iij iiiil i lili ii iiiiii ! i i i pi. mi ill lil I 11 liiSli i|5f Ijiii i il i i ! 1 WmB I m mm. i mm 1 te' I !Ji ii;! Ii i 1 hlHililj I !!i ill hi! ii HI litliHI;; IliiliipiiiiiHiiilii!:::; il iii ii Ii i llHi yiiliHliH^ I !! illiiliillH I ill Ii HI II lIHlHili? i iipiiii i ill iijii ?i i! 1 ii m ] iiiiiliui iiiiliiiili GIFT OF 1 J PLANE GEOMETRY A SERIES OF MATHEMATICAL TEXTS EDITED BY EARLE RAYMOND HEDRICK THE CALCULUS By Ellery Williams Davis and William Charles Brenke. PLANE AND SOLID ANALYTIC GEOMETRY By Alexander Ziwet and Louis Allen Hopkins. PLANE AND SPHERICAL TRIGONOMETRY WITH COMPLETE TABLES By Alfred Monroe Kenyon and Louis Ingold. PLANE AND SPHERICAL TRIGONOMETRY WITH BRIEF TABLES By Alfred Monroe Kenyon and Louis Ingold. THE MACMILLAN TABLES Prepared under the direction of Earle Raymond Hedrick. PLANE GEOMETRY By Walter Burton Ford and Charles Ammerman. PLANE AND SOLID GEOMETRY By Walter Burton Ford and Charles Ammerman. SOLID GEOMETRY By Walter Burton Ford and Charles Ammerman. PLANE GEOMETRY BY WALTER BURTON FORD JUNIOR PROFESSOR OF MATHEMATICS, THE UNIVERSITY OF MICHIGAN AND CHARLES AMMERMAN THE WILLIAM McKINLEY HIGH SCHOOL, ST. LOUIS EDITED BY EARLE RAYMOND HEDRICK !C Neto gork THE MACMILLAN COMPANY 1920 All righU reserved COPTEIGHT, 1913, By the MACMILLAN COMPANY. -Set dp und'electrotyped. Published September, 1913. J. S. Gushing Co. — Berwick & Smith Co. Norwood, Mass., U.S.A. • PREFACE While the preparation of a textbook on Geometry presents much the same problem whenever undertaken, the arrangement of its details and the balancing of its parts must be expected to change as new educational ideals become established, and as new conditions of society arise. It is no longer possible, for example, to expect favor for the typical lengthy text of a generation ago. Although such texts proceed in good logical fashion to deduce the traditional theo- rems one after another, yet they make little or no appeal to the world of common experience ; and they present far more mate- rial than the student can absorb in the course as it is now usually taught. Neither is the ideal book something so far removed from this traditional type that its chief feature is a large variety of illuminating diagrams dra\vn from the Arts. Between these two extremes lies the only proper course to-day, and this the authors have tried diligently to find. The traditional manner of presentation in a logical system is pre- served, but logical development is not made the sole purpose of the book. Thus, problems drawn from the affairs of prac- tical life are inserted in considerable number. The function of such problems is not to train the student in the technique of any of the Arts ; rather it is to illuminate the geometric facts, and to make clear their importance and their significance. Geometry is and is likely to remain primarily a cultural, rather than an informational subject. But the intimate con- nection of Geometry with human activities is evident ui)on every hand, and constitutes fully as much an integral part of the subject as does its older logical and scholastic aspect. 459955 vi PREFACE Teachers of Geometry throughout the country recognize this and appreciate strongly the value of the study, not only as subject-matter in which practice in logical processes is ideally simple and beautiful, but also as an instrument for meeting real human needs. It is certainly most desirable from the standpoint of effective instruction that the students also come to realize all of these merits, thus answering for themselves the persistent question as to why the subject is studied. The necessary attention and concentration of mind on the students' part will then become from the very outset easy and natural. Attention is called to the Introductory chapter. In it the student is acquainted with the use of the ruler and compasses in the more simple construction problems, and is thus brought early face to face with many of the fundamental notions of Geometry. Great care has been exercised at this point that the student may approach the more rigid demands of the de- ductive side of the subject gradually, rather than plunge into them at the outset. Even in Chapter I, in which strictly deductive work commences, no abrupt change of style occurs, but an easy transition is made into the parts that so often appear foreign and unintelligible to the beginner. Indeed, throughout the book, an effort has been made to soften the austere and unnatural style that has frequently proved a bar to ready comprehension, and to avoid an excess of sym- bolic characters and technical phrases that do not add to the reasoning. The book is distinguished by its acceptance of the principle of emphasis of important theorems laid down by the Committee of Fifteen of the National Education Association in their Re- port.* Thus, theorems of the greatest value and importance are printed in bold-faced type, and those whose importance is considerable are printed in large italics. * Printed as a separate pamphlet with the Proceedings for 1912. Reprinted also in School Science, 1911, and in The Mathematics Teacher, December, 1912. PREFACE vii The Report just mentioned has been of great assistance, and its principles have been accepted in general, not in a slavish sense but in the broad manner recommended by the Committee itself. A perusal of the Report will give, more fully and accurately than could be done in this brief preface, the consid- erations which led to the adoption of these principles, in par- ticular, the principle of emphasis upon important theorems, both by the Committee and by the authors of this book. The authors are indebted also to many another book and to many recent reports and papers for ideas and problems. In the Introduction, the effect of the excellent little book of Godfrey and Siddons * is sufficient to deserve explicit mention. Detailed description of the principles followed in the Solid Geometry may be omitted, since they are in spirit the same as those of the Plane Geometry. The great excellence of the figures, particularly that of the very unusual and effective * phantom' halftone engravings in the Solid Geometry, de- serves mention. These figures should go far toward reliev- ing the unreality which often attaches to the constructions of Solid Geometry in the minds of students. W. B. FORD. CHARLES AMMERMAN. E. R. HEDRICK, Editor. Godfrey and Siddons, Plane Geometry, Cambridge University Press. CONTENTS Introduction Part I. Drawing Simple Figures Part II. The Principal Ideas used in Geometry- Part III. Statements for Reference Supplementary Exercises Chapter I. Rectilinear Figures Part I. Triangles Parallel Lines . Angles and Triangles Quadrilaterals Polygons Locus of a Point Part II. Part III. Part IV. Part V. Part VI. Miscellaneous Exercises . Chapter II. The Circle Part I. Chords — Arcs — Central Part II. Tangents and Secants Part III. Measurement of Angles Part IV. Construction Problems Miscellaneous Exercises . Angl Chapter III. Part I. Part II. Part III. Part IV. Part V. Proportion. Similarity General Theorems on Proportion Proportional Line Segments . Similar Triangles and Polygons Proportional Properties of Chords, Secants and Tangents .... Similar Kight Triangles, Trigonomet tios Miscellaneous Exercises Ra PAOBS 1-33 1-9 10-25 26-29 30-33 35-88 35-48 49-57 58-67 68-75 76-78 79-82 83-88 89-125 89-98 99-106 107-116 117-119 120-125 126-161 126-129 130-138 139-147 148-151 152-156 157-161 CONTENTS IX PAGES Chapter IV. Areas of Polygons. Pythagorean Theorem . 162-186 Miscellaneous Exercises . . . . . . . 180-185 Chapter V. Regular Polygons and Circles . . . 186-205 Miscellaneous Exercises 201-205 Appendix to Plane Geometry: Maxima and Minima . 206-213 Tables i-xxvii Table I. Quantities Determined by a Given Angle . ii-vii Table IF. Powers and Roots viii-xxvi Table III. Important Numbers xxvii Index xxix-xxxi ?e This edition contains only the Chap- ters relating to Plane Geometry, as listed above. An edition that contains also the usual topics of Solid Geometry is published under the title Plane and Solid Geometry. The Solid Geometry is also published separately. PLAICE GEOMETRY INTRODUCTION PART I. DRAWING SIMPLE FIGURES 1. Geometry is the branch of mathematics that deals with space. In it we consider points, lines, triangles, circles, spheres, and so forth. Such questions as the following are considered : When do two triangles of different shapes have the same area ? How does the length around a circle compare with the diameter ? How can the volume of a sphere be calculated ? Geometry not only answers many questions of this kind, but it also pre- sents a systematic study of the general principles involved, and the reasons for all statements are given with care. 2. Ruler and Compasses. In studying geometry it is desir- able to draw accurate figures. The student will need a nder^ with which to draw (or extend) straight lines; and a pair of compasses, with which to draw circles and to transfer lengths from one position to another. Fig. 1 For distinctness, the curve that forms the circle is often called the circumference of the circle. The position of the fixed point of the compasses is called the center of the circle. Any portion of the circumference is called an arc of the circle. The distance from the center to the circumference, which is the same all the way around the circle, is called the radius. B 1 INTRODUCTION [§3 3. Problem 1. Draw with a ruler and compasses a triangle each of whose sides is one inch long. [A triangle is a figure bounded by three straight sides.] Solution. With a ruler and pencil draw a line and mark two points A and B on it one inch apart. Around each of these points as a center, with a radius one inch long, draw an arc of a circle above the line AB so that the two arcs cut each other in a third point, which we will call C. C Fig. 2 Connect A with by a straight line, and connect B and C by another straight line. The triangle ABC is the required figure ; state why each side is one inch long. EXERCISES 1. Show how to draw a triangle each of whose sides is a given length AB. [Hint. See Problem 1.] 2. Copy the adjoining figure. Draw all the triangles that have equal sides which you can by draw- ing straight lines connecting pairs of points marked in the figure. 3. Draw in separate positions on heavy paper or on cardboard two triangles, each of whose sides is three inches long. Cut out one of these. Will it fit precisely upon the other one ? Will it fit one drawn by another stude»i^ §4] DRAWING SIMPLE FIGURES 3 4. Problem 2. Draw with a ruler and compasses a triangle ichose three sides are equal, respectioely, to three given lengths. Solution. Let a, 6, c be the three given lengths. (Fig. 3.) Draw a line. Mark two points A and B on it so that the distance AB is equal to one of the given lengths, say c. This is done with the compasses. Around A as center draw an arc of a circle with a radius equal to another of the given lengths, say h. Around B as center draw another arc with a radius equal to the remaining given length a, and in such a position that the two arcs cut each other. Let the point where the two arcs cut each other be called C. Draw CA and CB. Then the triangle ABC is the one required. EXERCISES 1. Draw, with ruler and compasses, a triangle whose three sides are, respectively, two inches, three inches, and four inches in length. 2. Is there any difference between this triangle and one whose sides are, respectively, four inches, two inches, and three inches in length? Draw both triangles separately on card- board, cut one out, and see if it will exactly fit on the other one. Was it necessary to turn it over ? 3. What can you say about any two triangles if the three sides of one are equal, respectively, to the three sides of the other ? 4 INTRODUCTION [§5 5. Problem 3. Drav^, by means of ruler and compasses, a perpendicular* to a given straight line at a point in that line. fP I C Fig. 4 Solution. Let AB be the given line and C the given point in it. With C as center, and with any convenient radius, draw- arcs of the same circle cutting AB at two points which we will call F and Q. With a somewhat longer radius than before, draw two arcs of circles above AB with centers at P and Q, being careful that these arcs have the same radius. These two arcs will cut each other at a point that we will call D. Draw the straight line joining C and I). This is the required perpendicular to AB at O; and there is no other. EXERCISES 1. Draw with ruler and compasses a perpendicular to a line four inches long at a point one inch from the right-hand end. 2. Draw a triangle with two sides perpendicular to each other, making the perpendicular sides three inches long and four inches long, respectively. Measure the third side. [The third side should be five inches long. In making the drawing, any lines may be extended as desired.] * The perpendicular CD can be drawn, of course, with a carpenter's square or a drawing triangle ; but the problem is to draw it loith ruler and compasses only. The drawing may be tested by folding it in a crease along the line CD ; then one end CB of AB should fall on the other end CA of AB. If it does, CD is said to be perpendicular to AB at C. §6] DRAWING SIMPLE FIGURES 5 6. Problem 4. Draw ivith ruler and compasses a perpen- dicular to a given straight line from a given p>oint not on that line. ^ L_^ Fig. 5 Solution. Let AB be the given line and let P be the given point not on that line. With P as center and with any radius long enough to reach past the line AB, draw two arcs of the same circle so as to cut the line AB in two points, which we shall call R and S. Around the points R and S as centers, with the same radius for both, draw two arcs of circles below AB, and let C be the point where they cut each other. The straight line joining P and C is the desired perpendicii- lar. Show this by folding the figure on the line PC and con- vincing yourself that TSB falls along TRA. EXERCISES 1. Draw a triangle whose three sides are in the ratios 2:3:3. Then draw with ruler and compasses a perpendicular from each corner to the opposite side. [The accuracy of the drawing can be tested by the fact that these three perpendiculars should meet in one point.] 2. In every drawing of one side of a house, a line is drawn representing the horizontal (or base) line. Show how to draw with ruler and compasses, through any point in the figure, a vertical line, that is, a perpendicular to the base line. 6 INTRODUCTION [§ 7 7. Problem 5. Draw with ruler and compasses an angle equal to a given angle. [An angle is formed by two portions of straight lines that end at the same point. ] B Fig. 6 Solution. Given the angle at A, the problem is to draw an angle equal to A from a point P on another line MN. With A as center, draw an arc of a circle cutting the lines that form the angle J. in the two points B and C. With P as a center, draw an arc x with the same radius as that just used, cutting MN at Q and extending upwards quite a distance. With Q as center, and with a radius equal to the distance from B to G, draw an arc y cutting the arc ic at a point 0. Then the angle at P formed by the portions of straight lines PQN and PO is equal to the given angle at A, for the two figures will fit each other exactly if one is placed upon the other with AdX P and AB along PQN. EXERCISES 1. Draw two angles that are exactly equal to each other. [Hint. Draw any angle and then draw another equal to it.] 2. Draw a triangle with one angle equal to the angle in Fig, 6 and with the two sides that form that angle two inches long and three inches long, respectively. Cut out your tri- angle and compare it with those made by other students. Will it fit exactly on theirs ? 3. What can you say of any two triangles if an angle and the two sides that include it in one triangle are equal to the corresponding parts of the other triangle ? Why ? §8] DRAWING SIMPLE FIGURES 7 8. Problem 6. Divide a portion of a straight line into two equal parts. % Fig. 7 Solution. Let A and B be two points on a straight line. Draw around A and B as centers, with the same radius, two circles. These circles will cut each other in just two points P and Q if the radius is large enough. Only small arcs of the circles are shown in the figure. The straight line joining Pand Q cuts the line AB at a point 0. This point divides AB into two equal parts AO and OB, for the whole figure may be folded on the line PQ so that O remains at 0, and B falls on A. EXERCISES 1. Show how to divide a line into four equal parts. 2. How can you divide accurately the ruled page of a note- book into two columns of equal width ? 3. Show that the line PQ in Fig. 7 is also a perpendicu- lar to the line AB by noting the effect of folding the figure along PQ. 4. Draw any triangle; then find by ruler and compasses the middle point of each of the sides. Connect each corner of the triangle to the middle point of the opposite side by a straight line. [Do these three new lines meet in one point ? The accuracy of the drawing may be tested in this manner.] 8 INTRODUCTION [§ 9 9. Problem 7. Divide a given angle into two equal parts. Solution. Let the given angle be the angle at B between the portions of straight lines BC and BA. With B as center and with any radius draw a circle that cuts BC at L and BA at M. With L and M as centers, draw two arcs of circles with the same radius (of any convenient size) so that these two arcs cut each other at a point G. Then the straight line joining B and G divides the given angle into two equal parts. Convince yourself of this by think- ing of folding the figure on the line BG. EXERCISES 1. Show how to divide a given angle into four equal parts. 2. Draw any triangle and divide each of its angles into two equal parts. State anything that you notice about the way in which the three new lines meet each other. Try to see whether the same thing happens on another triangle of a differ- ent shape. State all your conclusions in one sentence. 3. Draw two perpendicular lines ; then divide the angle be- tween them into two equal parts. State the connection between this problem and the processes that occur, for example, in making the corner of a picture frame, or in mortising a joint. Can you mention any other manufacturing processes in which the same problem arises ? §9] DRAWING SIMPLE FIGURES 9 MISCELLANEOUS EXERCISES FOR PART I 1. A shelf is to be fitted into a corner near a window, and is to be triangular. The distance from the corner to the win- dow casing is 9 inches, the shelf is to be 12 inches long on the other wall of the corner, and the edge of the shelf is to be 15 inches long. Are these measurements enough to fix the shape of the shelf ? Why ? If a carpenter cuts out of a piece of board a triangle that has exactly these lengths of sides, why is it that it will fit in the place? See Problem 2, p. 3, and Ex. 3, p. 3. Will the shelf fit either side up? 2. Draw a triangle with each side just one third the length of the corresponding side of the triangle mentioned in Ex. 1. 3. A square is bounded by four lines of equal length, two of which are perpendicular to each other at each corner. Draw a square whose sides are each two inches long. [The drawing may be tested by the fact that the two straight lines joining the two pairs of opposite comers should be equal in length.] 4. If the Korth and South line is shown on any map of a city, show how to draw the East and West line through any point. 5. Show how to draw a triangle with any given angle and with the sides that form that angle of any given lengths. Will two triangles fit each other exactly if made with the same given angle and the same given lengths of the two sides ? Must one of the triangles be turned over before they will fit ? 6. If two triangles are drawn with two angles of one equal to two angles of the other, respectively, will one necessarily fit the other exactly ? What more is needed to insure that the triangles shall fit each other exactly ? 10 INTRODUCTION [§ 10 PART II. THE PRINCIPAL IDEAS USED IN GEOMETRY 10. Solids, Surfaces, Curves, Points. A limited portion of space, such as that bounded by a sphere or a cube, is called a geometric solid. In geometry we have nothing to do with the material of the solid. Thus we would speak of the space occupied by this book as a solid, without thinking that it is made of cloth and paper. Even very soft objects, such as a cloud or a drop of water, would be thought of in geometry only with respect to the form of the portion of space they occupy, and this portion of space would be called a geometric solid, even though the actual object is not solid in the sense of being hard to bend or warp. We often think of the actual object as taken away, and we consider the portion of space that it did or would occupy. For every solid there is an ideal boundary that separates the solid from the rest of space; this ideal boundary is called a surfox^e. Surfaces are either curved, as is the surface of a sphere ; or else they are flat in every direction, that is, plane. A plane surface is simply called a plane. Thus, each of the faces of a cube is a piece of a plane. When two surfaces cut each other, their common points form a curve. Thus, when a plane cuts a sphere, the curve formed by their common points is a circle. Again, when two planes cut each other, the curve formed is the simplest possible curve — a straight line. As here, we shall often use the word curve, in general, to include straight lines as special curves. A point is that which is common to two curves which cut each other; in particular, two lines cut each other in a point. § 12] FUNDAMENTAL IDEAS 11 11. Ideal Nature of Geometric Things. It should be noticed that the points and curves and surfaces just mentioned are ideal things which cannot be actually made out of materials. Thus a surface must be thought of as having no thickness whatever. A curve or a straight line must be thought of as having length but no breadth or thickness. A point has only a position. While we cannot manufacture such things as these, every one recognizes that they can be thought of in this ideal way. What we can do is to represent them very nearly : thus a point is represented by a dot, a curve by a line drawn with pencil or chalk, a surface by paper or tin or some other very thin substance. 