dimensions equal. Which are they ? 14. Point out upon a cube the intersection of two sur- faces. Also the intersection of two lines. 15. How many edges and how many corners has a square box? 16. In common language, a telegraph wire is called a line. Is it a geometrical line ? What is it ? Why ? 17. We call the trace of a pencil-point on paper a line. What is it in reality ? Why ? * />& 8 LESSONS IN GEOMETRY. Lesson 4. 1. Divide lines into two classes. Define each class. There are two kinds of lines, straight and curved. A straight line has the same direction from end to end. A curved line is a line whose direction is continually changing. Curved lines are often called curves. A moving point describes a straight line when the direction of its motion does not change, and a curve when the direction of the motion continually changes. In Figure 12, AB is a straight line ; ~ B CD is a curve ; EF is a broken line ; GH is a composite line. The figures on carpets and on _j. wall-papers furnish numberless Flo 12 examples of broken and com- posite lines. The word "line" used alone means straight line. 2. Draw and name with letters two straight lines which intersect. Also two straight lines which do not intersect. 3. The same exercise with two curved lines. 4. Draw three broken lines. Define a broken line. 5. Draw three composite lines. Define a composite line. 6. What kind of lines are the edges of a cube ? the edges of a prism? the edges of a pyramid? the edges of a cylin- der ? the edges of the room ? the edges of a ruler ? 7. What kind of lines are represented by the spokes of a wheel? the tire of the wheel? a telegraph wire? a watch- chain? the letters M, 0, S, T, V, Z? the digits 2, 4, 6, 8? 8. What is the path of a falling apple? a ball thrown straight upwards? a ball thrown sideways into the air? the end of a clock hand ? a raindrop? a fly ing bird? INTRODUCTION. 9 9. Classify surfaces, and define each class. There are two kinds of surfaces, plane and curved. A plane surface is a surface on which straight lines can be drawn in as many different directions as you please. Plane surfaces are also called planes. A curved surface is a surface which is not plane. 10. What kind of surfaces are the faces of a cube ? the faces of a prism? the faces of a pyramid? the base of a cone? the lateral surface of a cone? the surface of a sphere? 11. Classify the surf aces of the following bodies : a table, a room, a hat, a lamp-shade, a glass chimney, a mirror, a basin, the water in the basin, the ocean. 12. Hold a sheet of paper so that its surface is at first a plane, and then a curved surface. 13. How can you test whether a surface is a plane ? Apply to the surface the straight edge of a ruler in sev- eral different directions ; if the edge always touches the surface at all points, the surface is a plane. Test if the surface given you is a plane or not. 14. Classify bodies, and define each class. Bodies are of two kinds, polyhedrons and curved bodies. A polyhedron is a body bounded wholly by planes. A curved body is a body bounded partly or wholly by. curved surfaces. 15. What kind of body is a cube? a cylinder? a sphere? 16. What does the word figure mean in Geometry? In Geometry every line or group of lines in short, every- thing having shape or form is called a figure. 17. What is Geometry ? Geometry is the study of geometrical figures, their proper- ties, their construction, and their measurement, CHAPTER II. STRAIGHT LINES. Lesson 5. 1. How many directions has a straight line? A straight line AB (Fig. 13) has two directions, the direction from A towards -B, and the opposite direction from B towards A. If either direction be known or given, the other direction is of course also known. NOTE. In Geometry "given" is used very nearly in the sense of "known." 2. How is the position of a straight line determined 9 Different straight lines may be drawn, all having the same direc- ,. tion ; also different straight lines ^ > may be drawn through a given point P. But through a given point C in a given direction (as that of the FlG 13 arrow) only one straight line can be drawn. Also through two given points P and Q only one straight line can be drawn. Hence if we know either one point and the direction of a straight line, or two points of the line, we know enough about the line to be able to construct it, or to represent it by draw- ing it on paper. In other words, the line is determined in position. STRAIGHT LINES. 11 3. How are straight lines drawn on paper? Straight lines are drawn on paper by means of a ruler with a straight edge. -To draw a line through two given points, lay the ruler on the paper so that its edge lust touches the two ^ ^ FIG. 14. points ; then draw the line with a well-sharpened pencil, holding the pencil nearly vertical, and always touching the edge of the ruler. 4. What is freehand drawing ? If we draw a straight line without the aid of a ruler to guide the hand, we are said to draw it freehand. In general, freehand drawing is drawing without the aid of instruments, except the pencil or pen. 5. Draw four lines through a point, and name them. 6. Make two points, and join them by a straight line. NOTE. To " join AB " means to " draw a straight line from A to B" 7. Make three points, and join them by drawing as many straight lines through them as possible. 8. The same exercise with four points. 9. The same exercise with five points. 10. Draw freehand a straight line ; correct or rectify it with the aid of a ruler. Repeat several times. 11. Draw one of the faces of the body given you in three different ways, as follows : (1) Lay the face on paper and trace its boundary. (2) Lay the face on paper ; mark the corners ; join them. (3) Draw the face as well as you can freehand. 12. Test whether the edge of your ruler is straight. To do this draw a line through two points, turn the ruler end for end, draw a line through the same points along the same edge as before. The lines ought to coincide. 12 LESSONS IN GEOMETKY. Lesson 6. 1. How cure straight lines and plane surfaces classi- fied with respect to the surface of the earth? With respect to the surface of the earth straight lines and planes are either vertical, horizontal, or inclined. A vertical line is a line having the direction of a plumb line, or string held at rest in the hand, and sup- porting at its lower end a small weight (Fig. 15). Every plane in which a vertical line can be drawn is a vertical plane. A horizontal line is a line having the direction of a pencil, stick, or other object which is float- ing on the surface of still water. The surface of still water, and every plane sim- FIG is ilarly placed with respect to the earth's surface, is a horizontal plane. A line or a plane which is neither vertical nor horizontal is said to be inclined. 2. Give examples of lines and planes vertical, horizontal, and inclined. 3. Hold a pencil vertical, horizontal, inclined. 4. Hold a book vertical, horizontal, inclined. 5. Draw on the blackboard a line of each kind. 6. What direction has the mast of a ship ? the path of a falling apple ? a rail on a railway track ? the side of a house ? the roof of the house ? the trunk of a growing tree ? 7. Place the body given you on the table, and then exam- ine the position of its edges and its faces with respect to the surface of the earth. 8. At what time of day are the sun's rays most nearly vertical? When are they horizontal? STRAIGHT LINES. 13 9. When is the hour-hand of a clock vertical ? horizontal ? 10. When two vertical planes intersect, what kind of a line is the intersection ? Example : the walls of a room. 11. When a vertical plane intersects a horizontal plane, what kind of a line is the intersection? Give an example. 12. How are vertical and horizontal lines repre- sented on paper, when they are situated in a vertical plane which stands directly in front of the eye ? Vertical lines are represented by lines drawn on paper straight towards us or straight from us ; horizontal lines by lines drawn from left to right or from right to left. 13. Draw a vertical line, mark on it four points, and draw through each point a horizontal line. Draw freehand. 14. Draw a horizontal line, mark on it four points, and draw through each point a vertical line. Draw freehand. 15. Draw an inclined line, mark on it four points, and draw through each point, freehand, a horizontal and a vertical. N. horizontal N.W. Wr N.E. -E. S.E. FIG. 16. FIG. 17. 16. How are the cardinal directions of a horizontal plane represented on paper (Fig. 17)? The cardinal directions are North, South, East, and West. A line from north to south is drawn straight towards us. A line from east to west is drawn from right to left. 14 LESSONS IN GEOMETRY. 17. Draw, freehand, through a point, the four cardinal directions, and also the four directions which lie midway between them and are termed, northeast, northwest, south- east, southwest. Name the directions (as in Fig. 17), by affixing the letters N., S., E., W., N.E., N.W., S.E., S.W. 18. What direction is opposite to N.E. ? opposite to S.E. ? 19. How many horizontal lines can be drawn through a point? How many vertical lines? How many inclined lines ? 20. How many horizontal lines can be drawn in a hori- zontal plane ? How many in a vertical plane ? 21. How many vertical lines can be drawn in a vertical plane? How many in a horizontal plane? NOTE. The plumb line is used for testing vertical walls, and the spirit level for testing horizontal planes. On the ground the difference in level of two places is often found by means of a water level (Fig. 18). It is a metal tube mounted horizontally on a tripod stand with glass vials cemented vertically at its ends. It contains water colored red to distinguish it from the glass. FIG. 18. STRAIGHT LINES. 15 Lesson 7. 1. Explain how a straight line may change in direction. The direction of a straight line may be conceived to change. The hands of a clock represent straight lines which are continually changing in direction. Suppose a line OA (Fig. 19) to turn about the point O from the position OA to the position OB ; then the line is said to rotate, or revolve, about the point 0. If the motion con- tinue, the line will, after a time, return to its first position OA. It is then said to have made one revolution, If, during the motion, the length of the line does not change, the point A will describe a curve which returns into itself, and every point of which is equally distant from 0. 2. Define the terms circumference and circle. A circumference is a curve, every point of which is equally distant from a fixed point called the centre. A circle is a portion of a plane surface bounded by a circumference. 3. Define the terms radius, arc, chord, diameter. A straight line drawn from the centre of a circle to the circumference is called a radius (plural radii] . A part of a circumference is called an arc, A straight line joining the ends of an arc is called a chord. A chord passing through the centre is called a diameter. FIG. 19. 16 LESSONS IN GEOMETHY. 4. How are circles drawn or described on paper? Circles, or strictly speaking, circumferences, are described on paper with the aid of the compasses, or dividers. This instrument has two legs joined together by a pivot, about which they can turn. In the form best adapted for school work, one leg is provided with a metal point, and the other leg with a pencil point. In order to de- scribe a circle on paper, we open the legs and proceed as illustrated in Fig. 20. On the ground circles are made as FIG. 20. illustrated in Fig. 21. NOTE. The shorter word, "circle," is often used instead of the word " circumference," in cases where a misunderstanding is not possi- ble. For example : " describe a circle " ; " an arc of a circle." i FIG. 21. 5. Name in Fig. 19 a radius, an arc, a chord, a diameter. 6. Give an example of a circle or a circumference. 7. What does the tire of a carriage wheel represent? the spokes ? the axle ? the part of the tire between two spokes ? 8. What is true of all radii of the same circle? 9. What is true of all diameters of the same circle? 10. Compare a radius and a diameter of the same circle. STRAIGHT LINES. 17 11. Draw a circle, a radius, a diameter, and a chord. 12. Describe an arc ; draw its chord and radii to its ends. 13. Describe a circle ; then draw (1) a chord equal to the radius, (2) the longest possible chord. 14. Describe several circles freehand. 15. Describe three concentric circles, or circles having the same point for centre. 16. Describe two circles so that their circumferences shall touch each other, and one shall lie wholly outside the other. 17. Describe two circles so that their circumferences shall touch each other, and one shall be wholly within the other. 18. Describe two equal circles so that the circumference of each shall pass through the centre of the other. 19. Describe three unequal circles so that the centre of each shall lie on the circumference of one of the others. 20. Construct Figure 22. First describe the circle ; then the arcs whose centres are marked 1, 2, 3, 4, 5, 6 ; then find the centres of the remaining arcs by repeated trial. 4 FIG. 23. 21 . Construct Figure 23. The circles have the same radius. Begin by describing the circle in the middle of the figure. 18 LESSONS IN GEOMETRY. Lesson 8. 1. Define parallel lines, and give examples. Parallel lines are lines having the same direction. A B Examples. The lines AB, CD, _ D EF, and GH (Fig. 24) ; the legs of a table ; the rails of a railroad *- -F (Fig. 25). G H Abbreviation. The sign || is used for the word "parallel"; "AB is || to CD," is read " AB is parallel to CD." FIG. 25. 2. Parallel lines cannot meet however far produced. Show why this follows from the definition. Suppose that two parallel lines, for example, AB and CD (Fig. 24), when produced (prolonged), should at length meet, and call the point of meeting P. Then we should have two lines, AB and CZ>, drawn through P in the same direction. Now, two lines so drawn must coincide (p. 10, No. 2). But AB and CD plainly do not coincide, therefore we know that they could never meet, however far produced. 3. Are two lines, which would never meet if produced, necessarily parallel? Illustrate by an example. STRAIGHT LINES. 19 4. Problem. To draw a line parallel to a given line AB through a given point C. METHOD I. (Fig. 26). Instruments : ruler and compasses. Join C to any point D in AB. With centre D and radius DC describe an arc, cutting AB in E. With centre C and radius DC describe the arc DF. With centre D and radius CE cut the arc DF in.F. Through C and F draw a straight line. The straight line CF is the parallel required. n E D FIG. 27. B METHOD II. (Fig. 27). Instruments : ruler and compasses. With any point D in AB as centre and DC as radius, describe a semicircle AECB. With centre A and radius BC cut the semicircle in E. Draw CE. The line CE is the parallel required. METHOD III. (Fig. 28) . Instruments : ruler, and a piece of wood with three straight edges, called a triangle. Place the ruler and triangle as shown in the figure. Slide the triangle along till its edge touches the point C; then draw a straight line along the edge of the triangle. This line will be parallel to AB because the edge of the tri- angle during the motion re- mains parallel to AB. FIG. 28. 20 LESSONS IN GEOMETRY. 5. Draw a straight line ; then draw a parallel line through a point not in the line by Method I. 6. The same exercise, using Method II. 7. The same exercise, using Method III. 8. Draw a parallel to a given line through a given point by Method I. Test the accuracy of your result by Method III. 9. Draw by Method III. six parallels as nearly equidistant as you can. 10. Draw a vertical line, mark five points on it, and draw through the points parallel lines by Method III. 11. Draw six parallel lines, two vertical, two horizontal, and two inclined. 12. Draw a three-sided figure, and then draw through each of its corners a line || to the opposite side. 13. Describe a circle, and draw two parallel chords. 14. Draw a straight line ; then try to draw freehand a parallel to it. Begin by marking points which, as well as you can judge, must be in the required line. 15. Draw freehand a series of six parallel lines. 16. Hold two pencils (1) parallel; (2) so that they would intersect if prolonged ; (3) so that they are not parallel, and also would not intersect. 17. Point out on the bod}' given to you edges which are (1) parallel; (2) intersecting; (3) neither parallel nor in- tersecting. 18. The same exercise with the edges of the room. 19. When is a straight line parallel to a plane? A straight line is parallel to a plane when the line will never meet the plane however far they are both produced. 20. When are two planes parallel to each other? 21. Give an example of a line || to a plane ; also of two parallel planes. STRAIGHT LINES. 21 Lesson 9. 1. What has a straight line besides direction? Besides direction, a straight line has length. As regards length, two straight -lines are either equal or unequal. 2. Define equal and unequal straight lines. Two straight lines are equal if they can be so placed, one upon the other, that their ends coincide. If this cannot be done, the lines are unequal. 3. How is the equality of two lines expressed ? The equality of two lines AB and CD is expressed thus : AB = CD. A- -B This is called an equation, and is C D read, " AB is equal to CD." E _ F The sign = is the sign of equality. FIG. 29. 4. How is the inequality of two lines expressed ? The inequality of the two lines AB, EF is thus expressed : AB>EF, orEFa, d < b. 12. Read the following : m = n, AB < CZ>, x > y. 13. Draw four straight lines, a, 6, c, d ; then compare a with each of the other lines, writing the result with the proper sign. 14. Draw freehand two lines as nearly equal as you can, each longer than the greatest opening of your dividers. Can you test their equality by means of your dividers ? 15. Compare different lines drawn between two points. A straight line is the shortest line between two points ; hence the length of a straight line joining two points is taken as the distance of the points from each other. rr FIG. 30. 16. How do sign-painters make use of this truth (Fig. 30) ? They chalk a cord and stretch it tightly between the points through which the line is to pass ; then, seizing this cord by STRAIGHT LINES. 23 the middle, they draw it back 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. 17. Define an axiom, and illustrate the meaning. There are statements or assertions which are so obvious that they stand in no need of any explanation or proof. Every one sees at a glance that they are true. In Geometry, an assertion which is admitted to be true, without proof, is called an axiom. An instance of an axiom occurs in No. 5. We assume that if AB and CD are separately equal to the distance be- tween the points of the dividers, they are equal to each other. In general, if any two magnitudes are each equal to a third magnitude, we assume, as quite obvious, that they are equal to each other. Another example of an axiom is the assertion that a straight line is the shortest distance between two points. 18. State the most important axioms which are found useful in the study of Geometry. 1. Two magnitudes, each equal to a third, are equal to each other; or, for any magnitude, its equal may be substituted. 2. If equals are added to equals, the sums are equal. 3. If equals are taken from equals, the remainders are equal. 4. If equals are multiplied by equals, the products are equal. 5. If equals are divided by equals, the quotients are equal. 6. A ivhole is greater than any of its parts. 7. Through a given point in a given direction, only one straight line can be drawn. 8. Through two points, only one stmight line can be drawn. 9. A straight line is the shortest dis&mee between two points. 10. Through a given point only one parallel to a given straight line can be drawn* LESSONS IN GEOMETRY. Lesson 1C. 1. Explain the addition and subtraction of lines. In order to add two straight lines, let AB and- CD (Fig. 31) be two lines, draw a straight line and take on it EF= AB, FG=CD\ then the line EG will be the sum of AB and BC ; or, EG = AB + CD. A\ IB C\ \D El FIG. 31. If, instead of laying off CD towards the right, we lay it off from F to H towards the left, then we obtain a length EH which is the difference between AB and CD ; or, using signs, EH=AB-CD. 2. How are lines multiplied or divided by numbers ? To multiply a line by a number is to add the line repeat- edly to itself. Thus : produce AB (Fig. 32) by adding to it lines BC, CD, etc., each equal to itself ; then AC is twice as long as AB, AD three times as long, etc. A, _ L* _ ^ _ L5 _ ^ _ , _ ^ _ Jl _ l K_ tL FIG. 32. These results are usually written as follows : AC=2AB, AD = SAB, AE = AB, AL = 9AB. Division is the inverse of multiplication. To divide a line by a number is to separate the line into as many equal parts as there are units in the number, and take one of the parts as the quotient. The operation of division is usually ex- pressed in the form of a fraction, thus (see Fig. 32) : ete. STRAIGHT LINES. 25 3. Problem. To bisect a given straight line AB. With centres A and B (Fig. 33) , and a radius greater than half of AB, describe arcs intersecting at C and D. Join CD, which will cut AB at a point M. Then M is the middle point of AB. H K FIG. 34. 4. Problem. To divide a given straight line into any number (say five) equal parts. Draw through A (Fig. 34) any straight line AX. Upon AX lay off five equal lengths from A to G. Join GB, and draw through the points C, D, E, F, between A and G, lines || to GB (p. 19, No. 4). These parallels will divide AB into five equal parts. NOTE. The use of the lines CM, DN, etc., will appear hereafter. 5. Draw any two straight lines ; then construct a line equal to their sum, and a line equal to their difference. 6. In Fig. 32 name lengths equal to the following : AE + BD, AC + GL, AE - AD, AG - EH, AD - HL. AC, 2x (AB + BD), 3 x (DH-DE), x(AG-BF). AE, IAL, \BL, (AD + AH), (CL-CF). 7. Draw a broken line, then develop it ; that is, draw a straight line equal to the entire length of the broken line. 8. Draw a straight line equal in length to the sum of the edges of one face of the solid given to you. 26 LESSONS IN GEOMETRY. 9. Draw AB = CD, and MN= PQ. Then draw lines equal to AB + MN, CD + PQ, AB - MN, CD - PQ. Are the first two of these lines equal? Why? Are the last two of these lines equal? Why? 10. Produce a line AB to P, making AP= 5 AB. 11. Draw a line AB and then bisect it. 12. Divide a line into four equal parts. First bisect the line and then bisect each of the halves. 13. Divide a line into six equal parts. 14. Divide a line into seven equal parts. 15. Bisect a line freehand. Test and correct the result. 16. Trisect a line freehand. Test and correct the result. D A F B A. E B FIG. 35. FIG. 36. 17. Bisect a line by repeated trial with the dividers. 18. Draw a three-sided figure ABC. Bisect its sides at D, E, F. Join the middle points to the opposite corners. If your work is accurate, the lines AD, BE, CF will all pass through the same point (Fig. 35). 19. Draw a four-sided figure ABCD. Bisect its sides at E, F, G, H. Join the middle points, taken in order around the figure. If your work is accurate, EF will be || to GH, and FGl to EH (Fig. 36). NOTE. In performing Exercises 18 and 19 do not try to choose points so that your figures will resemble Figures 35 and 36, but mark their positions on your paper without reference to these figures. Do not choose your points too near together ; the farther they are from each other the better. STKAIGHT LINES. 27 Lesson 11. 1. What is a unit of length? Give examples. A unit of length is a length named and defined by law. Examples : the/oo, the yard, the mile, the meter, 2. What is meant by measuring a line? To measure a line is to find how many times it will con- tain a unit of length. The number of times a line will con- tain a unit of length, joined to the name of the unit, is called the length of the line. Examples : 6 feet, 9 miles. 3. What units of length are in common use? The units of length in common use in this country and in England are the inch, the foot, the yard, the rod, the mile. Abbreviations : in. = inch, ft. = foot, yd. = yard. 4. How are these units related to one another? They are related as shown in the following table : 12 inches = 1 foot. 5|- yards = 1 rod. 3 feet = 1 yard. 320 rods = 1 mile. 5. How are lengths less than an inch expressed? The inch is usually divided into halves, quarters, eighths, and sixteenths. If we wish to take into account very small parts of an inch, it is more convenient to divide it decimally into tenths, hundredths, etc. 6. How many inches in one rod? in one mile? 7. How many feet are there in one mile? 8. How many inches in 3 feet? 10 feet? 21 yds.? 9. What part of a foot is 3 inches? 4 inches? 8 inches? 9 inches? 10 inches? 10. Reduce 100 inches to feet. Also to yards. 28 LESSONS IN GEOMETRY. 11. You have a scale for measuring lengths. What is the length of the scale ? Describe how each inch is subdivided. Name one of the shortest parts. Name two of them taken together. Name three of them taken together, etc. 12. How many half-inches are there in 5^ inches? 13. How many quarter-inches are there in 4| inches? 14. Reduce 9J inches to quarters of an inch. 15. Reduce 3^ inches to quarters of an inch. 16. Reduce 2J inches to eighths of an inch. 17. Reduce 4J inches to sixteenths of an inch. 18. Reduce 24 quarters of an inch to inches. 19. Reduce 28 eighths of an inch to inches. 20. Find the difference between 6J in. and 6^| in. 21. How many eighths of an inch are there in one foot? 22. How many quarter-inches are there in one yard? 23. Draw straight lines having lengths as follows : 4fin. ; 3f in. ; 2| in. ; 1ft in. 24. Through a point A draw three straight lines. Upon one of them, mark off a length AB = 2| inches ; upon an- other, AC 3 inches ; upon the other, AD = 4 T 3 g- inches. 25. Draw four parallel lines (7, 4). Then lay off upon the first AB = 4J in. ; upon the second, CD = 2J in. ; upon the third, EF = AB + CD', upon the fourth, GH=AB- CD. Then measure EF and GH. Do not put the scale on the paper. Record the lengths of EF and GH thus : EF= , GH= 26. Draw a broken line, ABCDEF, making AB= 2 in., BC= 2f in., CD = l in., DE = Z^ in., EF=2% in. Then develop it. In what part does the middle point of the developed line lie ? Find the length of the developed line (1) by adding the parts, (2) by measuring it. STRAIGHT LINES. Lesson 2. DIRECTIONS TO BE FOLLOWED. Work out Exercises 1 and 2 before the hour of recitation, and bring the results with you. Do each of the other exercises twice; first, using no instruments but your pencil, and estimating all lengths as well as you can by your eye ; secondly, using rule, scale, and dividers, and measur- ing lengths correct to a sixteenth of an inch at least. Record neatly all your results, both the estimated lengths and the measured lengths. 1. Find the average value of your pace. To do this, take ten steps straight forward as you naturally walk, meas- ure the distance walked, and divide this distance by ten. Repeat, and if the two results differ, take their mean, which is equal to half their sum. 2. Measure by pacing the distance from the schoolhouse to 3*our home. 3. Draw a three-sided figure ABC. Then measure and record the lengths of the sides, AB, BC, AC. FORM OF RECORD. AB BC AC Estimated Length. Measured Length. Difference. 4. Measure the length (1) of one of the horizontal bars in the form of record above, (2) of one of the vertical bars. Make a neat form of record for yourself. 5. Draw AB= 1| in., AC = 2 in. Join BC. Produce AB to D making ED =\\ in. Draw DE II to BC and meet- ing AC produced at E. Measure BC, DE, CE. 6. Measure and record the dimensions of the body which is given you. 30 LESSONS IN GEOMETRY. Lesson 13. 1. Suppose three points A, B, C on the ground are in a straight line, and that by measurement you find AB = 1%5 feet, AC =300 feet. How can these lengths be represented on this page by a straight line drawn from left to right ? Sincfc the width of the printed matter on the page is only about 31 inches, it is clear that the line which I draw to rep- resent AC cannot be more than 31 inches long. Suppose I take 1 inch on paper to represent 100 feet or 1200 inches on the ground. Then T ^ 7 of an inch will represent 1 foot. Therefore 300 feet will be represented by f$$ inches, or 3 inches, and 125 feet by ij|- inches, or 11 inches. Hence, to represent the measured lines I draw on the paper AC ' = 3 inches, and take AB = 1 J inches. A B C _j | L_ Scale: 1 in. = 100 ft. The two measured lengths have now been drawn to scale; the reducing factor is 1200 ; the scale of reduction is y^Vo' or as sometimes written 1 : 1 200 ; a common way of expressing it is as follows : 1 inch = 100 feet. 2. What is meant by " drawing to scale " ? Drawing to scale means drawing lines on paper so that each line shall be the same fractional part of the line on the ground which it represents. 3. What is the reducing factor ? The Reducing factor is the number by which each line on the ground must be divided in order to obtain the reduced length to lay off on paper. STRAIGHT LINES. 31 4. What is the scale of reduction ? The Scale of reduction is the fraction having 1 for nume- rator and the reducing factor for denominator. In making a drawing, the scale of reduction should be so chosen as to leave a fair margin around the paper after the lines are all drawn, and should be neatly written by the side of the drawing, usually at the lower right-hand corner. 5. Give examples of drawing to scale. On maps distances of hundreds and even thousands of miles are represented by lines a few inches long. Plans of estates and buildings are always drawn to a reduced scale. In general, the same is true of all pictures and paintings. 6. Write in the form of a fraction the following scales : 1 in. 1 yd., \ in. = l ft., 2 in. = l mile, ^ in. = 10 miles. 7. If the scale of reduction be ^V^, what length on the ground will be represented by 1 inch on the paper ? 8. If 4 inches on paper represent a line 200 feet long, express the scale of reduction in a fractional form. "What length on paper will represent half a mile ? 9. The scale of a map is -^Vo- The distance between two towns on the map is 8.8 inches ; what is the real distance between the towns ? 10. Draw 5 parallels and lay off on them to the scale -^ the lengths : 100 feet, 150 feet, 70 feet, 50 feet, 30 feet. 11. Draw a three-sided figure, and find what lengths the sides would represent to the scale 1 inch = 1000 feet. 12. Find from a map of the United States the distance in a straight line from New York to Chicago. 13. Measure the line on the blackboard assigned to you, and record its length. Then represent the line on paper, choosing a suitable scale of reduction for this purpose. 82 LESSONS IN GEOMETRY. Lesson 14. 1. What is the nature of the Metric System ? The Metric System of units is a decimal system^ The fundamental unit of length is the meter. All the other units of length are derived from the meter by decimal multiplication and division. 2. Write out in tabular form the names, abbrevia- tions, and relative values of the units of length. THE METRIC UNITS OF LENGTH. Name. Abbreviation. Relative value. Meter m. Dekameter dkm 10 meters. Hectometer hm 100 Kilometer km 1000 Decimeter dm 0.1 of a meter. Centimeter ....... cm 0.10 " Millimeter mm 0.100 " The units most largely employed in practice are the meter, the kilometer, the centimeter, the millimeter. 3. How is one unit reduced to another unit? Reduction from a larger unit to a smaller unit is performed by multiplying by 10 as many times as may be necessary ; that is, by moving the decimal point towards the right the proper number of places. Reduction from a smaller unit to a larger unit is performed by dividing by 10 as many times as may be necessary ; that is, by moving the decimal point towards the left the proper number of places. Examples. 32 m = 320 dm = 3200 cm = 32000 mm ; 0.764 km = 7.64 hm = 76.4 dkm = 764 m = 764000 mm ; 376 mm = 37.6 cm = 0.376 m = 0.000376 km . STRAIGHT LINES. 4. In practice, only one unit is used in expressing a length. Suppose the length of a line is measured and found to be 7 dm , 6 cm . Reduce this to meters ; decimeters ; centimeters. 5. Reduce 12.4 km to each of the other units. 6. Reduce 1800 cm to decimeters ; to meters ; to kilometers. 7. What part of a kilometer is 1 millimeter? 8. What part of a meter is 5 cm ? 2 dm ? 10 mm ? 9. 41 _|_ 4 4. 4'm _j_ 4mm _ } 1OW many me t er s ? 10. What is the difference between 4 em and 28 mm ? 11. From 10 m take 10 mm . 12. How long a piece of wood is required to make 20 rulers, each ruler to be 30 cm long? 13. How many rails 7.5 rn long are required to build a rail- road track 300 km long? 14. Reduce the following values to English equivalents : height of a barometer, 76 cm ; height of Mt. Blanc, 4815 m . 15. Reduce 236 miles to kiloYneters. 16. Describe the divisions of your metric scale. 17. Measure to the nearest millimeter (1) the length of this page ; (2)' the length of a line of print upon the page. Record the results neatly in a tabular form. 18. Construct on paper a scale, l dm long, precisely like the decimeter scale printed below. 3 TTTTlTTTT 4 TTlTTTT TTTT 10 TABLE OF ENGLISH AND METRIC EQUIVALENTS. More exactly. 1 kilometer = 0.6214 mile. 1 meter 39.37 inches. 1 centimeter = 0.3937 inch. Approximately. 8 kilometers = 5 miles. 1 meter = 1 T J T yards. 2 centimeters = 1 inch. 34 LESSONS IN GEOMETRY. Lesson 15. 1. Express in fractional form the scale 4 CW = l m . 2. Express in fractional form the scale 2 mm = I*" 1 . 3. A certain map is drawn to the scale I em =l00 m . What length on the map will represent a distance of 1.2 km ? 4. The distance between two towns on a map drawn to the scale 1 : 40000 is represented by a line l dm in length. What is the actual distance between the towns? 5. A monument is 40 m high, and a man standing beside it is 2 m high. Draw to the scale 1 : 200 straight lines to represent the monument and the man. 6. Draw three parallel lines of any lengths, and then find what lengths they would represent to the scale 1 : 200. For example : a line 3 cm long would represent 3 cm x 200 = 600 cm = 6 m . 7. Draw AB=26 mm , AG=37 mm . Join BC. Produce AB to D, making BD = 50 mm . Draw DE \\ to BC, and meeting AC produced at E. Measure BC, DE, CE. 8. Draw a line by your eye 4 cm long ; measure it ; record your error in millimeters. Repeat several times. NOTE. In doing Exercises 9, 10, and 11 follow the directions given in Lesson 12, and measure lengths to the nearest ^//-millimeter. 9. Measure the dimensions of the following figure : I) 10. Draw a four-sided figure ABCD, and measure the lengths of the sides. 11. Measure the dimensions of the body given you. STRAIGHT LINES. S5 Lesson 16. Review. 1. Review the italicized exercises in Lessons 5-9. 2. What is the greatest possible number of straight lines that can be drawn through four points? Mark four points, and draw all the straight lines possible through them. 3. Place four points so that four and only four straight lines can be drawn through them. Draw the lines. 4. In how many points will five lines intersect if four of the lines are parallel ? if three are parallel ? if two are par- allel? if no two of the lines are parallel? 5. Through a vertical line how many vertical planes can be passed (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? 6. Through a horizontal line how many vertical planes can be passed? how many horizontal planes? how many inclined planes? 7. Through an inclined line how many vertical planes can be passed? how many horizontal planes? how many inclined planes? 8. Describe two concentric circles. Then describe two circles which shall touch, but not cut, the concentric circles. 9. Describe a circle with radius two inches, and draw in it a chord equal to one and one-half times the radius. 10. Describe a circle, and within this circle two smaller circles which shall touch each other and the large circle. 11. Describe a semicircle, and draw in it three chords parallel to the diameter. 12. Give an example of parallel lines ; of parallel planes. 13. Describe the situation of all the points in a plane which are one inch from a straight line drawn in the plane. 86 LESSONS IK GEOMETRY. Lesson 17. Review. 1. Review the italicized exercises in Lessons 10-13. 2. Which is the greater, 3.6 in. or 3| in. ? What is the difference ? 3. If a pace = 21 ft., how many paces are there in 1 mile? 4. Express as fractions the following scales : 1 in. = 1 ft. ; 1 in. = 1 yd. ; 1 in. = 1 rod ; 1 in. = 1 mile. 5. If 4 in. represent 66 ft., what will represent 44 yds. ? 6. Among the scales in use in the engineer service are : Plans of buildings, 1 in. 10 ft. Maps 1| miles square, 2 ft. = 1 mile. Maps 3 miles square, 1 f t. = 1 mile. Maps 4 to 8 miles square, 6 in. = 1 mile. Maps 9 miles square, 4 in. = 1 mile. Maps 12 to 24 miles square, 2 in.= 1 mile. Maps 50 miles square, 1 in.= 1 mile. Express these scales in the form of fractions. 7. Divide by your eye an inch into tenths. 8. Divide a line into 8 equal parts. 9. Draw a three-sided figure ; bisect its sides, and pro- duce (if necessary) the bisecting lines till they meet. They should all meet in the same point. 10. Draw a line AB, and divide it into three parts, such that the second part shall be double the first, and the third part double the second. 11. Draw a four-sided figure, two sides of which are par- allel ; then describe circles upon the sides taken as diameters. 12. Draw a four-sided figure ABCD, having its opposite sides parallel ; join AC and BD, and mark their point of intersection 0. Measure the lengths -4(7, BC, CD, DA, AO, CO, BO, DO. CHAPTER III. ANGLES. Lesson 18. 1. Define an angle, its sides, and its vertex. An angle is the difference in direction of two straight lines which meet, or would meet if produced. The lines are called the sides of the angle ; their point of meeting is called the vertex. 2. Illustrate these definitions. The lines AB and AC (Fig. 37) form the angle BAC. A is the vertex, AB and AC the sides. The lines DE and FG (Fig. 38) also form an angle. Its vertex is the point H, where the lines would meet if produced. A B D E II FIG. 37. FIG. 38. 3, How is an angle named or denoted? An angle may be named in either of two ways : (1) By a small letter placed just inside the vertex. (2) By writing the letter which stands outside the vertex between two letters which stand one on each of the sides. Thus, the angle in Fig. 37 is the angle m, or the angle BAC. Abbreviations : /, stands for k ' angle," A for " angles." 38 LESSONS IN GEOMETRY. D. II 4. Angles differ in magnitude. How can a clear idea of the magnitude of an angle ~be obtained ? To get a clear idea of angular magnitude we must conceive an angle to be described by the rotation of a line. ^ Suppose a line to start from the position OA (Fig. 39), and turn about the point in the direction of the arrow. When the line has reached the position OJ5, it has described the angle AOB; when it has reached the position OC y , it has described the angle AOC. If the motion continue, an angle larger than AOC will be described. It is quite obvious that the mag- nitude of the angle depends entirely upon the amount of ro- tation of the line, and not at all upon the length of the line. 5. Define the right angle and the straight angle. When a line makes one-fourth of a complete revolution it describes an angle which is called a right angle. Thus, the angler AOC (Fig. 39) is a right angle. When a line makes one-half of a complete revolution it describes an angle which is called a straight angle. The angle AOE (Fig. 39) is a straight angle ; its sides are oppo- site in direction, and form a straight line ; hence its name. 6. Can an angle be greater than a straight angle ? Yes. When the rotating line (Fig. 39) has made three- fourths of a revolution it takes the position 06r, and the angle described, AOG, is equal to three right angles. The side OG of this angle is opposite in direction to the line 0(7. When the rotating line has made exactly one revolution it arrives at the position OA from which it started, and has described an angle equal to four right angles. ANGLES. 39 7. Define an acute angle and an obtuse angle. An angle less than a right angle is called an acute angle, and an angle greater than a right angle is called an obtuse angle. 8. What kind of an angle (Fig. 39) is AOB? AOD? DOE? COE? FOG? EOG? DOG? GOH? HOC? 9. Is the angle a on the blackboard right, acute, or obtuse? the Z 6? the Z c? etc. 10. Mark three points ^4, -B, C. Join them by straight lines. How many angles are formed? Name them. 11. Draw two intersecting lines. How many angles are formed? Name them, using as few letters as possible. 12. Draw freehand an acute angle, an obtuse angle, a right angle. 13. What angle does a vertical line make with a hori- zontal line ? 14. What kind of an angle do the hands of a clock make with each other at three o'clock ? at six o'clock ? 15. Mention times of the day when the hands of a clock make a right angle, an acute angle, an obtuse angle. 16. How long a time does it take the minute-hand of a watch to describe a right angle ? How long does it take the hour-hand to do the same ? 17. What kind of an angle does the hour-hand describe in one hour? three hours? five hours? six hours? 18. A man is walking due north. If he suddenly change the direction in which he is moving through two right angles, in what direction will he now be going? In what direction if he makes a change from due north through one right angle to his left? through three right angles to his right? through two and one-half right angles to his right? 40 LESSONS IN GEOMETRY. Lesson 19. 1. Define perpendicular lines and oblique lines. If two lines form a right angle, they are said to-be perpen- dicular to each other ; if they form any other angle, they are said to be oblique to each other. Abbreviation : The sign _L means " perpendicular." 2. Problem. To erect a perpendicular at any point C of a given straight line AB. METHOD I. (Fig. 40) . Instruments : ruler and compasses. With centre (7, and any radius, cut AB in D and E. With centres D and E, and a greater radius than before, describe arcs intersecting at F. Join CF; OF is the perpendicular required. D\ C FIG. 40. E B FIG. 41. METHOD II. (Fig. 41). Instruments: ruler and triangle. Place the ruler and the triangle as seen in the figure. Draw a line along the edge CD. This line will be the perpendicular required. 3. Give examples of perpendicular lines. 4. Draw a line AB ; erect a perpendicular by Method I. 5. Draw a horizontal line, and erect (by Method II.) perpendiculars at six equidistant points along the line. 6. How many right angles are formed when two perpen- dicular lines cut each other? ANGLES. 41 7. Problem. To erect a perpendicular at the end of a line AB without producing the line (Fig. Mark any point C not in the line. With centre C and radius GB describe an arc greater than a semi- circumference, cutting AB at D. Join DC, and produce DC to meet the arc at E. Join EB ; EB is the perpendic- ular required. 8. Draw a line AB = 4 inches ; FlG . 42 . erect a perpendicular at each end, making each perpendicular, as nearly as you can judge by the eye, equal to AB. Test the result with dividers. 9. Draw a line AB ; erect at A a perpendicular equal to 4 AB ', at B a perpendicular equal to ^ AB. Work out the exercise (1) freehand ; (2) with instruments. 10. Draw a horizontal line AB four inches long ; at A erect a perpendicular AC = 1^ in.; at B erect a perpendicular BD = 4J- in. Join CD ; measure CD, and write its length along its side. Work the exercise (1) freehand; (2) with instruments. 11. Draw AB = 5 in., J3CJ_ to AB and = 3 in., CD || to AB and = 1 in. Join AD, and measure its length. Work (1) freehand; (2) with instruments. 12. Draw AB= 6 in. Upon AB&$ diameter, describe a semicircle. At a point C, two inches from A, draw CD A. to AB, and meeting the curve at D. Measure CD, and multiply its length by itself. Record the result thus : CD x CD= If your work were perfectly done, the product would be 8. The nearer your result is to 8, the better your work. 13. Erect a perpendicular freehand, and test the result with ruler and triangle. - \ 42 LESSONS IN GEOMETRY. Lesson 2O. 1. Problem. To drop a perpendicular from a given point C to a given line AB. METHOD I. (Fig. 43). Instruments : ruler and compasses. With centre C and radius greater than the distance from C to AB, cut AB at D and E. With cen- G _ x , tres D and E and the same .,-'' E B radius as before, describe arcs ,.- intersecting at F on the other ,/' side of AB. Join CF; CF is # -L to AB. METHOD II. Instruments : ruler and triangle. This method will be obvious from Fig. 41. 2. Drop by Method I. a perpendicular from a point C to a line AB. 3. The same exercise, using Method II. 4. Draw a four-sided figure ABCD, and drop perpendic- ulars from C and D to the side AB. 5. Draw a three-sided figure and drop perpendiculars from the corners to the opposite sides (produced when nec- essary). If your work is accurate, the three perpendiculars will all pass through the same point. 6. Draw an angle BAC. Upon AC take AM = 11 in. ; AN= 2^ in. ; AO = 3| in. ; AP= 5 in. Drop perpendicu- lars ME, NS, OT, PVto AB. Measure their lengths, and also AR, AS, AT, AV. Record your results in a table. 7. A path goes straight for 15 yards, then turns through a right angle and goes straight for 40 yards, then turns and runs straight to its starting-point. Find its entire length by drawing to scale (1 in. = 10 yds.). ANGLES. 43 Lesson 21. 1. Define equal angles. Two angles are equal if they can be placed so that their vertices coincide in position, and their sides coincide in direction. For example, the angles AOB, BOC, COD, etc. (Fig. 44), are equal angles. 2. What consequences follow from this definition ? (1) All straight angles are equal; for the sides of a straight angle form a straight line, and two straight lines can always be so placed that they coincide. (2) All right angles are equal; for a right angle is half of a straight angle, and the halves of equal things are also equal (see Axiom VIII.). (3) Angles, like lines, can be added, subtracted, etc. For example : in Fig. 44, AOC=AOB + BOC and AOB = AOC-BOC. Also, the angles being equal, AOC=2AOB, AOD=3AOB, etc. And it is also clear that FIG. 44. D = %AOD, etc. 3. Describe the units used for measuring angles. The right angle is divided into 90 equal parts called degrees, the degree into 60 equal parts called minutes, the minute into 60 equal parts called seconds. Abbreviations : the sign stands for "degree," ' for "minute," " for "second." 4. Reduce to degrees 4 rt. A, \ rt. Z, \ rt. Z, \ rt. Z. 5. Reduce 48 54' 36" to seconds ; 120,000" to degrees. LESSONS IN GEOMETRY. 6. Define equal arcs. Two arcs are equal if they can be placed, one upon the other, so that they coincide. Example : the arcs AB and CD (Fig. 45). 7. Suppose a circumference (Fig. 4$) to be divided into any number of equal arcs by drawing radii, and then compare the corresponding angles at the centre. Compare any two adjacent an- gles, as AOB and BOG. Imagine AOB folded upon the line OB till A falls on C. Why must A fall on O? Then OA will coincide with OC. Why must OA and OC coincide ? Therefore, Z AOB = Z BOG. In the same way all the angles at the centre can be proved equal. Therefore, if a circumference be divided by radii into any number of equal arcs, the corresponding angles at the centre are also equal. 8. How is this relation between arcs and the corre- sponding angles at the centre applied in order to measure angles? The circumferences of circles are supposed to be divided into degrees, minutes, and seconds precisely in the same way as angles. A circumference, therefore, contains 360 degrees of arc, each degree contains 60 minutes, and each minute contains 60 seconds. It follows that the arc, described from the vertex of an angle as centre, and with any radius, will contain the same number of degrees, etc., as the angle. And the angle is measured by finding how many degrees, etc., there are in this arc. FIG. 45. ANGLES. 45 9, Why may the arc employed to measure an angle be described with any radius? It is true that the greater the radius the greater will be the arc. Thus : the arc AB (Fig. 46) is larger than the arc CD ; but if the arc AB be divided into degrees, etc., and radii are drawn to all the points of division, these radii will divide the arc CD into exactly as many equal parts as there are in AB, only the parts will be smaller. Therefore, we shall get the same measure for the angle whichever arc we use. FIG. 47. 10. How are angles drawn on paper measured? Angles on paper are measured with an instrument called a protractor (Fig. 47). It is a semicircle, the circular edge of which is divided into 180 degrees. To measure an angle, place the centre of the protractor over the vertex and the zero line on one side of the angle ; then read on the divided edge the division through which the other side of the angle passes. 11. Make an angle BAG, and measure it with the pro- tractor. 12. Measure with your protractor the angle given you. 13. Make with your protractor the angles : 30, 60, 75, 120, 150. 14. Make with your protractor the angles : 45, 63, 107. 46 LESSONS IN GEOMETRY. Lesson 22. 1, Problem. At a point ^C in a line AB to con- struct an angle equal to a given angle DEF' (Fig. J8).. METHOD I. Instruments : ruler and compasses. With centre E and any radius describe the arc GH. With centre C and the same radius describe an arc LM, cutting AB at L. With centre L and radius GH, cut the arc LN at N. Join CN-, Z. BCN= Z DEF. METHOD II. Instruments : ruler and protractor. This method is obvious without explanation. Ji/F AC L B FIG. 48. FIG. 49. 2. Problem. To bisect a given angle BAC (Fig. With centre A describe the arc DE. With centres D and E, and the same radius as before, describe arcs intersecting at F. Draw AF-, the line AF bisects the Z. BAC. 3. Make an angle BAC, and then construct with ruler and compasses an angle DEF equal to BAC. 4. Make with your protractor an angle of 40 ; then con- struct an equal angle with ruler and compasses. 5. Construct a right angle, and then bisect it. What is the value of each of the equal parts ? ANGLES. 47 6. Problem. To construct an angle of 60 (Fig. 50). Draw a line AB. E With centre A and any radius describe the arc CD. With centre (7, and the same radius, cut the arc CD in E. Join AE ; Z BAE = 60. C FIG. 50. 7. Construct an angle of 60, and then bisect it. 8. Construct an angle BAG = 60 ; from C drop CD _L to ^4J5 ; and find the value of the angle ACD. 9. Construct an angle BAC= 60 ; join BC ; and find the values of the angles ABC and ACB. 10. Construct with ruler and compasses an angle of 30. Verify your result with your protractor. 11. Construct with ruler and compasses an angle of 221. Verify with your protractor. 12. Draw any acute angle BAC ; then construct an angle three times as large. 13. Draw freehand the angles 30, 45, GO . Correct your results with your protractor. Record your errors in degrees. 