12. Angles. Two portions of straight lines that end at the same point form an angle. The two lines are called the sides of the angle, and the common end-point is called the vertex of the angle. The length of the sides has nothing to do with the size of the angle, which depends only on the amount of the opening. Strictly speaking, when two portions of straight lines thus end at a common point, there are two angles formed : thus in the figure, the portions of straight lines BA and BC may be thought of as forming the angle marked x, or they may be thought of as forming the very much larger angle marked y which is all that is left of the j,^^ y plane if the angle marked x is taken away. We shall always understand that the smaller angle x is in- tended in such a case, unless the contrary is stated. An angle may be read in any one of three ways: (1) By the single letter at the vertex ; as the angle B. (2) By three letters, one on each side and one at the vertex ; as, the angle ABC. The middle letter, here J8, is always the one that stands at the vertex of the angle. {3) By a single letter placed in the opening of the angle,* as, the angle x. 12 INTRODUCTION [§13 13. Measurement of Angles. Units. The most usual imit of angle is the degree (°). It is formed as follows : Divide the circumference of any circle into 360 equal parts (or arcs), and then join the ends of one of these arcs to the center of the circle by straight lines ; the angle thus formed at the center is one degree. Thus one degree is one three hundred sixtieth of one complete revolution. The degree is subdivided into 60 equal parts, called minutes, ('). The minute is divided into 60 equal parts, called seconds ("). Thus an angle of 10 degrees, 20 minutes, and 15 seconds is writ- ten 10° 20' 15". 14. The Protractor. A protractor is an instrument for meas- uring angles. It is a half -circle made of cardboard, celluloid, or metal, with the center marked at (Fig. 10) and with the circum- Protractor. ference divided by fine lines into 180 equal arcs. Each of these arcs corresponds to 1°, if the vertex of the angle is placed at 0. To measure an angle, place the protractor upon it so that one side of the angle lies along the radius OA, with the vertex of the angle at 0. Then the other side of the angle will fall in some such position as OP, and the number of degrees and frac- tions of a degree can be read off directly from the scale §14] FUNDAMENTAL IDEAS 13 EXERCISES 1. Through how many degrees does the minute hand of a clock turn in fifteen minutes ? 2. Through how many degrees does the hour hand of a clock turn in one hour ? 3. The second hand of a watch turns on a circular dial that is divided into sixty equal parts. What is the angle between two successive marks ? What is the angle between the mark for 10 seconds and the mark for 15 seconds ? What is the angle between the mark for 10 seconds and that for 20 seconds ? 4. Ordinary scales for weighing small objects are often made with a circular face like a clock face ; the divisions of the scale indicate pounds ; if the entire face represents 24 pounds, what is the angle between two successive pound marks ? 5. There being 16 ounces in one pound, what is the angle between two successive ounce marks on the scale of Ex. 4 ? 6. How long does it take the minute hand of a clock to turn through 36° ? How long does it take the hour hand of a clock to turn through 36° ? 60° ? 75° ? 7. What weight will cause the hand of the scale described in Ex. 4 to turn through 15° ? 60° ? 75° ? 150° ? 8. Through how many degrees does a screwdriver turn in half a revolution ? 9. If a wheel makes ten revolutions per minute, through how many degrees does it turn in one second ? 10. Draw an angle of 150° with a protractor, and divide it into four equal parts by means of ruler and compasses. Meas- ure the resulting angles with the protractor: How much error did you make in each case ? 11. Draw an angle as nearly equal to 75° as you can judge by your eye. Then measure your angle with a protractor. How much error did you make ? What fraction of 75° is your error ? 14 INTRODUCTION [§15 15. Generation of Curves and Surfaces by Motion. We may think of curves and surfaces as formed, or generated, by motion. If a point moves, its path is a curve. If a curve moves, it generates a surface. If a surface moves, it generates a solid. Thus, the point of the compasses that draws a circle may be thought of as a moving point that is generating the circle. Again, if a circle is rotated about a line through its center, it generates the surface of a sphere. If a square that lies horizontally is lifted vertically, it generates a rec- tangular block ; the block becomes a cube when the height through which the square is lifted becomes equal to one of its sides. Notice, however, that a moving curve does not always gen- erate a surface. Thus, when a wheel turns on its axle, the curve formed by its circumference is moving, but no surface is being generated. Likewise, a surface that merely slides upon itself does not generate a solid. All such motions as these are exceptions to the general rules stated above. 16. Generation of Angles by Rotation. An angle may always be thought of as generated by a line which rotates about one of its extremities (re- garded as fixed). Thus, if the line AB rotates about the point A, it takes one after another the positions AC, AD, AE, AF, AG, and finally comes back to its original position AB. In each case it makes an angle with its Pj^ j^ original position AB, and it is to be noted that this angle increases as the rotation goes on. t^ X §17] FUNDAMENTAL IDEAS 15 |D Right Angle 17. Important Special Angles. An- other important unit angle is the right angle, which is the angle between two lines that are perpendicular to each other (see footnote, p. 4). Still another important angle is a complete revolution, the angle formed when a line turns around one of its extremities until it comes to its original position. A complete revolution is also called simply a revolution ; or sometimes a •perigon. ^ -^^^ A When the two sides of an angle lie along the same straight line, and in opposite directions from the vertex, the angle is called a straight angle. A straight angle is equal to two right angles, since a perpen- dicular to a straight line makes two equal right angles on one side of the line to which it is per- ^ ^. - Straight pendicular. Angif ^— ^180 B Fig. 12 One Revolution B Fig. 14 A revolution is equal to two C — straight angles, or to four right angles. Since a revolution is 360°, a straight angle is 180°, and a right angle is 90°. An acute angle is an angle less than a right angle. An obtuse angle is an angle greater than a right angle and less than a straight angle. Acute Angle Obtuse Anglr Fig. 15 If two straight lines cross each other the sum of the two angles formed on the same side of either of the lines is a straight angle. If one of them is acute, the other is obtuse. 16 INTRODUCTION [§ 17 EXERCISES 1. In Fig. 11, pick out a right angle ; an acute angle ; an obtuse angle; an angle of one revolution; a straight angle. Are there any other right angles in the figure ? 2. Show from the footnote on p. 4, regarding a perpen- dicular, that if two lines cross at right angles, any two of the four angles formed are equal to each other. 3. How many degrees are there in half a right angle ? in one third of a right angle ? in one fourth of a right angle ? 4. How many degrees are there in one and one half right angles ? in two right angles ? What is another name for two right angles ? 5. How many degrees are there in one revolution ? in one fourth of a revolution ? in one twelfth of a revolution ? in one fifteenth of a revolution ? 6. Is one fifth of a revolution an acute or an obtuse angle ? 7. Is two thirds of a straight angle acute or obtuse ? 8. Name two streets that you know which meet each other at right angles. Name two that do not. In the latter case describe the corner or corners at which there is an acute angle ; those at which there is an obtuse angle. 9. Does a rafter of the roof of a barn make an acute or an obtuse angle with an upright in the side wall ? What can you say of the angle at the peak of the roof where the rafters join? 10. Through what kind of an angle has a door turned on its hinges when it is said to be ajar? Can a door be opened through an obtuse angle ? 11. The earth turns on its axis once in 24 hours. How many degrees of longitude correspond to 1 hour ? 12. Apply the construction for dividing an angle into two equal parts (§ 9) to a straight angle. Show that the construc- tion that results is the same as that of § 6. §19] FUNDAMENTAL IDEAS 17 Fiu. 16 Thus the angles x and y C B Fig. 17 18. Relations between Two Angles. Two angles that have a common vertex and one common side between them are called adjacent angles. Thus the angles a and b in the hgure are adjacent angles. If the sura of two angles is equal to a right angle, the two angles are said to be comple- mentary to each other; or, either of them is called the complement of the other. in Fig. 17 are complementary to each other. If the sura of two angles is equal to two right angles, the two angles are said to be supplementary to each other ; or, either of them is called the supplement of the other. Thus the angles x and y in Fig. 18 are supplementary to each other. By the sum of two angles is meant the angle formed by placing the angles adjacent to each other; the sura is the total angle thus formed, as in the preceding figures. The measure of the sum in degrees is the sura of the nuraber of degrees in the two given angles. Fia. 18 19. Vertical Angles. If two lines AB and CD cross each other at a point 0, the angles that lie opposite each other across the common vertex are called vertical angles. Thus, in Fig. 19, the angles x and y are vertical angles ; and u and v also are vertical angles. Since the sura of u and a; is a straight angle, and the sum of u and ?/ is a straight angle, it is easy to see that x and y must be equal. That is, any two vertical angles are equal to each other. Fig. 19 18 INTRODUCTION [§ 19 EXERCISES [In this list, and hereafter, the sign Z is used for the word angle.l 1. Angle ^5(7 is a right angle. IfZa;=40°, C how many degrees in Z.y? What is the com- plement of 40° ? 2. What is the complement of 30° ? 50° 20' ? 45° 18' 20"? 74° 31' 14"? 3. The complement of a certain angle x is 2 X. How many degrees are there in a; ? 4. The complement of a certain angle is eight times itself. What is the angle ? Draw a diagram by means of a protractor. 5. Inthefiguve, Z ABC -hZCBD = 2 right angles, or the straight angle ABD. If Z ic=50°, how many degrees are there in Z y? What is D the supplement of 50° ? B " 6. What is the supplement of 80° ? 40° 15' ? 100° 30' 20'' ? 7. The supplement of a certain angle a; is 4 a;. How many degrees are there in a; ? Draw a diagram. 8. The supplement of a certain angle is eleven times itself. What is the angle ? Draw a diagram. 9. Compare the complements of two equal angles. 10. Compare the supplements of two equal angles. 11. What kind of an angle is equal to its supplement ? 12. Find two complementary angles whose difference is 36°. 13. Two supplementary angles are such that one is 40° more than the other. Find each of the angles. 14. One line meets another line so that one angle is five times its adjacent angle ; find each of the angles. 15. How do two angles u and v compare if they have che same complement ? if they have the same supplement ? §21] FUNDAMENTAL IDEAS 19 20. Contrast between Drawing and Construction of Fig- ures. One purpose of a part of our study will be to show how figures can be drawn without any other instruments than a ruler and compasses; for distinctness, we shall say that a figure has been constructed when only these instruments have been used. The direction to construct a figure will carry with it the direction that only these instruments are to be used. There is no objection whatever to the use of other instru- ments for quickly sketching a figure. Thus perpendiculars may be drawn by means of a fixed square, such as that used by carpenters or draftsmen. We shall continue to use the words "to draiv ajigure^' whenever we intend that other draw- ing instruments than ruler and compasses may be used. When a figure is to be drawn only in a very rough fashion, for example by free-hand without any instruments, we shall say that it is to be sketched. 21. Reduction or Enlargement of Drawings. It is often inconvenient or impossible to draw a figure its actual size, or, as is often said, life-sized. Thus a plan of a house cannot con- veniently be drawn the size of the house. In such cases, a figure is drawn in which every distance is reduced in the same ratio. Thus in a figure drawn half size, the distances in the figure are all half the actual distances. House plans are usually drawn on a scale which makes a distance of one quarter of an inch on the drawing represent one foot in the actual house ; that is, the scale is reduced in the ratio of one to forty-eight. The angles, of course, remain unchanged. An accurate record of the scale used should be written on the face of every drawing that is not life-size. Many geomet- ric figures do not change their properties when merely en- larged, and for the purpose of showing something about a figure, it may be drawn in any convenient size. Thus angles are not changed by reducing the size of the drawing. 20 INTRODUCTION [§ 21 Figures described in exercises in which small distances are used should be enlarged when they are drawn on the black- board. When large distances are mentioned in exercises, the size may be reduced for a drawing on paper or at the board. EXERCISES 1. What distance on the drawing represents 20 ft. in a house plan drawn to the scale mentioned in § 21 (^ in. to 1 ft.) ? 2. What actual distance does 6 in. represent in a house plan drawn to the scale mentioned in § 21 ? 3. What are the real dimensions of a room that appears on a house plan to be 21 by 3 in., if the plan is drawn on the scale described in § 21 ? What is the actual floor area of the room ? (The area is the length times the breadth.) 4. A table 2 ft. 6 in. wide and 4 ft. long is to be placed in one of the rooms. How large a spot will it represent in the drawing on the scale of § 21 ? 5. On a map whose scale is 37 mi. to the inch, the dis- tance between Chicago and Ann Arbor is 5f in. What is the actual distance ? 6. New York is 143 mi. from Albany. How far apart are they on the map referred to in Ex. 5 ? 7. A ship on leaving port sails N.W. 18 mi., then N. 15 miles. Draw a map showing her course, using a scale of 1 in. for 10 mi. In this manner find (by measurement on your map) her approximate distance and her bearing from port ; that is, how many degrees West of North. 8. When a vertical pole 20 ft. high casts a shadow 35 ft. long, what is the acute angle the sun's rays make with the hor- izontal ? [This angle is called the angle of elevation of the sun. Draw a map, scale 10 ft. to the inch, and use a protractor to measure the angle.] §231 FUNDAMENTAL IDEAS 21 Fig. 20 Fig. 21 22. Triangles. Notation. A figure bounded by three straight lines is called a triangle. The bounding lines are called the sides, and the points where the sides meet are called the vertices. Usually, the small letters, a, 6, c, are used to denote the sides, while the capital letters. A, B, C, are used to denote the vertices. The side a is then always placed opposite the vertex A, while b is likewise placed opposite to B, and c opposite to C, as indicated in the figure. The angle at A is called the included angle of the sides b, c. Similarly, B is the included angle of the sides a, c ; and C is the included angle of a, b. The angles at A, B, C are known as the interior angles of the tri- angle. Besides these, every tri- angle has what are known as exterior angles. In the figure, x represents the exterior angle at B. In general, an exterior angle is one which, like x, is formed between one side of the triangle and the prolongation (extension) of another side. 23. Circles. A circle is a closed curve, every point of which is equally distant from a fixed point within called the center. The distance from the center to any point on the circle is called the radius. A straight line through the center terminated by the circle, is called a diameter. It follows from this definition of a circle that all its radii are equal. How do the radius and the diameter compare in length ? When several circles have the same center, but different radii, they are called concentric. Draw three concentric circles. The curve that forms the circle is often called the circumference. The word circle is sometimes used to denote the space enclosed by the circumference. Fig. 22 22 INTRODUCTION [§ 24 24. Squares and Rectangles. Any figure bounded by four lines is called a quadrilateral. A rectangle is a quadrilateral each of whose angles is a right angle. The two sides that meet at any corner (vertex) of a rectangle are therefore perpendicular to each other. A line that joins opposite corners (ver- tices) of a rectangle is called a diagonal. Thus ^O in Fig. 23 is a diagonal of the Fig. 23 rectangle A BCD. If all the sides of a rectangle are of equal length, it is called a square. 25. Areas. To measure an area of any sort, a unit is usually chosen which is a square, any side of which is equal to the unit of length. The most common unit of area is one square foot ; that is, a square each of whose sides is 1 ft. long. Any given area is measured by comparing its size with that of the unit square. In particular, the area of any rectangle is found to be the product of the number of units of length in its base times the number of units of length in its height. This rule is usually learned in Arithmetic. Areas that are bounded by curved lines or by pieces of straight lines are usually measured by supposing them filled up with little squares, each of whose areas we can find. If the area bounded by the figure cannot be precisely filled in this way, at least it is greater than the sum of the areas of those squares that lie entirely within it ; and it is less than the sum of the areas of squares that entirely cover it. A good practical way to estimate the area of any figure is to draw it on paper that is ruled into little squares of known size. Such paper (called squared paper, or cross section paper) can usually be bought at any stationers, the ruling being into squares one tenth of an inch on each side. There are, of course, one hundred such squares in one square inch. §25] FUNDAMENTAL IDEAS 23 Beneath, in Fig. 24, several figures are drawn on a sheet of squared paper. Estimate the areas of each of them in the manner just d& scribed. ^ C iV? B\ 7^ \ t K ^^ i -^ ^ 7 ^ S L ^ fa) S, _ -bl ^ (a; ^ ^ ^s^i. / ^n I ^K r 3D — , M ■^M . .^ V 7J 1 ^ A t t ^ ,^ H '^ H^4 ^ i- t t ^^ \ e: t\ ^^^ \ L ^^ ^S .--^^ z s ^ ^^ T \ J I - •. ^ - - ^^ ^ - V £ - / ^^tx . J 5 L S^ ^Z S^ ^2 ^>,^-^^ _t — M ' 1 One Inch 1 r n Fig. 24 The very best conception of area is that which results by imagining the rulings on the squared paper just mentioned to be made finer and finer, so that the estimated area becomes more and more nearly the cor- rect area. The ideal or exact area bears the same relation to these esti- mates that the drawings made by human beings do to the ideal figures in geometry. (See § 11.) In the same way, lengths of curved lines can be estimated by first re- placing each small bit (arc) of the curved line by a straight line joining the ends of the arc, and then taking the sum of the lengths of all these pieces of straight lines. The smaller the arcs, the more accurate this result 24 INTRODUCTION [§25 EXERCISES 1. How many exterior angles has a triangle ? 2. In a certain triangle ABC the interior angle at A is 49°. What is the exterior angle at the same vertex ? 3. Draw three triangles of different shapes, and then, using the protractor, determine the sum of the three interior angles for each triangle. Are the three sums equal, and if so, to what ? 4. The end of the minute hand of a clock always travels in a circle. Why ? 5. Draw on a piece of squared paper a rectangle ^ in. wide by 1 in. high. Draw its diagonal. Estimate the area in each triangle into which the rectangle is divided. Are the areas of the two triangles equal ? 6. Draw on squared paper a triangle with one right angle, and with the perpendicular sides 1.5 in. and .8 in. long, re- spectively. Estimate its area. 7. Draw on squared paper a rectangle whose diagonal is the longest side of the triangle mentioned in Ex. 6. Find its area. 8. Draw on squared pa^^er two concentric circles, with the radius of one twice that of the other. Estimate their areas. 9. Construct, on squared paper, a triangle whose sides are, respectively, 2 in., 1.5 in., and 1.7 in. Estimate its area. [The area can also be found by dividing the triangle into two triangles that have one right angle in each, by a perpendicular from one corner to the opposite side, and then completing each of these smaller triangles into rectangles, as in Exs. 5 and 7.] 10. What is the sum of the four angles of a rectangle ? 11. A courtyard is 25 yd. long by 15 yd. wide. Draw a plan of it on squared paper, scale 10 yd. to the inch. What area is represented by one of the ruled squares on your paper ? Find the area of the courtyard. §25] FUNDAMENTAL IDEAS 25 MISCELLANEOUS EXERCISES FOR PART II 1. In Fig. 19, p. 17, the angles a and 2/ are vertical angles. If Z.X = 40°, what is the value of Z w, its supplement ? Since Z w is also the supplement of Z?/, what is the value oi Z.y? Compare Z x and Z y. 2. If Z a; in Fig. 19 is 38°, what is the value oiy? If Z ?/ is 78°, what is the value of Za;? 3. Name pairs of vertical angles in the adjoining figure. What is the value ofZic-h Zy -\- /.z? ~ — '^^^ — ^ ^ 4. Draw a diagram showing the complement and the supplement of ^ an acute angle ABC. What is their difference ? 