14. Make an angle, and bisect it freehand. Correct your result with your protractor. Record your error. 15. Draw a three-sided figure ABC. Find the values of its angles, (1) by estimating them by the eye, (2) by meas- uring them with your protractor. Record your results in tabular form, thus : BAC ABC ACB Sum. Estimated Value. Measured Value. Difference. 48 LESSONS IN GEOMETRY. Lesson 23. 1. Define complementary and supplemental^ angles. Two angles are said to be complementary if their sum is equal to a right angle, or 90, and each is called the comple- ment of the other. Example : 67 and 23. Two angles are said to be supplementary if their sum is equal to a straight angle, or 180, and each is called the supplement of the other. Example : 150 and 30. 2. Define adjacent angles. When are they supple- mentary ? Two angles which have the same vertex and a common side between them are called adjacent angles. If a line OO (Fig. 51) is drawn from any point in AB, two adjacent angles AOC, BOO are formed ; and whatever be the direction of OC it is obvious that AOC+COB=180. The two adjacent angles formed when one straight line meets another are supplementary. BOA FIG. 51. FIG. 52. 3. What is the complement of 35? 60? 1? 89? 45? 4. What is the supplement of 60? 90? 1? 135? 62 12'? 5. Draw two angles adjacent but not supplementary. 6. If OD (Fig. 51) is _L to AB, what is the complement of Z .400? 7. If Z .400= 48, find Z COD and Z COB. ANGLES. 49 8. Define vertical angles. Two angles are called vertical angles if one of them is formed by producing the sides of the other from the vertex. When two straight lines intersect (Fig. 52) two pairs of vertical angles, a and c, b and c?, are formed. 9. Two vertical angles are equal. How can this be shown to be true? We may use three different methods : (1) Measure them directly with a protractor. (2) Cut them out, and place one upon the other. (3) Observe that the vertical angles a and c (Fig. 52) both have the angle b for their supplement, and then reason as follows : Since a=180-&, and c = 180 - 6, therefore a = c. (Axiom 1). 10. What advantages has the third method? The third method requires no instruments ; it employs only reasoning. Moreover, it is easy to see that this reason- ing applies equally well to the vertical angles b and d, or to any pair of vertical angles whatever. Therefore it convinces us immediately that vertical angles are always equal. 11. If Z a = 37 (Fig. 52), find the A b, c, and d. 12. Two lines intersect so that one of the angles formed is 90. What are the values of the other angles? 13. Draw any two intersecting lines, and find the values of the four angles which are formed. 14. Write out the proof by the third method employed in No. 9 that Z b = Z d (Fig. 52) . 50 LESSONS IN GEOMETRY. Lesson 24. 1. Name certain pairs of angles formed when a straight line EF cuts two parallel lines AB and CD (Fig. 53). The line EF forms with AB four angles, a, b, c, d, and with CD four angles, m, n, o, p. The pair of angles, a and m, and the other pairs similarly placed with respect to AB and CD, are called exterior-interior angles. The pairs, c and m, d and n, are called alternate- interior angles. 2. Draw two parallels. Cut them by a third line. Mark the angles about the points of intersection 1, 2, 3, 4 and 5, 6, 7, 8. Name the exterior-interior angles and the alternate- interior angles. E A y^ B n/m . C = 40 mm ; join DB and DA, and find in millimeters the difference between DB and DA. 3. Describe a circle with centre and radius = 60 mm . Draw any radius OA, bisect it at C, and let D, E be the points where the line of bisection meets the circumference. Join OD, OE. Measure the length DE, and also measure with the protractor the angles ODE, OED, DOE. 4. Draw AB = 80 mm . Bisect it at G. Erect perpendic- ulars at A, B, C. Draw an}' line cutting these perpendicu- lars at D, E, F, respectively. Measure AD, BE, CF. Find with protractor the sum of the angles ADE and BED. 5. Draw any line AB. Erect the bisecting perpendicular. Join any two points D, E in this perpendicular to A and B. Measure AD, BD, AE, BE. Is any general truth sug- gested by your results? 6. Three ships start from the same port, sailing N., E., S., at the rates 6, 8, 10 kilometers an hour, respectively. How far are they from one another at the end of 4 hours ? Find the distances correct to half -kilometers (scale, 2 mm = l km ). 7. Two cities are connected by a road which forms a broken line ABODE (Fig. 56). AB= 5 km , BC.= 3.75 km , OD=3 kin , DE = 4 km ; Z ABC = 90, Z BCD = 90, Z ODE = 135. A new road is made, going- straight from A to C. Find by drawing to scale the dis- tance from A to E by the new road. CHAPTER IV. TRIANGLES. Lesson 28. 1. Define a triangle, Us sides, perimeter, vertices. A triangle is a plane figure bounded by three straight lines. The straight lines are called its sides; their sum, its perimeter ; their points of intersection, its vertices. The sides form three angles ; the three sides and the three angles are called the parts of the triangle. The sign A stands for " triangle," A for " triangles." 2, How are triangles divided with respect to their sides ? There are three classes ; namely : Equilateral triangles, having all the sides equal. Isosceles triangles, having two sides equal. Scalene triangles, having no two sides equal. EQUILATERAL. ISOSCELES. SCALENE. 3. Draw, freehand, a triangle, and name its six parts, using only the letters placed at the vertices. 4. Draw, freehand, two equilateral triangles, two isosceles triangles, and two scalene triangles. 56 LESSONS IN GEOMETRY. 5, How are triangles divided with respect to their angles ? As regards angles there are three classes ; namely : Acute triangles, having all the angles acute. Obtuse triangles, having one obtuse angle. Right triangles, having one right angle. In a right triangle the side opposite the right angle is called the hypotenuse, and the other sides are called the legs. ACUTE. OBTUSE. RIGHT. 6. Draw, freehand, two acute triangles, two obtuse tri- angles, and two right triangles. 7. How are the dimensions of a triangle named ? The two dimensions are named base and altitude. In the isosceles triangle the side not equal to the others is called the base, and the equal sides are called the legs. In other kinds of triangles any side may be taken as base. In all triangles the alti- tude is the perpendicular dropped to the base from the opposite vertex. In Fig. 57 the three altitudes are all drawn : notice that they meet in one point. N B M FIG. 57. In an obtuse triangle two of the altitudes lie oulside the figure, and meet, not the base, but the base produced. Thus, in the obtuse triangle MNO (Fig. 57) the altitude OP meets the base MN produced at P. TRIANGLES. 57 8. Draw an acute triangle, and construct the altitudes. 9. Draw an obtuse triangle, and construct the altitudes. 10. Draw a right triangle, take the hypotenuse as base, and construct the altitude. What is the altitude if one of the legs is taken as the base? 11. Draw any two lines a and 6; then make a triangle having a base equal to a and an altitude equal to b. Can you make more than one such triangle? 12. The perimeter of an isoceles A =17 in. Find (1) the base if one leg = 5 in. ; (2) each leg if the base = 5 in. 13. Draw a A ABC. Which is the greater, AB or AC+ CB? Why? BCorAB+AC? ACorAB + BC? What is true o'f the sum of two sides of any triangle ? 14. Problem. To construct a right triangle having given the lengths of the hypotenuse and one leg. Let a and b be the given lengths. Draw AB equal to b. At B erect a _L ED. With centre A and radius a cut BD at C. *A. jj Join AC ; then the A ABC is FlG 58 the A required. 15. Construct a right triangle having given the hypotenuse 3f in., one leg 2J in. Measure the other leg. 16. Construct right triangles having for legs (1) 3 in. and 4 in. ; (2) 11 in. and 2 in. Measure the hypotenuses. Compare your results. 17. On the ground in front of a wall is a flower-bed 15 ft. wide. It requires a ladder 25 ft. long to reach from the outer edge of the bed to the top of the wall. How high is the wall? Solve by drawing to scale. 58 LESSONS IN GEOMETRY. IJesson 29. 1. Problem. To construct a triangle, having given the lengths of the three sides (Fig. 59). Let a, 6, c be the lengths. Draw AB equal to a. "With centre A and radius b describe an arc DC. With centre B and radius c describe an arc EC, cutting the arc DC atC. Join AC and BC. A ABC is the A required. B FIG. 59. 2. Construct a triangle, with sides 4 in., 3^ in., 2-|- in. 3. Construct a triangle, with sides 4 in., 3 in., 3 in. 4. Construct an isosceles triangle, given : base = 3 in. ; one leg = 5 in. 5. Construct an equilateral triangle, having given one side. 6. Construct an equilateral triangle with a perimeter equal to 8 in. 7. Problem. To construct a triangle, having given two sides and the included angle (Fig. 60). Let a, 6, be the sides, and Z m the given angle (Fig. 60) . a Draw a straight line, and at any point A of the line con- struct an angle equal to m. Upon its sides lay off AB = a, AC=b. Join BC. A ABC is the A required. B FIG. 60. NOTE. In Fig. 60 the auxiliary lipes used for constructing the angle are not given. Draw auxiliary lines dotted and very fine. TRIANGLES. 59 8. Construct a triangle, given AB = 3 in., AC = 3^ in., Z A = 45. 9. Construct an isosceles right triangle, having given one leg. 10. Explain how the distance between two points, situated on opposite sides of a pond, can be measured. Let A and B (Fig. 61) be the points. First choose a point C from which A and B are both visible ; then measure AC, BC, and Z ACB ; then construct to scale on paper a triangle acb, mak- ing Z ac& = Z ACB, and taking ac of the proper length to rep- resent AC, and be to repre- sent BC. Measure ab, and multiply it by the reducing factor you have used. The product will be the distance from A to B. FlG . 61 . Instead of constructing to scale on paper, if the ground is level and clear, we may produce AC to D, making CD = CA, and BC to E, making CE = CB. Then measure ED. The distance from E to D will be the same as the distance from A to B. 11. Suppose your notes of measurement made to find AB (Fig. 61) to be: AC =1200 yds., (7=1500 yds., Z ACB= 110. Find the distance from A to B. 12. Two villages A and B are separated by a river. There is a straight road going from each village to a bridge over the river at C. If AC= 5 miles, BC= 6 miles, Z ACB = 30, find how far A is from B. 13. Newton is 7 miles west of Boston ; Arlington is 6 miles northwest of Boston. How far is Newton from Arlington? 60 LESSONS IN GEOMETRY. Lesson 3O. 1. Problem. To construct a triangle, having. given one side and the two adjacent angles (Fig. 62). Let a be the given side ; w and n the given angles. Draw AB equal to a. At A construct /. BAC=m. At B construct Z. ABCn. Produce the lines AC and FIG. 62. BC till they meet at C. A ABC is the A required. 2. Construct a triangle, given AB=3 in., BAC=0, ABC =80. 3. Define angle of elevation and angle of depression. If a line is drawn from a point to the eye of an observer, the angle which this line makes with the horizontal plane passing through the eye is the angle of elevation or the angle of depression of the point, according as the point is above or below the horizontal plane. For example (Fig. 63) : if the eye is at (7, the angle of elevation of A is the angle ACB ; if the e} T e is at A, the angle of depression of C is the angle DAG. D A B 4. What is the angle of depression of E as seen from A? 5. What pairs of angles in Fig. 63 are equal? Why? TRIANGLES. 61 6. Construct a right A, given one leg = 2 J in., one angle = 50. 7. Construct 'a right A, given hypotenuse = 5 in. , one angle = 50. 8. Construct an isoceles A, given the base, and angle at the base. 9. Find the height of a church spire, if the angle of ele- vation of the top, 100 yds. from the bottom, is 45. 10. At a distance of 150 ft. from the foot of a tree, I find the angle of elevation of the top to be- 30. How high is the tree ? 11. Wishing to find the height AB of a tower (Fig. 63), I observe the angle of elevation of the top at a point E, and find it to be 30. At a point F, 120 ft. nearer the tower, I find that the angle of elevation of A is 45. What is the height of the tower? 12. Wishing to find the breadth AB of a river, I measure on a line _L to AB along the bank a distance of 400 ft., and find that Z ACB = 70. What is the breadth of the river ? 13. A lighthouse A stands on an island. To find its dis- tance from the shore a man runs a base line BC equal to 1000 yds. along the shore, and measures the angles ABC, ACB. He finds /. ABC =80, Z ACB =70. Find the distance from B to the lighthouse. NOTE. The learner who has carefully worked out the exercises in Lesson 29 and in the present lesson will be able to understand what is meant by the indirect measurement of a line. To measure any consid- erable distance by yard-stick, chain, or other direct means, would be a very laborious process. In many cases, also, direct measurement is practically impossible on account of obstacles, such as water, swamps, etc. In such cases, Geometry teaches us how to measure the line indi- rectly, by treating it as a part of a triangle which we can construct after we have found the values of three other parts by direct measurement. 62 LESSONS IN GEOMETRY. Lesson 31. 1. How can it be shown to the eye that the sum of the angles of a triangle is equal to a straight angle, or 180? Draw a triangle ABC (Fig. 64) and the altitude CD. Cut out the triangle, and fold the corners over the dotted lines till they come together at D. The three angles of the tri- angle will just make a straight angle at D. C D C E B 2. Prove that the sum of the angles of every tri- angle is equal to 180. Draw any triangle ABC (Fig. 65) ; let a, 6, c, denote its angles. We wish to prove that a + b + c = 180. Through C draw DE || to AB. Let m denote the angle DC A, and n the angle ECB. The A a and m are equal, because they are alt. -int. angles. The A b and n are equal for the same reason. Therefore, a + 6 + c = ra + n + c. (Ax. 2.) But m + n -f c == a straight angle, or 180. Therefore, a + b + c = 180. (Ax. 1 .) This reasoning applies equally well to any triangle ; there- fore the sum of the angles of every triangle is equal to 180. TRIANGLES. 63 3. State three truths which follow immediately from what has just been proved. (1) If two angles of a triangle are equal, respectively, to two angles of another triangle, then the third angles must also be equal. For the third angle in each case is found by subtracting the sum of the other two from 180 ; and the sum of the other two in each case is the same (Ax. 2) ; and if equals be taken from equals, the remainders will be equal (Ax. 3). (2) A triangle can have only one right angle or one obtuse angle. For two right angles would make 180, and two obtuse angles would make more than 180. (3) The two acute angles of a right triangle are comple- mentary ; that is, their sum is always 90. For their sum is found by subtracting the right angle, or 90, from 180, and 180- 90= 90. 4. If one angle of a triangle is 45, what is the sum of the other two angles ? 5. If two angles of a triangle are 37 and 75, find the third angle. 6. If two angles of a triangle are 64 47' 33" and 77 18' 41", find the third angle. 7. If one of the acute angles of a right triangle is 30, what is the other? If it is 27 5', what is the other? 8. If the acute angles of a right triangle are equal, what is the value of each? 9. Find the value of each angle in an equiangular triangle. 10. Draw a triangle (the larger the better) , measure its angles, and find their sum. If the sum is not exactly 180, what is the reason ? NOTE TO THE TEACHER. The pupils should work No. 10 together, and the results should be put on the board and compared with 180. 64 LESSONS IN GEOMETRY. Lesson 32. 1. Compare two triangles with respect to size and shape. There are four cases, as follows : (1) Two triangles may have the same size (magnitude). They may have the same shape (form). They may agree both in size and shape. (4) They may differ both in size and shape. For example (Fig. 66) : the triangles ABC and DEF have the same size ; ABC and MNO have the same shape ; ABC and PQR agree both in size and shape ; and DEF and MNO differ both in size and shape. (2) (3) .D E M FIG. 66. 2. Define equivalent triangles; similar triangles; equal triangles. Two triangles are equivalent, if they have the same size. Two triangles are similar, if they have the same shape. Two triangles are equal, if they agree in size and shape. 3. To what extent do these definitions apply to other magnitudes than triangles? They apply, in general, to lines, to surfaces, and to solids. Thus, a curved line may have the same length as a straight line ; a field bounded by curved lines may enclose the same extent as a field bounded by straight lines ; a cubical vessel may hold the same quantity of water as a vessel shaped like TRIANGLES. 65 a cylinder. In all these cases, the size is the same, the shape different. They are examples of equivalent magnitudes. Again, two straight lines have the same shape whatever be their lengths ; so have two circles, two cubes, or two spheres, however much they differ in size. They are similar magnitudes. If two straight lines have the same length they are equal magnitudes ; so also are two circles, or two spheres, which have the same size. Of these three kinds of agreement, equivalence, simi- larity, and equality, the last is the simplest, and should be studied first. 4. The study of equal triangles raises two questions. What is the first question , and what is the answer ? If we know that two triangles are equal, what can we infer respecting their sides and their angles ? The answer is obvious. Since two equal triangles agree in size and in shape, they can differ only in position; and if placed one upon the other, they must coincide in all their six parts. In other words, their sides and angles must be equal, each to each. In the equal A ABC and PQR (Fig. 66), AB=PQ, AC=PR, BC=QR; ^ACB = ZPRQ, Z.ABC=^PQR, Z. BAG ' = Z QPR. In any two equal triangles, the sides opposite tivo equal angles are equal; two such sides are called homologous sides. 5. Draw, freehand, two equal triangles. Name the equal angles, pair by pair. Name the equal or homologous sides. 6. What is the other question about equal triangles ? How many, and what parts, of two triangles must be known to be equal, in order that we may conclude, without error, that the triangles themselves are equal? To answer this question, special cases must be examined. 66 LESSONS IN GEOMETKY. Make two unequal triangles : 7. Each having a given length for one side. 8. Each having two given lengths for two sides. 9. Each having a given angle. 10. Each having two given angles. What is true of the remaining pair of angles ? Why ? 11. Each having a given length for one side, and also a given angle adjacent to this side. 12. Each having a given length for one side, and a given angle opposite to this side. Make the angle first. 13. Is a triangle determined if one part is given? If two parts are given ? If the three angles are given ? 14. Theorem. Two triangles are equal, if two sides and the included angle of one are equal, respectively, to two sides and the included angle of the other, HYPOTHESIS. A ABC and DEF, AB=DE, AC=DF, CONCLUSION. -^ B D A ABC = A DEF. FlQ ' 67 ' PROOF. We may divide the proof into five steps. (1) Imagine A ABC placed on A DEF so that A falls on D, and the equal angles coincide. (2) Then E will fall on B, and F on C. Why? (3) Therefore EF will coincide with BC. Why? (4) Therefore the two triangles will coincide. Why ? (5) Therefore A ABC = A DEF. Since this reasoning applies equally well to any two tri- angles, the theorem is proved. 15. Name all the equal parts in the A ABC and DEF. TRIANGLES. 67 Lesson 33. 1, What is a theorem, and what are its parts? A theorem is a statement to be proved. A theorem may be separated into two parts : (1) The hypothesis, or what is assumed as true. (2) The conclusion, or what is to be proved. The proof of a theorem is the chain of reasoning by which the conclusion is drawn from the hypothesis. 2. Theorem. Two triangles are equal if a side and the two adjacent angles of one are equal, respectively, to a side and the two adjacent angles of the other. HYPOTHESIS. Two A ABC, DEF. AB = DE. A B D E FIG. 68. CONCLUSION. A ABC = A DEF. PROOF. Use the method of superposition. (1) Place A ABC on DEF so that the equal parts coin- cide. Mention these parts, pair by pair. (2) AC will coincide in direction with DF. Why? (3) BC will coincide in direction with EF. Why? (4) Therefore C, which is common to AC and BC, must fall on F, which is common to DF and EF. (5) Therefore the triangles will coincide. (6) Therefore A ABC = A DEF. The theorem is now proved, because the same reasoning will hold good for any two triangles that can be drawn. 3. Name all the equal parts in the A ABC and DEF. 68 LESSONS IN GEOMETRY. 4. Theorem. In an isosceles triangle the angles opposite the equal sides are equal. HYPOTHESIS. ABC an isosceles triangle. AC and BC the equal sides. CONCLUSION. Z ABC =Z BA C. A D B PROOF. Successive steps : FIG. 69. (1) Draw CD so as to bisect Z ACB. (2) Prove that A DBC = A DAC. They have three parts equal, each to each. CD is common. BC = AC by hypothesis ; Z BCD = Z ACD by construction. Why, then, are the two triangles equal ? (3) The two triangles being equal, their six parts are equal, pair by pair. Therefore Z ABC = Z BAG. 5. What method of proof is made use of in proving the above theorem? The method of equal triangles. This method may often be employed for the purpose of proving two lines equal, or two angles equal. We begin by finding or making two tri- angles in which the two lines, or the two angles, appear as corresponding parts. We then prove these triangles to be equal. Then it follows immediately that the two lines, or the two angles, must also be equal. 6. Apply this theorem to an equilateral triangle. In an equilateral triangle the angles, taken in pairs, must be equal, because in each case they are opposite equal sides. Therefore all three angles are equal. An equilateral triangle is also equiangular. TRIANGLES. 69 7. Theorem. If two angles of a triangle are equal, the opposite sides are also equal. HYPOTHESIS. ABC a triangle. ^ABC=Z.B and 7. The hypothesis of one is the conclusion of the other. When two theorems are thus related, either one is called the converse of the other. 9. How is the converse of a theorem formed ? By changing the hypothesis to the conclusion, and the conclusion to the hypothesis. 10. Apply the theorem of No. 7 to a triangle, all of whose angles are equal; that is to say, an equi- angular triangle. If all the angles are equal, then the sides taken in pairs must be equal, because each pair stands opposite equal angles. Therefore all the sides are equal. An equiangular triangle is also equilateral. 11. What is the value in degrees of each angle of an equilateral triangle ? 70 LESSONS IN GEOMETRY. Lesson 34. 1. Theorem. Two triangles are equal if the sides of one are equal, respectively, to those of the other. HYPOTHESIS. In the A ABC, DEF, AB = DE, AC=DF, BC=EF. CONCLUSION. A ABC '= A DEF. FIG. 71. PROOF. Successive steps : (1) Place A DEF so that the longest side DE coincides with AB, and F and C lie on opposite sides of AB. (2) Join OF. What two isosceles A are thus formed? (3) Prove that Z ACF= Z AFC (p. 68, No. 4). (4) Prove that Z.BCF=^BFC. (5) Prove that Z ACB= /.AFB. (6) Prove that A ACB= A AFB (p. 66, No. 14). (7) Therefore AACB^ADEF. 2. Theorem. Two right triangles are equal if the hypotenuse and a leg of one are equal, respectively, to the hypotenuse and a leg of the other. HYPOTHESIS. Two right A ABC, DEF; AC=EF, BC=DF. CONCLUSION. ABC = A DEF, c FIG. 72. TRIANGLES. 71 PROOF. Successive steps : (1) Place A DEF so that the leg DF coincides with BC, and A and F are on opposite sides of BC. What isosceles A is thus formed ? Why isosceles ? (2) Show that Z ACB = Z FCB (p. 63, No. 3) . (3) And that &ABC=AFBC; (4) That is, A ABC = A DEF. 3. Theorem.^? perpendicular is the shortest dis- tance from a point to a straight line. HYPOTHESIS. A straight line AB. A point P, not in AB. PC-LtoAB. A PD any other line drawn from P to AB. CONCLUSION. PC < PD. "C PROOF. Successive steps : FIG. 73. (1) Produce PC, making CQ = PC, and join DQ. (2) Prove that A PDC = A QDC. Give all the reasons. What two parts are equal by construction ? (3) Then show that PQ < PD + DQ (p. 23, Ax. 9). (4) Hence show that 2 PC < 2 PD. (5) And therefore PC < PD. 4, Define the distance from a point to a straight line. By the distance from a point to a line is meant the shortest distance ; therefore the perpendicular dropped from the point to the line. 72 LESSONS IN GEOMETRY. Lesson 35. 1. Theorem. Every point in the perpendicular which bisects a straight line is equidistant- from the ends of the line. HYPOTHESIS. D the middle point of AB. CD _L to AB. P any point in CD. CONCLUSION. PA = PB. B FIG. 74. PROOF. Prove that A ADP = A BDP. What follows ? 2, Theorem. Every point equidistant from the ends of a straight line lies in the perpendicular which bisects the line. HYPOTHESIS. D the middle point of AB. P any point such that PA = PB. CONCLUSION. PD is _L to AB. PROOF. Prove that A ADP = A BDP. FlG - 75 - What follows with respect to the angles ADP and BDP? Then show that Z ADP= 90 (see p. 48, No. 2). 3, What truth follows from the last theorem? Since two points determine a straight line, any two points equidistant from the ends of a straight line determine the per- pendicular which bisects the line. This perpendicular is called the bisecting perpendicular. 