5. Show that the bisectors of two adjacent supplementary angles are perpendicular to each other. 6. By use of squared paper deter- mine (approximately) the number of square inches in the adjoining figure, taking AB — 4 in., BC = 3 in., and (7^ = 2 in. 7. The figures (a), (6), (c) of Fig. 24, p. 23, have the same area. Is the length of the boundary therefore the same for each of them ? Esti- mate these lengths. 8. In what ratio is the drawing of a house reduced from the actual house if a distance of 12 ft. is represented by a line 1^ in. long ? What actual distance is represented on the same drawing by a line 2 in. long ? 9. Find the difference in longitude at two places on the earth if the difference in sun time is 2 hr. 30 min. 10. Find the difference in the sun time between two places that differ in longitude by 30° ; between two places that differ in longitude by 20°. 26 INTRODUCTION [§ 26 PAET III. STATEMENTS FOR REFERENCE 26. Assumptions. In Chapters I-Y, which we are about to study and in which many of the principles thus far used are more carefully considered, we shall make use of certain self- evident general statements. These statements are those upon which Geometry is based. They are divided into two classes, known respectively as Axioms and Postulates. Axioms refer to quantities in general, that is, without special regard to geom- etry; postulates refer especially to geometry. The following lists (§§ 27, 28) contain axioms and postulates that will be clear at this time. 27. Axioms. 1. If equals are added to equals, the sums are equal. Thus, if a = 6 and c= d, then a-\-c = b-{-d. 2. If equals are subtracted from equals, the remainders are equal. Thus, if a = 5 and c = d, then a—c=h — d. 3. If equals are multiplied by equals, the products are equal. Thus, ii a — b and c = d, then ac = bd. 4. If equals are divided by equals, the quotients are equal. Thus, if a = 6 and c = d, then - = - • In applying this axiom c d it is supposed that c and d are not equal to zero. 5. If equals are added to unequals, the results are U7iequal and in the same order. Thus, if a = 6 and c> d, then a-\- c > b-hd. 6. If equals are subtracted from unequals, the results are un- equal and in the same order. Thus, it a > b and c = d, then a — c>b — d. 7. If unequals are added to unequals in the same sense, the results are unequal i7i the same order. Thus, if a > 6 and c > d, then a-\-c>b-{- d. § 29] STATEMENTS FOR REFERENCE 27 8. If unequals are subtracted from equals, the results are uiia equal in the opposite order. Thus, if a = 6 and c > d, then a — cKh — d. 9. Quantities equal to the same quantity, or to equal quanti- ties, are equal to each other. In other words, a quantity may he substituted for its equal at any time in any expression. 10. The ichole of a quantity is greater than any one of its parts. 11. Tlie whole of a quantity is equal to the sum of its parts. 28. Postulates. 1. Only one straight line can he drawn joining two given points. 2. A straight line can he extended indefinitely. 3. A straight line is the shortest curve that can he drawn be- tween two points. 4. A circle can he described about any point as a center and with a radius of any length. 5. A figure can be moved unaltered to a new position. 6. All straight angles are equal. Hence, also, all right angles are equal, for a right angle is half of a straight angle. § 17. 29. Names of Statements. Aside from the above axioms and postulates, the words Theorem, Problem, Proposition, and Co7'- ollary will hereafter be used in the following special senses : Theorem. A statement of a fact which is to be, or has been, proved is called a theorem. Problem. A statement of a construction (see § 20) which is to be made is called a problem. Proposition. Either a theorem or a problem is known as a proposition. Corollary. A theorem which follows immediately as a conse- quence of some other theorem is called a corollary of it. 28 INTRODUCTION f§ 30 30. Summary of Construction Problems. In this Intro duction we have shown how to make the following construc- tions (with ruler and compasses alone) : 1. To construct a triangle, each of whose sides is equal to a given length. § 3, p. 2. 2. To construct a triangle, whose three sides are, respec- tively, equal to three given lengths. § 4, p. 3. 3. To construct a perpendicular to a given straight line at a given point in that line. § 5, p. 4. 4. To construct a perpendicular to a given line from a given point not on that line. § 6, p. 5. 5. To construct, at a given point in a given line, another line that makes an angle equal to a given angle with the given line. § 7, p. 6. 6. To divide a portion of a straight line into two equal parts. (To bisect a line.) § 8, p. 7. 7. To divide a given angle into two equal parts. (To bisect an angle.) § 9, p. 8. 31. Facts or Theorems now Known. We have also either assumed or proved the following geometrical facts : 1. All radii of the same circle are equal. § 2, p. 1 ; and § 23, p. 21. 2. Circles whose radii are equal can be placed upon each other so that their centers and their circumferences coincide (lie exactly upon each other). 3. Equal angles may be placed upon each other so that their vertices coincide and their corresponding sides fall along the same straight lines. This is, in fact, what we mean by equal angles. 4. Two straight lines have at most one point in common. See postulate 1, p. 27. 5. Two circles have at most two points in common. See § 8. § 31] STATEMENTS FOR REFERENCE 29 6. A straight line and a circle may have at most two points in common. 7. At a given point in a given line only one perpendicular can be drawn to that line. (A consequence of Problem 3, § 5.) 8. Complements of the same angle, or of equal angles, are equal. Ex. 9, p. 18. 9. Supplements of the same angle, or of equal angles, are equal. Ex. 10, p. 18. 10. Vertical angles are equal. § 19. 11. If two adjacent angles have their exterior sides in a straight line, they are supplementary. § 18. 12. If two adjacent angles are supplementary, they have their exterior sides in a straight line. See Fig. 18. 13. If each of two figures can be placed upon a third figure so as to coincide with it, they can be placed upon each other so that they coincide. 14. Any desired angle may be drawn, and any angle may be measured, by the use of a protractor. (But the use of this instrument is not permitted when a figure is to be constructed. See § 20.) 15. A perpendicular to a given line through any given point may be drawn by means of a set square or a drawing triangle. (But the use of these instruments is not permitted when a figure is to be constructed. See § 20.) 16. The area of a rectangle (in terms of a unit square) is equal to the product of its width and its height, measured in units of length equal to one side of the unit square. 17. The area of any given figure is greater than the area of any figure that is drawn completely within it. 18. The areas of two figures are equal if they consist of cor- responding portions that can be made to coincide. 30 INTRODUCTION [§3] SUPPLEMENTARY EXERCISES 1. Draw (using protractor) an angle of 50°. Construct (using ruler and compasses) its complement. Measure your new angle with the protractor and see if it has the proper num- ber of degrees. 2. Draw an acute angle and then an obtuse angle. In each case estimate as nearly as you can without using the pro- tractor how many degrees there are in the angle. Then, check your estimate by measurement, note your errors, and find what fraction of the correct amounts each of these errors is. 3. Find the angle whose complement and supplement are in the ratio 4 : 13. 4. Show how to construct an angle of 45°; an angle of 22° 30'. 5. Construct two lines that bisect (divide into two equal parts) each other at right angles. 6. Draw the patterns shown below. Your drawings should be twice the size of the copies. The curves are formed by joining arcs of certain circles. 7. A traveler wishes to go due north but finds his way barred by a swamp. He therefore goes five miles northeast, then five miles north, then five miles northwest, and he now finds himself due North of his starting point. Draw a map (scale one mile to the inch) and determine by measurement how many miles he lost by going out of his course. §31] SUPPLEMENTARY EXERCISES 31 LIGHT 8. A tower is observed from a point 500 ft. distant from its foot, and the angle the line of sight makes with the hori- zontal is found to be 15°. What is the height of the tower ? 9. It is a principle of physics that when a ray of light is re- flected from a mirror, the re- flected ray makes the same angle with a line perpendicu- lar to the mirror that the origi- ^>^^^^% ^ ^^ ^^ ^ nal ray makes with the same line. Show that the original ray and the reflected ray also make equal angles with the mirror itself. 10. Two forts defending the mouth of a river, one on each side, are 10 mi. apart. Their guns have a range (possible shooting distance) of 4| mi. Draw a plan (scale 1 mile to the inch) showing what part of the river is exposed to fire from the two forts. [Godfrey and Siddons.] 11. Construct two triangles each with the ^ sides a, b, and c, as indicated in the adjacent figure. See § 4, p. 3. Cut them out and place ^ ^ one on the other so that they coincide. What C a B conclusions do you draw concerning such triangles ? 12. Construct two triangles each with the base AB = 2 in., angle ABC = angle DEF, and angle ACB = angle DFE. Cut one of these triangles from the paper and place it upon the other so that the corresponding parts coincide. What conclu- sions do you draw concerning such triangles ? 2 IN. 2 IN. 13. The length of a rectangular field is 50 yd. and the length of the entire boundary is 180 yd. What is its breadth and its area ? 32 INTRODUCTION l§31 14. Construct two equal angles, ABC and DEF. On the sides of these angles lay off the distances BA and ED, each 2 in. ; also the distances BC and EF, each IJ in. Join MAKE 2 IN. MAKE 2 IN. AC and DF, thus forming two triangles ABC and D^i^. What conclusions do you draw concerning such triangles ? 15. Draw a triangle with two sides AC = 3 in. and AB = 5 in. and their included angle = 35°. (Use protractor.) Draw another triangle in which AC = 3 in., AB = 5 in., but their included angle = 20°. Do two sides alone fix (deter- mine) a triangle ? 16. Is a triangle determined if two sides and their included angle are given ? DF^^ Show the relation of the last question to the following fact from everyday life : Two pieces of board AB and BC hinged at B can be held rigid by nailing a crosspiece DE to both sides. 17. Many designs may be made by emphasizing a part of the lines on squared paper, or the diagonals of those squares. Copy and complete the following; and also invent others. §31] SUPPLEMENTARY EXERCISES 33 18. To construct the plan for a Gothic window, proceed as follows : Take any line AB and divide it into four equal parts. With A and B as centers and a radius equal to AB, draw arcs intersecting at C. What radius and what centers are used to de- scribe the small arcs ? is found by tak- ing A and B as centers and AK as a radius. Complete the figure. 19. Construct (by ruler and compasses) patterns like the following : <^ ^ (b) (e) SYMBOLS AND ABBREVIATIONS The following symbols and abbreviations will be used for the sake of brevity throughout the present book : = equal, or is equal to Ax. Axiom Tt not equal, or is not equal to Cons. Construction, or by con- > greater than struction < less than Cor. Corollary ^ is congruent to Def. Definition _L perpendicular, or is perpen- Hyp. Hypothesis, or by hy dicular to pothesis II parallel, or is parallel to Iden. being identical ~ similar, or is similar to Prop. Proposition Z angle rt. right A angles st. straight A triangle Th. Theorem A triangles Prob. Problem O parallelogram Fig. Figure or diagram UJ parallelograms .... and so on O circle hence or therefore CD circles ^^ arc The signs +, — , X, -5-, are used with the same meanings as in algebra. The following agreements are also made : a -5- & = a/b = a:b CHAPTER I RECTILINEAR FIGURES PART I. TRIANGLES 32. Definitions. It is desirable to distinguish between several kinds of triangles as follows: c c Equilateral Triangle Isosceles Triangle Fig. 25 Scalene Triangle An equilateral triangle has all three of its sides equal. A triangle that has any two of its sides equal to each other is called an isosceles triangle. A scalene triangle is one that has no two of its sides equal. fC C " c Equiangular Triangle Acute Trla.ngle Fig. 26 Obtuse Triangle An equiangular triangle has all three of its angles equal. An acute triangle is one whose angles are all acute. An obtuse triangle is a triangle that has one obtuse angle. 36 36 RECTILINEAR FIGURES [I, § 32 A triangle one of whose angles is a right angle is called a right triangle. In such a triangle, the side opposite the right angle is called the hypotenuse, and the word "side" is used only for the other two sides, as indicated in Fig. 27. The side upon which any triangle appears to rest is called its base. The vertex opposite to the base is called the vertex of the triangle, and the angle opposite to the base is called the angle at the vertex. The perpendicular distance from any vertex to the opposite Right Triangle Fig. 27 c M P B c M B . Fig. 28 side (extended if necessary) is called an altitude of the tri- angle. The distance from any vertex to the middle point of the opposite side is called a median of the triangle. Any portion of a straight line between two points is called a segment of that line. Thus the sides, altitudes, and medians of any triangle are line segments. Any figure composed wholly of points and straight lines is called a rectilinear figure. 33. Congruent Figures. Two triangles that can be made to fit each other exactly (coincide) by properly placing the one upon the other are called congruent. More generally, any two geometric figures are congruent if they can be made to coincide exactly. I, § 34] TRIANGLES 37 When two congruent figures are made to coincide, any cor- responding parts coincide : corresponding angles of congruent figures are equal ; corresponding lengths are equal; any portion of one of two congru- ent figures is congru- ^ ^ P ent to the correspond- ^ " ' ing portion of the ^ \ 3 ^^ other. An illustration of two congruent fig- ures, each broken up into certain corresponding parts which are also congruent, is given in Fig. 29. Fig EXERCISES 1. How many altitudes has a triangle? How many me- dians ? Draw figures illustrating your answers. 2. Can a right triangle be isosceles ? Draw figures. 3. Are all right triangles isosceles ? Draw figures. 4. Can a right triangle be equilateral? Explain your answer. 5. What kind of triangle was drawn in Problem 1, p. 2? In Problem 2, p. 3 ? 6. Investigate the following questions : (1) Is there any kind of triangle for which the medians coincide with the alti- tudes, and also with the bisectors of the three angles ? If so, describe it. (2) Will these lines usually all be different for a triangle ? Illustrate your answer by drawings. 34. Congruence of Triangles. We now proceed to state and prove certain theorems regarding triangles. The pupil should first reread carefully § 22. Our first topic of study will be the following question: " When are two triangles congruent ? " Note again the defini- tion of congruent figures as given in § 33. 38 RECTILINEAR FIGURES [I, § 35 35. Theorem I. If two triangles have two sides and the included angle of the one equal, respectively, to two sides and the included angle of the other, the triangles are congruent. Fig. 30 In the figure, let ABC and A'B'C be the two triangles, and let us suppose that we know (as the theorem says) that AB = A'B', AC= A' a, and that the angle A = the angle A\ We are now to prove that these two triangles are congruent, which means that we have to show that the one may be fitted on to the other so that they will exactly coincide in all their parts. Kow, since AB=:A'B', we can place the triangle A'B'C on the triangle ABO so that A'B' will coincide with its equal AB, making the point C" fall somewhere on the same side of AB as a Then, A'C will lie along AO, because angle A' = angle A, this being also one of the given (supposed) facts. Moreover, C will fall exactly at (7, since A'C = AC, which is another of the given facts. It thus follows that the side C'B' will fit exactly upon CB, for, according to Postulate I, only one straight line can be drawn through the two points, C, B. Therefore, the two triangles are congruent. Note. The fact stated in this theorem was indicated in Exs. 2 and 3, p. 6 ; and in Ex. 14, p. 32. In those exercises, however, the truth was only suggested. What we do in a proof such as that just given is to make certain what was previously simply plausible. I, § 36] TRIANGLES 39 36. Corollary I. Two right triangles are congruent if the two sides of the one are equal respectively to the two sides of the other. Reasoning from Theorem I, the student should convince himself of the truth of this corollary. [Hint. First, note the form of all right triangles, as illustrated in Fig. 27. Note also from the same figure how the word "side " is used for such triangles. Now draw two such triangles having the "sides" of the one equal respectively to the " sides " of the other, and see how Theo- rem I applies to them.] EXERCISES "^1. Write out the proof of Theorem I for two triangles shaped as in the adjoining figure, in which we suppose AB^A'B', AC = A'C, and angle A = angle A'. 2. Using a protractor, draw a triangle in which one angle is A"*^-- 'B A-^ 'b' 30° and the two sides that in- clude that angle are respectively 3 and 4 inches long. Show, by Theorem I, that if another triangle is drawn which has these parts, it is congruent to the one first drawn. 3. In the figure, L represents a lake. It is required to find its length, i.e. the distance between A and B. Show that this may be done as follows : .^^^^^^^ (1) Fix a stake at some convenient point ^^^ ^^ C, and measure the distances AO and BC. \y^ (2) In line with.^C set a stake D, such /C\^ that CD = AC. Then in line with BC, set a / \ stake E, such that CE = BC. E-- -^D The distance from stake D to stake E will be the required distance between A and B. Show that this is a consequence of Theorem I. 4. Show how the method of Ex. 3 could be applied to de- termine the greatest length of your school building. 40 RECTILINEAR FIGURES [I, § 37 37. Theorem II. If two triangles have two angles and the included side in the one equal, respectively, to two angles and the included side in the other, the triangles are congruent. In the figure, let ABC and A'B'C be the two triangles, and let us suppose that we know that ^A = ZA',ZB = ZB',^dAB = A'B'. [For brevity we here use the symbol Z for the word angle (see p. 34). Also, mstfad of " side AB = side A'^ " we write simply AB = A'B'.^ We shall now prove that the triangles ABC and A'B'C are congruent by showing that it is possible to place the one upon the other so that they shall coincide. To carry out the proof, place the triangle A'B'Cf upon the triangle ABC so that A'& coincides with its equal AB, and so that O and C fall upon the same side of AB. Then A'C will fall along AC because ZA = Z A', which was one of the given facts. Moreover, B'C wiU fall along BC, since Z B = Z B', which was also given. It follows from this that C will falT exactly at C, for, by Statement 4, § 31, the two lines AC and BC c&n intersect in only one point. Thus, the two triangles can be made to coincide completely, and they are, therefore, congruent. Using the symbols of p. 34, this result is written in the following form : A ABC ^ A A'B'C I, § 39] TRIANGLES 41 38. Corollary 1. Two right triangles are congruent if an acute angle and its adjacent side in one are equal, respectively, to an acute angle and its adjacent side in the other. Thus, the two right tri- angles ABC and A'B'C are congruent if AB = A'B' and Z.A = Z.A'. This corollary, like all others, should be proved by the student. EXERCISES 1. Draw a right triangle having one angle 30° and the adja- cent side (not the hypotenuse) 3 in. long. Measure the other angle with a protractor, and measure each of the other sides. Repeat this with an angle of 45° in place of 30°. 2. Theorem II was used by Thales (640-546 B.C.) to determine the distance of a vessel V from the shore, by measur- ing the angles w and v and then con- structing the congruent triangle ABX on shore, this can be done. 3. A drawing triangle is usually made of celluloid, with one angle (Z C in the figure), a right angle. One other angle A is usually made 30°. Show that if the length AC is chosen, the form of the triangle is completely determined. Explain how b C Drawing Triangle 39. Form of Proofs. Hereafter we shall use symbols when- ever possible, thus enabling us to give proofs in a more con- densed form. See the list of symbols, p. 34. Xote that in the arrangement of the proof the different steps are in the left column, while the reason for each step is in the right column. 42 RECTILINEAR FIGURES [I, § 40 40. Theorem III. In an isosceles triangle the angles opposite the equal sides are equal. Given the isosceles triangle ABC in which AC — BO, To prove that ZA=ZB. Proof. Draw CD bisecting Z ACB. Then, in the A ACD and BCD we have CD = CD, Men. AC=:BC, Given ZACD = ZBCD. Cons. Therefore A ACD ^ A BCD ; . § 35 and hence ZA = ZB. § 33 41. Corollary 1. If a triangle is equilateral., it is also equi- angular. State carefully the reasons for this conclusion, in the form of the proof of Theorem III. EXERCISES 1. Construct (using ruler and compasses) an isosceles triangle and measure by means of the protractor the angles opposite the equal sides. Do your results conform to what Theorem III says ? I, § 40] TRIANGLES 43 2. An ordinary gable roof has a form (cross section) such as indicated in the accompanying figure. How does this illus- trate Theorem III ? Ridge 3. Mention some other familiar object in which an isosceles triangle occurs, draw a figure to represent it, and state how it illustrates Theorem III. 4. Prove that the medians drawn through the base angles of an isosceles triangle are equal. Write out the proof in the precise form used in the proof of Theorem III, using two columns, one for state- ments of fact and the other for the reasons ; and use the symbols of p. 34. [Hint. Given the isosceles A ABC in which AC = BC and let AD and BE be the medians drawn through the base angles A and B. To prove AD = BE. In the proof, direct your attention to the k^EAB and DBA. They have Z EAB = Z DBA by Theorem III ; at the same time they have AE = BD because each is one half of the equal sides AG and BC. Also AB = AB. From these facts, reason by means of Theorem I that A ABE ^ A ABD, and hence AD = EB, which was to be proved.] 5. Prove that the bisectors of the base angles of an isosceles triangle are equal. [Hint. Draw a figure similar to that of Ex. 4, but make AD bisect Z A, and BE bisect Z B. Direct your attention to the A EAB and DBA. They have Z EAB = Z ABD by Theorem in ; at the same time Z EBA = Z DAB, being halves of equal angles. Whence, show by Theorem II that A EAB ^ A DBA. Then, AD = BE, as was to be proved.] 44 RECTILINEAR FIGURES [I, § 42 42. Nature of Proofs in Geometry. Hypothesis. Conclu- sion. Every proof, as arranged in the form which we have now adopted and as illustrated by the proof of Theorem III, consists of four parts, as follows : (1) A careful statement of the theorem, accompanied by an appropriate figure. (2) A condensed statement of what we have to work with, or what is given. This part of the statement of the theorem is called its hypothesis, or thing supposed. Thus, in Theorem III the hypothesis is that the triangle is isosceles. (3) A condensed statement of what is to be proved. This is called the conclusion. Thus, in Theorem III the conclusion is that "the angles opposite the equal sides are equal." (4) The details of the proof. This consists of a series of statements, each accompanied with its reason. EXERCISES 1. What is the hypothesis of Theorem I? What is its conclusion ? Answer the same questions for Theorem II. 2. Hypotheses and conclusions occur in all careful argu- ments in other subjects as well as in Geometry. Pick out the hypothesis and the conclusion in each of the following state- ments : (a) If he is guilty, he should be punished. (5) If a piece of iron is magnetized, it will attract other pieces of iron. (c) A body heavier than water will sink in water. (d) Corollaries of §§ 36, 38, 41. 3. Make some conditional statement similar to those of Ex. 2, and give the hypothesis and the conclusion. 4. Make some statement that you think is true about some geometric figure. Pick out the hypothesis and the conclusion. Draw a figure to illustrate the statement. I, § 44] TRIANGLES 45 43. Theorem IV. The hisectoi' of the angle at the vertex of an isosceles triangle is perpendicular to the base and bisects the base. isosceles A ABC in bisects the vertical Given the which CD za To prove that CD ± AB and that AD = DB. Proof. In the A ADC and DBC we have AG= CB, Given and Z ACD = Z DCS. Given Moreover CD = CD. Therefore A ADC ^ A DBC ; whence Z ADC = Z BDC, so that CD±AB. Also AD = DB. § 17 §33 44. Corollary 1. In any isosceles triangle (a) The bisector of the angle at the vertex divides the triangle into two congruent right triangles. (h) The bisector of the vertical angle coincides with both the altitude and the median drawn through the vertex. (c) The perpendicular bisector of the base passes through the vertex^ and divides the triangle into two congruent right triangles. EXERCISES 1. If a plumb line (a string with a heavy bob attached at one end) be let down from the ridge of the roof represented in Ex. 2, p. 43, the bob strikes the floor in a way that illustrates Theorem IV. How ? 2. Prove that if the bisector of the angle at the vertex of any triangle is perpendicular to the base, the triangle is isosceles. [HiKT. Apply Theorem II to the &.ADC, BDC (Fig. 33), and show that^C= CB.] 46 RECTILINEAR FIGURES fl, § 45 45. Theorem V. If two triangles have three sides of the one equal respectively to the three sides of the other, they are con- gruent c Fig. 34 Given the j^ABG and A^B'O in which AB = A^B\ BO = B'C, and CA = C'A', To prove that A ABC ^ A A' B'C. Proof. Place AA'B'G' in the position ABG", thus making the side A'B' coincide with its equal AB and causing the point C to take up the position marked C". Join G and C" by a straight line. Then, in AAGG", we have AG=AG". Given Therefore Z AGG"=Z AG"G. § 40 Likewise, in A BGG", we have BG=BG"; Given hence Z BGG" = Z BG" G. Why ? Therefore ZAGG"-\-ZBGC = ZAG"G-\-ZBG"G, Ax. 1 (§ 27) that is, ZAGB=ZACB; whence A AGB^A AG"B, § 35 that is, A ABG ^ A A' B'C. 46. Corollary 1. Three sides determine a triangle; that is, if the three sides are given, the triangle is thereby fixed. The statement means that if the three sides of a triangle are known, any triangle made with these sides is congruent to any other triangle that has the same sides. See Problem 2, p. 3 ; Exs. 2, 3, p. 3 ; Ex. 11, p. 31. I, § 46] TRIANGLES 47 EXERCISES 1. If three boards are nailed together in the form of a tri- angle, with one nail at each corner, will the frame thus made be quite stiff? Show how the corollary just stated is related to your answer. 2. Two circles whose centers are at and 0' intersect in A and B. Prove thsit A OAO' ^ A OBO'. 3. Prove that two triangles ABC and A'B'C are congruent if AB = A'B', BC = B'C, and the median through A equals the median through A'. [Hint. Draw the two triangles. Let D be the point of intersection of the median through A with the side ^C in A ABC, and D' the cor- responding point in A A' B'C. First show, by Theorem V, that AADB ^ A A'D'B', remembering that BD = ^ BC and B'D' = \ B'D'. Hence show that ZADC = ZA'D'C' by using 9, p. 29. Finally, show that AADCc^AA'D'C. Then AABC^AA'B'C by § 33.] 4. The framework of bridges, scaffolding, and other struc- tures consists usually of a network of triangles whose sides are stiff pieces of iron or wood. Show, by means of Theorem V and Corollary 1, why the entire structure is stiff. 5. Show that the temporary bracing of a window frame in a building during its erection by a board nailed to the frame and to the floor, is an illustration of § 46. 6. Point out the triangle in the roof construction of Ex. 2, p. 43, which makes the roof structure rigid. Note. These Exercises illustrate the great importance of Theorem V in structures of all kinds. The practical value of the study of triangles arises principally from the frequent application of Theorems I, II, and V, §§ 35, 37, 45, both in practical affairs and in the proofs of other geometric theorems. These three theorems are printed in boldface type to indicate their especial importance ; they should be studied most carefully. 48 RECTILINEAR FIGURES [I, §47 47. Theorem VI. An exterior angle of a triangle is greater than either of the opposite interior angles. Given the A ABC hav- ing the exterior Z DBC. ^ To prove that Z DBC > A A; alsothatZi>J5(7>Z(7. Proof. Through E, the mid-point of BC, draw AE and prolong it to F, making EF— AE. Also, draw BF. Then, in the A AEC and BFE we have AE = EF, and BE = EC, Cons, and Z CEA = Z JS^i/'. 10, § 31 Therefore A AEC ^ A BFE ; Why ? whence Z O = Z i^^J^. Why ? But Z i)5^ > Z i^^^;; Ax. 10 hence Z D5^ > Z C, Ax. 9 that is, Z.DBC > ZC, which was to be proved. [Similarly, by bisecting AB and proceeding as above it can be proved that Z. ABK, which equals Z DBC (why '?), is greater than Z A.^ EXERCISES 1. By drawing a series of triangles, determine what form a triangle tends to take as one of its exterior angles comes closer and closer in size to either of its opposite interior angles. 2. Prove that no triangle can have two right angles. [Hint. If Z O, Fig. 35, were equal to 90", then ZDBC would be greater than 90° (§47) ; hence Z ABC <90°, since it is the supplement of Z DBC. Write out the proof in full. ] 3. Prove that no triangle can have two obtuse angles. 4. Prove that the bisector of an exterior angle of a triangle is perpendicular to the bisector of the adjacent interior angle. I, §50] PARALLEL LINES 49 PAKT II. PARALLEL LINES 48. Definitions. Lines that lie in the same plane but do not meet however far they may be extended are said to be parallel A B Parallel Lines Fig. 36 If two lines are cut by a third line, this third line is called a transversal, and the angles at the points of intersection are named as follows : tf u, X, and w are interior angles. Sj V J y, and z are exterior angles. s and z, or v and y, are alternate ex- terior angles. t and IV, or u and x, are alternate interior angles. V and w, or s and x, or ti and z, or t and y, are corresponding angles. 49. Parallel Postulate. Besides the postulates of §^ 28, the following is necessary for the study of parallel lines : Parallel Postulate. Only one line can be drawn through a given point parallel to a given line. That is, any other line so drawn will coincide with the first one. This postulate was stated in a somewhat different form by Euclid (about 300 b.c). It was recognized by him and by later mathematicians that it is peculiarly important. It is often called the Euclidean postulate. 50. Corollary 1. Lines parallel to the same line are parallel to each other. For, if two lines parallel to one and the same line should meet, there would be more than one parallel through a point, thus contradicting the preceding postulate. 50 RECTILINEAR FIGURES [I, §51 51. Theorem VII. When two lines are cut by a transversal, if the alternate interior angles are equal, the two lines are parallel. T ■— --. K S Fig. 38 Given any two lines AB and CD cut by the transversal. ^iST at E and F, so that the alternate interior angles x and y are equal. To prove that AB is parallel to CD. Proof. Let us suppose for the moment that AB and CD are not parallel. In this case they would meet in some point which we will call K, and a triangle EFK would be formed. Then the exterior angle x of this triangle would be greater than its interior angle y, by Theorem VI. But this is impossi- ble, since our hypothesis is that Z.x = /.y. In other words, the conclusion that AB and CD are not parallel cannot be true if our hypothesis that /.x = Zyis true. The only other possible conclusion is that AB and CD are parallel. 52. Corollary 1. Lines perpendicular to the same line are parallel. EXERCISES 1. If several strips are nailed to a board at right angles to it, show that the strips are parallel. 2. If another board is nailed perpendicular to the strips of Ex. 1, show that it is parallel to the first board. 3. Draw any line CD (Fig. 38) and select any point E not on it. Draw any line through E to cut CD at some point, F. Lay off /.x = Z-y with a protractor, or as in § 7, p. 6. Show that this process gives a line parallel to CD through E. I, §53] PARALLEL LINES 51 53. The Indirect form of Proof. Argument by Reduction to an Absurdity. The student may have observed that the proof just given in § 51 differs essentially from other proofs we have given. In the proof of Theorem VII we meet for the first time what is called an indirect proof, otherwise known as a reductio ad absurdum, or reduction to an absurdity. This form of proof is of frequent occurrence. In substance, an indirect proof of a theorem consists in first supposing something differ- ent from the theorem's conclusion, and then showing that an absurdity results, thus leaving the theorem itself as the only possible statement of fact. The use of the indirect proof is common, not only in geom- etry, but also in many of the familiar experiences of everyday life. Suppose, for example, that on a certain night a robbery is committed in a certain store and the next day Mr. A is sus- pected of having done it. He succeeds in proving, however, that throughout the night in question he was in another town. This is sufficient to free him of suspicion. Why ? — Because the supposition that he is guilty is thus made to lead to the absurd conclusion that he was in two different places at the same time. The very strongest kind of argument is to show that some contention of an opponent leads to absurd conclusions. EXERCISES 1. A closed wooden box is known to contain a piece of metal. A magnet is brought near and found to be attracted. The conclusion is drawn that the metal within the box is either iron or steel. Show that this conclusion is drawn by a process of indirect proof. 2. Using Theorem III, give an indirect proof of the follow- ing theorem : If no two of the three angles of a triangle are equal, the triangle cannot be isosceles. 3. Show that the reasoning used in § 50 is really a reduction to an absurdity. 52 RECTILINEAR FIGURES [I, §54 54. Theorem VIII. If two parallel lines are cut by a trans< versal, the alternate interior angles are equal. V d___ A D c s^ / A Fig. 39 d' Given the two parallel lines AB and CD cut by the trans- versal ST Sit E and jP and making the alternate interior angles X and y. To prove that the angles x and y are equal. . Proof. Suppose A x ^ /.y (For the symbol ^, see p. 34.) Now draw CD' through F, making Z D'FT = Z.t. Then CD'WAB', §51 but CD li AB. Given Thus we should have two different lines through the same point F parallel to the line AB, which is impossible, according to the postulate of § 49. Therefore Z x = Zy, which was to be proved. 55. The Converse of a Theorem. The student may have observed that Theorems YII and VIII, though not the same, are very closely related. Careful examination of the two will show that the precise nature of this connection lies in the fact that the hypothesis and conclusion of the one have become, respec- tively, the conclusion and hypothesis of the other. In other words, the one is the other simply turned about. In general, when any two theorems are related in this way, the one is said to be the converse of the other. Aside from geometry, there are many instances of state- ments and their converse. The following will illustrate this I, §55] PARALLEL LINES 53 fact, and also the fact that because a theorem or other state- ment is true we cannot always be certain that its converse is true. {Statement) If a boat is made of wood, it floats. {True) {Converse) If a boat floats, it is made of wood. {False) An instance in Geometry in which the converse of a true statement is false is : {Statement) If a figure is a square, it is a rectangle. {True) {Converse) If a figure is a rectangle, it is a square. {False) An instance in Geometry in which both the original and the converse theorems are true is : {Statement) If two sides of a triangle are equal, the angles opposite them are equal. Theorem III, § 40. {Converse) If two angles of a triangle are equal, the sides opposite them are equal. Theorem XVI, § 72. In case both a theorem and its converse are true, they may be stated together in one sentence by properly using the clause " if and only if." Thus, Theorems VII and VIII may be combined into one as follows : " When two lines are cut by a transversal, the alternate interior angles are equal if and only if the lines are parallel." The phrases "and conversely" or "and vice versa'' are often used for the same purpose. Thus we might say, " If a body is lighter than water, it will float; and conversely.'' Or, "If a body is lighter than water, it will float ; and vice versa." EXERCISES 1. Make a true statement whose converse is false. 2. Make a true statement whose converse is true. 3. Write out the illustration you used in Ex. 2 in the form of a single sentence, using the clause " if and only if." 4. Rewrite each double statement on this page, using the phrase, (a) "if and only if"; {b) "and conversely," (c) "and vice versa." Which are true and which are false ? 54 RECTILINEAR FIGURES fl,§56 Fig. 40 56. Theorem IX. If two lines are cut by a transversal and the corresponding angles are equal j the lines are parallel. Given AB and CD cut by the transversal ST m. such a way that the corresponding angles x and y are equal. To prove that AB is parallel to CD. Proof. Z.x = Zz, 10, §31 but . Z.x= /.y, Given therefore Z.y = Z.z, Ax. 9 and hence also AB \\ CD. § 51 57. Corollary 1. If two lines are cut by a transversal and the two interior angles on the same side of the transversal are supple- mentary, the lines are ixtrallel. 58. Corollary 2. From a given poi^it only one perpendicular ca7i be drawn to a given line. For, if there were two perpen- diculars through the same point, they would be parallel (Why?) but parallel lines cannot meet (§ 48). See also Ex. 2, p. 48. EXERCISES 1. Two parallel pipes for hot and cold water lie flat along the same wall ; at the end of each of them an elbow is screwed on which turns the pipe through a right angle. If the pipes connected to these elbows also lie flat against the same wall, will they be parallel ? Connect your answer with §§ 56-58. 2. A rectangle (§ 24) has all its angles right angles ; show that the opposite sides are parallel. I, §60} PARALLEL LINES 55 59. Theorem X. {Converse of Theorem IX.) If two paral- lel lines are cut by a transversal^ the corresponding angles are equal. Given the two parallel lines AB and CD cut by the transversal ST^ making corresponding angles X and y. To prove that /.x=/.y. Proof. /.x=Z.z, Why? also /.y=/.z\ Why? therefore Ax=Ay. Why ? Fio. 41 60. Corollary 1. If a line is perpendicular to one ofttuo par- allelsy it is perpendicidar to the other also, EXERCISES 1. State and prove the converse of Corollary 1, Theorem IX. 2. The crosspieces (arms) which are put on a telephone pole to carry the wires are usually all perpendicular to the pole. How does this illustrate Theorem X or its corollary ? 3. Prove that the diagonals of all of the squares on a sheet of squared paper form continuous lines. 4. Prove that the bisectors of any pair of corresponding angles formed when a transversal cuts two parallel lines are themselves parallel. Is this true also for bisectors of alternate interior angles ? 5. Prove that the bisectors of any two interior angles formed when a transversal cuts two parallels are either par- allel, or else perpendicular to each other. 6. Prove by §§ 56, 59 that lines perpendicidar to the same line {or to parallel lines) are parallel, (See also § 52.) 56 RECTILINEAR FIGURES [I, §61 61. Problem 1. To construct a line parallel to a giveii line and passing through a given point. T/ / o/ K ? Fig. 42 Given the line AB and the point 0. Required to construct a line through O parallel to AB. (See Ex. 2, p. 50.) Construction. Draw any line OT through 0, cutting AB at some point, as D. At construct Z TOK = Z ODB. Prob. 5, p. 6. Then the line OK (extended) will be the desired line through parallel to AB. Proof. Since Z TOK= Z ODB, the lines AB and OK are parallel. § 56 Note. The fact that in the problem of § 61, above, we have not only shown how to draw the desired line, but have afterwards proved that our method is correct, illustrates the principle that every construction problem should, in its solution, contain not only the construction, but also the proof of its correctness. This will be done hereafter in such problems as are worked out in the text, and the student should do the same in all construction problems that occur in exercises. EXERCISES 1. Show how to construct a line parallel to a given line and passing through a given point, by means of § 52. 2. Show how to construct a line that makes one half a given angle with a given line at a given point. [Hint. Mrst bisect the given angle.] 3. Show how to construct a line that makes an angle of 45° with a given line at a given point. I, §61] PARALLEL LINES 57 4. (a) In order to draw a parallel to a line I through a point P, a draftsman will place a drawing triangle, or other object with two fixed edges, so that one side of the triangle coincides with Z, and the other side passes through P. He will then lay a ruler against the side of the triangle that passes through P, and finally slide the triangle along the edge of the ruler, until one corner of the triangle comes to P. Show that a line drawn along the side of the triangle, originally in coincidence with I, will be the parallel to I through P. (b) Show how to draw, by means of a drawing triangle, a per- pendicular to a given line AB through a given point. (Fig. b.) Notice that Z i? = 90°. See Ex. 3, p. 41. 5. Draw Fig. 38, omitting the portion dotted in the figure. Through some point C on CD construct a line parallel to ST. (The quadrilateral formed is called a parallelogram.) 6. A parallelogram (Ex. 5) is formed when one pair of parallel lines cuts another pair. Show that the sum of the two interior angles that have one side in common is 180°. 58 RECTILINEAR FIGURES [I, §62 PART III. ANGLES AND TRIANGLES 62. Theorem XI. The sum of the three angles of any tri' angle is equal to two right angles, or 180^. Given any triangle ABC. To prove that /. A+Z.B + ^ (7 = 2rt. A. Proof. Prolong AB to D and through B draw BE \\ AC. Denote Z DBE by x, and Z EBC by y. Then Zx-^ Ay + Z ABC = 2 rt A-, Why? moreover A A = Ax, § 59 and AC = Ay. §54 Therefore Z ^ -f- Z O + Z ^.BO = 2 rt. A. Ax. 9 Note. This famous theorem is of great practical importance. It was known by Pythagoras (about 500 e.g.). This figure was used by the Greek philosopher Aristotle (384-322 b.c.) and by the famous Greek geometrician Euclid (about 300 e.g.). 63. Corollary 1. The sum of the two acute angles of any right triangle is one right angle, or 90°. f 64. Corollary 2. An exterior angle of any triangle is equal to the sum of its opposite interior angles. That is, in Fig. 43 ACBD=ZA + ZC. 65. Corollary 3. Each angle of aii equilateral triangle is equal to 60°. 66. Corollary 4. If tico angles of one triangle are equal re- spectively to two angles of another triangle, then the third angles are likewise equal. I, §661 ANGLES AND TRIANGLES 59 EXERCISES 1. Two angles of a triangle are 10° 30' and 85° 15', respec- tively. What is the size of the third angle ? 2. If the rafters of the roof represented in Ex. 2, p. 43, make an angle of 35° with the horizontal, show that the total angle at the ridge of the roof is 110°. 