4. How are the two theorems on this page related ? 5. By what method are both theorems proved? TRIANGLES. 73 6. Theorem. Every point in the bisector of an angle is equidistant from the sides of the angle. HYPOTHESIS. AD the bisector of Z BAC. P any point in AD. CONCLUSION. P is equally distant from AB and from AC. FIG. 76. PROOF. Successive steps : (1) Draw PE _L to AB, PF _L to AC. (2) Prove that A APE = A APF (see p. 63, No. 3) . (3) What follows? 7. Theorem. Every point within an angle, and equidistant from its sides, lies in the bisector of the angle. HYPOTHESIS. P any point within Z. BAC equi- distant from AB and AC. CONCLUSION. PA bisects /. BAC. FIG. 77. PROOF. Successive steps : (1) Draw PE JL to AB, PF A. to AC. Join PA. (2) Prove that A APE = A APF. (3) What follows? 8. What truth follows from the last theorem? One point equidistant from the sides of an angle determines the bisector of the angle. Why is one point in this case sufficient ? 9. What relation is there between the theorems given on this page? Why? 74 LESSONS IN GEOMETRY. Lesson 36. Review. 1. Review the italicized exercises in Lessons 28-31. 2. Upon a given length as base, how many isosceles tri- angles can be constructed ? How many equilateral triangles ? 3. If the acute angles of a right triangle are 30 and 60, what angle is formed by their bisectors ? 4. If one angle of a triangle is 40, and the other two angles are equal, what are their values? 5. Find the angles of a triangle, if the second angle is twice the first, and the third three times the second. 6. Draw a A ABC, produce AB to any point Z>, and show that Z. DBC = Z BAG + Z ACS. 7. Show that a A with 1, 2, 3 for sides cannot be made. 8. Show that a A with 1, 2, 4 for sides cannot be made. 9. Two towns, A and _B, are 10 miles apart. A place is described as 7 miles from A, and 5 miles from B. Draw a plan showing the situation of (7. Is its position determined? 10. Construct a right triangle having one acute angle 20 larger than the other, and hypotenuse = 4J in. 11. Construct a A ABO ; given : AB = 2 in., Z. A = 30, ZC=45. 12. Construct a A ABC', given: AB=5 in., BC=3% in., /. BAC = 30, and show that there are two solutions. 13. How high is a tower if the angle of elevation of the top, 300 yds. from the base, is 20? 14. A, J5, (7, are the corners of a triangular field. A is 80 ft. west of B, and 800 ft. southwest of C. Find the cost of fencing the field at 50 cents a foot. TRIANGLES. 75 Lesson 37. Keview. 1. Review the italicized exercises in Lessons 32-35. 2. If one side of an equilateral triangle is produced, what is the value of the exterior angle thereby formed ? 3. Find the values of the angles at the base of an isos- celes triangle if the angle at the vertex is 24. 4. Find the angles of an isosceles triangle if each angle at the base is double the angle at the vertex. 9 5. What are the angles of an isosceles right triangle? 6. Can you devise an easy way to find the height of a tree by means of an isosceles right triangle ? 7. Construct an isosceles triangle, having given the base and the angle at the vertex. First find, by construction, the base angles 8. Construct the A ABC, given AB= 2 in., AC 3 in., BC 4 in. Then find a point D which is 2 in. from C and equidistant from A and B. 9. Prove that the lines joining the middle points of the sides of an equilateral triangle divide the triangle into four equal equilateral triangles. 10. Prove that the altitude of an isosceles triangle bisects the base and the angle at the vertex. 11. A monument AB on a horizontal plane subtends the angle 30 when seen from C (that is, Z ACB= 30) ; 150 ft. nearer, the angle it subtends is doubled. Find the height of the monument. 12. A man 75 ft. from the bank of a stream finds that a tree on the opposite bank subtends an angle of 45. At the bank, this angle is 60. How high is the tree? 76 LESSONS IN GEOMETEY. Lesson 38. Keview. 1. Prove that the construction in Fig. 26, on page 19, is correct ; that is, prove that it follows from the construction that CF is II to AB. For this purpose, show that A DCF= ADCE, and then apply No. 5, p. 51. Prone the following constructions correct. In each case first state the problem ; then describe how the construction was made ; then state what must be proved ; then prove it. 2.* Fig. 33, page 25. 5. Fig. 48, page 46. 3. Fig. 40, page 40. 6. Fig. 49, page 46. 4. Fig. 43, page 42. 7. Fig. 50, page 47. 8. Draw a A ABC and produce each side in both direc- tions. Name with letters all the angles that are formed, and find their values, if Z ABC = 70 and Z BAG '= 35. 9. In general, three parts determine a triangle. Can you think of an exception to this rule ? 10. Find a point in a given straight line AB which is equidistant from two given points C and D. Prove that your construction is correct. 11. Ju Fig. 78, AB is given, and an equilateral tri- angle is constructed, having AB for altitude. The con- struction is quite obvious from the figure. CD is drawn _L to AB ; all the arcs have the same radius. Make the construction, and prove it to be correct. TKIANGLES. 77 Lesson 39. Review. 1 . Through each vertex of a triangle a line parallel to the opposite side is drawn. Prove that the triangle formed by these lines is equal to four times the given triangle. 2. Find the values of all the angles of an isosceles tri- angle, if the angle formed by producing the base is 130. 3. In a A ABC the side AB is produced to any point D. Prove that the Z CBD = Z BAG + Z ACB. 4. Construct an isosceles right triangle whose hypotenuse shall be 6 cm long. 5. Describe the position of all points which are situated 2 cm from a given point A ; 3 cm from a given point B ; 2 cm from A, and 3 cm from B. 6. A surveyor erects two flagstaffs, A, B, on opposite sides of a pond, chooses a favorable point (7, and makes the following measurements: .4O=660 m , BC= 1380, Z C ' = 45. Find the distance AB. 7. In a triangular field ABC a man measures AB= 600 m , Z B = 70. He then walks over BC without measuring it. He finds, however, that Z (7=45. Draw a plan, and find BC. 8. Two roads meet at J., forming the angle 60. How far apart are two houses, one of which is situated 700 m from A along one road, and the other 900 m from A along the other road? 9. The distances between three towns A, J5, C are : AB = 9 km , AC= 7 km , BC= G km . A straight railroad runs from A to B. It is desired to have a station placed as near as* possible to C. Draw a plan, and find at what distance from A the station should be built. 78 LESSONS IN GEOMETRY. Lesson 4O. Review. 1. Re\iew all the italicized exercises in this chapter. 2. Find in- one side of a triangle a point equidistant from the other sides. Prove your construction correct. 3. Find a point within a triangle equidistant from the three vertices. Prove your construction correct. 4. Find a point within a triangle equidistant from the three sides. Prove your construction correct. 5. Three towns, J., jB, (7, are situated at the vertices of an isosceles triangle, whose base is the line BC. /. ABC= 30. The distance from A to the base BC is 6 1 " 11 . Find the dis- tances between the towns. X FIG. 79. 6. In order to find the distance XY (Fig. 79) , without any means for measuring angles, I chose four points, A, B, (7, J5, and measured their distances with the following results : AB=140 m , AD=18Q m , u4(7=115 m , .RD=65 m , CD=85 m . Draw a plan, and find XY. CHAPTER V. POLYGONS. Lesson 41. 1. Define a polygon: its sides; perimeter; vertices. A polygon is a plane figure bounded by straight lines. The bounding lines are called its sides ; their sum, its perime- ter ; their intersections, its vertices. 2. Name the most important polygons. Polygons are named according to the number of sides. A triangle has three sides ; a quadrilateral, four sides ; a pentagon, five sides ; a hexagon, six sides ; an octagon, eight sides ; a decagon, ten sides ; a dodecagon, twelve sides. 3. Define a diagonal, and explain how a polygon may be divided into triangles by drawing diagonals. A diagonal is a straight line joining two vertices not on the same side. A polygon may be divided into triangles by drawing all the diagonals possi- ble from one vertex. The num- ber of diagonals that can be drawn from a vertex is always three less than the number of sides ; and the number of tri- angles into which the polygon is divided is always two less than the number of sides. 80 LESSONS IN GEOMETRY. 4. Theorem. The sum of the angles of a polygon is equal to 180 X the number of sides less two. We prove this theorem for a hexagon as follows : HYPOTHESIS. A hexagon ABCDEF. CONCLUSION. The sum of the A of the hexagon is equal to 180 x 4, or 720. PROOF. Divide the hexagon into 6 2, or 4, triangles, by drawing the diagonals AC, AD, AE. Obviously, the sum of the A of the hexagon is equal to the sum of the A of these triangles. The sum of the A of the A = 180 x 4. Why? Therefore the sum of the A of the hexagon = 180 x 4. Now suppose that in place of 6 sides the polygon has any number n of sides ; then there would be n 2 triangles in place of 4 ; and, therefore, the sum of the A of the polygon would be 180x(n-2). 5. Prove, without assuming the above theorem to be true, that the sum of the angles of a quadrilateral is equal to 360. 6. If the angles of a quadrilateral are all equal, what is the value of each one? 7. Three angles of a quadrilateral are 40, 70, and 120. What is the value of the other angle ? 8. Find the sum of the angles of a pentagon. 9. Find the sum of the angles of an octagon. 10. Find the sum of the angles of a decagon. 11. Find the suin of the angles of a dodecagon. POLYGONS. 81 11. Classify quadrilaterals. Quadrilaterals are divided into three classes : parallelo- grams, trapezoids, and trapeziums. Parallelograms have their opposite sides parallel. Trapezoids have only two sides parallel. Trapeziums have no parallel sides. PARALLELOGRAM. TRAPEZOID. TRAPEZIUM. 12. Classify parallelograms. There are four kinds of parallelograms : the square, the rectangle, the rhombus, and the rhomboid. A square has four equal sides and four right angles. A rectangle has unequal adjacent sides and four right angles. A rhombus has four equal sides and no right angles. A rhomboid has unequal adjacent sides, and no right angles. SQUARE. RECTANGLE. RHOMBUS. RHOMBOID. 13. Make, freehand, a parallelogram ; a trapezoid ; a trapezium. 14. Make, freehand, a trapezium having two right angles. 15. Give examples of parallelograms. What kind is each ? 16. Make, freehand, a square ; a rhombus ; a rectangle ; and a rhomboid. 82 LESSONS IN GEOMETRY. Lesson 42. NOTE. In working the exercises in this lesson, construct perpen- diculars and parallels with the ruler and the triangle, and angles with the protractor. 1. How are the dimensions of a parallelogram named ? They are called base and altitude. Any side may be taken as the base, and then the altitude will be the perpendicular dropped to the base from any point of the opposite side. Thus, in Fig. 82, AB is the base, DE the altitude. A E B A E B FIG. 82. FIG. 83. 2. How are the dimensions of a trapezoid named ? They are also named base and altitude. One of the par- allel sides is always taken as the base, and the altitude is the perpendicular drawn from one parallel side to the other. Thus, in the trapezoid ABCD (Fig. 83) DE is the altitude. 3. How are the sides of a trapezoid named ? The parallel sides are called the bases, and the other two sides are called the legs. 4. Construct a parallelogram with a base of 3 in. and an altitude of 2 in. ; also a trapezoid with the same dimensions. 5. Draw a triangle ABC. Through D, any point in AC, draw DE II to AB, and DF II to BC. What kind of a figure is ABED ? Why ? What kind of a figure is BEDF ? Why ? POLYGONS. 83 6. Construct a square having a side 2 in. long. Draw its diagonals. Measure their lengths and the angle between them. 7. Construct a rhombus with each side 2 in. long, and one angle equal to 60. Draw its diagonals. Measure their lengths and the angle between them. 8. Construct a rectangle having for two adjacent sides l^in. and 2 in. Draw its diagonals. Measure their lengths and the angle between them. 9. Construct a rhomboid ABCD, having given AB= 3^ in., <7=lin., Z. ABC =55. Find the values of the other three angles. 10. Draw a line AB ; upon one side construct an equi- lateral A ABC, upon the other side a square ABDE. Join CD, CE. Prove that A CDE is isosceles. 11. Construct two rectangles, one within the other, and make their adjacent sides everywhere -J- of an inch apart. 12. Construct two squares. Divide one into four equal squares, and the other into four equal rectangles. 13. Construct Fig. 84 (scale, 2:1). Then erase the dotted lines, and shade the portions of surface at the corners and at the middle of the sides. 14. Through a point draw two per- pendicular lines. With centre and any radius, describe a circle, cutting the lines in A, B, C, D. Join AB, BC, CD, DA. What kind of a figure is ABCD? Can you prove your answer to be correct? 15. A man walks 5 miles due north, then 3 miles north- west, then 4 miles due south, and then he goes straight home. What figure does his path enclose? Find, by drawing to scale, how far he has walked. FIG. 84. LESSONS IN GEOMETRY. Lesson 43. 1. Theorem. The opposite angles of a parallelo- gram are equal. HYPOTHESIS. A parallelogram ABCD. Also let its angles be de- noted by a, 6,.c, d, as shown in the figure. CONCLUSION, a = c, b = d. E FlG . 85 . PROOF. Produce AS to 2, DC to F, and let We know that a = m (p. 50, No. 3). And that c = m (p. 51, No. 4). Therefore a = c (p. 23, Ax. 1). Also, we know that d = n. Why? And that b = n. Why? Therefore b = d. Why? 2. Theorem. The opposite sides of a parallelogram are equal. What is the hypothesis? D What is the conclusion ? PROOF. Successive steps : (1) Draw the diagonal AC. (2) Prove that A ABC A ADC. What follows? FIG. 86. 3. What truths follow immediately from No. 2? (1) A diagonal divides a parallelogram into two equal A. (2) Parallels between parallels are equal. POLYGONS. 85 4. Theorem. Two parallels are everywhere equi- distant. HYPOTHESIS. ? AB is II to CD. E, Gr, any points in AB. EF is _L to AB. GH is _L to AB. ^ ^ 5- CONCLUSION. EF=GH. FlGt 87> PROOF. ^FEG = /.HGB. Why? Then JS^ is II to <7# (p. 51, No. 5). Hence EF=GH. Why? 5. One angle of a parallelogram = 50. Find the others. 6. What is the angle which the diagonal of a square forms with one of the sides? 7. In what kinds of parallelograms is the altitude equal to one of the sides ? 8. How many trees 10 yds. apart can be set out around a rectangular park 800 yds. long and 480 yds. wide? 9. Upon a line AB as side, construct with ruler and compasses a square. Draw BC _L to AB ; take BC AB ; and with centres A and 0, and radius AB, describe arcs meeting at D. ABCD is the required square. Prove that it is a square. 10. Construct a square having its perimeter equal to a line MN. 11. Construct a square having its diagonal equal to a line MN. 12. Make a plan of a wall 20 ft. high and 64 ft. long, with a passage-way in the middle 10 ft. high and 16 ft. wide. 13. Make a plan of two vertical posts 18 ft. high and 24 ft. apart, with a cross-bar 10 ft. from the ground. 86 LESSONS IN GEOMETRY. Lesson 44. 1, Theorem. The diagonals of a parallelogram bisect each other. HYPOTHESIS. Let ABCD be a parallelo- gram, and let the diagonals AC, BD, intersect at 0. CONCLUSION. FIG. 88. PROOF. Prove that A ABO = A DCO, and keep in mind that in equal triangles equal sides are opposite equal angles. 2. Prove that the diagonals of a rectangle are equal. Begin by stating the hypothesis (Fig. 89) ; then state the conclusion ; then prove by showing that the triangles ABD, ACD, are equal. 3. Prove that the diagonals of a rhombus are perpen- dicular to each other (Fig. 90) . State hypothesis and con- clusion, and prove by showing that A AOB = AAOD, etc. C FIG. 89. 4. Prove that the construction, on page 25, for dividing a line into equal parts, is a correct one. Draw CM, DN, etc., II to AB, prove the A AHC, CMD, etc., equal, and apply No. 3, p. 84, and Ax. 1, p. 23. POLYGONS. 87 5. Problem. To construct a parallelogram, having given two adjacent sides and the included angle. CONSTRUCTION. Instruments : ruler and compasses. Let a and b be the given sides ; ra the given angle. Make Z BAD = m. Lay off AB = a, AD = b. ' j~~ With centre D and radius a describe an arc EF. With centre B and radius b describe an arc cutting EF at (7. Join BC and DC. ^ 9i ABCD is the figure required. PROOF. (1) The figure obviously has the three given parts. (2) Join .4(7. A ABC = A ADC. Why? (3) Therefore Z BAC = Z DC A, and Z .4<7= Z C^Z>. (4) Therefore AB is II to DC, and C is II to AD. Why? (5) Therefore ABCD is a parallelogram. Why? NOTE. In the following exercises, use ruler and compasses only. First make the construction ; then describe it ; then prove it to be correct. 6. Construct a square, having given a side. 7. Construct a square, having given the diagonal. 8. Construct a rectangle, given two adjacent sides. 9. Construct a rectangle, given a side and a diagonal. 10. Construct a rhombus, given the two diagonals. 11. Construct a rhombus, given a side and an angle. 12. Construct a parallelogram, given two sides and a diagonal. 13. Construct a parallelogram, given one side, one angle, and one diagonal. 14. Construct a trapezoid, given the bases, the altitude, and one of the legs. LESSONS IN GEOMETRY. Lesson 45. 1. Problem. To construct a polygon equal to a given polygon ABCDEF (Fig. 92). CONSTRUCTION. Instruments : ruler and compasses. Draw the diagonals AC, AD, AE. Make A MNO = A ABC (explain fully how) . Then make A MOP = A ACD. Also A MPQ = A ADE. And A MQR = A AEF. Then the polygon MNOPQ R is the polygon required. ED Q P B N FIG. 92. PROOF. By construction, the A, pair by pair, are equal. They are also, pair by pair, similarly placed. Therefore the polygon MNOPQR=the polygon ABCDEF. NOTE. Use only ruler and compasses for the following exercises. First make the construction ; then describe it ; then prove it to be correct. 2. Construct a square, given the perimeter. 3. Construct a rectangle, given one side and the angle between the diagonals. 4. Construct a rectangle, given a diagonal and the angle between the diagonals. 5. Construct a rectangle, given one side and the perimeter. 6. Construct a rhombus, given one side and the altitude. POLYGONS. 89 7. Construct a rhombus, given a side and a diagonal. 8. Construct a parallelogram, given the base, the alti- tude, and one angle. 9. Construct a parallelogram, given the two diagonals and the angle between them. 10. Construct a parallelogram, given the sides, and know- ing that one angle is double the other. 11. Construct a trapezoid, given the legs, one base, and the altitude. 12. Construct a trapezoid, given one leg, one of the bases, and the angles at the base. 13. Construct a trapezium, given the four sides and one of the angles. 14. Construct a trapezium, given the four sides and one of the diagonals. 15. Make a pentagon, and then construct an equal pen- tagon. 16. Make a hexagon, and then construct an equal hexagon. 17. Make an octagon, and then construct an equal octagon. 18. Draw any triangle ABC. Upon AB describe the square ABEF, and upon AC the square ACGH. Drop a perpendicular from A to BC, and join EC, GB. If the figure is correctly drawn, EC and GB will intersect on the perpendicular. 19. Draw a cross vertical section of a ditch whose depth 'is 6 ft., breadth at bottom 4 ft., at top 8 ft., and whose sides are equal. Find the length of the side. In this exercise, besides ruler and compasses, the scale must be used. 20. Construct a square, and on the four sides construct equilateral triangles ; join their vertices, and show by meas- urement that the figure so formed is also a square. Can you prove that it is a square ? 90 LESSONS IN GEOMETRY. Lesson 46. 1. Define an equilateral polygon. An equilateral polygon is a polygon whose sides are equal. 2. Define an equiangular polygon. An equiangular polygon is a polygon whose angles are equal. 3. Define a regular polygon. A regular polygon is a polygon which is both equilateral and equiangular. 4. How is an angle of a regular polygon found ? First find the sum of all the angles (p. 80, No. 4) ; then divide this sum by the number of angles ; in other words, by the number of sides of the polygon. 5. What are the usual names of the regular triangle and the regular quadrilateral ? 6. What is the perimeter of a regular octagon, if one side is 6 in. long? 7. What is the length of one side of a regular decagon, if the perimeter is 35 in. ? 8. What quadrilateral is equilateral, but not equiangular? also what quadrilateral is equiangular, but not equilateral? 9. Make a pentagon which shall be equilateral, but not equiangular. 10. Make a hexagon which shall be equiangular, but not equilateral. 11. Find the angle of a regular dodecagon. 12. Find the angles of the regular polygons from three sides to ten sides, and arrange the results in tabular form. POLYGONS. 91 As the number of sides increases, in what way does the angle change? 13. Stone pavements furnish examples of the use of regu- lar polygons. Only those polygons can be used which will fill up all the space about a point. Show that equilateral triangles may be used, and make a pattern like that in Fig. 93. 14. Make a pattern arranged in squares, like that shown in Fig. 94. FIG. 93. FIG. 94. FIG. 95. 15. Show that a pattern can be made of regular hexagons arranged in groups of three about a point, and make a pat- tern like that in Fig. 95. 16. Make a pattern consisting of a combination of equi- lateral triangles and hexagons (Fig. 96). FIG. 96. FIG. 97. 17. Show that a pattern cannot be made of regular octa- gons alone, but that one can be made of regular octagons combined with squares, as shown in Fig. 97. 92 LESSONS IN GEOMETRY. Lesson 47. Review. 1. Review the italicized exercises in Lessons 41-43. 2. A diagonal divides a parallelogram into two'equal tri- angles. What kind of triangles are they if the parallelogram is a rectangle ? a rhombus ? a square ? 3. The perimeter of a rectangular lot is 360 ft., and the street side is 80 ft. Find the depth of the lot. 4. Make, freehand, a hexagon having as many right angles as possible. 5. Find the sum of the angles of a polygon of 15 sides. 6. Prove that the figure formed by joining the middle points of the sides of a square, taken in order, is also a square. 7. Draw a triangle ~ABC. Through B and C draw par- allels to the opposite sides, meeting at D. What kind of a figure is ABCD? Why? 8. A man walks 8 miles east; then 4|- miles southwest; then 8 miles west ; then straight home. Draw a plan, and find how far he walked. 9. Construct a trapezoid having an altitude of 1 in., and equal legs each 2 in. long. 10. Construct a quadrilateral ABCD, given AEAD = 2 in., BC= CD = 1 in., Z BAD = 35. 11. A man walks 3 miles ; turns to his right through 60, and goes 2 miles ; turns again to his right, through 60, and goes 2 miles ; and then goes straight home. Draw a plan, and find how far he has walked. 12. Construct a square ABCD, and upon its sides equi- lateral triangles ABE, BCF, CDG. Join E, F, G, H, taken in order. What kind of a triangle is AEH? What kind of a figure is EFGH1 Prove your answers, POLYGONS. 93 Lesson 48. Review. 1. Review the italicized exercises in Lessons 44-46. 2. Into what figures is a rhombus divided by drawing both its diagonals? 3. Construct a rhombus, one side of which shall be 2 in. long, and one angle of which shall be double the other. 4. The intersection of the diagonals of a parallelogram is often called its centre. Prove that any straight line which can be drawn through the centre will divide the perimeter into two equal parts. 5. Construct a square, given one side and the position of its centre. 6. Construct a trapezoid, given one base, the altitude, and the two diagonals. 7. A stone dam is 20 ft. high, 34 ft. wide at the bottom, and 4 ft. wide at the top. The slant height of each side is the same. Draw to scale a cross section of the dam, and find the slant height. 8. Construct a parallelogram, given one side and the two diagonals. 9. A man walks 5 miles straight ; then turns through 30, and walks 2 miles ; then takes a direction parallel, but oppo- site, to his first direction. Draw a plan of his route. How far is he from his starting-point? 10. Draw two lines making an angle of 60, and find a point which shall be 1 in. distant from one line, and 2 in. distant from the other. 11. A certain town is 3 miles from a straight railroad. A man lives 5 miles from the town, and 1 mile from the rail- road. Find, by construction, how many locations satisfy these conditions. 94 LESSONS IN GEOMETBY. Lesson 49. Review. 1. Construct a square, given the diagonal = 7 cm . 2. Construct a rhombus ABCD, given AB=5 cm , Z ABC = 45. 3. Construct a rectangle ABCD, given AB = 4 cm , and the diagonal AC= 9 cm . Construct a parallelogram ABCD, having given : 4. AB = 3 cm , AD = 5 cm , Z ABC =70. 5. AB = 4 cm , AD=7 cm , AC=9 cm . 6 AC = 7 cm Z ACD 30 Z DAC 1 = 50 7. AB=6 cm , BD=8 cm , ZABD=30. 8. ^B=4 cm , Z CAB = 80, Z DAB =100. Construct a trapezoid ABCD (AB II to CD) , having given : 9. AB=3 cm , CD5 cm , AD=4: cm , ZADC= 70. 11. BD=Q cm ,ZDBC=QO,ZBDC= 30, ZADB= 30. 12. AB=4: cm , BD=5 cm , BC=3 cm , Z.CBD= 80. Construct a quadrilateral ABCD, having given : 13. AB=3 cm , AC=5 cm , BD= cm , 14. AB=3 cm , AC=4: cm , BD=5 cm , 15. Draw a plan of a rectangular floor 15 m long and 8 m wide, and find the distance between the opposite corners. 16. ABCD is a rectangle. JLB=1200 m , BC= 600 m . What is the distance from A to the middle point M of the side CD? What kind of a triangle is ABM? Prove your answer. 17. A barn is 12 m wide, 8 m high to the eaves, and 14 m high to the ridgepole. Draw a plan of a vertical cross section. How long are the rafters? CHAPTER VI. THE CIRCLE. Lesson 5O. 1. Review the italicized exercises in Lesson 7. 2. Define an angle at the centre, a sector, a seg- ment, a quadrant, a semicircle. An angle at the centre is an angle whose vertex is at the centre of a circle. (For example : Z. AOB, Fig. 98.) A sector is a portion of a circle bounded by two radii and the intercepted arc. A segment is a portion of a circle bounded by an arc and its chord. A quadrant is a sector whose radii form a right angle. A semicircle is a sector whose radii form a straight angle, or segment whose chord is a diameter of the circle. 3. Define a semi-circumference. 4. Compare, as regards size, a circle, a semicircle, and a quadrant. 5. Draw a circle, and make in it an angle at the centre ; a sector ; a seg- ment ; a quadrant. Also name them with letters. 6. Into what figures is a sector divided by the chord of its arc? 7. Construct a sector whose angle is 60. 8. Construct a segment whose chord = radius of the circle. 90 LESSONS IN GEOMETRY. 9. Define equal circles and equal circumferences. Two circles, or two circumferences, are equal if they can be so placed that they coincide. 10. What truth follows immediately from the defi- nitions which have been given ? If two circles have equal radii, or equal diameters, they are equal ; and their circumferences are also equal. 11. Define equal arcs (p. 44> No. 6). 12. What conditions must two equal arcs satisfy? (1) They must have equal radii. (2) They must correspond to equal angles at the centre. 13. Draw two unequal arcs having equal radii. 14. Draw two unequal arcs that correspond to equal angles at the centre. (The radii must be unequal.) 15. Theorem. JT& equal circles, equal angles at the centre intercept equal arcs (Fig. 99). What is the hypothesis ? What is the conclusion ? PROOF. Place /. AOB on Z CRD so that they coincide. A will fall on C. Why? ^ B will fall on D. Why? Arc AB will coincide with arc CD because all their points are at equal distances from R. 16. Theorem. In equal circles, equal arcs corre- spond to equal angles at the centre (Fig. 99). What is the hypothesis ? What is the conclusion ? Prove by the method of superposition (as in No. 15) . 17. In what relation does No. 16 stand to No. 15? THE CIRCLE. 97 18. In what way is the term " subtend" used? A chord AB (Fig. 100) is said to subtend the arc AB, and the arc AB is said "to be subtended by" the chord AB. To subtend means to stretch across. 19, Theorem. In equal circles, equal chords sub- tend equal arcs (Fig. 100). What is the hypothesis ? What is the conclusion ? PROOF. Apply No. 1, p. 70, and No. 15, p. 96. FIG. 100. 20. Theorem. /^ equal circles, equal arcs are sub- tended by equal chords (Fig. 100). What is the h}*pothesis? What is the conclusion ? PROOF. Apply No. 16, p. 96, and No. 14, p. 66. 21. What relation is there between Nos. 19 and 20? 22. Problem. To make an arc equal to a given arc AB whose centre is at a given point (Fig. 101). CONSTRUCTION. Draw OA, OB, AB. With radius OA, and any point R as centre, describe an arc CE. / \ \ With centre (7, and radius / \ \ / equal to AB, cut CE at D. ^ ^ V Then arc CD = arc AB. FIG. 101. PROOF. Apply No. 19. 23. Make an arc equal to three times a given arc. LESSONS IN GEOMETRY. Lesson 51. 1. Theorem. The radius perpendicular to a chord bisects the chord, the arc subtended by the chord, and the corresponding angle at the centre. HYPOTHESIS. Let the radius 00 be J_ to the chord AB, and meet AB at 0. CONCLUSIONS. AD=BD, arc AC '= arc J50, PROOF. Compare A AOD, BOD, and apply No. 16, p. 96. 2. Theorem. The perpendicular erected at the middle point of a chord passes through the centre of the circle. HYPOTHESIS. Let AB be a chord. Let D be its middle point. Also let DE be J_ to AB. CONCLUSION. DE passes through 0, the centre of the circle. FIG. 103. PROOF. DE passes through all points which are equally distant from A and B. (No. 1, p. 72.) is a point equally distant from A and B. Why ? Therefore DE passes through 0. 3. Bisect an arc, and prove your construction correct. 4. Construct an arc containing 22 30'. THE CIRCLE. 99 5. Problem. To describe a circumference through three points, A, B, C, not in a straight line. CONSTRUCTION". Draw AB, EG. Draw the bisecting perpendiculars to AB and BC. Let these _k meet at 0. With centre and radius OA de- scribe a circle. This circle will pass through the points A, B, and C. PROOF. Apply No. 1, p. 72, or No. 2, p. 98. 6. Can a circle be described through three points which are in the same straight line ? 7. Describe a circle and erase the centre. Then find the centre by a construction, and prove that your construc- tion is correct. (Apply No. 2, p. 98.) 8. Draw a triangle, and then describe a circle which shall pass through its vertices. 9. Make a square of 2 in. side, and describe a circle which shall pass through its vertices. 10. Describe three circles, all passing through two points A and B. How many circles can be described through two given points ? In what line are all the centres located ? 11. Describe a circle with a radius of 2 in. which shall pass through two points A and jB, 3 in. apart. 12. Mark two points A and B, and draw a line CD. Then describe a circle whose centre shall be in the line OZ>, and whose circumference shall pass through both A and B. Can you draw CD so that the problem cannot be solved ? 100 LESSONS IN GEOMETRY. Lesson 52. 1. Define an inscribed angle. An inscribed angle is an angle whose vertex is" in the cir- cumference of a circle, and whose sides are chords. For example : Z SAC (Fig. 105) is an inscribed angle. 2. Theorem. An inscribed angle is equal to half the angle at the centre, which intercepts the same arc. HYPOTHESIS. Circle with centre 0. An inscribed Z BAG. CONCLUSION. BAC = | BOO. Case 1 (Fig. 105). The centre in the line AB. PROOF. ZBAC + ZOCA + Z^OC=180. Why? Also Z BAG = Z OCA. Why ? Therefore 2 Z BAG + Z AOC =180. Why? But ^BOC + Z AOC = ISO . Hence 2 ZJB.4C + Z .40(7 = ZBOC + ZAOC. Why? Therefore 2^ BAG = /.BOG. (Ax. 3.) Therefore Z BAG =|Z BOG. (Ax. 5.) A A A FIG. 107. Case 2 (Fig. 10G). The centre between AB and AC. Draw the diameter AD. Apply Case 1. Add the results. Case 3 (Fig. 107). The centre outside the angle BAG. Draw AD. Apply Case 1. Subtract the results. THE CIRCLE. 101 3. When is an angle inscribed in a segment? When its vertex is in the arc of the segment, and its sides pass through the ends of the arc. 4. Make a segment, and inscribe in it three angles. 5. What truths follow immediately from JVa 2 ? (1) Inscribed angles that intercept equal arcs are equal. (2) All angles inscribed in the same segment are equal. For they all intercept the same arc. Thus, the A ACB, ADB, AEB (Fig. 108), all intercept the arc AB. (3) Every angle inscribed in a semicircle is a right angle. For the intercepted arc is a semi-circumference. Thus, the A ACB, ADB, AEB (Fig. 109) are right angles. D n FIG. 108. O FIG. 109. 6. What is the value of an inscribed angle if the inter- cepted arc is 60? 153? 320? 118 35'? 7. What value has an angle inscribed in a segment whose arc contains 60? 90? 135? 180? 270? 300? 8. Prove the construction in Fig. 27, p. 19, to be correct. Construct a right triangle, having given : 9. Hypotenuse = 3 in., one leg = If in. 10. Hypotenuse = 3 in., altitude upon hypotenuse = 1 in. 11. Hypotenuse = 3 in., one acute angle = 60. 12. A point A is 2 miles from a line BC ; ; BC=3 miles, / BAC= 90. Find AB and AC (by drawing to scale) . 102 LESSONS IN GEOMETRY. Lesson 53. 1. Define a tangent to a circle. A tangent to a circle is a straight line which touches the circumference at a point, but does not cut it. A tangent is said to touch the circle. The point where it touches the circle is called the point of contact. 2. Theorem.,^ perpendicular ; erected at the end of a radius is a tangent to the circle. HYPOTHESIS. A circle with centre 0. A radius OA. [ o The line MAN JL to OA. CONCLUSION. The line MN is a tangent to W the circle at the point A. PROOF. Join to any other point in MN, as B. Then we know that OA < OB (No. 3, p. 71). Therefore B must lie outside the circle. In the same way we can show that every point in MN except A must lie outside the circle. Therefore, by definition, the line MNia a tangent, and A is the point of contact. 3. Mark a point A in a line MN, and then describe three circles which shall touch MN at the point A. Take for the radii of the circles 1 in., 2 in., and 3 in. How many circles touching MN at A can be described ? In what line do their centres lie ? 4. Draw a line MN, and then describe three circles with the same radius, all of which shall touch MN. How many circles, all having the same radius and all touching MN, can be described ? In what lines do their centres lie ? THE CIRCLE. 103 5. Problem. To draw a tangent to a given circle through a given point A. Case 1 (Fig. 111). Let A be in the given circumference. CONSTRUCTION. Draw the radius OA, and through A draw the line BAG _L to OA. BAG is the required tangent. PROOF. See No. 2. B FIG. 111. FIG. 112. Case 2 (Fig. 112). Let A be outside the given circle. CONSTRUCTION. Join AO. LTpon AO as diameter describe a circle. This circle will cut the given circle at two points, B and G. Draw the lines AB and AC. Then AB and AC are tangents to the given circle. PROOF. The angles ABO, AGO, are right angles. Why? Also B and G are the ends of radii. Therefore AB and AC are tangents. Why? 6. Prove that the tangents AB, AC (Fig. 112), are equal. 7. Describe a circle with a radius of 2 in., and draw tan- gents to it from a point 3J in. from the centre. 8. Describe a circle passing through a given point, and touching a given line at a given point in the line. Describe a circle, and then draw a tangent : 9. Parallel to a given straight line. 10. Perpendicular to a given straight line. 104 LESSONS IN GEOMETKY. Lesson 54. 1. When are two circles said to touch each other? Two circles are said to touch each other, or to be tangent to each other, when they both touch the same straight line at the same point. If the circles lie on opposite sides of the straight line, they are said to touch each other externally (Fig. 113). If they lie on the same side of the straight line, they are said to touch each other internally (Fig. 114). In both cases, the line is a common tangent to the circles. 2. What truth easily follows from No. 7 ? If two circles touch each other, the radii drawn to the point of contact are both perpendicular to the common tan- gent (No. 2, p. 102). Therefore they must lie in the same straight line. In other words, the centres and the point of contact lie in the same straight line. FIG. 113. FIG. 114. 3. Construct an equilateral triangle, side 2 in., and by means of it describe three circles, so that each circle shall touch the other two circles. 4. Describe the smallest circle which can touch a given circle and pass through a given point exterior to the given circle. 5. Describe a circle touching two given parallel lines, and passing through a given point between the lines. THE CIRCLE. 105 6. Draw an angle, and describe three different circles, each touching the sides of the angle. How many circles touching the sides of a given angle can be described? In what line do their centres lie ? 7. Draw a triangle, and describe a circle having its centre in one side and touching the other two sides. 8. How many common tangents can be drawn in Fig. 113? 9. Define circumscribed figures and inscribed figures. A circle is circumscribed about a polygon if its circum- ference passes through all the vertices of the polygon ; and the polygon is, in this case, inscribed in the circle. A circle is inscribed in a polygon if it touches all the sides of a polygon ; and in this case the polygon is circumscribed about the circle. 10. Illustrate by figures the definitions in No. 9. 11. Draw a triangle, and then circumscribe a circle about the triangle (see p. 99, No. 5). 12. Inscribe a circle in a given triangle. (Bisect any two angles. Prove your construction correct. See No. 6, p. 73.) 13. Circumscribe a circle about an equilateral triangle, and also inscribe a circle in the same triangle. 14. Inscribe a circle in a square whose side is 2 in. 15. Construct,, a square of side 3 in., inscribe a circle in the square, and form an octagon by cutting off the corners with tangents to the circle. 16. Construct a square of side 4 in., and then describe four equal circles, so that each circle shall touch two of the other circles, and also two sides of the square. 17. Draw, a rhombus, and inscribe in it a circle. 106 LESSONS IN GEOMETRY. Lesson 55. 1. Define the centre of a regular polygon. The centre of a regular polygon is a point equidistant from all the vertices of the polygon. 2. Theorem. The point where the bisectors of any two angles of a regular polygon meet, is the centre of the polygon. HYPOTHESIS. Let ABCDE be a regular polygon. Let the bisectors of the A A and B meet at 0. CONCLUSION. AO, BO, CO, DO, EO, are all equal. PROOF. Successive steps. Supply all the reasons. (1) A OAB is isosceles, and OA= OB. (2) A OBC = A OAB (No. 14, p. 66). (3) A OCD = A OBC (No. 14, p. 66). (4) A ODE= A OCD (No. 14, p. 66). FIG. 115. What follows? What follows? What follows ? 3. What truths follow immediately from No. 2 ? (1) The lines draivn from the centre of a regular polygon to the vertices bisect all the angles. (2) A circle can be circumscribed about every regular polygon. 4, Theorem. The centre of a regular polygon is equidistant from all the sides. What is the hypothesis (use Fig. 115) ? What is the conclusion ? PROOF. Compare the right triangles OFB, 0GB ; and then OGC, OHC, etc. (see p. 63, No. 3; p. 67, No. 2). THE CIRCLE. 107 5. What truths follow immediately from JV0. 4 ? (1) Perpendiculars dropped from the centre of a regular polygon to the sides bisect the sides. (2) A circle can be inscribed in every regular polygon. 6. Define the radius and the apothem of a regular polygon. The radius of a regular polygon is a line drawn from the centre to any one of the vertices. The apothem of a regular polygon is a perpendicular dropped from the centre to any one of the sides. 7. Show how to find the centre of a regular polygon. 8. Inscribe a square in a given circle. To do this, draw two perpendicular diameters, and join their ends, taken in order (Fig. 116). Prove that the figure thus obtained is a square. 9. Inscribe in a given circle a regular octagon. FIG. 117. 10. Inscribe in a given circle a regular hexagon. To do this, apply the radius as a chord six times (Fig. 117). Then prove that the figure thus formed is a regular hexagon (show that Z AOB = 60) . 11. Inscribe in a given circle an equilateral triangle. 12. Inscribe in a given circle a regular dodecagon. 13. Construct a regular hexagon with a side equal to 3 in. 108 LESSONS IN GEOMETRY. Lesson 56. 1. What is the relation between the circumference of a circle and the diameter ? The diameter of a circle is contained in the circumference a little more than three times. The exact value of the quotient cannot be expressed by a number, but it is known to be the same for all circles, and, for convenience, is represented by the Greek letter TT (pronounced like pie) . The value of TT, accurate enough for most purposes, is 3* or f - The value of ?r, correct to five decimal places, is 3.14159. If we take TT = ^ 2 -, the relation between circumference and diameter is expressed by either of the following equations : Circumference = diameter x 3?-. Diameter = circumference x -^V The relation is often expressed in the following general form, in which c denotes circumference, and r radius : NOTE. The method by which the value of TT is shown to be the same for all circles, and the process of computing its approximate values, require the aid of theorems which are not given in this book. The value has been computed to over 700 decimal places. Find the circumference of a circle, having given : 2. Radius 7 in. 6. Diameter 77 ft. 3. Radius 21 in. 7. Diameter 49 ft. 4. Radius 40 ft. 10 in. 8. Diameter 18 yds. 2 ft. 5. Radius 6 yds. 1 ft. 3 in. 9. Diameter 2 T \ in. 10. The circumferences of two concentric circles are 16J feet and 18 ft. Find the width of the ring between them. w THE CIRCLE. 109 11. If the radius of a circle is 3^ in., what is the length of an arc of 30 ? The circumference = 2 X f X ty in. = 22 in. The circumference is divided into 360. Therefore an arc of I = ^j- in. And an arc of 30 = 22x8 in. = lf in. Find the length of an arc, having given : 12. Circumference 100 in., angle at centre 30. 13. Radius 14 in., angle at centre 45- 14. Radius 42 ft., angle at centre 11 15'. 15. Diameter 70 in., angle at centre 36. Find the radius of a circle, having given : 16. Circumference 132 ft. 17. Circumference 198 ft. 18. Arc of 60= G6 ft. 19. Arc of 15= 3 ft. 8 in. 20. A wheel makes 220 revolutions in going half a mile. What is its diameter ? 21. A wheel whose diameter is 3J ft. made 1200 revolu- tions in going a certain distance. Find the distance. 22. How deep is a well if the wheel, whose diameter is 2 ft. 4 in., makes 30 revolutions in raising the bucket? 23. What angle at the centre will intercept an arc of 6 ft. 5 in. if the radius of the circle is 8 ft. 2 in. ? Circumference 2 X - 2 7 - X 98 in. 616 in. Since 616 in. contain 360, 1 in. of arc will contain -, and 77 in. will contain 77 * 3600 , or 45. 616 24. If the radius of a circle is 7 in., what angle at the centre will intercept an arc 7 in. long ? 110 LESSONS IN GEOMETRY. Lesson 57. Review. 1. Review all the italicized exercises in Lessons 50-53. 2. Make an arc of 225. 3. Make an arc equal to twice a given arc. 4. Make a sector equal to twice a given sector. 5. Describe a circle with a radius of 2 in. and passing through two given points A and _B. When is the solution of the problem impossible? 6. If airy number of parallel chords are drawn in a circle, upon what line are all their middle points located ? 7. Through a given point within a circle draw that chord which is bisected at the given point. Prove that your construction is correct. 8. A segment is made by joining the ends of the arc of a quadrant. What is the value of any angle inscribed in this segment? 9. If a series of circles are made, all of them touching two parallel lines, upon what line will all their centres lie? 10. Describe a circle touching two given parallel lines, one of them at a given point A. 11. A man wishes to locate his house so that it shall be 2 miles from a church, and equidistant from the homes of two of his friends. Show by a construction how the proper location may be found. 12. A man walks 3 miles in a straight line ; and then walks in the arc of a circle through whose centre he has passed, and whose radius is 1 mile, till he has made the chord of the arc equal to the radius. Find how far he is from home. THE CIRCLE. Ill Lesson 58. Review. 1. Review all the italicized exercises in Lessons 54-56. 2. Describe a circle, and upon one of its radii OA as a diameter describe another circle. Through A draw any straight line, meeting the circumferences again at B and C. Prove that AB = BC. 3. Make two equal circles touching each other externally, and draw all the common tangents. 4. Make two equal circles, exterior to each other and not touching. Then draw all the common tangents. 5. Inscribe a regular hexagon in a circle, and then cir- cumscribe a regular hexagon about the same circle. This is done by drawing tangents through the vertices of the in- scribed hexagon. Can you prove that the figure formed by these tangents is a regular hexagon ? 6. Inscribe four equal circles in a given square so that each circle shall touch two other circles and one side only of the square. 7. A circular park is laid out having a radius of 350 ft. What will it cost to build a stone wall around the park at $8.00 per yard? 8. The latitude of Leipsic is 51 21', that of Venice is 45 26'. Venice is due south of Leipsic. How many miles are they apart? (Take as the radius of the earth 4000 miles.) 9. If the radius of a circle is 4 ft. 8 in., what is the perimeter of a sector whose angle is 45? 10. Draw four intersecting lines so as to make four tri- angles, and circumscribe a circle about each of these triangles. If your work is correct, the four circumferences will be found to pass through the same point. 112 LESSONS IN GEOMETRY. Lesson 59. Review. NOTE. In the exercises of this lesson take 7r = 3.1416. 1. Review all the italicized exercises in Chapter VI. 2. Find the circumference of a circle if the radius = 10 m . 3. Find the radius of a circle if the circumference = 10 m . 4. What must be the diameter of a round dining-table for 12 persons, if 60 cm is allowed to each person? 5. A wheel 75 cm in diameter makes 3000 revolutions in going a certain distance. What is the distance ? 6. How many trees 10 m apart can be set out around the edge of a circular pond 500 m in diameter ? 7. The radii of two concentric circumferences are 8 cm and 10 cm . Find the length of the circumference situated just half way between them. 8. Two toothed wheels work together. One has 24 teeth, the other has 144 teeth. The distance between the centres of the teeth is the same on each wheel. How many revolu- tions will the second wheel make while the first is making 600? If the radius of the first wheel is 4 cm , what is the radius of the other? 9. Describe a circle with a radius of 2 cm , and then cir- cumscribe about the circle a square. Find (to the nearest millimeter) the difference in length between the circumference of the circle and the perimeter of the square. Diameter of a circle = 10 cm . Find the length of the arc : 10. Intercepted by a side of the inscribed square. 1 1 . Intercepted by a side of the inscribed regular pentagon. 12. Intercepted by a side of the inscribed regular hexagon. 13. Intercepted by a side of the inscribed regular octagon. . CHAPTER VII. AREAS. Lesson 6O. 1. How are surfaces measured? Surfaces, like lines, are measured by choosing a unit, and then finding how many times this unit is contained in the surface which we wish to measure. The number of times the unit is contained in the surface to be measured, followed by the name of the unit, is called the area of the surface. 2. What units of area are in common use ? The square inch, the square foot, the square yard, the acre, and the square mile. All these units, except the acre, are squares whose sides are equal to the units of length. Their abbreviations are formed by prefixing to those of the corresponding units of length the letters " sq." (standing for " square"). 3. How are those units related ? These units are related as shown in the following table : 144 sq. in. = 1 sq. ft. 9 sq. ft. = 1 sq. yd. 43,560 sq. ft.= l acre. 640 acres = 1 sq. mile 4. How many square feet make 1 square mile? 5. How many square feet in J of an acre? 6. Reduce 16 sq. in. to the fraction of a square foot. 7. Reduce -| of a square foot to square inches. 8. A man bought an acre of land for $5000, and sold it for 20 cents per square foot. How much money did he make ? 114 LESSONS IN GEOMETRY. FIG. 118. 9. How can a square ~be divided into smaller squares ? Let ABCD (Fig. 118) be a square. Divide AB into any number of equal parts (here, six). Draw through the points of division lines parallel to AD. These lines will divide the square into six equal rectangles. Divide also AD into six equal parts, and draw through the points of division lines parallel to AB. These lines will subdivide each rectangle into six equal squares. Therefore the entire square is now divided into 6x6 or 36 equal squares. In general, the number of equal squares into which a square is decomposed by this process is found by multiplying the number of equal parts in one of the sides by itself. 10. Problem. To find the area of a square. Let ABCD (Fig. 119) be a square. Measure one side AB. Let, for example, AB = 4 in. By proceeding as in No. 9, we see that the area of the square will be 4x4 or 16 sq. in. Next, let JJ3 = 4.25 in. The same A reasoning will still apply, only we shall now have in AB 425 equal parts, each a hundredth of an inch long, and in ABCD, 425x425 or 180,625 equal squares. Now one inch contains 100 of these equal parts. Therefore^ one square inch will contain 100 x 100 or 10,000 of the equal squares. Therefore the area of the square ABCD, in square inches, will be \Vimr or 18.0625, a result at once found by multiplying 4.25 by itself. AREAS. 115 In general, by multiplying by itself the number of units of length in a line AB, we obtain the number of units of area in the square constructed upon AB as a side ; the unit of area being always the square whose side is equal to the unit of length. Multiplying a number by itself is called squaring it, and the square of the number of units of length in a line AB is called, for brevity, " the square of AB," and written Aff. Hence the result may be briefly expressed as follows : Area of a square = square of one side. Find the area of a square if one side is equal to : 11. 9 in. 16. 1 ft. 8 in. 12. 14 ft. 17. 2 ft. 4 in. 13. 18 yds. 18. 4 yds. 1 ft. 2 in. 14. 1\ in 19. 98 ft. 6 in. 15. 2i in. 20. 