3. A crank AB is operated by means of a rod DVB, which slides through a ring at C. Show that g the angle ACB is always half the angle ^^^^y XAB, provided AC = AB. - ^^"^^^^ { X 4. The exterior angles at A and C of a triangle ABC are 71° and 140°, respectively ; how many degrees in the angle B ? 5. Eind the three angles of an isosceles triangle when one of the angles at the base is equal to one half the angle at the vertex. 6. Draw an equilateral triangle, and draw three other equilateral triangles, placing one on each of the sides of the first one as a base. Show that if this process is repeated continu- ally, the plane is divided into equilateral tri- angles that completely fill it. This fact is the basis for many interesting designs, some of which are shown below. l/f^plq i>Ki>< A/\/\A/\/yV\A/\7 KAA/Ox^VAAAkX ><>< N^i^lcTi Orchard Plan Tiled Floor Pattern Linoleum Pattern 7. Show by § 62 that a triangle can have no more than one obtuse angle. Can it have more than one right angle ? See Exs. 2, 3, p. 48. 60 RECTILINEAR FIGURES [I, §67 67. Theorem XII. Two angles whose sides are re- spectively parallel are either equal or supplementary, ^^ ^<.'A__. ^ X c There are two cases, i/ / / X^ as indicated by Figs. ^ ^ ^ B 44 a and 44 6. ^^«- ^« Fig. 44& Case 1. Given A B and B' with AB \\ A'B' and BC || 5'0', as in Fig. 44 a. To prove ZB = ZB\ Proof. Prolong BC and B'A' until they meet, thus form- ing the angle x. Then ZB= Zx, Why ? and ZB'=Zx. Why? Therefore ZB = ZB'. Why ? Case 2. Given Zs^ and B' with ^5 II A'B' and 50 1| J5'C", as in Fig. 44 b. To prove that A B and B' are supplementary. [The details of the proof for this case are left to the student.] EXERCISES 1. When in Theorem XII will the two angles be equal? when supplementary? 2. Show that Theorem XII is illustrated by the angles at the intersection of any two straight streets of uniform width. 3. Show that Theorem XII is illustrated by the angles at the intersection of two straight railroads. 4. A parallelogram (Ex. 5, p. 57) is a figure formed when one pair of parallel lines cuts another pair. Show by means of § 67 that the opposite interior angles are equal, and that the adjacent interior angles are supplementary. I, §681 ANGLES AND TRIANGLES 61 68. Theorem XIII. T%t)o angles ivhose sides are re- sjjectively jyerpendicular to each other are either equal or supplementary. There are two cases, as indicated by Figs. 45 a and 45 b. Case 1. Given AB and B' with AB±B'a and BC±A'B\ as in Fig. 45 a. To prove that ZB = ZB'. Proof. Prolong AB and B'C until they meet at some point such as T, and through T draw a line TT II BC, meeting A'B' pro- longed at K. Then Zx==Zy. Why? Now Zz is the complement of Zy. ' Moreover, TKB' is a right angle ; hence Zz is the complement of Z B'. Therefore Zy = Z B', 8, § 31 or ZB=ZB'. Ax. 9 Case 2. Given zi S and B' with ^S±5'C' and BC±A'B', as in Fig. 456. To prove A B and B' supplementary. [The details of the proof for this case are left to the student.] 'B' Fig. 45 a Def. § 18 §60 §a3 EXERCISES 1. When in Theorem XIII will the two angles be equal ? when supplementary ? 2. An object lies at a point K on an inclined plane AB. Show that the angle between the vertical line through K and a perpendicular to '^ AB at K is equal to the angle CAB which the inclined plane makes with the horizontal. 62 RECTILINEAR FIGURES [I, §69 69. Theorem XIV. Two right triangles are congruent if the hypotenuse and an acute angle of the one are equal respec- tively to the hypotenuse and an acute angle of the other. Ftg. 46 Given the rt. A ABC and A'B'C with the hypotenuse BC = hypotenuse B'C srnd Z B=Z B'. To prove A ABC ^ A A'B'C. Proof. We have Z:A = ZA'3indZB=ZB'. Given Hence ZC=ZC'. ^66 Moreover BC= B'C, Given and therefore A ABC ^ A A'B'C. Why ? Note. Theorem XIV is frequently stated in the following form : A right triangle is determined by its hypotenuse and one acute angle. EXERCISES 1. Draw an acute angle and then construct the right triangle containing this angle and having a hypotenuse 2 inches long. 2. Show that if two right triangles have one acute angle of one equal to one acute angle of the other, all of the angles of the' one are equal to the corresponding angles of the other. 3. If one angle of a right triangle is 45°, show that the tri- angle is isosceles. 4. State and prove the converse of Ex. 3. 5. How could you construct (using only ruler and compasses) the right triangle one of whose acute angles is 60° and whose hypotenuse is a given length AB ? (See § 65.) I, §71] ANGLES AND TRIANGLES 63 Fig. 47 70. Theorem XV. Two right triangles are congruent if the hypotenuse and a side of the one are equal respectively to the hypotenuse and a side of the other. Given the rt. A ABC and A'B'C with hyp. BC= hyp. BfC and side AB = side A'B'. To prove that rt. A ABC ^ rt. A A'BC. Proof. Place A A'B'C in the position ABC" so that A'B^ coincides with its equal AB and C falls at C'\ opposite to C. Then, the A CAB and CAB being rt. A (why?), the line CAC" will be a straight line. 12, § 31 Now, in the A CC"B, we have BC=B'C"', Given therefore A CC"B is isosceles ; hence AACB = AAC"B, or AACB=:AA'C'B'. Therefore rt. A ABC ^ rt. A A!BfC', 71. Corollary '1. If two oblique lines of equal length are drawn from a point C in a pei-pendicular CD to a line AB (Fig. 32, p. 42), they cut off equal distances from the foot of the per- pendicular, and conversely. EXERCISES 1. In the figure a mast is being held in a vertical position by means of a number of ropes (guy ropes) at- tached to the mast at the same distance from the ground. Show that the ropes will all have the same length if they are anchored at equal dis- tances from the foot of the mast. 2. Construct a right triangle whose hypotenuse is 4 inches long and one of whose sides is 2 inches long. §40 Ax. 9 §69 64 RECTinXEAR FIGURES [I, §72 72. Theorem XVI. IConverse of Theorem IIL~\ If two angles of a triangle are equal, the sides opposite are equals and the triangle is isosceles. C Given the A ABC in which ZA=ZB. To prove that AC = BC Proof. Draw CD bisecting Z C. Then, in the A ACD and BCD we have Zx = ZyajidZ.A = ZB. Why? Therefore Zu = Zv: § 66 Fig. 48 but CD=CD. Iden. Hence A ACD ^ A BCD, Why? and therefore AC=BC. Why? 73. Corollary 1. -4n equiangular triangle is also equUaterdl. 74. Corollary 2. If two oblique lines are draion from a point C in a perpendicular CD to a line AB (Fig. 48), so as to make equal angles with AB, they are equal. 75. Theorem XVII. If two angles of a triangle are unequal, the sides opposite them are unequal and the greater side is oppo- site the greater angle. r Given the A ABC in which ZB> ZA. ^^-^^ \ To prove that AC > BC. ^ ^ -^B Fig. 49 Proof. Draw BD meeting AC at D and making Zx= ZA. Then, in the A ABD AD = BD, Why? and BU + DC > CB. Post. 3 Therefore AD + DC > CB ; Ax. 9 that is, AC>CB. 76. Corollary 1. If two triangles have two sides of the owe eqiuil to two sides of the other, hut the included angle of the first I, §77] ANGLES AND TRIANGLES 65 greater than the included angle of the second, then the third side of the first is greater than the third side of the second. [Hist. In ^ABC and A'B'C, \etAB=A'B', BC=B'C', And Z ABC > Z A'B< a. Suppose AB < BC. Place A'B' C on ABC with A' B> on AB. Join CC. Then ABCC is isosceles, and ZBC'C = ZBCC'. Hence, in HAC'C, ZAC'C>ZACC\ whence (§ 76) AC>AC'.] Note. Corollary 1 is sometimes stated in the following brief form wherein it finds numerous illustrations in physics and mechanics. " The growth of an angle in a triangle means the growth of the side opposite it.'' It is to be understood, of course, in this statement, that as the angle is allowed, to grow, the lengths of its including sides remain fixed. 77. Corollary 2. If two oblique lines are drawn from a point C in a perpendicular CD to a line AB (Fig. 48), and if the base angles at A and B are unequal, the oblique line opposite the greater base angle is the greater; in particular, the perpendicular CD is itself the shortest line from C to any point of AB. EXERCISES 1. If, on account of some obstruction, one of the guy ropes mentioned in Ex. 1, p. 63, must be anchored nearer the foot of the mast than the others, show why that rope will be the shortest. 2. Is the string attached to a kite usually \ ^^^ equal in length to the height of the kite above the ground ? Connect your answer with § 75. 3. A simple form of crane consists of a beam AB hinged at ^ to a vertical mast AC and con- trolled by a wire rope attached at B and run- C""*"® ning over a pulley at C. When the rope is let out, the beam AB descends. Connect this fact with § 76. 66 RECTILINEAR FIGURES [I, §78 78. Theorem XVIII. If two sides of a triangle are unequal, the angles opposite them are unequal and the greater angle is op- posite the greater side. C Fig. 50 Given A ABC in which BC > AC. To prove that Z CAB > Z.B. Proof. On CB take CB = AC', draw AD, Then, in the triangle ADC we have ZCAD=ZCDA. Why? But ZCDA>ZB', §47 therefore Z CAD > Z B. Why ? Moreover ZCAB> ZCAD-, Ax. 10 therefore Z CAB > Z B. Why ? 79. Corollary 1. If two triangles have tivo sides of the one equal to two sides of the other, but the third side of the first greater than the third side of the second, then the included angle of the first is greater than the included angle of the second. [Hint. Prove by reduction to absurdity. Sliow first that equality of the included angles leads to a violation of § 35. Show that if the included angle of the second triangle is the greater, § 76 is violated.] Note. Corollary 1 is sometimes stated in the following- brief form wherein it finds numerous illustrations in physics and mechanics : " The growth of a side of a triangle means the growth of the angle opposite." 80. Corollary 2. If from a point C in a perpendicular CD to a line AB (Fig. 48) unequal oblique lines are drawn to the base AB, the longer of the oblique lines is opposite the larger of the two base angles. I, §80] ANGLES AND TRIANGLES 67 MISCELLANEOUS EXERCISES 1. Prove that the hypotenuse of a right triangle is its longest side. 2. By use of Theorem XVIII, prove that the perpendicular is the shortest line that can be drawn from a point to a straight line. (Compare § 77.) 3. If the crank AB mentioned in Ex. 3, p. 59, is so arranged that AC > AB (see the figure, p. 59), show that the angle ACB will always remain less than half the angle XAB during the rotation. 4. If the pans of a balance of the ordi- nary form shown in the figure are not pre- cisely on the same level, show that each pan is nearer to the middle post than when the balance is level. Show also that the pans are always at equal distances from the middle post. Balance Scales King.Post Truss 5. A simple piece of bridge work consists of a frame like that shown in the adjoining figure, AM and MB being stiff pieces of steel merely hinged together at M, but the hinge resting on a plate P, which at some lower point C is connected to A and B by strong flexi- ble wires. Show that a heavy weight may safely be put at Jf, even though the bridge is supported only by buttresses at A and B. (An arrangement of this kind is called an inverted King-Post truss.) 6. Determine the angles of a triangle when they are in the ratio 3:4:5. 7. If the exterior angle at A in a triangle ABC is 115°, and C is three times B, find B and C 68 RECTILINEAR FIGURES [I, §81 PART IV. QUADRILATERALS 81. Definitions. A plane figure bounded by four straight lines is called a quadrilateral. It is desirable to distinguish be' tween several kinds of quadrilaterals as follows : If two sides of a quadrilateral are parallel, it is called a trapezoid. If each pair of opposite sides of a quadrilateral are parallel, it is called a parallelogram. B H Quadrilateral Trapezoid Fig. 51 Parallelogram A parallelogram all of whose angles are right angles is called a rectangle. A quadrilateral all of whose sides are equal is called a rhombus. A rectangle all of whose sides are equal is called a square. Rectangle Rhombu Fig. 52 Square The side upon which a quadrilateral appears to rest is called its base. Trapezoids and parallelograms, however, are consid- ered as having two bases, one being the side upon which the figure appears to rest, and the other being the parallel side opposite it. Thus, in Fig. 53, AB and CD are bases. I, §81] QUADRILATERALS 69 The perpendicular distance between the bases (prolonged if necessary) of a trapezoid or parallelogram is called its altitude, as the line h in the figures below. B A D \ 1 '-" f X / ^ \ \ 1 \ /X N \ \ I _J Fig. 53 A line joining opposite corners (vertices) of a quadrilateral is called a diagonal, as the line AG in Fig. 53. In the figures above, the angles ABC and BCD are said to ije adjacent to the side BC. A pair of angles such as A and C are said to be oj^osite angles of the quadrilateral. EXERCISES 1. Prove that the angles adjacent to any side of a parallelo- gram are supplementary. 2. Prove that opposite angles of a parallelogram are equal. 3. If two adjacent angles of a parallelogram are in the ratio 17 : 1, how large is each angle of the parallelogram ? 4. Construct the rhombus each of whose sides equals 2 inches and one of whose angles is 30". 5. Show that two equilateral triangles that have a common side, together form a rhombus. This is popularly called a " diamond," and is the basis of many designs. 70 RECTILINEAR FIGURES [I, §82 82. Theorem XIX. Either diagonal of a parallelogram divides it into two congruent triangles. Given the parallelogram ABGD in which the diagonal AC has been drawn. To prove that A ABC ^ A ADC. Proof. IniliQ A ABC, ADC we have AC = AC, and /.z = Z.w\ therefore A ABC ^ ADC. Fig. 54 83. Corollary 1. side opposite it. §54. Why? Why? Any side of a parallelogram is equal to the 84. Corollary 2. The segments of par- allel lines included between parallel lines are equal. [Thus, in Fig. 55, AB and CD are one pair of parallels, while A'B' and CD' are another pair ; they form the quadrilateral whose sides are represented by a, &, c, and d. Then, the corollary states that a = c and h = c?.] EXERCISES 1. How does the ruled paper used in drawings in the Intro- duction (see § 25) provide an illustration of Corollary 2 ? 2. In the parallelograms that occur in the framework of bridges, a crosspiece is usually inserted along at least one of the diagonals. Why will this make the whole parallelogram stiff ? 3. Cut a piece of paper in the form of a parallelogram and then cut it in two along one diagonal. Will the two triangles thus formed fit exactly upon each other ? Why ? I. §86] QUADRILATERALS 71 85. Theorem XX. If a quadrilateral has both pairs of opposite sides equal, it is a parallelogram. Why? Why? Why ? Why? §81 Fia. 56 Given the quadrilateral ABCD in which AB = DC and BC = AD. To prove that ABCD is a EJ. Proof. In the A ABC, ADC we have AC=AC, AB = DC, and BC = AD. Therefore A ABC ^AADC-, hence Z x = Z. y, and Z z = Zw. It follows that AB II DC, and BCWAD-, hence ABCD is a O. 86. Theorem XXI. If a quadrilateral has one pair of sides equal and parallel, it is a parallelogram. Given the quadrilateral ABCD in which ^B is equal and parallel to DC. To prove that ABCD is a O. Proof. Draw the diagonal AC. ADC, we have AC=AC,AB = DC, Zx hence A ABC ^ A ADC Therefore Z z = Zio, and hence AD But since AB II DC, ABCD is a O. Fio. 57 Then in the A ABC and ^y, Why? Why? BC. Why? Why? 72 RECTILINEAR FIGURES [I, §87 87. Theorem XXII. The diagonals of a parallelogram bi- sect each other. Given the O ABCD, and let its /^\m diagonals intersect at M. To prove that AM= MG and that . DM=MB. t. .« ^ Fig. 58 [Hint. Prove that A AMB ^ A DMC. Since the proof is easily carried out it is left to the student.] 88. Theorem XXIII. Two parallelograms are congruent if two sides and the included angle of the one are equal respectively to two sides and the included angle of the other. ■^B Fig. 59 Given the n ABCD and A'B'C'D' in which AB = A'B\ AD = A'D', and Z DAB = Z D'A'B'. To prove O ABCD ^ O A'B'C'D'. [Hint. Draw the diagonals DB, D'B', and prove A ADB ^ A A'D'B'; also prove that A DCB ^ A D'C'B'. Then apply § 33.] EXERCISES 1. Prove that the diagonals of a rectangle are equal. 2. Prove that the diagonals of a rhombus are perpendicular to each other. 3. Show that if each of the diagonals of all of the squares on a piece of squared paper are drawn, two new sets of continuous straight lines at right angles to each other are formed. This is the basis of many designs. I, §90] QUADRILATERALS 73 89. Theorem XXIV. The line joining the middle points of the two sides of a triangle is parallel to the base and equal to half the base. Given the triangle ABC and the line DE joining the mid- points of the sides AC and BC. To prove that DE II AB, and that DE = AB/2. Proof. Draw BB' II AC meeting DE prolonged at F. Then, in the A DEC and EBF, we have CE = EB,Zr = Zs,Zx = Zy. Why? Therefore, A DEC ^ A EBF-, hence, DC = BF. Why ? But DC = AD', hence, BF = AD, and ABFD is a O. Why ? It follows that DE II AB. § 81 The fact that DE = AB/2 may now be established as follows : Since, as just shown, ABFD is a O, we have DF=AB, Why? But DF=DE + EF. Moreover DE = EF, since A DEC ^ A ^Bi^. It follows that DF=2 DE, ov AB = 2 DE, that is DE = AB/2. Ax. 4 90. Corollary 1. (Converse of % 89.) The line draivn through the middle jyoint of one side of a triangle parallel to the base bisects the other side. [Hint. Draw the parallel; and draw the line connecting the middle points of the two sides. If these do not coincide, show by § 89 that § 49 is violated. For another proof, see Ex. 1, p. 75.] 74 RECTILINEAR FIGURES [I, §91 91. Theorem XXV. If three parallel lines cut off two equal portions of one transversal, they cut off two equal portions of any other transversal, T V E R'/ n\ F C A Given the three parallel lines AB, CD, and EF', let >ST be a transversal of which the two portions PQ and QR cut off by the three parallels, are equal. To prove that the three parallels cut off equal portions LM and MN on any other transversal UV. Proof. Through It draw a line RK II UV cutting CD at J. Then KJ:= LM, and JR = MN-, § 84 but KJ=JR\ §90 hence LM— MN. Ax. 9 92. Corollary 1. If a series of parallel lines cut off equal portions of one transversal, they cut off equal portions of any other transversal. [Hint. Show that the portions cut off on any trans- versal are equal, taking two at a time, by Theorem s U XXV.] Fig. 62 93. Corollary 2. If three parallel lines cut off two portions of one transversal, one of which is double the other, j V they cut off two portions of any other transversal, -6^ — ly one of which is double the other. " "" 7 V K V J y I V P / \ [Hint. First draw a fourth parallel through the mid- dle point of the larger portion of the first transversal, and then use Cor. 1, § 92.] I,§W1 QUADRILATERALS 75 94. Corollary 3. If three parallel lines cut off two portions oj one trayisversal, one of which is n times the other, they cut off two portions of any other transversal, one of which is n iimes the other. EXERCISES 1. Prove that the line DE drawn through the middle point of one side of a triangle ABC parallel to the base bisects the other side by drawing BF || AC and showing that A DCE ^ A FBE. Compare § 90. 2. Prove that the lines joining the middle points of the three sides of a triangle divide it into four congruent triangles. / 2 [Hint. Prove A1^A2^A3^A4.] .C 3. Prove that perpendiculars drawn from the middle points of two sides of a triangle to the third side are equal. 4. Prove that the lines joining the middle points of the sides of any quadri- lateral form a parallelogram. [Hint. Draw the diagonals of the original quadrilateral, and use § 89.] 5. A long board 7^ in. wide is to be cut into 4 equal parallel strips. Show that it can be marked ready for sawing in the following manner : Lay the corner (heel) of a carpen- ter's square on one edge of the board (see figure) and turn until the 12 v mark is on the other edge. With an awl make dents at 3, 6, and 9. Move the square and repeat the operation. Then draw parallels through the dents thus made. Verify this process. 76 RECTILINEAR FIGURES [I, §95 PART y. POLYGONS 95. Definitions. A plane figure bounded by any number of straight lines is called a polygon. The bounding lines are called sides ^ an angle between two adjacent sides, as the angle ABG in the figure, is called an interior angle of the polygon, while an angle between any one side and the adjacent side pro- longed, as the angle CBK in the figure, is called an exterior angle of the polygon. In what follows, we assume that all polygons mentioned are convex, i.e. that each interior angle is less than 180°. E A line joining any two non-consecutive vertices is a diagonal, as AC in the figure. 96. Kinds of Polygons. A triangle is a polygon of three sides. A quadrilateral is a polygon of four sides. A pentagon is a polygon of five sides. A hexagon is a polygon of six sides. An octagon is a polygon of eight sides. A decagon is a polygon of ten sides. An equilateral polygon is one all of whose sides are equal. An equiangular polygon is one all of whose interior angles are equal. A regular polygon is one which is both equilateral and equi- angular. I, §97] POLYGONS 77 97. Theorem XXVI. The sum of the interior angles of a polygon is two right angles taken as many times as the figure has sides, less two. ^ Given the polygon ABCD • • • having n sides. [In Fig. 65, n = 6.] To prove that the sum of its inte- rior angles = (n — 2) 2 rt. A. Proof. Draw the diagonals, AC, AD, ••', dividing the polygon into (n-2)A. Fig. 65 The sum of the angles of the polygon is equal to the sum of the angles of these triangles. Ax. 11 But, the sum of the angles of any triangle = 2 rt. zi, § 62 ; hence the sum of the angles of ABCD — is (n — 2) 2rt. A. EXERCISES 1. "What is the sum of the interior angles of a pentagon ? a decagon ? an octagon ? 2. How many degrees in one angle of a regular pentagon ? Answer the same question for a regular hexagon ; regular deca- gon ; regular octagon. 3. How many sides has the polygon each of whose exterior angles equals 30° ? 4. Show that a regular hexagon can be made by placing six equilateral triangles with one vertex of each at the same point. (See Ex. 6, p. 59.) 5. Show that if a regular hexagon is Irawn on each side of a given regular hexagon, the space in the plane around the given hexagon is just tilled. If the process is continued, show that the entire plane is divided into regular hexagons, in the manner of a honeycomb. 78 RECTILINEAR FIGURES [I, §98 98. Theorem XXVII. Tlie sum of the exterior angles of a polygon formed by producing the sides in succession is equal to four right angles. Given the polygon ABCD —. ""dTo c\ To prove that the sum of its exterior / \ angles = 4 rt. A. /'eV ^V^'^ Proof. Denote the interior A by A, B, \L__A^__ C, D, ••• and the corresponding exterior \ angles by a, h, c, d, •••. " Fig. 66 Then, A A + A a =^2 rt. A, AB-\-Ab = 2vtA. In like manner the sum of each pair of angles at a vertex = 2 rt. A. Therefore, the sum of both interior and exterior angles about the whole polygon will be 2 rt. A taken as many times as the polygon has sides ; that is, it will be « • 2 rt. A. But the sum of the interior angles alone is (n — 2) • 2 rt. A. § 97 Therefore, the sum of the exterior angles alone is n • 2 rt. Zs - (n - 2) . 2 rt. Zs = 2 . 2 rt. Zs = 4 rt. Z. EXERCISES 1. What is the sum of the interior angles of a square? What is the sum of the exterior angles ? 2. What is the sum of the interior angles of any quadri- lateral ? What is the sum of the exterior angles ? Compare with Ex. 1. 3. What is the sum of the exterior angles of a pentagon ? of a hexagon ? 4. How large is each of the exterior angles of a regular pentagon ? hexagon ? decagon ? 5. How many sides has the polygon the sum of whose interior angles equals the sum of its exterior angles ? I, §99] LOCUS OF A POINT 79 PART VT. THE LOCUS OF A POINT 99. Locus of a point. In § 23 (Introduction) it is stated that a circle is a curve every point of which is equally distant from a point within (center). This definition may be stated in the following language: A circle is the locus (position) of all points equidistant from a given point. Using the same lan- guage, it may be said that the locus of all points equidistant from two parallel lines ^' is the line lying midway between them, as ^ the line EF in the figure. ^ ~ Similarly, the locus of all points common to two intersecting lines is simply one point; namely, their point of intersection. Note. It is important to observe that in each of the pre- ceding illustrations, the locus not only (1) contains all points that satisfy a certain given condition, hut it is also true that (2) there are no points on the locus that do not satisfy this condition. Thus, in the figure above we can make the following two statements about EF: (1) EF contains all points equidistant from AB and CD ; (2) there are no points on EF that are not equidistant from AB and CD. Every true locus ]?ossesses the properties (1) and (2) ; hence in all locus problems two things are to be proved. This will be illustrated presently. EXERCISES 1. What is the locus of all points 2 inches distant from a given straight line ? [Hint. Note that there are such points on either side of the given line.] 2. What is the locus of points 1 in. from a fixed point ? 3. What is the locus of all points 4 inches distant from each of two given points which are 6 inches apart ? 80 RECTILINEAR FIGURES [I, § 100 100. Theorem XXVIII. The locus of all points equidistant from the extremities of a line is the perpendicular bisector of that line. E Given the line whose extremities are A and B. Given also the line EC drawn _L AB at its middle point C. To prove that EC is the locus of all points equidistant from A and B ; that is (see § 99), to prove that (1) any point D which is equidistant from A and B lies on EC, (2) there is no point on EC not equidistant from A and B. Proof. From D draw DA and DB. Then A ADC ^ A BDC. § 45 Therefore Z ACD = Z BCD, Why ? so that DC±AB. Why? Hence D must lie on EC, which is the property (1) to be proved. T, § 31 Again, let D' represent any point on the perpendicular bisector CE. Then AAD'C^A BCD'. § 35 Hence D'A = D'B, which is the property (2) to be proved. EXERCISE 1. What is the locus of the vertices of all isosceles triangles constructed on a given base ? I, § 101] LOCUS OF A POINT 81 101. Theorem XXIX. The bisector of an angle is the locus of all points equidistant from its sides. Given the angle ABC and its bisector BE. To prove that BE is the locus of all points equidistant from AB and BC\ that is, to prove that (1) any point D equidistant from AB and BC lies on BE, (2) any point i>' on BE is equidistant from AB and BC. Proof. For the proof of (1) show that A BDF^A BDG. For the proof of (2) show that A BDF' ^ A BD'G\ [The details of the proof are left as an exercise.] EXERCISES 1. By means of Theorem XXVIII prove the correctness of the construction given in § 5, p. 4. Similarly, prove the cor- rectness of the constructions indicated in §§ 6, 8. 2. By means of Theorem XXIX prove the correctness of the construction given in § 9, p. 8. 3. What is the locus of a point that is equidistant from three given points ? Show how to construct the locus. 4. A carpenter bisects an Z ^ as follows : Lay off AB = AC. Place a steel square so that BD= CD as shown in the figure. Mark D and then draw the line AD. Show that AD bisects Z A. Would this method be correct if the square were not a right angle at D ? 82 RECTILINEAR FIGURES [I, § 102 102. Supplementary Theorems on Altitudes, Medians, etc. Theorem XXX. The perpendiculars erected at the middle points of the sides of a triangle meet in a point. C Outline of proof. Let be the point where the perpendicular bisectors EI, FH, of the sides AB, BO meet. Join O to the middle point D of AC. Prove that OD is then the perpendicular bisector of AC. Theorem XXXI. The bisectors of the angles of a triangle meet in a point. C D Fig. 71 Outline of proof. Draw the bisectors of A A and B and suppose that they meet at 0. Join O to the third vertex and prove that rt. A OCD ^ rt. A OCE, thus making Z ECO =:ZDCO. Theorem XXXII. TJie altitudes of a triangle meet in a poijit. Outline of proof . GivenAvl^C. Through its vertices draw lines parallel respectively qI to the opposite sides, forming A A'B' C ^\ Then A is the mid-point of B'C, since v BO AC and BOB' A are parallelograms. Similarly, O and B are mid-points of A'B' and A'C. Then, AD, BE, and OF are per- pendicular to the sides of A'B'C at their mid-points and therefore meet in a point (Theorem XXX). FiQ. 72 I, § 102] MISCELLANEOUS EXERCISES 83 Theorem XXXIII. The medians of a triangle meet in a point which is tico thii'ds of the distance from any vertex to the middle point of the opposite side. Outline of proof. Draw the medians CF and BE and suppose they intersect at 0. Draw AO and extend it to cut BC at D. Now draw BG parallel to FC, meeting AO (prolonged) at G. Johi G and C. Then ^*^^^. in the triangle ABG we have AF = FB. """-A-''" Therefore, AO = OG. (§ 90, p. 73.) ^ Next prove OE II GO. Then BOCG is a ^^^' '^^ parallelogram. Then BD = DC (§ 87) which proves that the medians meet in a point. Moreover, A0 = OG, while 0G = 2 0D (§ 87). Whence, A0z=2 OD. But AD=A0+ OD=S OD so that AO/AD = 2/3 ; that is, AO = i AD. MISCELLANEOUS EXERCISES ON CHAPTER I 1. Given two angles of a triangle. Construct the third angle. 2. Prove that the bisectors of two supplementary adjacent angles are perpendicular to each other. 3. If a weight is hung from a small ring that slips freely on a cord, and if the cord is tied fast to two sup- ports at equal heights, the ring will come to rest at the middle of the cord. Prove that the string by which the weight is attached then bisects the angle between the two portions of the cord. 4. A simple form of carpenter's level consists of three pieces of board nailed together in the shape of a capital letter A. A plumb bob is hung on a hook screwed at B. Show that any object upon which the feet A, C are set will be level in case the plumb line passes through the middle point Q of DE. 84 RECTILINEAR FIGURES [I, § 102 5. An angular joint for water pipes is to be constructed for two pipes that meet each other at an angle q \ h / of 40°. If the seam FII is to make equal angles with the lines of centers ; that is, if angle 50i^ = an gleJOF, show that each of these angles must be taken equal to 70°. 6. Prove that if two angles of a quadrilateral are supple- mentary, the other two are supplementary also. 7. What is the size of the obtuse angle formed between the bisectors of the acute angles of any right triangle ? 8. Given a diagonal, construct the corresponding square. 9. Given the diagonals of a rhombus, construct the rhombus. 10. A so-called T joint for pipe is a piece made in the form of ^ a letter T ; the angle between the arms is accurately a right angle. Show, by § 52, that two pipes along the same wall joined to the same main pipe by T joints, are parallel. 11. Prove that the bisector of the exterior angle at the vertex of an isosceles triangle is parallel to the base. Given the A ABC with AC = BG and Z DCE = Z.BCE. To prove CE \\ AB. 12. Prove (using Theorem I), that in an equilateral triangle the bisector of any angle forms two congruent triangles. 13. To cut two converging timbers by a line AB which shall make equal angles with them, a carpenter proceeds as follows : Place two squares against the timbers, as shown in the figure, so that AO==BO. Show that AB is then the required line. ozrt EH I, § 102] MISCELLANEOUS EXERCISES 85 14. Let ABO and EST be two congruent triangles. Prove that the medians drawn through A and R are equal. Prove also that the altitudes through A and B are equal. C 15. In the figure, CA = CB, AD = BE. Prove A ADB ^ A ABE. D ^E 16. In a right-angled triangle, if one of the acute angles is 30°, prove that the side opposite is half the hypotenuse. 17. Prove that the bisectors of the interior angles of a rec- tangle form a square. C 18. If the bisector of an angle of a triangle is / perpendicular to the side opposite the angle, the / triangle is isosceles. Prove this statement. / ADB 19. Prove that the line joining the mid-points of the non- parallel sides of a trapezoid is equal to half the sum of the bases. 20. i? is a river, and it is required to find the distance be- tween the points B and A situated on the opposite shores. Show that this may be done (without crossing the stream) as fol- lows: (1) Set a stake at some point C in line with AB. (2) Set a second stake at some point D from which all three of the points A, B, C can be seen. (3) Set stake at E in line with DC and such that ED = DC, and similarly a stake at F which shall be in line with DB and such that FD = DB. (4) Set stake at G in line with both EF and AD. Then FG will be the required distance AB. 86 RECTILINEAR FIGURES [I, § 102 c p D A 2 ' B given 21. Prove that the angle between the bisectors of two angles of an equilateral triangle is double the third angle. 22. In the triangle KLN, NM is per- ^ pendicular to KL, and KM = MN = ML. Prove that KLN is an isosceles right triangle. K 23. Show how to obtain (as a crease) a parallel to a given line through a given point on a piece of paper, by- folding the paper, and prove that the result is correct. 24. A man wishes to measure the distance between two points R and T on opposite sides of a stream. He takes a line VR and measures the angle TRV. He then walks along VR prolonged until he reaches a point S where Z TSR = \^TRV. He then concludes RS=RT. Is he right ? Why ? Note. The sailor uses this principle when he " doubles the angle on the bow '' to find his distance from any object on shore. Thus if he is sailing in the direction ABC, and if Z is a lighthouse, he measures the angle A, and if he notices when the angle that the lighthouse makes with his course is just twice the angle noted at Af then BL = AB. He knows AB from his log; hence he knows the distance BL. 25. Show that if the two ends of one side of one square on squared paper are connected by straight lines to the two ends of any parallel side of any other square of the same size, taken in the same order, a parallelogram is formed. D C / / / / / / 1 5 5 I, § 102] MISCELLANEOUS EXERCISES 87 ■?rr".^.^Vrr.^r^.^rVr^- ^^^^^^^^^^^^^J-^. 26. If a jointed frame of the form of a parallelogram is moved so that the angle at A grows less, show that (1) the angle at B in- creases ; (2) the diagonal DB decreases ; (3) the diagonal AC increases. 27. Can the frame of Ex. 26 be braced effectively by flexible wires along the diagonals ? Why does this make the frame quite stiff? This principle is used in bridge building. 28. An isosceles trapezoid is one whose sides (other than the bases) are equal. Prove that the diagonals of an isosceles trapezoid are equal ; also that its base angles are equal. Given the trapezoid ABCD in which BC = AD. A F G B To prove (1) that AO = BD and (2) that Z ABC = Z BAD. [Hint. Draw the altitudes DF, CG, and provj (2) first.] 29. Show that if a rectangular door suspended on hinges hangs out of a vertical line, the bottom edge of door is out of a horizontal line. If the hinged edge leans away from the vertical line by 2 in. in every 3 ft. of its length, show that the bottom edge rises by 2 in. in every 3 ft. of its length. 30. A quadrilateral of which two pairs of ad- jacent sides are equal is called a kite. Show that one of the diagonals divides a kite into two congruent triangles ; show that the other diago- nal divides the kite into two triangles, each of which is isosceles. 31. Show that one pair of opposite angles of a kite are equal. If these angles are both right angles, show that the other two angles are supplementary. 88 RECTILINEAR FIGURES [I, § 102 32. Construct a triangle having given the mid-points of its sides. See Theorem XXXII, § 102. 33. What is the locus of the middle points of all straight lines drawn from a fixed point to a fixed line of unlimited length? 34. Prove that if the diagonals of a parallelogram are equal and perpendicular to each other, the figure is a square. 35. Construct an equilateral triangle, having given its altitude. 36. If in A ABC, AC = BC, and if AC is extended to D so that CD = AC, prove that DBA-AB. Hence show how to draw a perpendicular to a line AB at one end B without extending AB. 37. Show how to trisect (divide into three equal parts) a straight angle ; a right angle ; an angle of 45°. 38. ABC is a right-angled triangle. BD is drawn from the right-angle to the mid-point of the hy- potenuse. Prove that the triangle ABC is thus divided into two isosceles triangles. 39. Prove that the sum of the sides of a quadrilateral is greater than the sum of its diagonals, but less than twice their sum. 40. Prove that the difference between the diagonals of a quadrilateral is less than the sum of either pair of opposite sides. 41. A line is terminated by two parallel lines. Through its mid-point any line is drawn terminated by the parallels. Prove that the second line is bisected by the first. 42. Prove that the perpendiculars drawn from the extremities of one side of a triangle to the median upon that side are equaL CHAPTER II THE CIRCLE PART I. CHORDS. ARCS. CENTRAL ANGLES 103. Definitions. The circle (§§ 2, 23) is a curve, all points of which are equally distant from a point within, called the center; or (§ 99), it is the locus of all points equally distant from a given point. By the definition of a circle, all its radii must be equal. (See §§2, 23.) Any portion of the circumference is called an arc (§ 2). One quarter of a circumference is called a quadrant. A chord is a straight line joining the extremities of an arc. A diameter is a chord that passes through the center. The angle between any two radii is called a central angle. In Fig. 74, the central angle AOB is said to intercept (cut off) the arc AB (written AB ) ; while the arc AB is said to subtend the angle AOB. An area bounded by two radii of a circle and the arc between them is called a sector. An area bounded by a chord of a ciicle and its arc is called a segment of the circle. Two circles are said to be equal when the radius of the one is equal to the radius of the other. Two circles that have the same center are said to be concentric 89 Fig. 75 90 THE CIRCLE [II, § 104 104. Postulates. In what follows we shall use the follow- ing facts as postulates : (1) As a central angle increases, its intercepted arc increases, and vice versa ; and as a central angle decreases, its intercepted arc decreases, and vice versa. (2) In the saine circle {or equal circles), equal central angles iiitercept equal arcs; and equal arcs subteyid equal central angles. Fig. 76 Thus, in Fig. 76, the equal central angles AOB and A'O'B' intercept the equal arcs AB and A'B', and the equal arcs AB and A'B' subtend the equal central angles AOB and AOB'. 105. Rotation. In considering the relations between the angles at the center of a circle and their intercepted arcs, it is helpful to think of the rotation of a wheel about its axle. During such a rotation, any spoke of the wheel turns through a constantly increasing angle. The end of the spoke describes the arc of the circle that forms the rim of the wheel, while the angle described by the spoke intercepts on the rim precisely the arc described by the end of the spoke. Thus, any two spokes of the same wheel describe equal angles in equal times. The arcs described by the ends of the two spokes are also equal. [(2), § 104.] As the angle a spoke describes increases, the arc that its end describes also increases, that is, the greater of two angles at the center intercepts the greater arc. [(1), § 104.] II, § 105] CHORDS, ARCS, CENTRAL ANGLES 91 £X£RCIS£S 1. What is the central angle between the hands of a clock when it is three o'clock ? Answer the same question for four o'clock, eight o'clock, and half past nine. 2. Show that the diameter of a circle is its q greatest chord. [Hint. In the figure, AC = AO -\-0C = AO + OB. But AO-h OB>AB.;\ B 3. Through a point within a circle to construct the longest possible chord. 4. Show that if one arc of a circle is double another arc of the same circle, the angle at the center sub- p tended by the first is double that subtended by the second. [Hint. § 104.] First bisect the larger angle ; then apply (2), 5. Show that if one arc of a circle is n times another arc of the same circle, the angle at the center sub- tended by the first is n times that subtended by the second. 6. Show that if a circumference is divided into 360 equal arcs, the central angles subtended by these arcs are all equal (one degree) ; show that the number of degrees in any central angle is equal to the number of these small arcs contained in its intercepted arc. 7. Prove that two intersecting diameters divide a circumfer- ence into four arcs each of which is equal to one of the others. ^ 8. The diameter AB and the chord CD are prolonged until they meet at E. Prove that EA > EC and EB < ED. 92 THE CIRCLE [II, § 106 106. Theorem I. In the same circle or in equal circles, equal arcs subtend equal chords. /^ Given the two equal © / and 0' in which j3= I ^ )k :t [O is the symbol for circle; -^ is the symbol for arc.'] ^ " Fig. 78 ^ To prove that chord AB = chord A'B'. Proof. Draw the radii OA, OB, O'A', 0'B\ Then in the A AOB, A'O'B', we have OA = 0A\ OB = O'B', § 103 and ZO = Z 0'. § 104 Therefore A AOB ^ A A'O'B' ; § 35 hence AB = A'B'. 107. Theorem II. (Converse of § 106.) In the same circle or i7i equal circles, equal chords subtend equal arcs. Given the two equal (D and 0' (Fig. 78) in which the chord AB = the chord A'B'. To prove that AB = A'B'. Outline of proof. Draw the radii OA, OB, O'A', O'B'. Then show, by § 45, that A AOB ^ A A' O'B' and apply § 104. EXERCISES 1. Show that in the same or equal circles the greater of two arcs subtends the greater chord, and vice versa. [Use § 104 (1) and § 76.] 2. Show that the construction of an angle equal to a riven angle (§ 7) illustrates § 107. Hence prove that the construc- tion in § 7 is correct. 3. State Theorems I and II, using the phrase (a) " and con- versely " ; (b) " and vice versa '' ; (c) ^' if and only if." See § 55. II, § 108] CHORDS, ARCS, CENTRAL ANGLES 93 108. Theorem III. A diameter perpendicular to a chord bisects the chord and the arc subtended by it. Given the diameter DF ± chord AB at K. To prove that AK= KB and that AF = FB. Proof. Draw the radii OA and OB Then 6k= ok, and OA=OB; Why ? hence rt. A OKA ^ rt. A 0KB. Why ? Therefore AK=KB, and Z AOK=Z. BOK; Why ? hence AF= FB. (2) § 104 Note 1. In the figure above the chord AB may be regarded as subtending not only the arc AFB, but also the larger arc ADB. It is customary to speak of the two as tae minor arc and the major arc corresponding to the given chord. Similarly, any central angle subtends both a minor and a major arc. Unless otherwise stated, the minor arc is the one to be under- stood hereafter in any statement where both might play a part. Note 2. This extremely impoHant, though simple, figure (Fig. 79), occurs in the greatest variety of practical affairs, and in many geometric theorems and constructions. (See §§ b,Q, 8, 9, 40, 43, 44, 72, 100, 102, and Exs. 2, p. 43; 1, p. 45, etc.) J EXERCISES 1. Prove that a diameter perpendicular to a chord bisects the major arc subtended by it. 2. What is the locus of the mid-points of a system of parallel chords ? •* 3. Prove that the perpendicular bisector of a chord passes through the center of the circle and bisects the arcs (major and minor) subtended by the chord. 94 THE CIRCLE [II, § 109 109. Theorem IV. In the same circle or in equal cir- cles, equal chords are equally distant from the center; and, conversely, chords that are equally distant from the center are equal. Fig. 80 Given the equal chords AB and A'B^ in the equal (D and 0'. To prove that AB and A'B^ are equally distant from the centers and 0', respectively. Proof. Draw OD J_ AB and O'D' ± A'B') draw also the radii OB and 0'^'. Prove that A DOB ^ A D'O'B', and hence OD = O'D'. In the converse, it is given that the chords AB and A^B^ are equally distant from the centers of the equal © and 0'. To prove that AB = A 'B'. Proof. Show that A DOB ^ A D'O'B' and hence that DB = D'B'. But AB = 2DB and ^'5' = 2 Z>'5' ; § 108 whence AB:==A'B\ Why? EXERCISES 1. Show that two boards sawed from the same log, or from equal logs, at equal distances from the center, are equal in width. 