56 ft. 4 in. 21. Find the area in acres and square feet of a square field each side of which is 178 yds. long. 22. How many tiles 8 in. square are required to cover a square court one side of which is 33 ft. 4 in. long? 23. What will it cost, at 20 cents per square foot, to paint a square floor the side of which measures 18J ft. 24. Find the side of a square whose area is 2000 sq. ft. SOLUTION. We must take the square root of 2000. V2000=44.72 +. Hence the side (correct to inches) = 44.72 ft. 44ft. 9 in. 25. Find the side of a square whose area is 2J sq. yds. 26. Find the side of a square whose area is 6000 sq. ft. 27. What will it cost, at $8 per yard, to enclose with an iron fence a square park containing 1000 acres? 23. Explain why it requires 144 sq. in. to make 1 sq. ft. 116 LESSONS IN GEOMETRY. Lesson 61. 1. Problem. To find the area of a rectangle. Let ABCD (Fig. 120) be a p FIG. 120. B rectangle. Measure AB and AD. Let AB = 7 in. , AD = 4 in. We can show by reasoning as in No. 9, p. 114, that the area of the rectangle is equal to 7 X 4 or 28 sq. in. In general, the number of units of area in a rectangle is found by multiplying the number of units of length in one side by the number of units of length in the adjacent side. The adjacent sides are equal to the dimensions of the rec- tangle, and are usually called length and breadth ; hence we have the following condensed formula : Area of a rectangle = length x breadth. Find the area of a rectangle, having given : 2. Length 12 in., breadth 9 in. 3. Length 15 J in., breadth 6 in. 4. Length 3^ in., breadth 2^ in. 5. Length 200 ft., breadth 60 ft. 6. Length 5 ft. 6 in., breadth 1 ft. 6 in. 7. Length 4 yds. 1 ft. 6 in., breadth 6 ft. Find the other dimension of a rectangle, having given : 8. Area 100 sq. in., one dimension 8 in. 9. Area 288 sq. in., one dimension 6 in. 10. Area 22^ sq. in., one dimension 2 in. 11. Area 180 sq. ft., one dimension 4 yds. AREAS. 117 12. How many bricks 9 in. by 4 in. are required to cover a floor 34 ft. long and 17 ft. wide? 13. A street ^ a mile long has on each side a sidewalk 7J ft. wide. Find the cost of paving the sidewalk with stones 2 ft. 9 in. by 1 ft. 8 in., and costing, laid down, $1 each. 14. A lawn measures 144 yds. by 98 yds. ; the turfs which are used measure 18 in. by 14 in., and cost, laid down, 75 cents per dozen. Find the cost of turfing the lawn. 15. A floor is 27 ft. 6 in. long and 8 yds. wide. If the planks to cover it measure 11 ft. by 9 in., how many are needed, and what will they cost at 10 cents per square foot? 16. How many yards of paper 27 in. wide are required to paper a room 18 ft. long, 12 ft. wide, and 11 ft. high? 17. What will it cost to line with lead, at 2 cents per square inch, the inside of an open cistern 10 ft. long, 6 ft. wide, and 4 ft. deep? 18. A man wants 4 sq. ft. of wood, and has only a plank 18 in. wide from which to cut it off. Find the length of the piece he must cut off. 19. The dimensions of a rectangle are 36 ft. and 20 ft. If the length be diminished by 6 ft. , how much must be added to the breadth in order that the area may remain the same ? 20. A rectangular field, half a mile long, contains 100 acres. Find the width of the field in feet. 21. A room is 39 ft. by 23 ft. and 17 ft. high, with three windows each 9^ ft. by 8 ft. ; two doors each 10J ft. by 6 ft. ; and two fireplaces each 6 ft. by 4 ft. The carpet is 2 ft. wide, and costs $1.50 per yd. ; the paper is 2 ft. wide, and costs 75 cents per roll of 10 yds. Find the cost of carpet- ing and papering this room, the carpet running lengthwise, and the labor costing $12, LESSONS IN GEOMETRY. * Lesson 62. 1. Problem. To find the area of a parallelogram. Let ABCD (Fig. 121) be a parallelogram, having the base. AB and the altitude BE. F D EC Draw AF _L to AB, and meeting AB produced at the point F. The figure ABEF is a rectangle having the same FlG - 12L base AB and altitude BE as the parallelogram. ABCE = AADF. Why? Entire figure ABCF-A ADF = parallelogram ABCD. Entire figure ABCFA BCE = rectangle ABEF. Therefore, parallelogram ABCD = rectangle ABEF. But rectangle ABEF= ABxBE(p.ll6,No.l). Therefore, parallelogram ABCD = AB x BE. Hence, the area of a parallelogram is found by measuring its dimensions and multiplying their values together. Area of a parallelogram = base x altitude. 2. Does this formula also hold true for the square and for the rectangle ? Find the area of a parallelogram, having given : 3. Base 16 in., altitude 5 in. 4. Base 2J in., altitude 14 in. 5. Base 1 ft. 9 in., altitude 4 ft. 6. Base 15 ft., altitude equal to half the base. 7. The base of a parallelogram = 18 in. What must be its altitude in order that it may contain exactly 1 sq. ft. ? AREAS. 119 8. What conclusion may be drawn from JV0. 1 re- specting all parallelograms that have equal bases and equal altitudes? Parallelograms having equal bases and equal altitudes have also equal areas; in other words, they are equivalent. For by No. 1 each one of the parallelograms is equivalent to a rectangle having an H N G F M D E C equal base and an equal altitude. For example (Fig. 122) : the parallelo- grams ABCD, ABEF, and ABGH are equiva- lent, each being equivalent to the rectangle ABMN. 9. Construct a set of four equivalent parallelograms. 10. Make two parallelograms, differing very much in shape, but having the same area. 11. Construct a parallelogram, and then find its area. 12. The bases of two parallelograms are each 9| in. long. The altitude of one is 1 in., and the altitude of the other is 3 in. How much greater is the area of one than the area of the other? 13. The bases of two parallelograms are equal ; the alti- tude of one is three times that of the other. Compare their areas. Draw a figure to illustrate your answer. 14. The altitudes of two parallelograms are equal ; the base of one is three times that of the other. Compare their areas. Draw a figure to illustrate your answer. 15. Find the area of a rhombus, the perimeter of which is 156 ft., and the altitude 9 ft. 4 in. 16. The area of a field having the shape of a rhombus is exactly 10 acres. The distance from any one side to the opposite side is 400 ft. Required the cost of fencing the field at $2.50 per rod. 120 LESSONS IN GEOMETRY. Lesson 63. 1, Problem. To find the area of a triangle. Let ABC (Fig. 123) be a triangle, AB the base, CD the altitude. Draw AE \\ to BC, and CE II to AB ; the figure ABCE is a parallelogram having AB for base, CD for altitude. Therefore, A ABC = A AEC (p. 84, No. 3) . Hence, area of A ABC % area of parallelogram ABCD. Now, area of parallelogram ABCD = AB x CD. Hence, area of A ABC = %AB x CD ; or, Area of a triangle = % x base x altitude. FIG. 123. 2. What follows immediately from this formula? A having equal bases and equal altitudes are equivalent. Thus (Fig. 124) the A ABC, ABD, ABE, are equivalent. Find the area of a triangle, having given : 3. Base 11 in., altitude 10 in. 4. Base 2^ in., altitude 1 in. 5. Base 3 ft. 6 in., altitude 2 ft. 6. Base 20 ft. 6 in., altitude 10 ft. 4 in. 7. Find the area of a right triangle if the lengths of the two legs are 4J in. and 3^ in. 8. A garden is laid out in the shape of a right triangle. One leg = 150 yds., the other leg = 200 yds. Find the value of the garden at 12 J cents per square foot. AKEAS. 121 9. Apply the formula of No. 1 to a rhombus. Let ABCD (Fig. 125) be a rhombus. The diagonals AC and BD bisect each other at right angles (p. 86, No. 3). AC divides the rhombus into two equal isosceles triangles, each having AC for base, and half of BD for altitude. The area of each A = \AC x %BD ; therefore the area of the rhombus = \ X AC X BD ; or, in general : Area of a rhombus = half the product of the diagonals. E FIG. 126. B 10. Apply the formula of No. I to a trapezoid. Let ABCD (Fig. 126) be a trapezoid, AB and CD the bases, DE the altitude. Draw the diagonal DJ5, dividing the trapezoid into the two triangles ABD and BDC. Area of A ABD = %ABx DE. Why ? Area of A BDC = \DC x DE. Hence area of trapezoid ABCD = (AB + DC) X DE. Putting this result into general terms, we have the formula : Area of a trapezoid = J (sum of bases) x altitude. Find the area of a rhombus, having given : 11. The diagonals 9| in. and 6J in. 12. The diagonals 26 yds. and 34| yds. 13. Find the area of a trapezoid if the two bases are 37 ft. and 25 ft., and the altitude 19 ft. 14. Find in square feet the area of ABCD (Fig. 126) if AB = i mile, CD = J mile, DE = 1 mile. 122 LESSONS IN GEOMETRY. Lesson 64. 1. Problem. To find the area of a regular polygon. Let ABCDE (Fig. 127) be a regular polygon, the centre, OF the apothem. Draw the radii OA, OB, etc. The radii divide the polygon into the equal isosceles A AOB, BOG, etc. Altitude of each A = OF. Area of A AOB = ABx OF. Area of A BOG = %BC x OF, etc. Adding these areas, we obtain : FI- 127. Area of polygon ABCDE = %(AB+BC+etc.) x OF', or, Area of a regular polygon = j- x perimeter x apothem. If one side of a regular polygon and the number of sides are given, the apothem may be found (correct enough for all practical purposes) by multiplying the given value of the side by the proper number in the following table : No. of sides. Apothem. 3 one side X 0.288 4 one side X 0.500 5 one side X0.688 6 one side X 0.866 No. of sides. Apolhem. 7 ...... one side X 1-038 8 one side X 1-207 10 one side X 1.539 12 . . one side X 1.866 Find the area of: 2. An equilateral triangle, if one side = 5 in. 3. A regular pentagon, if one side = 6 in. 4. A regular octagon, if one side =15 ft. 5. A regular decagon, if one side = 19 ft. 6. A park has the shape of a regular hexagon ; each side is 1000 ft. long. Find the value of the park, at 8 cents per square foot. AREAS. 123 7. How may the area of any polygon be found ? The area of any polygon may be found by dividing it into triangles, or into' triangles and trapezoids, computing the areas of these parts, and then adding the results. 8. Find the area of the polygon ABCD (Fig. 128), having given : AC, 400 yds. ; BE, 120 yds. ; DF, 80 yds. FIG. 128. FIG. 129. 9. Find the area of the polygon ABODE (Fig. 129), having given : BE, 108 yds. ; EC, 96 yds. ; AF, 49 yds. ; HD, 35 yds. ; CG, 67 yds. 10. Find the area of the polygon ABCDE (Fig. 130), having given : AC, 475 yds. ; BG, 160 yds. ; AF, 175 yds. ; AG, 320 yds. ; AH, 420 yds. ; EF, 160 yds. ; DH, 90 yds. 11. Find the area of the polygon ABCDEF (Fig. 131), having given : AC, 300 yds. ; BK, 115 yds.; FG, 63 yds. ; EG, 54 yds. ; FH, 283 yds. ; DH, 60 yds. ; AF, 125 yds. ; CF, 325 yds. Z AFC= 90. 124 LESSONS IN GEOMETRY. Lesson 65. 1. Problem. To find the area of a circle. Circumscribe about a circle a regular polygon r then Area of the polygon = J X perimeter X apothem ; or, since the apothem is equal to the radius of the circle, Area of the polygon = -J- x perimeter x radius. The greater the number of sides of the polygon, the nearer will its perimeter approach to the circumference of the circle, and its area to the area of the circle. v If the polygon had 1000 sides, it would not differ sensibly from the circle ; if it had 10,000,000 sides, the difference would of course be still less. But however many sides the poly- gon may have, its area may always be found by the formula above given ; and it can be proved that this formula will still hold true if we substitute the area of the circle for the area of the polygon, and the circumfer- ence of the circle for the perimeter of the polygon. Making these substitutions, the formula becomes : Area of a circle = X circumference x radius. To change this formula into a more practical form, we must substitute for circumference the value which we have previously found (see p. 108) ; namely, -^ X diameter, or 4^ X radius. After reducing to the simplest form, we have : Area of a circle = - 2 ^ X (radius) 2 ; or, if one prefers to use the symbols TT and r, Area of a circle =^1^. AREAS. 125 2. From the formula for the area of a circle find the value of the radius in terms of the area. Take the equation given in No. 1 : %f- X (radius) 2 = area. Multiply both sides by -^ , and extract the square root. Indicating by the radial sign y' the square root of the second side, we obtain for the result : radius = ^- X area. Find the area of a circle, having given : 3. Radius 7 in. 4. Radius 14 ft. 5. Radius 40 ft. 10 in. 6. Radius 6 yds. 1 ft. 3 in. 7. Diameter 49 ft. 8. Diameter 18 1 yds. 9. Diameter half a mile. (Find the answer in acres.) 10. Find the radius, if the area = 24G4 sq. ft. 11. Find the radius, if the area= 6 acres. 12. Find the diameter, if the area = 1 sq. mile. 13. The radii of two concentric circles are 3^ in. and 7 in. Find the area of the ring bounded by their circumferences. 14. Find the area of a circular ring 4 ft. wide, if the radius of the larger circle is 32 ft. 15. Out of a square piece of wood 5 ft. 10 in. long is cut the largest possible circle. Find its area. 16. Find the area of a sector, if the radius of the circle = 14 in. and the angle at the centre = 45. The sector is the same part of the circle that its angle is of 360. 17. Find the area of a sector, if the radius = 10 in. and the angle at the centre = 18. 126 LESSONS IN GEOMETRY. Lesson 66. 1. Theorem. In every right triangle the square of the hypotenuse is equal to the sum of the squares of the legs. HYPOTHESIS. Let ABC be a right triangle, and let Z ACB = 90. CONCLUSION. E FIG. 133. PROOF. Construct the squares ABDE, ACFG, BCHK. Draw CM _L to AB, and pro- duce it to meet DE at N. Join BG and CE. In the A ACE, ABG, AC = AG, AE = AB. Why ? Also, Z CAE = Z BAG. Why ? Therefore, A ACE = A ABG (p. 66, No. 14). Now, the A ACE and the rectangle AMNE have the same base, AE, and the altitude of each is the distance between the parallels AE and CN. Therefore, A ACE = % rectangle AMNE (p. 120, No. 1) . Similarly, A ABG = % square ACFG. Therefore, rectangle AMNE = square ACFG. Therefore, rectangle AMNE = square ACFG. Similarly (joining CD and AK) , we can prove that rectangle MBDN= square BCHK. Adding the last two equations, we have : square ABDE = square ACFG + square BCHK, or NOTE. This very useful theorem is called, from the name of the discoverer, the " Theorem of Pythagoras." AREAS. 127 Find the hypotenuse of a right triangle, having given : 3. The two legs 3 in. and 4 in. 4. The two legs 8 in. and 15 in. 5. The two legs 81 ft. and 108 ft. 6. The two legs 237 ft. and 310 ft. Find one leg of a right triangle, having given : 7. Hypotenuse 13 in., one leg 5 in. 8. Hypotenuse 40 ft., one leg 24 ft. 9. Hypotenuse 51 ft., one leg 24 ft. 10. Hypotenuse 185 ft., one leg 111 ft. 11. Find the area of a right triangle, if the hypotenuse is 25 in. and one leg is 15 in. 12. Find (correct to two decimal places) the hypotenuse of a right triangle, if each leg is 10 in. 13. Find (correct to two decimal places) the legs of an isosceles right triangle, if the hypotenuse is 10 in. 14. The side of a square is 20 ft. Find the diagonal. 15. The diagonal of a square is 20 ft. Find one side. 16. A ladder 34 ft. long just reaches a window when placed with its foot 16 ft. from the side of the house. How high is the window above the ground? 17. Two vessels start at the same time from the same port, and sail one north, the other east. Their rates are G and 8 miles per hour. How far apart are they after 3 hours ? 18. The perimeter of an isosceles triangle is 64 ft. ; the base is 24 ft. Find the area of the triangle. 19. The side of an equilateral triangle is 14 yds. Find the area of the triangle. 20. The radius of a circle is 1 ft. 3 in. Find the distance from the centre to a chord 2 feet long. 128 LESSONS IN GEOMETRY. Lesson 67. 1. Define the mean proportional between two lengths. The mean proportional between two lengths is -the length whose square is equal to their product. Thus : since 6x6 = 4x9, a line 6 in. long is the mean proportional between a line 4 in. long and a line 9 in. long. 2. Problem. To find the mean proportional be- tween two given lines AB and CD (Fig. 1$4)> (1) By calculation. Measure the two lines, multiply their lengths together, and find the square root of the product. (2) By construction. Draw a straight line, and upon it lay off AB a, and BC=b. Upon AC as a diameter, describe a semicircle. Erect at B a perpendicular, cutting the semicircle at D. Then BD will be the mean pro- portional required. NOTE. This construction can be proved to be correct with the aid of the Theorem of Pythagoras, but the shortest and easiest proof is founded on the properties of similar triangles (see page 151, Ex. 8). 3. Draw a line 2 in. long and a line 3J in. long, and then find (1) by calculation, (2) by construction, the mean pro- portional between them. 4. Find the mean proportionals between the following numbers : 1 and 9, 9 and 25, 16 and 25, 1 and ^, % and %. 5. What is meant by transforming a figure ? To transform a figure is to change its shape without chang- ing its size. 6. Construct a parallelogram, and then transform it by a construction into a rectangle (see p. 118, Ex. 1). AREAS. 129 FIG. 135. 7. Problem. To transform a rectangle into a square. (1) By calculation. Measure the dimensions of the rec- tangle ; find the square root of their product ; draw a line equal to this root in length ; and upon this line as a side, construct a square. (2) By construction. Denote the two dimensions by a and b. We must find a length x such that x 2 = a x b. Hence the side x of the required square is the mean proportional between a and b. Therefore the construction is the same as that in No. 2. 8. Construct a rectangle 4 in. long and 2\ in. wide. Then find (1) by calculation, and (2) by construction, the side of the equivalent square. 9. How can a triangle be transformed into a square ? Answer. The area of a triangle = |- the base X altitude. Therefore, the side of an equivalent square is the mean proportional between half of the base and the altitude. 10. Construct a triangle with base 4 in. and altitude 3^ in., and transform it to a square (1) by calculation ; (2) by con- struction. 11. Construct an equilateral triangle, and then transform it by construction into a square. 12. Construct a rhombus, and then transform it by con- struction into a square (see p. 121, No. 9). 13. Two fields are equal in area. One is a square, and the other a rectangle 625 ft. long and 324 ft. wide. Find the cost of fencing each field at 10 cents a foot. 14. One leg of an isosceles right triangle is 40 ft. long. Find (correct to two decimals) a side of the equivalent square. 130 LESSONS IN GEOMETRY. Lesson 68. 1, Problem. To transform any polygon into a tri- angle. Let ABODE (Fig. 136) be any pentagon. Draw the diagonal BD. Draw through C a line parallel to BD, and meeting AB produced at F. Join DF. The figure AFDE has one side less than the given polygon, and is equivalent to it. Repeat this construction by joining AD, drawing through E a line parallel to AD, and joining DG. The A DFG will be equivalent to the given pentagon ABODE. The repetition of this process will reduce a polygon of any number of sides to an equivalent triangle. E G A. B F FIG. 136. PROOF. The A BDF, BDC, have the same base BD and equal altitudes (their vertices C, F, being in a line II to BD) . Therefore A BDF is equivalent to A BDC (p. 120, No. 2) . Hence, if A BDC be taken from the given polygon, and A BDF substituted for it, the polygon will suffer no change in area, but it will be transformed into the figure AFDE. For like reasons, A ADG is equivalent to A ADE, and when substituted for it, the figure AFDE will be reduced to the A DFG. AREAS. 131 NOTE. The next seven exercises are to be done by construction. 2. Draw a trapezium, and transform it into a triangle. 3. Draw a pentagon, and transform it into a triangle. 4. Draw a hexagon, and transform it into a triangle. 5. Draw a hexagon, and transform it into a square. 6. Draw a triangle, and then upon one side of this tri- angle as a base, construct an equivalent isosceles triangle. 7. Draw a square, and upon a side of the square as a base, construct an equivalent isosceles triangle. 8. Draw a triangle, and upon one of its sides construct an equivalent rectangle. NOTE. The remaining exercises are to be done by calculation. 9. The legs of a right triangle are 4 in. and 8 in. Find the side of the equivalent square. 10. Find the side of a square equivalent to a triangle whose base is 18 ft. and altitude 16 ft. 11. Find the side of a square equivalent to a trapezoid whose bases are 30 ft. and 34 ft., and altitude 8 ft. 12. Find, correct to two decimals, the side of a square equivalent to a circle whose radius is 7 in. 13. Find, correct to two decimals, the radius of a circle equivalent to a square whose side is 11 in. 14. A rectangle is 320 ft. long and 100 ft. wide. If the length be reduced by 50 ft., how much must the breadth be increased in order that the area may remain the same as before ? 15. The hypotenuse of an isosceles right triangle is 10 in. Find (correct to two decimals) a side of the equivalent square. 132 LESSORS IN GEOMETRY. Lesson 69. Review. 1. Review the italicized exercises in Lessons 60-65. 2. What will it cost, at $2 per rod, to fence a square field containing 2^ acres? 3. How many five-acre lots can be made out of a field containing 5 square miles? 4. How many planks, each 10 ft. long and 8 in. wide, will cover a floor 14 ft. 3 in. long and 13 ft. 4 in. wide? 5. Find the side of a square having an area equal to that of a regular hexagon whose side is 8 ft. 6. The circumferences of two concentric circles are 440 ft. and 330 ft., respectively. Find the width of the ring. 7. Out of a circular piece of wood whose radius is 3 ft. 4 in. is cut the largest possible square. Find its side. 8. The radius of a circle is 4 ft. Find the area of the inscribed regular hexagon (see p. 107, No. 10). 9. The side of a square is 42 yds. Find the areas of the inscribed and the circumscribed circles. 10. The diagonals of a diamond-shaped pane of glass are 12 in. and 16 in. How many panes will cover an area of 400 sq. ft. ? 11. A square and a rectangle have the same perimeter, 100 yds. The length of the rectangle is four times its breadth. Which has the greater area, and by how much? 12. What will it cost to turf a lawn 35 yds. by 27 yds., with turfs 21 in. by 18 in., and costing 3 cents apiece? 13. A path 8 ft. wide surrounds a rectangular court 60 ft. long and 36 ft. wide. If tiles are 9 in. by 4 in., and cost 10 cents each, find the cost of paving the path with tiles. AREAS. 133 Lesson 7O. Review. 1. Review the italicized exercises in Lessons 66-68. 2. Show by a figure the difference between half a square foot and half a foot square. 3. Can you make a rectangle, 9 in. by 4 in., and then divide it into two parts of such a form that when placed together in a certain way they will make a square? 4. A man has a lot of land 127 yds. square. What will it cost him, at 30 cents per square yard, to make a gravelled walk 1 yd. wide around his lot? 5. How high is a window from the ground if a ladder 30 ft. long will just reach the window when placed with its foot 18 ft. from the side of the house? 6. The width of a house is 47 ft. ; height to ridge, 67 ft. ; height to eaves, 39 ft. Find the cost of painting the side of the house up to the ridge, at 18 cents per square yard, de- ducting 240 sq. ft. for the windows. 7. If a right triangle has an acute angle of 30, the leg opposite this angle is always equal to half the hypotenuse. Knowing this truth, find the legs of a right triangle if the hypotenuse = 20 ft. and one angle = 30. 8. If the distance from the centre of a circle to a chord 10 in. long is 1 ft., find the distance to a chord 2 ft. long. 9. Transform a square into a right triangle. 10. Transform a rectangle into an isosceles triangle. 11. Transform a parallelogram into a square. 12. Construct a square equal to the sum of two given squares. 13. Construct a square equal to the difference of two given squares. 134 LESSONS IN GEOMETRY. Lesson 71. Review. 1. Show by a figure that the square of is equal to J. 2. Show by a figure that the square of 1J is" equal to 2^. 3. Find the side of a square equal in area to the sum of the squares constructed upon the lengths 2 in., 3 in., and 4 in. 4. In a railway curve the radius of the circle is 300 ft., and the centre of the circle is located at a point 780 ft. from a station. Draw a plan showing the location of the station, the centre of the circle, and the point where the track begins to curve. Find the distance from -this point to the station. 5. How many bricks 9 in. by 4J in. are required to cover a square yard? 6. Transform a triangle into a rectangle. 7. A circular grass-plot has a diameter of 400 ft., and a walk 3 yds. wide around it. Find the area of the grass-plot, and also the area of the walk. 8. A square, whose side is 4 in., is to be transformed into a rhombus in which one diagonal equals twice the other. Find the lengths of the diagonals and of one side of the rhombus. Can you solve this problem also by construction? 9. Find the radius of a circle equivalent to a square whose side is 11 ft. long. 10. Find the side of a square equivalent to a circle whose radius is 7 ft. long. 11. Find the area of a trapezium, if one diagonal is 54 ft., and the perpendiculars to this diagonal from the opposite vertices are 23 ft. 9 in. and 36 ft. 6 in. 12. Compare the areas of a circle, a square, and an equi- lateral triangle, if the perimeter of each figure is 132 ft. AREAS. 135 Lesson 72. Review. 1 . Review all the italicized exercises in this chapter. 2. Find the area of a right triangle, if the hypotenuse is 8J in. and one of the legs is 4 in. 3. Find the area of an isosceles triangle, if the base is 20 ft. and each of the other sides is 26 ft. 4. Draw any pentagon, and transform it into a square. 5. Construct a square whose diagonal shall be 3 in., and then find its area. 6. Make a square half as large as a given square. 