2o What is the locus of the mid-points of a system of equal chords in a circle ? II, § 111] CHORDS, ARCS, CENTRAL ANGLES 95 110. Theorem V, In the same circle, or in equal circles, if two unequal chords are drawn, the longer one is nearer the center. Given the O with the two chords AB and CD such that AB > CD. Also, let OE and Oi^be the perpendicular distances from to AB and CD, respectively. To prove that OE < OF. Proof. From A lay off the chord AB' = CD. Then draw the perpendicular OK, and finally join K to E by the line KE. Since AE^^AB and AK= \ AB', it follows that AE > AK, for AB > AB'. Hence Zb > Z.c, and consequently Z a < Z d. Therefore OE < OK. But 0F= OK. Therefore OE < OF. FiQ. 81 § 108 Given § 78 Ax. 8 §75 §109 Ax. 9 111. Corollary 1. {Converse of § 110.) In the same circle or in equal circles, if two chords are unequally distant from the center, the more remote is the less. EXERCISES 1. Show that a board sawed from a circular log is wider than another board sawed from the same log at a greater dis- tance from the center. 2. Show that the least chord that can be drawn through a given point within a circle is the chord per- pendicular to the radius through that point. [Hint. Draw any other chord through P and draw OC perpendicular to it. First prove 0P> OC.J 3. Show that a diameter is longer than any other chord. 96 THE CIRCLE [II, § 112 Fig. 82 112. Chords. The relations between chords, arcs, and cen- tral angles of the same circle appear vividly **^ D in connection with ro- tation (§ 105). Thus the chord that sub- tends the arc grows as the angle grows ; that is, as the wheel rotates, until the chord reaches its greatest possible size, the diameter of the circle. At this time, the angle the spoke has described is 180°, or a straight angle. As the rota- tion goes on, the chord shrinks again to less than the length of the diameter, until the wheel has made one complete revolution, when the chord has shrunk to zero. If a circle of known radius is drawn, any central angle subtends a chord of some definite length. Taking the radius as 1 unit, the lengths of the chords corresponding to various central angles for every degree from 0° to 90° are given in the table of chords, Tables, pp. iii-vii. By means of this table, any angle can be laid off from any point as vertex, by drawing a circle of unit radius about it. Thus, the chord that corresponds to a central angle of 29"" is very nearly equal to \. To lay off 29° at a point P on a line MN^ draw a circle X of unit radius about P as center, cutting MN at §, as in § 7, p. 6. Then draw an arc y about Q as center with radius \. If x and y intersect at 0, Z OFQ = 29°, approximately. Let the student lay off the angles of 31°, 69°, etc., in a similar manner, using the Tables. II, § 113] CHORDS, ARCS, CENTRAL ANGLES 97 113. Angular Speed. The facts about the rotation of wheels or other parts of machinery are often clearly expressed in terms of the speed with which the wheel (or other part) is rotating. This speed of rotation means the amount of angle through which the wheel turns in one unit of time ; for example, we say that a certain wheel is turning at the rate of four revolutions per minute, or that some other wheel is rotating 50° per second. Such a speed of rotation of a wheel is called its angular speed. Which of the two wheels just mentioned is moving the faster ? The answer to this question is found by changing revolutions per minute into degrees per second as follows : The first wheel is rotating four revolu- tions per minute. Since four revolutions means 1440^, that wheel is making 1440° per minute, or 1440° every 60 seconds. Dividing by 60, we find that it is going 24° each second. Hence the second wheel is rotating over twice as fast. EXERCISES 1. Prove that in the same circle, or in equal circles, equal chords subtend equal central angles, and vice versa. 2. Prove that the greater of two chords of a circle subtends the greater central angle, provided the central angle less than 180° is understood in each instance. 3. Draw a circle of radius one inch. Measure approximately the length of the chord of each of the following angles : 45°, 60°, 75°, 90°, 120°, 180°, 210°. Check, when possible, by means of the tables, pp. iii-vii. 4. If an angle is doubled, is its chord doubled? Compare the lengths of chords of angles of 45° and 90°. Compare the lengths of chords of angles of 60° and 180°. Check your results by means of the tables on pp. iii-vii. 5. If one wheel is rotating through 5 revolutions per minute, and another through 30° per second, which is rotating the more rapidly ? 98 THE CIRCLE [11, § 113 MISCELLANEOUS EXERCISES. PART I 1. Show that a chord equal to the radius subtends an angle of 60°. 2. If a circle is drawn through all four vertices of a square, show that the arcs in- tercepted by its sides are equal. Prove that each arc subtends a central angle of 90°. 3. Show that if the vertices of an equilateral triangle all lie on a circle, each side intercepts an arc of 120°. 4. Show that if the vertices of an equilateral hexagon all lie on a circle, each side intercepts an arc of 60°. Show how to construct such a figure. 5. If a wheel is rotating two revolutions per minute, what part of a revolution does it make in one second ? How many degrees does it turn through in one second ? 6. The earth revolves once in 24 hours. How many degrees correspond to one hour ? Ans. 15°. 7. A jointed extension rod — such as that used on desk lights — is made by having a piece in the form of a circular ring fit over the end of another circular piece, in the manner illustrated in the figure. An excessive motion is prevented by projections at the points B and R' and at /S and S'. Show that when the stop S has reached E, the line AB has turned away from the line DC by half the difference between the angles BOS and EOF. n,§ii4] TANGENTS AND SECANTS 99 PART II. TANGENTS AND SECANTS CIRCUMSCRIBED AND INSCRIBED TRIANGLES 114. Definitions. A line of indefinite length wliich cuts a circle is called a secant. A line of indefinite length which touches a circle in but one point is called a tangent; this point P is then called the point of con- tact, or point of tangency. A triangle or other polygon is said to be inscribed in a circle when its vertices all lie on the circumference. Under the same conditions, the circle is said to circumscribe the polygon. Fig. 8;i Inscribed Triangle Circumscribed Triangle Fig. 84 A triangle or other polygon is said to be circumscribed about a circle when its sides are all tangent to the circle. Under the same conditions, the circle is said to be inscribed in the polygon. A good idea of a line tangent to a circle is obtained by placing a coin (representing a circle) against the edge of a ruler (rep- resenting a straight line). A tangent to a circle may be roughly drawn by placing a ruler so that its edge just meets, but does not cut across, the circumference. If several tangents are drawn to the same circle in this manner, and extended to meet, a circumscribed polygon results. 100 THE CIRCLE [II, § 115 115. Theorem VI. A line perpendicular to a radius at its extremity is tangent to the circle. Given the O and the line AB ± to the radius OE at its extremity. To prove that AB is a tangent to the circle. Proof. Take any point D on AB except E and draw OD. Then OD > OE. § 77 Therefore the point D is not on the circle ; § 103 that is, no point of AB except E lies on the circle. Whence AB is a tangent. § 114 116. Corollary 1. A tangent to a circle is perpendicular to the radius drawn to the point of contact. [Note that we know that OjE'< OD. Then apply § 77.] 117. Corollary 2. A perpendicular to a tangent at its point of contact passes through the center of the circle. [Hint. Draw the radius to the point of contact and apply Corollary 1, together with § 58.] 118. Theorem VII. Two tangents drawn to a circle from a point outside are of equal length. [The proof is left to the student, with the aid of Fig. 86.] Fig. 86 n,§ii8] TANGENTS AND SECANTS 101 £X£RCISES 1. Prove that two tangents drawn to a circle at the extremi- ties of a diameter are parallel. 2. Construct a tangent to a circle parallel to a given line. 3. Draw two concentric circles (§ 103) and prove that all chords of the greater circle that are tangent to the smaller circle are equal. 4. Let and 0' be two circles which are tangent to each other ; that is, which have but one point in common. Prove (1) that the line joining and 0' (line of centers) passes through the point common to the two circles (see 12, § 31) ; and (2) that a perpendicular to the line OO at the common point P will be a tangent to both circles. Consider the case in which the circles are tangent externally, and also the case in which they are tangent internally, as indicated in the figures. 6. A railroad curve joining two pieces of straight track AB and CD, is usually a circular arc tangent at B and C to AB and CD, respectively. Show that the center O of the cir- cular arc is the intersection of the perpendiculars- to AB and CD at B and C, respectively. 6. If AB and CD (Ex. 5) are extended to meet at E, show that EO bisects the angle at 0. Hence show that F, the intersection of EO and BC, is the center of Sb. 7. Show that the two radii and the tangents at their ex- tremities (Fig. 86, or figure for Ex. 5) form a kite ; that is (Ex. 30, p. 87), two pairs of its adjacent sides are equal 102 THE CIRCLE [II, §119 119. Theorem VIII. Tivo parallel lines intercept equal arcs on a circle. Case I. WJien the parallels are a tangent and a secant. Given the tangent AB II the secant EF (Fig. 87 a) ; also, let C be the point of contact of AB. To prove that EQ=6f. Proof. Draw the diameter CD. Then CD1.AB', § 116 hence OD _L EF, Why ? and therefore EC = CF. § 108 Case II. When the parallels are both secants. Given the parallel secants AB and EF (Fig. 87 b). To prove BF= AE. Proof. Draw DC II AB and tangent to the circle. Let Jf be the point of contact. Theni>CMI^F. Why? Whence FM= EM, and BM= AM. Case I Therefore BF= AE. Ax. 2 Case III. When the parallels are both tangents. Given AB and EF parallel tangents touching the circle at M and K, respectively (Fig. 87 c). To prove that MLK= MRK. Outline of proof. Draw CD II AB (Fig. 87 c) ; and prove MC=Mb, and KC = KD, Then apply Ax. 1, n, § 119] TANGENTS AND SECANTS 103 EXERCISES 1. Show that a tangent parallel to any chord of a circle has its i^oint of tangency in the center of the arc that the chord intercepts. 2. State and prove the converse of Theorem VITI, § 119. [Hint. (Case I) : Given GF= CE\ \jo prove AB II EF. Draw the diameter CD, also the radii OE, OF. Then CD ± AB. ' (Why ?) Now, since CF = CE, we have Z COF = Z COE. (Why ?) Therefore A FGO ^ A EGO (Why?), so that ZFGO = ZEGO. Therefore CD±EF. Since CD±AB and CD ± EF, we have ABW EF as desired.] 3. If every vertex of a trapezoid lies on a circle, prove that it is isosceles; that is, that BC=AD. 4. If a thick board is sawed out of a circular log, the circular arc that forms one edge is equal to that which forms the other edge. Connect this fact with § 119. 5. Let ABC be an isosceles triangle, with Z. B = Z C; and let AD be its altitude from A to BC. Draw a circle with center at A and with radius AD, cutting the sides ^B and AC in F and J5J, respec- tively. Show that EF II BC. 6. Draw a secant intersecting two concentric circles and prove that the portions intercepted between the two circles are equal. 7. Prove that if every vertex of a parallelogram lies on a circle, any two opposite sides are equidistant from the center. 8. Prove that if a polygon is inscribed in a circle, the per- pendicular bisectors of the sides meet in a point. 104 THE CIRCLE [II, § 120 120. Theorem IX. Through three given points not all on the same straight line, one and only one circle can he draivn. Given the three points A, B, and not all on the same straight line. To prove that one and only one circle can be passed through A, B, and C. Proof. The locus of all points equally distant from A and B is the perpendicular bisector DE of the line AB, while the locus of all points equally distant from B and C is the perpendicular bisector FG of the line BO. § 100 Therefore the intersection of DE and FG is equally dis- tant from all three of the points A, B, and C. Hence, the circle drawn with as center and with a radius equal to the line AO will pass through A, B, and C. That this is the only such circle follows from the fact that the lines DE and FG can intersect in but one point. 4, § 31 Note. Theorem IX is frequently stated in the following brief form : Three points determine {fix) a circle. The proof also shows how to construct the circle passing through three given points. Thus, given A, B, and G (Fig. 88); draw AB and BC and erect their perpendicular bisectors, DE and FG. The intersection of DE and FG is the center of the desired circle ; its radius is one of the equal distances OA, OB, or 00. II, § 123] TANGENTS AND SECANTS 105 121. Corollary 1. A cirde may be dravjn to circumscribe any tnangle. [Draw the three perpendicular bisectors of the sides of the triangle.] 122. Corollary 2. TTie perpendicular bisectors of the sides of a triangle meet in a point. (Compare Theorem XXX, § 102.) This point is called the circumcenter, because it is the center of the circumscribed circle. 123. Corollary 3. A circle may be completed if any arc of it is given, [Hint. Take three points on that arc and draw a circle through them.] EXERCISES 1. Try to prove Theorem IX, § 120, for three points A, B, C, that lie on a straight line. At what place does the proof break down, and why ? 2. Draw a triangle of any shape and then construct the circle which passes through its vertices. 3. How many circles can be drawn through two given points ? What is the locus of the centers of all such circles ? 4. How many circles having a given roilius can be drawn through two given points ? Is it always possible to have one such circle ? Why ? 5. Show how to construct the center of a given circle. [Hint. Erect the perpendicular bisectors of any two chords.] 6. Show that a circle can be circumscribed about a given quadrilateral if (and only if) the perpendicular bisectors of the sides all meet in a single point. 7. Show that a circle may be circumscribed about any square ; any rectangle. 8. Show that the perpendicular bisectors of the opposite sides of a parallelogram are parallel unless they coincide. Hence show that an inscribed parallelogram is a rectangle. 106 THE CIRCLE [II, § 124 124. Theorem X. A circle may he inscribed in any triangle. C Given the triangle ABC. ^ y^j\ To prove that a circle may be y/\~i\- inscribed in it. ^/f "^^je^ rva Proof. Draw the bisectors of >'''''0^'\^''^^^~>C \ the angles A, B, C: these three . j^^^ \^ I ^^^^A p bisectors meet in a point I. § 102 ^ Draw the perpendiculars /L, IM, INy to the three sides a, b, c, respectively ; then IL = IM=m. §101 Hence a circle with radius r = IL is tangent to a, b, and c at Lj Mj and 2^, respectively. § 115 Note. The preceding proof involves also the proof of the construction of the inscribed circle in any triangle. Compare § 120. 125. Corollary 1. A circle drawn from any point on the bisector of an angle, ivith a radius equal to the distance from that point to one side, is tangeyit to both sides of the angle. EXERCISES 1. How many circles can be drawn tangent to each of two given intersecting lines ? What is the locus of their centers ? 2. What is the locus of the centers of the circles of Ex. 1, in case the two given lines are parallel ? 3. Show that a circle can be inscribed in a given quadrilat- eral if the bisectors of the four angles all meet in a single point. 4. Show that a circle can be inscribed in any square. 5. Show how to round off the vertices of any triangle by circular arcs. (See § 125.) This process is used in rounding off the corners of triangles that occur in a triangular-shaped piece of ground, so as to build a fence or a sidewalk or a build- ing without sharp corners. (See also Ex. 5, p. 101.) II, § 127] MEASUREMENT OF ANGLES 107 PART III. MEASUREMENT OF ANGLES 126. Numerical Measure. In order to measure any quantity, say a line of fixed length, we must first select the unit which we are to use. In the case of a fixed line, the customary unit would be either the inch, the foot, the centimeter, the yard, or any one of several others. In the case of an area, the unit might be a square inch or a square foot or an acre, or any one of several others. Having once selected our unit, the process of measuring consists in obtaining some idea of the relative size of the given quantity as compared to that of the chosen unit. Thus, in the case of the fixed line, we lay a yardstick along- side of it, and read off by means of the scale provided for the purpose, a number which, at least with some degree of ac- curacy, tells us the relative size of the line in question to that of the inch. Even though we cannot usually determine in this way the exact length of a line, owing to imperfections both in our in- struments and our eyesight, still we suppose, and it is indeed an axiom of measurement, that there always exists in every case just one number which does express exactly the length in terms of the unit. This number is called the numerical measure of the given line, corresponding to the unit selected. In general, the numerical measure of any quantity of any kind is a number, obtained as above, which expresses the relative size of the quantity to some unit of the same kind selected in advance. 127. Ratio. The numerical measure of one quantity divided by the numerical measure of a second quantity of the same kind, provided the same unit has been used in each case, is called the ratio of the first quantity to the second. Thus, the ratio of 6 feet to 8 feet is 6/8 or 3/4; again, the ratio of 6 inches to 1 yard is 6/36 or 1/6. 108 THE CIRCLE [II, § 128 128. Commensurable and Incommensurable Quantities. Let AB and CD be two straight lines of different length and let it be supposed that a certain unit ^, , , , . ^ of length, as MN, is contained an exact (integral) number of times (that is, with- ' ' ' ' ' ^ out any remainder) in AB. For example, M— ^N let MN be contained 4 times in AB. ^^"- ^ Then MJSF may also be contained an exact number of times in the other line CD, but more often this will not be the case; ordinarily the unit JOT" will be contained in CD a certain exact number of times plus a remainder x, which will be less than MJSr. This is illustrated in Fig. 90, in which MN is con- tained 5 times in CD, with a remainder x. Now, if we select a very small unit MN that is contained an exact number of times in AB as before, we obtain a very small remainder x when the same unit is applied to CD. However, it may happen that we can never take MN so small that the corresponding remainder x will turn out to be exactly zero. In this case the two lines AB and CD (Fig. 90) are said to be incommensurable. Examples of incommensurable lines occur frequently in geometry. Thus, it will be seen later that in any ^ isosceles right triangle a side and the hypotenuse constitute two incommensurable lines; that is, in Fig. 91, AB and BC are incommensurable. BC may be thought of as the diagonal of a square of which AB and AC are two sides. ^^^' ^^ Two other interesting incommensurable quantities are the diameter of any circle and the length of its circumference. These will be considered in detail in Chapter Y. If, on the other hand, it is possible to choose a unit which is contained an exact number of times in AB (Fig. 90), and also an exact number of times in CD, then the lines AB and CD are said to be commensurable. IT, § 129] MEASUREMENT OF ANGLES 109 129. Other Cases. Limits. Thus far we have spoken only of commensurable and incommensurable lines, but similar definitions apply to angles, arcs, or any other sorts of quantities. Thus, two angles are commensurable when a sufficiently small unit will be contained in each an exact number of times, and they are incommensurable when no unit which is contained an exact number of times in the one, will at the same time be contained an exact number of times in the other. In any case, the remainder x can be made as nearly equal to zero as we please by taking the unit sufficiently small : we often say that x may be made to approach zero. If a variable quantity approaches a fixed quantity as nearly as we please, the fixed quantity is called the limit of the variable one. Thus, in the preceding paragraph, we might say that the limit of x is zero. EXERCISES 1. State accurately what is meant by two incommensurable arcs ; two incommensurable areas. 2. Show that the remainder a; of § 128 can always be made less than \ in. 3. Show that the remainder a; of § 128 can be made smaller than one thousandth of an inch ; one millionth of an inch. Hence show that any two lengths can be measured in terms of a common (small) unit, except for a remainder that is less than the human eye can see. 4. What is the smallest unit shown on your ruler ? Can you see lengths less than this smallest division ? If a line whose length you are measuring is not an exact number of these smallest units, how can you estimate its exact length ? 5. Estimate the exact width of one line of type on this page. 6. The length Of a material object will change on account of expansion due to changes in temperature. Show that a unit can be chosen so small that the remainder a; (§ 128) will be less than the expansion due to a temperature change of one degree. 