7. Find the side of a square equivalent to a rectangle 400 ft. long and 100 ft. wide. 8. One angle of a rhombus is 60, and the shorter diag- onal is 8 ft. Find the area (see p. 133, No. 7). 9. The angle of a sector is 60, and the chord of the arc is 14 ft. Find the area of the sector. 10. Two ships, each making 6 miles an hour, sail from the same port. One sails at noon, directly east ; the other sails at 3 P.M., directly south. How far apart will they be at 6 P.M. ? 11. The top of a centre-table has the shape % of a regular octagon. One side measures 2 ft. What will it cost to stain and polish the surface at $2.25 per square foot? 12. A room is 31 ft. by 17 ft., and 12 ft. high, with 3 windows each 5^ ft. by 6 ft., 2 doors each 7 ft. by 4 ft., and a fireplace 7 ft. by 6 ft. The carpet is 27 in. wide, costs $2.50 per yard, and is laid lengthwise along the floor. The paper is 27 in. wide, and costs $1.80 per roll of 10 yds. Find the cost of the carpet and the paper. 136 LESSONS IN GEOMETRY. Lesson 73. Review. 1. Define and name the metric units of area. The metric units of area are the squares of- the units of length. They are named by placing the word "square" before the name of the corresponding unit of length. Thus : square meter, square decimeter, etc. The square dekameter is also called an ar, and the square hectometer a hectar. The abbreviations are formed by placing the letter q be- fore those of the corresponding units of length. Thus : qm for square meter ; qcm for square centimeter, etc. 2. How do they stand related to each other? Each unit is 100 times larger than the next smaller one. For example : l qm = 100 qdm ; l qdm = 100 qcm , etc. 3. Show by a figure that 1 square meter = 100 qdra . 4. Reduce to square meters 4 dkm ; l,840,000 qmm . 5. How many square meters are there in a hectar? 6. Show that l qm = 10.764 sq. ft. (see bottom of p. 33). 7. Show that there are about 15^ sq. in. in l qdm . 8. A square park contains 64 hectars. How many trees, 20 m apart, can be set out around it ? 9. If glass is worth $1.40 per square meter, what will be the cost of 1000 panes of glass each 48 cm square? 10. Find the area of the largest rectangle which you can make under the condition that its perimeter is 60 cm . 11. A table 2.3 m by 1.2 m is covered with cloth which over- laps 3 cra on all sides. How much cloth was required? 12. Find the side of a square equivalent to a triangle whose base is 90 ra and altitude is 20 m . AREAS. 137 13. A ladder 13 m long leans against the top of a wall, its foot being 5 m from the bottom. How high is the wall? 14. Find the area of an equilateral triangle if the length of one side is 20 m . 15. Find the cost of covering a round table with cloth 60 cm wide, and costing $2 per meter, if the radius of the table is equal to 75 cm . 16. Find the cost of making a stone walk l m wide around a circular reservoir 64 m in diameter, the price paid being $5 per square meter. 17. At a point E in one side of the rectangular field ABCD there is an artesian well. The owner divides the field into three unequal parts by lines drawn from E to D and C. Find the value of each part, if AD = AE = 81 m ,. EB = 144 m , and a hectar is worth $3400. FIG. 137. 18. The radius of a circle is 4 in . Find the areas of the circumscribed and the inscribed squares. 19. In order to find the area of a swamp, ABCDEF, it was first surrounded by a rectangle, as shown in Fig. 138. The following measurements were then made : G B CH=75.o m , DI =68.5 m , IE = 68.5 KF= 29. 5 m , FA = 58.8 n Find the value of the swamp, K one hectar being worth $1500. 20. The diagonals of a rhombus are 4 m and 6 m . Find its area, its perimeter, and its altitude. OF THE^ " { UNIVERSITY 138 LESSONS IN GEOMETRY. Lesson 74. Review. 1. Review all the italicized exercises in this chapter. 2. Find the area of the figure which is given to you. 3. If a triangle and a parallelogram have equal bases and equal areas, what is true of their altitudes? 4. Find the radius of a circle three times as large as a circle whose radius is 7 m . 5. Show by a figure that the square of | is equal to 1. 6. The three dimensions of a room are each 6. 4 m . Find the cost of papering the walls at 30 cents per square meter. 7. Through the middle of a rectangular garden, 180 m by 76 m , there are two paths _L to each other and II to the sides. Find the area of the remainder of the garden. 8. If a rhombus and a square have equal perimeters, which has the greater area? Prove your answer correct. 9. Find the side of a square equivalent to a trapezoid whose bases are 60 m and 32 m , and altitude is 124 m . 10. Find the side of a square equivalent to a regular hexa- gon, one side of which is 8 m . 11. The perimeter of a regular octagon is 16 m . Find the area. 12. One leg of an isosceles right triangle is 32 m . Find the area. 13. The diagonal of a square is 12 m . Find the area. 14. Find the length of the shortest chord that can be drawn through a point 9 cm from the centre of a circle whose radius is 15 cm . 15. What is the mean proportional between 9 and 16? 16. The angle of a sector is 120, and the length of its arc is 66 cm . Find its area. CHAPTER VIII. RATIOS. Lesson 75. 1. Define the numerical measure of a quantity. The number of times one quantity contains another quan- tity of the same kind is called the numerical measure of the first quantity referred to the second as a unit. ABCDEFGH i i i i i i i i For example : if AB = BC = CD = DE = EF = FG = GH, the numerical measure of AH referred to AB as a unit is 7, and the numerical measure of AB referred to AH as a unit is | ; in other words, AH= 1 AB, AB = i of AH. 2. If in the above figure AB is taken as the unit, what is the numerical measure of AG? AF? AE? AD? AC? AB? 3. If AC is taken as the unit, what is the numerical measure of AH? AG? AF? AE? AD? AC? AB? 4. Define the ratio of two numbers. The ratio of one number to another number is their rela- tive magnitude, and is found by dividing the first number by the second number. A ratio is written either as a fraction or by placing a colon ( : ) between the two numbers. For example : the ratio of 4 to 7 is ^, or 4 : 7 ; the ratio of 3 to 1 is f (that is, 3), or 3 : 1. The two numbers which form a ratio are called its terms. 140 LESSONS IN GEOMETRY. 5. Define the ratio of two quantities. The ratio of two quantities of the same kind is the ratio of their numerical measures referred to the same unit. (1) The two quantities must be the same in kind. Two quantities which differ in kind cannot have a ratio to each other. For example, we cannot compare, as regards magni- tude, a line with an angle, or dollars with days. (2) The two quantities must be referred to the same unit. We may compare, for example, 6 in. with 2 ft., but their ratio is not 6 : 2, because 1 ft. = 12 in. ; it is 6 : 24, or 1 : 4. ABCDEFGH 6. In the above figure what is the ratio of AC to AD? AD to AC? AC to AH? AD to AG? AE to AG? etc. 7. What is the ratio of 2 in. to 6 in. ? 2 in. to 6 ft. ? 8. What is the ratio of 15 to a right angle? 9. Problem. To divide a straight line AB into two parts having a given ratio ; say, 3 : 4- Draw through A a line AC, making an acute angle with AB. Begin at A and lay off upon AC 3 + 4, or 7, equal parts. Mark the end of the 7th part D, and the end of the 3d part E. Join BD, and draw EF II to ^~ ~~p~ B BD, and cutting AB at F. FlG 139 PROOF. Draw parallels to BD through all the points of division in AC, and prove that they divide AB into 3+4, or 7, equal parts (p. 25, No. 4; p. 86, No. 4). Therefore AF : FB = 3 : 4 . 10. Divide a line into two parts, having the ratio 2 : 5. RATIOS. 141 11. Compare two triangles whose altitudes are equal. Let the triangles ABC, ADC (Fig. 140), have the same altitude CE, but 'let their bases AB, AD, be unequal. The areas of the two triangles have the values, If we divide one of these equations by the other, we obtain the ratio of the areas written as a fraction ; this fraction may then be reduced to its lowest terms by cancelling the common factors ^ and CE. We have as the result, A ABC = $ABx CE^AB A ADC ^ADxCE AD Hence we see that triaixjles having equal altitudes are to each other as their bases. c FIG. 141. 12. Show that two triangles having equal bases are to each other as their altitudes (Fig. 14D- 13. Compare two parallelograms having equal alti- tudes. 14. Compare two parallelograms having equal bases. 15. Two parallelograms have equal altitudes, but the base of one is ten times that of the other. Compare their areas. 16. A triangle and a parallelogram have equal bases, but the altitude of the triangle is four times that of the parallelo- gram. What is the ratio of their areas? 17. Construct a square and a rhombus so that their perim- eters shall be equal, and their areas be to each other as 3 : 2. 142 LESSONS IN GEOMETRY. Lesson 76. 1. Compare the two parts into which a triangle is divided Inj drawing a line from one vertex, to the opposite side. In the triangle ABC (Fig. 142) let CD be drawn from the vertex C to any point D of the base AB. The two triangles thus formed ADC, BDC, have the same altitude CE, but unequal bases AD, BD. Therefore (p. 141, No. 11), A ADC = AD A BDC ~ BD A ADC: A BDC = AD : BD, or Hence a line draivn from the vertex to the base of a tri- angle divides the triangle and its base into parts that have the same ratio. C D E FIG. 142. B E FIG. 143. 2. How must the line CD (Fig. 142) be drawn in order to divide A ABC into two equivalent parts? What would be the ratio of these parts ? What would be the ratio of one of them to the whole triangle ? 3. Draw a triangle and divide it into two parts which shall be to each other as 2:3. What part of the whole triangle is each one of the parts ? 4. It is desired to run a division line AE across a rec- tangular field ABCD (Fig. 143) so that the triangle cut off, ADE, may be equal in area to ^ of the whole field. How would you determine the point E? What part of DC is DE? What is the ratio of DE to EC? KATIOS. 143 5. In Fig. 143, if DE = $DC, what is the ratio of DE to DC? DC to DE ? DE to EC? What part of the rectangle ABCD is the A ADE? How should AE (Fig. 143) be drawn in order that 6. A ADE may be \ of the rectangle ABCD? 7. A ADE may be 1 of the rectangle ABCD? 8. A ADE : rectangle ABCD = 1:12? 9. A ADE : rectangle ABCD = 3:7? 10. Draw a triangle and divide it into three parts which shall be to each other as the numbers 1, 2, 3. 11. Draw a rectangle and then divide it into four equiva- lent parts by lines drawn from one of the vertices. 12. Draw a rectangle and then divide it into three equiva- lent parts by lines drawn from one of the vertices. 13. Draw a rectangle and then divide it by a line II to one side into two parts having the ratio 2:3. What fraction of the rectangle is each part? 14. In a triangle ABC (Fig. 144) the middle points Z>, E, of the sides AB, AC, are joined. What is the ratio of the triangle ADE to the triangle ABC? Join BE. Since A D = M B, A A ADE =A ABE (p. 141, No. 11). Also, since AE= I AC, Substituting this value of A ABE in the first equation, we have : FIG. 144. or, written as ratios, A ADE : A ABC=l : 4. 15. In Fig. 144, if AD = $AB, and AE = $AC, show that A ADE : A ABC = 1:9. 144 LESSONS IN GEOMETKY. Lesson 77. 1. Define a proportion. Four quantities are said to be proportional, or to form a proportion, if the ratio of the first to the second is equal to the ratio of the third to the fourth. For example: since T \ = f, the numbers 4, 10, 2, 5, are proportional. A B E F D G If Again, the lines AB, CD, have the ratio 3 : 5, and the lines EF, GH, have the same ratio 3:5. Therefore the four lengths AB, CD, EF, GH, form the proportion AB : CD = EF : GH, which is read "AB is to CD as EF is to GH." The four quantities that form a proportion are called its terms ; the first and fourth are called the extremes ; the second and third the means. 2. Give an example of four proportional numbers. 3. Give examples of proportions on pp. 142, 143. 4. Prove that, if four numbers form a proportion, the product of the extremes is equal to the product of the means. If the four numbers a, 6, c, d, are proportional, then Multiply both sides of this equation by the product ltd. abd cbd Then T--5" Cancelling common factors, we have left ad = be. RATIOS. 145 5. Find the first term of a proportion if the other three terms are 9, 28, 7. Let x denote the first term ; then x : 9 = 28 : 7. Hence (No. 5), 7z = 9 x 28. q v 90 Therefore x = ^4^ = 9 x - = 36. In general, either extreme is equal to the product of the means divided by the other extreme. 6. Find the first term if the other terms are 3, 7, 84. 7. Find the fourth proportional to 13, 65, 120. 8. Compare two circumferences and their radii. Let c, d, denote the circumferences ; r, s, their radii. Then c = 27rr, d = '27rs (p. 108, No. 1). Therefore &ACD 90 r s RATIOS. 151 4. Six different proportions may be formed from the homologous sides of the similar triangles in Fig. 150, and it is very easy to form them after the table in No. 3 has been completed. Write out all these proportions in a vertical column, thus : c: a = a : e, c : a = b : h. 5. Transform the six proportions obtained in No. 4 into c'(iuations by placing in each case the product of the means equal to the product of the extremes. 6. Define a mean proportional (p. 128, No. 1). The proportion c : a a : e gives the equation a 2 = ce. Between what two lines is a a mean proportional? 7. What two other examples of mean proportionals are there among the equations obtained in No. 5, ? 8. Prove that the construction in No. 2, p. 128, is correct. HINTS. First show that the triangle ACD (Fig. 134) is a right triangle ; then make use of one of the proportions which can be obtained from the table in No. 3. 9. Among the equations obtained in No. 5 are the two following : a 2 = ce, b 2 = cd. Hence show that a 2 -f- b 2 = c 2 . State this result as a theorem. Compare with No. 1, p. 126. HINT. Observe that d -f e = c. 10. If (Fig. 150) c = 9 in., e = 4 in., find a, h, and b. 11. If (Fig. 150) a = 6 in., b = 8 in., find c, d, and e. 12. If (Fig. 150) e = 3 in., h = 4 in., find a, 6, and c. 13. One of the equations obtained in No. 5 is the follow- ing : ab = ch. State this equation as a theorem. 152 LESSONS IN GEOMETRY. Lesson 81. Review. 1. Review the italicized exercises in LCSSODS 75-77. 2. What is the ratio of 3 inches to 2 feet ? 3. What is the ratio of 6 feet to 9 yards ? 4. What is the ratio of of an inch to ^ of an inch ? 5. What is the ratio of \ of an inch to J of a foot? 6. Produce a straight line AB to a point C such that AB : BO = 4 : 5. What is the ratio of AC to BC? 7. Divide a given line AB into three parts proportional to the numbers 2, 3, 5. 8. Divide a given line AB into three parts proportional to the numbers 1, ^, ^. 9. Construct a triangle whose sides shall be to each other as the numbers 3, 4, 5. 10. Construct an isosceles triangle so that the ratio of the base to one leg shall be as 2:3. 11. Compare the areas of a triangle and a parallelogram if they have equal bases and equal altitudes. 12. A triangle and a parallelogram have equal areas and equal altitudes. What is the ratio of their bases ? 13. Divide a triangle into two parts which are as 3 to 7. 14. A triangular lot is worth $6000. Show by a figure how you would run a line from one corner to the opposite side so as to cut off a piece worth $2400. 15. Divide a parallelogram by a line drawn from one vertex into two parts having to each other the ratio 3:8. 16. Find the fourth proportional to the numbers 9, 117, 10. 17. If the radii of two circles are as 1 : 4, and the area of the first circle is 1 square inch, what is the area of the other circle ? 18. Given a circle ; construct a concentric circle \ as large. RATIOS. 153 Lesson 82. Review. 1. Review the italicized exercises in Lessons 78-80. 2. Prove that two isosceles triangles are similar if an angle of one is equal to the homologous angle of the other. 3. Are two equilateral triangles similar? Why? 4. The sides of a triangle are 5, 6, 7. Find the sides of a similar triangle if its ratio of similitude to the given tri- angle is 10 : 7. 5. Draw an isosceles triangle, and then construct a similar triangle whose ratio of similitude to the first triangle is 3 : 1 . What is the ratio of their areas ? 6. Compare the areas of the two parts into which a parallelogram is divided by the line joining the middle points of two adjacent sides. (See p. 84, No. 3; p. 143, No. 14.) 7. The ratio of similitude of two similar triangles is 3 : 5. The first triangle has an area of 27 sq. ft. What is the area of the other triangle ? 8. A vertical rod 6 ft. high throws a shadow 4 ft. long at the same time that a church spire casts a shadow 120 ft. long. How high is the spire ? 9. In order to find the distance from my position A to the enemy's fort B, I run a line AC J_ to AB and equal to 2000 ft. Through a point D in AC 100 ft. from C I run a line DE II to AB, meeting BC at E. If DE is found to be 800 ft., how far am I from the fort? 10. The radius of a circle = 17 ft. Through a point upon a diameter 9 ft. from the circumference a chord _L to this diameter is drawn. Find the length of this chord. 11. How many miles is the light of a lighthouse 200 ft. high visible at sea? (Radius of earth = 4000 miles.) 154 LESSONS IN GEOMETRY. Lesson 83. Review. 1. Review all the italicized exercises in this chapter. 2. What is the ratio of 4 m to l km ? 3. What is the ratio of 5 cm to 5 m ? 4. What is the ratio of 8 cm to 24 dm ? 5. What length is to 60 m as 5 is to 6 ? 6. The areas of two triangles are 480 qm and 300 qm . If their bases are equal, compare their altitudes. 7. If the altitude of the larger triangle in No. 6 is 30 m , find its base and the altitude of the other triangle. 8. Compare the areas of two similar triangles if the ratio of similitude is equal to ^. 9. From one corner of a field in the shape of an equi- lateral triangle I wish to cut off a similar triangle equal in area to y^ of the entire triangle. How must I run the division line? 10. A rectangle 36 cm by 20 cm is so divided into three parts by lines drawn from one vertex that the parts are as the numbers 2, 3, 4. How great is each part? How are the sides divided by the division lines? 11. Two sides of a field in the shape of an isosceles right triangle are each 400 m long. It is desired to cut from the field a lot containing one hectar of land by a straight fence starting from the vertex of the right angle. What portion of the hypotenuse will be cut off by this fence ? 12. What is the mean proportional between 49 and 121 ? 13. How high is a tower if it casts a shadow 28 m long at the same time that a vertical staff 4 m high casts a shadow 1.45 ra long? CHAPTER IX. SOLIDS. Lesson 84. 1. Define a plane (see p. 9, No. 9). A plane is named by three or more letters written on its boundary lines, or by a single letter written upon its sur- face ; thus the plane in Fig. 151 is the plane ABCD, or the plane ABC, or simply the plane M. 2. When is a straight line parallel to a plane ? A straight line and a plane are parallel when they will not meet, however far both are produced. 3. What is the foot of a line ? If a straight line meets a plane, the point where it meets the plane is called its foot. 4. When is a straight line perpendicular to a plane ? When it is perpendicular to every straight line that can be drawn through its foot in the plane. 5. Name in Fig. 151 lines II to the plane J/, _L to Jf, oblique or inclined to M. 6. Are two lines necessarily parallel if they are J_ to the same plane? If they are II to the same plane? FIG. 151. Illustrate your answers by reference to Fig. 151. 156 LESSONS IN GEOMETRY. 7. When are two planes parallel to each other? Two planes are parallel if they will not meet, however far both are produced. Two parallel planes are everywhere equally distant, and their distance apart is equal to the length of a perpendicular dropped from any point in one of the planes to the other plane. 8. What is the intersection of two planes ? The intersection of two planes is a straight line. 9. Define a dihedral angle. If two planes intersect, the amount of rotation about the intersection required to make one of the planes coincide with the other is called the dihedral angle of the two planes. The two planes are the faces of the angle ; their intersec- tion is the edge of the angle. Dihedral angles are expressed in degrees, etc., like ordi- nary angles. H G D FIG. 152. FIG. 153. 10. Define the plane angle of a dihedral angle. The dihedral angle formed by two planes ABC, ABE (Fig. 152), is obviously equal to the angle formed by two lines PQ, PR, drawn one in each face from any point P of the edge and _L to the edge. The angle of any two lines so drawn is called the plane angle of the dihedral angle. Thus, the angles RPQ, CBE, DAF, are plane angles of the dihedral angle in Fig. 152, SOLIDS. 157 11. W*hen are two planes perpendicular to each other? AVlien their dihedral angle is a right angle. 12. Name in Fig. 153 two parallel planes, two perpen- dicular planes, two inclined planes. In the last two cases what are plane angles of the dihedral angles? 13. If two parallel planes are cut by a third plane, what is true of their intersections? Illustrate by Fig. 153. 14. If two planes are each _L to a third plane, what is true of their intersection? Illustrate by Fig. 153. 15. How are planes represented on paper? The part taken for representation is usually a rectangle. If the plane is to be represented as coincident with, or parallel to, the plane of the paper, we simply draw the rec- tangle in its true shape. In all other cases we must substi- tute a parallelogram for the rectangle. The proper way to draw the parallelogram depends on the position of the plane with respect to the eye and the line of sight. By a little practice any one can learn to draw with tolerable accuracy planes in various different positions. 16. Draw two parallel planes joined by a line JL to them. 17. Draw two perpendicular planes. 18. Draw two planes forming a dihedral angle of about 45. 19. Draw three planes each J_ to the other two. 20. Draw three parallel planes, and a fourth plane inter- secting them. 21. Draw a horizontal plane and three vertical planes, all passing through a line J_ to the horizontal plane. 158 LESSONS IN GEOMETKY. Lesson 85. 1. Define a solid (see p. 9, No. 9). 2. Define a cube (Fig. 154). A cube is a solid bounded by six equal squares. The six squares are called the faces of the cube, and their intersections are called its edges. 3. Name in Fig. 154 the faces and the edges that are (1) II to the face ABCD ; (2) _L to the face ABCD. H G FIG. 154. FIG. 155. 4. How is a model of a cube made 9 Draw six equal squares on stiff cardboard, in one group, as shown in Fig. 155. Cut out the entire group, and then cut half-way through the cardboard, along the interior lines of division. Then fold the squares over these lines into the shape of a cube, and fasten the edges with slips of thin paper and mucilage (or paste) . 5. What is meant by developing a surface? To develop the surface of a solid is to construct upon a plane surface (as a sheet of paper) the surfaces, plane or curved, which form the boundaries of the solid. The figure thus obtained is called the development of the surface. Fig. 155 shows the development of the cube. 6. Make a model of a cube (edge 3 in.). SOLIDS. 159 7. Define the volume of a solid. The volume of a solid is the amount of space it fills. Volumes, like surfaces and lines, are measured by finding how many times a unit of volume is contained in the solid to be measured. The common units of volume are the cubic inch, the cubic foot, and the cubic yard. Each of these is a cube whose edge is equal to the corresponding unit of length. 1 cub. ft. =1728 cub. in. ; 1 cub. yd. = 27 cub. ft. 8. How is the volume of a cube found ? Suppose, for example, that one edge =10 in. (Fig. 156). The lower face ABC may be di- vided into 10x10, or 100, square inches (p. 000, No. 0). Upon each square inch a pile of 10 cubic inches may be formed (like those seen along the edge DC} . Hence the entire cube contains 10 x 10 x 10, or 1000, cubic inches. A similar result is obtained, what- A ever value the edge may have. Finding the product of a number, when used three times as a factor, is called cubing the number ; hence the result may be expressed by a formula, thus : Volume of a cube = cube of one edge. Find the volume and the entire surface, having given : 9. Edge 8 in. 11. Edge 1 ft. 2 in. 10. Edge 9 in. 12. Edge 2 ft. 3 in. 13. Find one edge if the volume is 8000 cubic feet. 14. How many square feet of lead are needed to line the bottom and sides of a cubical tank 44 feet deep, and how many cubic feet of water will the tank hold? FIG. 156. 160 LESSONS IN GEOMETKY. Lesson 86. 1. Define a rectangular solid. A rectangular solid is a solid bounded by six rectangles. The rectangles are called its faces, and their intersections are called its edges. The upper and lower faces are also called the bases, and the distance between them the height. 2. In the rectangular solid shown in Fig. 157, name the edges equal to AB ; to BC ; to BF. Also name the faces and edges II to AB ; 1_ to An. Q H G -'" ^" ^B / i T, /i nun E FIG. 157. 3. Name the three dimensions of the solid in Fig. 157. 4. Name in Fig. 157 three pairs of equal and parallel faces. 5. Draw two rectangular solids differing in shape. 6. Make a model of a rectangular solid 6 in. long, 4 in. wide, and 3 in. high (see p. 158, No. 4). 7. How is the volume of a rectangular solid found ? Suppose that the dimensions are : length, 5 in. ; breadth, 3 in. ; height, 7 in. (see Fig. 158). The base may be divided into 3x5, or 15, square inches. Upon each square inch there may be formed a pile of 7 cubic inches ; hence the solid must contain 3 x 5 x 7, or 105, cubic inches. In general : Volume of a rectangular solid = length x breadth x height. SOLIDS. 161 Find the entire surface and the volume of a rectangular solid, having given : 8. Length 9 in., breadth 7 in., thickness 3 in. 9. Length 6f ft., breadth 5J ft., depth 4J ft. 10. Length 4 ft., breadth 1| yds., height 108 in. 11. Length 4 ft. 8 in., breadth 3 ft. 10 in., height 3 ft. 12. How many tons of coal will a bin 20 ft. by 16 ft. by 8 ft. hold, allowing 40 cubic feet to a ton? 13. A cellar which measures 12 ft. by 6 ft. is flooded to a depth of 4 in. Find the weight of the water, supposing that 1 cub. ft. of water weighs 1000 oz. (62 Ibs.). 14. What weight of water will a rectangular cistern hold, its length being 4 ft., breadth 2 ft. 6 in., depth 3 ft. 3 in., and 1 cub. ft. of water weighing 1000 oz.? 15. How many bricks 9 in. by 4^ in. by 3 in. are needed to build a wall 90 ft. long, 18 in. thick, and 8 ft. high? 16. A book is 8 in. long, 6 in. wide, and l\ in. thick. Find the depth of a box 3 ft. 4 in. long, and 2 ft. 6 in. wide, that it may hold 400 such books. 17. Marble is 2.716 times as heavy as water, and 1 cub. ft. of water weighs 1000 oz. Find the weight of a block of marble 9 ft. 6 in. long, 2 ft. 3 in. wide, and 2 ft. thick. 18. Iron weighs 7.2 times as much as water, and 1 cub. ft. of water weighs 1000 oz. What will an open cistern made of iron 1 in. thick weigh when empty, if its external dimen- sions are 5 ft., 4 ft., and 3 ft. ? 19. A cistern is 5 ft. 6 in. long, 3 ft. 9 in. wide, and 1 ft. 3 in. deep. How many gallons of water will it hold ? What weight of water will it hold ? (1 gal. of water = 231 cub. in. ; 1 cub. ft. of water weighs 1000 oz.) 162 LESSONS IN GEOMETRY. Lesson 87. 1. Define a right prism, and terms related to it. A right prism is a solid bounded by two equal and parallel polygons called bases, and by three or more rectangles called the lateral faces. The intersections of the lateral faces are called the lateral edges. The distance between the bases is called the height of the prism. A prism is called triangular if the bases are triangles, hexagonal if they are hexagons, etc. A regular prism is a right prism whose bases are regular polygons. The prism in Fig. 159 is a regular hexagonal prism. 2. What positions with respect to the bases do the lateral faces of a right prism have? the lateral edges? 3. Name on a right prism the edges that are parallel. 4. Draw a triangular prism. 5. Draw a hexagonal prism. FIG. 159. FIG. 160. 6. Make a model of a regular triangular prism (Fig. 1GO). 7. How is the volume of a right prism found ? It can be proved that the volume of a right prism is found by multiplying the area of the base by the height. Volume of a right prism = base x JieiyJit. SOLIDS. 163 NOTE. The word "prism" used alone means here "right prism." Find the entire surface and the volume of a prism, having given : 8. Area of square base 49 sq. in., height 8 in. 9. Square base, side of base 4 in., height 10 in. 10. Rectangular base 6 ft. by 3 ft., height 12 ft. 11. Base an equilateral A, side 2 ft., height 7 ft. 12. Base a regular hexagon, side 4 ft. 9 in., height 12-J ft. 13. Base a regular hexagon, side 10 in., height 10 ft. 14. Is a cube a right prism? Why? Is a rectangular solid a right prism? Why? 15. What is the volume of a regular four-sided prism if the height is 6 in. and one side of the base is 2 in. ? What would be the volume if the height were doubled? if the side of the base were doubled ? if both were doubled ? 16. The bases of a prism are trapezoids whose parallel sides are 12 ft. and 8 ft., and the altitude is 6 ft. Find the volume of the prism if the height is 32 ft. 17. How many cubic feet of stone are required to build a dam 1000 ft. long, 20 ft. high, 10 ft. wide at the bottom, and 4 ft. wide at the top? (Consider the dam to be a trapezoidal prism like that in No. 16.) 18. The distance around a reservoir in the shape of a regular hexagon is 360 ft. If the average daily loss from evaporation amounts to a layer of water 2 in. deep, how many cubic feet of water must be supplied daily to maintain the water at a constant level ? 19. Pencils are often made in the shape of regular hexa- gonal prisms. Find the volume of the pencil given you (cor- rect to a hundredth of a cubic inch). 164 LESSONS IN GEOMETRY. Lesson 88. 1. Define a right cylinder, and the related terms. A right cylinder is the solid generated by a rectangle revolving about one of its sides. A right cylinder is also called a cylinder of revolution. If the rectangle ABCD (Fig. 161) revolve about the side CD, the sides AD, BO, will describe equal and parallel circles, and the side AB will describe a curved surface. A right cylinder, therefore, is a solid bounded by two equal and parallel circles and a curved surface lying between the circles. The circles are called the bases of the cylinder, and the curved surface is called its lateral surface. The axis of a right cylinder is the line joining the centres of the bases ; it is perpendicular to the bases. The height of a cylinder is the length of its axis. 2. What is the development of the lateral surface? Illus- trate by means of a sheet of paper. FIG. 161. FIG. 162. 3. Draw two right cylinders differing in shape. 4. Make a model of a right cylinder (see Fig. 162) 5. What are the general formulas for finding the lateral surface and the volume? Lateral surface = circumference of base x height. Volume = base X height. SOLIDS. 165 NOTE. The word "cylinder" is here used meaning "right cylinder." Find the lateral surface and the volume of a cylinder, having given : 6. Radius of base 7 in., height 10 in. 7. Radius of base 1 ft. 2 in., height 5 ft. 8. Diameter of base 9 ft. 4 in., height 12 ft. 9. Circumference of base 7 ft. 4 in., height 10 ft. 10. How large a cylinder can be made by rolling up a rectangular sheet of tin 88 in. by 66 in., so that the height of the cylinder is 88 in.? How large, if. rolled up so that the height is 66 in.? 11. Find the height of a cylinder if the volume is 114 cub. yds. 2 cub. ft., and the radius of base 7 ft. 12. How many cubic yards of earth must be dug out to make a well 3 ft. in diameter, and 20 ft. deep ? 13. The diameter of a well is 4 ft. 8 in., its depth is 30 ft. Find the cost of digging it at $3.75 per cubic yard. 14. How many cubic yards of earth must be dug out in making a tunnel 100 yds. long, whose section is a semicircle with a radius of 10 ft. ? 15. What change in the volume of a cylinder is produced by doubling its height ? by doubling the diameter of its base ? by doubling both ? 16. Two cylinders have the same height, but the radius of the base of one cylinder is six times that of the other. Com- pare their volumes ; compare also their lateral surfaces. 17. Find the cost of cementing the side and bottom of a cylindrical tank 20 ft. deep and 18 ft. in diameter, at 32 cents per square foot. 18. How many gallons of water are there in a cylindrical well 7 ft. in diameter, if the water is 10 ft. deep and there are 1\ gals, in each cubic foot? 166 LESSONS IN GEOMETRY, Lesson 89. 1, Define a right pyramid, and related terms. A right pyramid is a solid bounded by a polygon and three or more isosceles triangles which have a common vertex. The polygon is called the base, the triangles the lateral faces, and their common vertex the vertex of the pyramid. The intersections of the lateral faces are called the lateral The distance from the vertex to the base is called the height of the pyramid. A regular pyramid is a right pyramid whose base is a regular polygon. The lateral faces are equal isosceles tri- angles, and their common altitude is called the slant height of the pyramid. Thus, AB is the slant height of the regular pyramid represented in Fig. 163. 2. Draw a triangular pyramid (a pyramid with a A for the base). 3. Draw a hexagonal pyramid. FIG. 163. FIG. 164, 4. Make a model of a square pyramid (Fig. 164). 5. How is the volume of a pyramid found ? It can be proved that the volume of any pyramid may be found by means of the formula : Volume of a pyramid = \ X base x height. SOLIDS. 167 NOTE. The word "pyramid," used alone, means "right pyramid." 6. Find the volume of a pyramid, having given : area of the base = 64 sq. in., height of pyramid = 15 in. 7. Find the volume of a pyramid if the height is 30 ft., and the base is a regular hexagon whose side is 6 ft. 8. Find the slant height of the pyramid in No. 7. 9. Find the total surface of the pyramid in No. 7. Find the volume, and also the total surface of a square pyramid, having given : 10. Side of base 40 ft., height 48 ft. 11. Side of base 22 ft., height 60 ft. NOTE. The frustum of a pyramid is the portion of the pyramid contained between the base and a plane parallel to the base. The base and the section made by the plane are called ^ the bases of the frustum. II ! \\ The distance between the bases is the height. To find the volume of a frustum, add together the areas of the bases and the square root of their product, and multiply the sum by | the height. FlG - 165 - 12. What kind of figures are the lateral faces of a frustum ? 13. Find the total surface of a frustum of a square pyr- amid (Fig. 165) if the sides of the bases are 12 in. and 4 in., and the slant height (or altitude of each face) is 5 in. 14. Find the volume of the frustum in No. 13 if the height of the frustum is 3 in. 15. Find the volume of the frustum of a square pyramid if the sides of the bases are 21 yds. and 15 yds., and height 84 yds. 16. A church spire has the shape of a frustum of a regular hexagonal pyramid; each side of the base is 5 ft., and of the top 2 ft. ; the altitude of each trapezoidal face is 20 ft. How many square feet of tin roofing are required to cover the lateral faces and the top ? 168 LESSONS IN GEOMETRY. Lesson 9O. 1. Define the right cone, and related terms. A right cone is the solid generated by a right triangle revolving about one of its legs. A right cone is also termed a cone of revolution. If the right triangle ABC (Fig. 166) revolve about the leg BC, the other leg AB will describe a circle called the base of the cone, and the hypotenuse AC will describe a curved surface called the lateral or convex surface of the cone. The point C is called the vertex of the cone. The leg jBC, about which the triangle revolves, is called the axis of the cone. The length of the axis is called the height of the cone. The length of the hypotenuse AC of the generating tri- angle is called the slant height of the cone. 2. Draw two right cones differing in shape. FIG. 166. FIG. 167. 3. What kind of a figure is the development of the lateral surface of a right cone (see Fig. 167)? 4. Make a model of a right cone. 5. What are the formulas for finding the lateral surface and the volume? Lateral surface = X circumference of base x slant height. Volume = 1 X base x height. SOLIDS. 169 NOTE. The word "cone," if used alone, means a "right cone." 6. The height of a cone is 6 in., the radius of the base is 2 in. Find the slant height (p. 126, No. 1). 7. Find the volume and the lateral surface of a cone if the height is 40 ft., and the radius of the base is 9 ft. 8. Find the volume and the lateral surface if the height is 28 ft., and the radius of the base is 21 ft. 9. A right triangle whose legs are 3 in. and 4 in. gene- rates a cone by revolving about its shorter leg. Find the volume and the lateral surface of this cone. 10. Solve No. 9 if the triangle revolve about the other leg. 11. How much canvas is required to make a conical tent 80 ft. high, and 70 ft. in diameter at the base? NOTE. The frustum of a cone is the part contained between the base and a plane II to the base. The bases and height are defined, and the volume found, as in the case of the frustum of a pyramid (p. 167). The lateral surface = half the product of the circumference ... j i ; . T. ' FIG. 168. of base and the slant height. 12. How many square feet of tin will be required to make a funnel with the radii of top and bottom 14 in. and 7 in. respectively, and the height 24 in. ? 13. Find the expense of polishing the curved surface of a marble column in the shape of a frustum of a right cone, slant height 12 ft., radii of bases 3 ft. 6 in. and 2 ft. 4 in., at 60 cents per square foot. 14. A round stick of timber is 20 ft. long, 3 ft. in diameter at one end, 2.6 ft. at the other. How many cubic feet does it contain ? 15. A bucket is 16 in. deep, 18 in. wide at the top, and 12 in. wide at the bottom. How many gallons of water will it hold, reckoning 7 gallons to the cubic foot? 170 LESSOMS IN GEOMETRY. Lesson 91. 1. Define "a sphere. A sphere is the solid generated by a semicircle revolving about its diameter. If the semicircle ACBO (Fig. 169) revolve about the diameter AB, the semicircle generates a sphere, and the semi-circumference ACB generates a curved surface forming the boundary of the sphere. The centre of the semicircle is the centre of the sphere. All points in a spherical surface are equidistant from the centre. 2. Define a radius and a diameter of a sphere. In Fig. 169 OA is a radius, and AOB is a diameter. A 3. Define great circles and small circles. Every section of a sphere made by a plane is a circle. If the plane pass through the centre, the circle is called a great circle ; in all other cases the circle is called a small circle. 4. Define the axis and poles of a circle. The axis of a circle (great or small) is the diameter which is perpendicular to the plane of the circle. The poles of a circle are the ends of its axis. 5. Illustrate the nbovc definitions by referring to Fig. 1 70. SOLIDS. 171 6. State three truths respecting great eircles. (1) A great circle divides a sphere into two equal parts. These two equal parts are called hemispheres. (2) Two great circles bisect each other. (3) The poles of a great circle are at the distance of a quadrant from every point in its circumference. 7. Define parallels. If a sphere is cut by a series of parallel planes, the cir- cumferences of the circles thus formed are called parallels. Example : parallels of latitude on the earth's surface. 8. Define a zone and its altitude. A zone is the portion of the surface of a sphere contained between two parallel planes. The distance between the two planes is the altitude of the zone. Example: the torrid zone, etc., on the earth's surface. 9. Define meridians* If a sphere is cut by a series of planes all passing through the same diameter, the circumferences of the great circles thus formed are called meridians. Example : meridians of longitude on the earth's surface. 10. Define a lune. A lune is the portion of the surface of a sphere contained between two semi-circumferences of great circles. 11. Point out in Fig. 170 parallels, meridians, a zone, its altitude, a lune. 12. Draw Fig. 170 (to a larger scale if you prefer). 13. Draw a sphere cut by three planes passing through the centre, and each perpendicular to the other two. 14. If (Fig. 170) PEQ represent the meridian through Greenwich, point out the latitude and longitude of the point M. 172 LESSONS IN GEOMETRY. Lesson 92. 1. When is a sphere inscribed in a cylinder? A sphere is inscribed in a cylinder when its surface touches the bases and the lateral surface of the cylinder. Also, the cylinder is circumscribed about the sphere (Fig. 171). 2. Compare the radius and diameter of a sphere with the dimensions of the circumscribed cylinder. P FIG. 171. 3. How is the surface of a sphere found ? It can be proved that the surface of a sphere is equal to the lateral surface of the circumscribed cylinder. Let r denote the radius of a sphere. Then r is also equal to the radius of the base of the circumscribed cylinder, and 2r is its height; therefore its lateral surface is equal to 2-n-r x 2r, or irr (p. 164, No. 5). Hence if the value of TT be taken as \ 2 -, Surface of a sphere = - 8 T 8 - r 2 . 4. How is the volume of a sphere found ? The volume of a sphere is found by multiplying the sur- face by one-third of the radius. Therefore if r denote the radius, the volume = \r X 4^= | Trr*. Hence if the value of TT be taken as , SOLIDS. 173 Find the surface and the volume of a sphere, given : 5. Radius 1 in. 10. Diameter 7 in. 6. Radius 2 in. 11. Diameter 16 in. 7. Radius 3 in. 12. Diameter 21 in. 8. Radius 4 in. 13. Diameter 42 ft. 9. Radius 3 ft. 6 in. 14. Diameter 5 ft. 10 in. 15. A regulation base-ball is 9J in. in circumference. How many square inches of leather are required in order to make 1000 base-balls? 16. The circumference of a dome in the shape of a hemi- sphere is 66 ft. How many square feet of tin roofing are required to cover it? 17. If the ball on the top of St. Paul's Cathedral in London is 6 ft. in diameter, what would it cost to gild it at 7 cents per square inch? 18. The area of a zone is equal to the product of its alti- tude and the circumference of a great circle. Find the area of the upper zone on the sphere in Fig. 172 if the altitude PR is 15 ft. and the radius is 35 ft. 19. The altitude of the torrid zone on the earth is about 3200 miles. What is its area in square miles, assuming the radius of the earth to be equal to 4000 miles ? 20. If one cubic inch of iron weighs 4J oz., what will an iron ball 10J in. in diameter weigh? 21. Find the weight of a spherical shell 10 in. in diameter and 2 in. thick, composed of a substance 1 cub. ft. of which weighs 216 Ibs. 22. How much rubber is there in a tennis ball whose diameter is 3^ in. if the thickness of the rubber is J in. ? How much cloth is needed to cover the ball? How much pasteboard is needed to make a cylindrical box, open at the top, which will just hold the ball? 174 LESSONS IN GEOMETEY. Lesson 93. Review. NOTE. 1 cub. ft. of water weighs 1000 oz. ; 16 oz. = 1 Ib. 1. Review the italicized exercises in Lessons 84-88. 2. Draw two parallel planes cut by a third plane. What is true of the two lines of intersection ? 3. Draw two planes each perpendicular to a third plane. What is true of their intersection ? 4. Find the volume of the cube given you. 5. Find the volume of the rectangular solid given you. 6. Find the volume of the prism given you. 7. Find the volume of the cylinder given you. 8. What will be the cost of digging a cellar 36 ft. long, 20 ft. wide, and 8 ft. deep, at 5 cents per cubic foot? 9. What will it cost to sink a well 100 ft. deep and 3 ft. in diameter, at $2 per cubic foot? 10. If a cubic foot of brass is drawn into a wire -^y of an inch in diameter, what will be the length of the wire? 11. How many cylindrical pieces of lead f of an inch in diameter and ^ of an inch thick must be melted, in order to form a cube whose edge is 3 in. long? 12. The piston of a pump is 14 in. in diameter, and moves through a space of 3 ft. How many tons of water will be thrown out by 1000 strokes? 13. A body is placed under water in a right cylinder 60 in. in diameter, and the level of water is observed to rise 30 in. Find the volume of the body. 14. How much will a brass cylinder weigh under water, if the height is 10 in. and the diameter of the base is 7 in. ? Brass is 7.8 times heavier than water, and a body when immersed in water loses a weight equal to the weight of the water displaced. SOLIDS. 175 Lesson 94. Review. 1. Review the italicized exercises in Lessons 89-92. 2. Find the volume of the pyramid given you. 3. Find the volume of the cone given you. 4. Find the volume of the sphere given you. 5. Make a model of a regular pyramid bounded by four equal equilateral triangles. Make each edge 4 in. long. Then find the volume and the .\ ^ ///// / \ / \ m/t entire surface. (The apothem of the base may be FIG. 173. found from the table on p. 122.) 6. If the height of a cone and the radius of its base are known, how is the slant height found? 7. Can the surface of a sphere be developed? 8. How much asphalt varnish will be needed to coat 1000 spherical bombs, each 10 in. in diameter, if 1 pint of varnish will cover 300 square inches of surface ? 9. If a cylinder and a cone have equal bases and equal volumes, compare their heights. 10. Find the volume of a cylinder circumscribed about a sphere whose radius is 7 in. 11. A cylindrical pail is partly filled with water. The radius of its base is 3^- in. If a bullet 3^ in. in diameter be dropped in, how much will the level of the water rise? 12. The chimney of a factory has the shape of the frustum of a square pyramid; its height is 180 ft., and the sides of its upper and lower bases are 16 ft. and 10 ft. respectively ; the section of the flue is throughout the entire length a square whose side is 7 ft. How many cubic feet of brick does the chimney contain? 176 LESSONS IN GEOMETRY. Lesson 95. Review. 1. What are the chief metric units of volume ? The cubic meter, having the abbreviation cbm. The cubic decimeter, having the abbreviation cdm. The cubic centimeter, having the abbreviation com. The cubic meter, when used for measuring wood, is called a stere ; and the cubic decimeter, when used for measuring fluid substances, is called a liter. 2. How are the metric units of volume related ? Since 10 centimeters =1 decimeter, it follows that 10 x 10 x 10, or 1000, cubic centimeters = 1 cubic decimeter. And since 10 decimeters = 1 meter, it follows that 10x10x10, or 1000, cubic decimeters =1 cubic meter. The reason is evident from a study of Fig. 156, p. 159. NOTE. l cbm = 35i cub. ft., nearly; 1 liter =1^ qts., nearly. 3. How many cubic centimeters make 1 cubic meter? 4. Define the gram and the kilogram. A gram is the weight of l ccm of pure cold water. A kilogram is 1000 grams, therefore equal to the weight of 1000 ccm , or 1 liter, of pure cold water. NOTE. 29 grams =1 oz., nearly; 1 kilogram 2 Ibs., nearly. 5. What is the weight of l cbm of water? 6. A bottle has a capacity of half a liter. How many grams of water will it contain? 7. The volume of a vessel is 2.4 cbm . What is the volume in liters? What weight of water will it hold? 8. A gallon jug contains 231 cub. in. Taking 1 inch :is equal to \ of a centimeter, find what weight of water in gnuns can be put into the jug. SOLIDS. 177 NOTE. In the following exercises take TT equal to 3.1416. 9. Find the total surface of a square pyramid if a side of the base is 3 m and the slant height is 15 m . 10. Find the total surface of a cone if its height is 12 m and the radius of the base 5 m . 11. If the radius of a sphere is 8 m , find the area of a zone whose height is equal to half the radius. 12. How high must a box 5 dm long and 2 dm wide be in order that it may hold exactly 30 liters ? 13. A cylindrical pail holding just 1 liter is 18 cm high. What is the diameter of its base? 14. A vessel has the shape of a frustum of a cone. The height is l m , and the inside circumferences of the bases are 5 m and 4 m . How many kilograms of water will the vessel hold? 15. What is the volume of the largest sphere that can be turned from a wooden cube whose edge is l dm ? 16. What is the volume of the largest sphere that can be turned from a wooden cylinder, if the height of the cylinder and diameter of the base are each equal to 12 cm ? 17. Marble is 2.7 times as heavy as water. What is the weight of a block of marble 20 dm by 16 dm by 8 dm ? 18. An immersed body is buoyed up by the weight of its own volume of water. Iron is 1\ times heavier than water. What will an iron ball whose diameter is 10 cm weigh under water ? 19. Eighty bullets, equal in size, are dropped into a cylin- drical vessel 68 cm in diameter containing water. The level of the water rises 2 cm . Find the diameter of a bullet. 20. Lead is about 11 J times heavier than water. Find the weight of a lead pyramid if the base is a square whose side is 40 cm , and the lateral faces are equilateral triangles. 178 LESSONS IN GEOMETRY. Lesson 96. Review. NOTE. When ?r is used, take IT equal to 3.1416. 1. How many square meters of sheet iron are needed to make a cylindrical pipe 12 m and 40 cm in diameter? 2. How many liters of water are there in a cubical tank if one edge is 24 m , and the water is 4 m deep? 3. A square field contains 4 hectares. What will it cost to dig all around the field a ditch 2 m deep and l m wide, the terms being 25 cents per cubic meter of earth thrown out? 4. A Dutch windmill, in the shape of the frustum of a cone, is 12 high. The outer diameters of the bases are 16 m and 12 m ; the inner diameters, 12 m and 10 m . How many cubic meters of stone were needed to build it? 5. The piston of a pump is 36 cm in diameter, and moves through a space 50 cm . How many kilograms of water are thrown out by 1000 strokes? 6. A body is placed under water in a cylinder 60 cm in diameter. The level of the water is observed to rise 30 cm . Find the volume of the body. 7. How much will a brass cylinder weigh under water if its height is 64 cm , and the diameter of its base 40 cm ? Brass is 8.4 times heavier than water. (See p. 177, No. 20.) 8. When a body floats in water, the weight of the water displaced is just equal to the weight of the body. Find the weight of a sphere 20 cm in diameter, which floats half under water and half above. 9. A cone of loaf-sugar is l m high, and the diameter of its base is 40 cm . How far from the base must the cone be cut by a plane parallel to the base, in order that the two parts may be equal in volume? VERSITY DRAWING EXERCISES. i. 3. 4. 6. 180 DRAWING EXERCISES. - 7. 8. n n 9. DRAWING EXEECISES. 181 11. 12. 182 DRAWING EXERCISES. 13. f '-: 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. NOV 18 NOV 18 SEP 141937 LD 21-50ni-8, D 17294- YB 17295 / B' 183667-