110 THE CIRCLE [II, § 130 130. Theorem XI. In the same circle^ or in equal cir- cles, two central angles have the same ratio as their inter- cepted arcs. Given the two equal circles O and 0' ; also let AOB be any central angle in 0, and let A'O'B' be any central angle in 0'. To prove that the ratio of Z. AOB to Z. A'O'B' is the same as that of arc AB to arc A'B'. Proof, (a) When arc AB and arc A'B' are commensurable. In this case a unit of arc may be found (see § 128) which is contained an exact number of times in both the arc AB and the arc A'B'. (Compare Ex. 5, p. 91.) Let m be such a unit, and let us suppose that the number of times it is contained in arc AB is r, while the number of times it is contained in arc A'B' is s. Then a.YG AB _r s (1) arc A'B' Now divide the arc AB into its r divisions, each of length 771, and through the points of division draw radii to the center 0. Likewise, divide the arc A'B' into its s divisions, each of the same length m, and draw radii through the points of divi- sion to the center 0'. Then Z AOB is divided into r equal angles, while Z A'O'B' is divided into s equal angles of the same size (§ 104) ; therefore ZAOB^r s (2) Z A'O'B' Erom (1) and (2) it follows that, ZAOB SiTcAB Z A'O'B' ^ic A'B' II, § 130] MEASUREMENT OF ANGLES 111 It would remain to prove Theorem XI when arc AC and arc A'C are incommensurable ; but this proof is interesting only theoretically. Instead of giving a proof, we may assume as a postulate that if any two geomet- ric ratios are equal whenr ever their terms are com- mensurable, they are equal also when their terms are incommensurable. The following proof may then be omitted at the discretion of the teacher : Proof. (6) When arc AB and arc A'B' are incommensurable. In this case if we take any unit of arc m which is contained an exact number of times in the arc AB and apply it to the arc A'B', there will re- main after the last point of division a certain arc B"B' less than m. § 128 But whatever the choice of m, we shall have ZAOB SLTC AB Fig. 93 (3) Case (a) ZA'O'Bf' Arc A'B" Now, as m is taken smaller and smaller, this equation (3) continues true at every step. At the same time, the individual membere of the same equation are changing, but only to the extent that ZA'O'B" comes closer and closer to ZA'O'B', while arc A'B" comes closer and closer to A'B'. Thus, by taking m sufficiently small, we can bring the first and second members of (3) as near as we please to the respective values ,^x ZAOB arc AB ^ ^ ZA'O'B'' SiYcA'B'' These last ratios, then, differ by as little as we please from the equal ratios in (3) ; hence they themselves differ from each other by as little as we please. But this is the same as saying that these ratios (4) are actually equal, for, if they were unequal the difference between them could not by any method of reasoning be shown to be as small as we please, but would always remain greater than some definite amount ; that is, in fact, what unequal means. Hence, ZAOB _ arc AB ZA'O'B' axe A'B'' 112 THE CIRCLE [II, § 131 Note. Theorem XI shows that the number of angulai units in a central angle is equal to the number of units of arc which the angle intercepts, if a unit of arc subtends a unit angle. Thus, the number of degrees in any central angle is the same as the number of degrees in the arc it ^ intercepts. This fact is expressed by saying that A central angle is measured by its intercepted arc. 131. Definition. An angle formed by the intersection of two chords in the circumfer- ence of a circle is called an inscribed angle ; or, the angle is said to be inscribed in the circle. 132. Theorem XII. An inscribed angle is measured by one half of its intercepted arc. B B B Fig. 95 (a) Fig. 95 (c) Given the inscribed Z ABC intercepting AO in the circle O. To prove that Z ABC is measured by one half of the arc AC. Case 1. When the center of the circle lies on one of the sides of the angle, as AB. (Fig. 95 a.) Proof. Draw DC ; then OC==OB; hence But therefore Since it follows that ZOBC=ZOCB. Z OBC+ Z OCB = Z AOC) 2ZABC:=ZA0C Z AOCis measured by AC, Z ABC is measured by ^ AC. Why? Why? Why? Ax. 9 Note, § 130 Ax. 4 II, § 135] MEASUREMENT OF ANGLES 113 Case 2. ^VJlen the center of the circle lies within the angle. Proof. Draw the diameter BD. (Fig. 95 6.) Then Z ABD is measured by ^ AD, Case 1 and Z DBC is measured by |-D0; Case 1 hence Z ABD 4-Z DBC is measured by |(AD + DC) ; that is, Z ^J56' is measured by ^ ^C. Case 3. Wheii the center of the circle lies outside the angle. Proof. Draw the diameter BD. (Fig. 95 c.) Then Z DBC is measured by \ DC, Case 1 and Z DJ5JL is measured by \ DA ; hence Z i)^0- Z Z)B^ is measured by ^{DG-DA) ; that is, Z ^5(7 is measured by \ AC. C. 133. Corollary 1. Any angle inscribed in a semicircle is a right angle ; as the angle BCA in the figure. 134. Corollary 2. Any angle inscribed in a segment (see § 131) greater than a semicircle is acute f ivhile any angle inscribed in a segment less than a semicircle is obtuse. Thus, in the figure, ABC is acute, while AB'C is obtuse. 135. Corollary 3. All angles inscribed in the same segment are equal. Thus, in the figure, the angles ABiC, AB^Cj AB^C, are all equal. FiQ. 98 114 THE CIRCLE [II, § 135 EXERCISES 1. A thin elastic band is stretched along tha diameter AB of a circle and then pinned firmly to the circumference at the two points A and B. If it be now stretched aside by means of a pencil point so that it takes the position indicated by the dotted line ; that is, so that a third point C of the band lies on the circumference, what can be said of the angle ACB? As the pencil is allowed to move about the circumference, the band meanwhile sliding over the pencil point, how does the angle ACB change ? 2. A corner of a piece of cardboard (of the usual rectangular shape) is pressed tightly against two pins E and F stuck into a board below. The cardboard is now turned in all possible positions, keeping it flat against the board. What is the locus of the point A ? 3. Prove by means of § 132 that the sum of the three angles of a triangle is two right angles. 4. If a circumference be divided into four equal arcs (quadrants), show that the chords which join the extremities form a square. 5. Show by § 132 that any parallelogram inscribed in a circle is a rectangle. 6. What is the locus, of all the vertices of right-angled tri- angles erected on a common hypotenuse ? 7. Prove that the opposite angles of any inscribed quad- rilateral are supplementary. 8. Prove that any equilateral polygon inscribed in a circle i? also equiangular. II, § 137] MEASUREMENT OF ANGLES 115 136. Theorem XIII. An angle formed by two chords intersecting loithin a circle is measured by one half of the sum of the intercepted arcs. Fig. 99 Outline of proof. Draw AD. Then /AKB=AD-{-^^A. § 64 But ZD is measured by ^ AB § 132 and ZAis measured by ^ CD ; § 132 hence Z AKB is measured by i {AB + CD). Why? 137. Theorem XIV. An angle formed by a tangent and a chord drawn through the point of tangency is measured by one half of the intercepted arc. C B. Outline of proof. Draw AE II CD. Then AB = BE §119 and ZABC=ZA. §54. But ZAis measured by | BE; Why ? hence Z ABC is measured by ^ AB. 116 THE CIRCLE [II, § 138 138. Theorem XV. An angle formed by two secants, or by a tayigent and a secant, or by two tangents that meet outside a circle is measured by one half the difference of the intercepted arcs. Fig. 101 [The proofs are left to the student. Draw DE parallel to BC and note how the angle ADE, which is equal to C, is measured.] EXERCISES 1. Prove Theorem XIII by drawing a line through A (Fig. 99) parallel to BD. 2. If, in Fig. 99, the arc BC contains 130° and the arc ABCD contains 170°, how many degrees are there in the angle AKB ? Ans. 20°. 3. Two angles of an inscribed triangle are 70° and 91°. Find in degrees the arcs subtended by each of the sides. 4. A chord that divides a circumference into arcs one of which contains 75° is met at one extremity by a tangent. At what (acute) angle does the meeting take place ? 5. A chord is met at one extremity by a tangent, making with it an angle of 61°. Into what arcs does the chord divide the circumference ? 6. If a tangent is drawn at the vertex of an inscribed square, •how many degrees are there in the angle included between the tangent and a side of the square ? Answer the same question for an inscribed equilateral triangle. II, § 139] CONSTRUCTION PROBLEMS 117 PART IV. CONSTRUCTION PROBLEMS 139. Problem 1. Through a given point to draw a tangent to a circle. Case 1. When the point is on the circumference. ,A i?"lQ. 102 Given the O and the point P on the circumference. Required to draw a tangent to the O through P. Construction. Draw the radius OP. At P draw AB ± OP. Then AB is the required tangent. Why ? Case 2. Wien the given point is not on the circumference. Outline of construction. On OP as diameter draw a circle cutting the given circle at A and B. ^ Pass a line through the points P, A, and another line through the points P, B. Either of these lines is a tangent such as desired. Why ? In answer- ing, note that Z PAO and Z PBO ^^-^^^_ ^^. are right angles and apply § 115. fiq. io3 EXERCISES 1. Show how to construct a tangent to a given circle parallel to a given line. 2. Show how to draw a tangent to a given circle perpen- dicular to a given line. 118 , THE CIRCLE [II, § 140 140. Problem 2. To circumscribe a circle about a given tri- angle. (See § 120. The student should carefully state the construction and the proof.) C Fig. 104 141. Problem 3. To inscribe a circle in a given triangle. (See § 124. State carefully the construction and its proof.) C EXERCISES 1. Can a circle always be drawn tangent to three lines, no two of which are parallel ? If so, how ? 2. Construct an equilateral triangle each of whose sides equals 2 inches, and then construct its inscribed circle. Prove that in this case the center of the circle lies at the inter- section of the three altitudes of the triangle. Can the same statement be made for any other triangle? 3. In Fig. 105, would the lines BI and AI (extended) ever pass through the points of contact M, L between the inscribed circle and the sides AC and BCj respectively ? If so, when would they, and why ? II, § 142] CONSTRUCTION PROBLEMS 119 142. Problem 4. On a given straight line to construct a seg- ment of a circle that shall contain a given angle. Given line AB and Z x. Required to construct on AB a segment of a circle that shall contain the Z x. Construction. Construct Z ABC= Z x. Draw ED ± AB at its middle point D. Draw FB ±GBsitB and meeting ED in O. Draw a circle with as center and OB as radius. Then AliB is the segment required. Proof. Z ABB = Z ABC = Z x. Why ? Fig. 106 EXERCISES 1. On a line 2 inches long construct a segment of a. circle that shall contain an angle of 60°. Do the same for 30° ; 135°. 2. Construct the locus of the vertices of all triangles hav- ing a common base 2 inches long and a common vertex angle of 30°. 3. In Ex. 2, p. 114, what can be said of the locus when the angle A is 60° instead of 90° ? 4. Construct the triangle whose base is 2 inches, whose alti- tude is 3 inches, and whose vertex angle is 30°. [Hint. First, draw the segment of a circle on a base of 2 inches, and such that 30° is contained in it. Now, draw a parallel to the same base at a distance of 3 inches above it. Where this parallel cuts the circle bounding the segment, will be the vertex of the triangle desired. Why ? The desired triangle is now easily drawn.] 5. How many solutions will there be to any problem similar to Ex. 4 ? May there be only one solution ? May there be no solution ? Describe all possible cases. 120 THE CIRCLE [II, § 142 MISCELLANEOUS EXERCISES. CHAPTER II 1. Prove that the construction described in § 5 for drawing a perpendicular to a straight line at a point within it is correct. 2. Prove that the constructions of §§ 6 and 8 are correct. 3. Through a given point in a circle construct the chord that is bisected by that point. 4. If a right triangle be inscribed in a circle, show that its hypotenuse will be a diameter. 5. Prove that an exterior angle of an inscribed quadrilateral equals the opposite interior angle. 6. Prove that the angle between two tangents to a circle is double the angle between the chord joining the points of con- tact and the radius to a point of contact. 7. Construct a circle such that it shall pass through two given points and shall have its center on a given line. 8. If four points A, B, C, D lie on a circle, and if AB — CD and AD=BC, show that the lines ^(7 and BD meet at the center. 9. Prove that the shortest line from a point to a circum- ference is along the radius through the point, extended if necessary, (a) when the point is outside the circle ; (&) when the point is inside the circle. 10. Prove that the angle between two tangents to a circle is the supplement of the angle between the radii drawn to the points of tangency. VI 1. If the sides of an inscribed angle are parallel to the sides of a central angle, how do their arcs compare ? 12. Show that two tangents to the same circle are parallel if and only if their points of tangency lie at opposite ends of the same diameter. 13. Show that a circle whose diameter is one side of any given triangle passes through the feet of the altitudes drawn to each of the other sides. II, § 142] MISCELLANEOUS EXERCISES 121 14. Following the suggestions of the ad- joining figures, show how to construct four common tangents to two given circles if the circles do not intersect. Note. The shadows thrown by a round body illuminated by a round source of light (e.g. earth and sun) illustrate this exercise. 122 THE CIRCLE [II, § 142 15. How many common tangents can be drawn to two given circles in each of the cases illustrated in the following figures ? A what Find approximately, from an 16. A common railroad turn is made by an arc of a circle tangent to two straight portions (Ex. 5, p. 101). In the adjoining figure, if OG = OF, and OKI. BF, show that : (1) ZOGB=Z.FOB/2=ZCOB/4:, (2) OJTis parallel to BG. (3) BK = BC/4., and Z 0GB = Z FOH. (4) ZEBF=ZBOK, and Z.EBC=ZBOF. 17. If in Ex. 16, ZCOB=60°, is the length of the chord BC ? accurate figure, or from the table of chords, p. iii, the length of the chord BF. Prove that BF = FC > BC/2. Note. In practice the entire curve BC can be (and frequently is) laid off by means of a surveying instrument located at B, by knowing accurately the lengths of a number of such chords through B, and their directions. I -* 18. Construct a circle through a given point tangent to a given line at a given point. p 19. Construct the triangle whose base is 2 inches, whose me- dian to the same base is 3 inches, and whose vertical angle is 30°. [Hint. The vertex must lie at the intersection of two circles, one bounding the segment mentioned in the Hint to Ex. 4, p. 119, and the other described about the middle point of the base with a radius of 3 inches.] TI, § 142 MISCELLANEOUS EXERC ISES 123 20. Show that a quadrilateral can be in- scribed in a circle if the sum of one pair of its opposite angles is 180°, 21. Show that a kite (see Ex. 30, p. 87) that is inscribed in a circle has two right angles; hence show that one diagonal is a diameter. 22. A simple turnout off of a straight track on street railways is of the form of a circular arc, as shown in the figure. The length i2=0^=0C= the radius of the outer rail of the circular track, is called the radius of the turnout. The distance L = BC from the point of the switch B to the crossing C is called the lead, and the angle TCS between the tangent to the circular track at G and the straight track, is called the angle of crossing. Show that if R is given (as it usually is), the angle SCT of crossing (which must be known accurately for the purpose of ordering the proper rail cross) can be found from the figure by measuring the angle COA. 23. A double opposite turnout from the end of a straight track is composed of two circular portions, the radii of the outer rails being given lengths r and R, respectively, and these outer rails being tangent to the straight track from which they start. Show that the angle of crossing SCT is the sum of the two angles COA and CO'B. 24. Let D be any point on the side AC of a right triangle ABC, and let E be the intersection of the hypotenuse AB with a circle whose diameter is AD. Show that the quadri- lateral DCBE is inscribed in the circle whose diameter is DB, 124 THE CIRCLE [H, § 142 25. If from one corner ^ of a paral- lelogram ABCD as center, a circle of radius AB is drawn, meeting AB pro- duced at L, and the sides AC and BD at M and N, respectively, show that LM= MN. 26. The usual form of carpenter's level contains a glass tube filled with liquid, except for a small bubble of air. The inner surface of the tube is cut in the form of a circular arc LHL', whose center is at 0; and it is set into a wooden or metal frame whose top edge is paral- lel to the tangent at H. Show that when the level is tilted through the angle A, the bub- ble moves through an arc HN which subtends the same angle A at the center 0. level be more sensitive if the radius R (= OH) is large or small ? 27. If two circles intersect, show that the obtuse angle between the two tan- gents to the two circles at either point of intersection is the same as the obtuse angle between the two radii drawn to that point. Hence show that the angle between the two tangents at one point of intersection is the same as that between the two tangents at the other point. 28. Prove that the sum of two opposite sides of a circum- scribed quadrilateral equals the sum of the other two sides. 29. Show how to construct the common tangents to the pairs of circles represented in (1), (2), (3), Ex. 15, p. 122. 30. Construct a rectangle having given a side and the diagonal Will the II, § 142] MISCELLANEOUS EXERCISES 125 31. Construct an isosceles triangle having given the base and the length of the altitude from one end of the base to one of the equal sides. 32. Construct a circle whose center lies on one side of a given triangle, and which is tangent to each of the other two sides. 1.33. Construct a triangle, having given the base and the altitudes to the other two sides. 34. Construct a triangle, 'having given the base, the altitude to the base, and a second altitude. 35. Construct a right triangle, from the suggestion contained in the figure, when the hypotenuse AB and the sum AD of the two sides are given. ^6. Show that the bisectors of opposite angles of a parallelo- gram are parallel or coincide. Hence show that the only kind of parallelograms in which a circle can be inscribed is a rhombus. 37. Show that the figure formed by two tangents PA and PB to a circle O and the radii OA and OB forms a kite which is in- scribed in a circle 0' whose diam- eter is OP. Let 00' meet the circle at (7 and D. Show that Z CO A is meas- ured by half the arc AP ; and Z CBA by one fourth the arc AP. 38. In the figure for Ex. 37, draw the radius O'A. Show that any change in Z CDA causes four times as much change in Z PO'A 39. Show what becomes of the figure of Ex. 37 if the angle COA is equal to 90°. CHAPTER III PROPORTION SIMILARITY PART I. GENERAL THEOREMS ON PROPORTION 143. Definitions. An equality of two ratios is called a proportion. Thus, the ratio 2/3 being equal to the ratio 4/6 gives us the proportion 2/3 = 4/6. More generally, if the ratio ajh is equal to the ratio c/d, then the equality ajh = c/d is a proportion. The form - = - is identical with a/& = c/d, since aJh is only a con- 6 d yenient way of printing a fraction. Besides writing a proportion in the form aJh = c/rf, either of the following forms may also be used, a : & = c : in., and CE IS 3* parallel in., Ans. If in. what is the length of EB? 2. If, in the figure of Ex. 1, we have CA = % ft., 05 = 12 ft. 3 in., and CE = ^ ft., what is the length of DA ? 3. A thread DE is stretched in any direction across a tri- angle ABC cutting two of its sides in the C points D, E. The thread is now moved y^\£. up and down over the surface of the tri- angle, but always in such a way as to be parallel to its original direction. How A B does the ratio CD/CE change during the motion ? Why ? 4. Prove that any line drawn par- allel to the base of a trapezoid cuts the other two sides proportionally. 5. If a common exterior tangent to two circles (see figures for Ex. 14, p. 121) meets the line of centers 00' in a point G, show that the lengths along this tangent from G to the two circles are in the ratio OG/O'G, 6. A channel for water has a cross-section of the form of a trapezoid ABCD. If BC and AD are horizontal, show that the wetted por- tions of the side walls BE B C and CF are proportional to the dry portions AE and DF. 7. What other proportions can you write down for Ex. 6 ? 8. In Ex. 6, if AB = 15 ft., CD = 12 ft., and the vertical height H from BC to AD is 10 ft., find the lengths of the dry portions of the side walls when the water is 4 ft. deep. Ill, § 148] PROPORTIONAL LINE-SEGMENTS 133 147. Theorem II. {Converse of Theorem I.) If a line divides two sides of a triangle proportionally, it is parallel to the third side. c Fig. 109 Given the A ABC and the line DE dividing the sides pro- portionally ; that is, such that CD/DA=CE/EB. To prove that BE is parallel to AB. Proof. Suppose that DE is not II AB. Then draw DE' II AB. We should now have (1) CA/CD = CB/CE'. § 146 Also, since by hypothesis CD/DA^CE/EB, it follows that (2) CA/ CD = CB/CE. Th. E, § 144 From equations (1) and (2) we should have CE' = CE ; Th. A, Cor. 1, § 144 whence DE' would coincide with DE. Post. 1 This is impossible unless DE II AB. Why ? Therefore, DE II AB. 148. Corollary 1. If a line cuts two sides of a triangle in such a ivay that either side is to one of its segments as the other side is to its corresponding segment, then the line is parallel to the third side [Hint. Show that under the given conditions the hypothesis of § 147 is satisfied. Compare with § 89 and § 146.] 134 PROPORTION SIMILARITY [III, § 149 149. Theorem III. The bisector of an angle of a triangle divides the opposite side into segments ivhich are proportional to the sides of the angle. Given the A ABC and the bi- sector CD of Z C To prove that AD/DB = AC/BG. Proof. Draw BE il CD and meet- ing AC prolonged at E. Then AD/DB = AC/ CE. Why? Moreover, Zx = Zyj Z x= Zw, Zy = Zz. Why ? Therefore, Zz = Zw; whence CE = BC ; Why ? hence AD/DB = AC/BC. Ax. 9 EXERCISES 1. In the triangle ABC of Fig. 110, CA = 11 in., CB=S in., and AB = 14 in. What are the lengths of the segments AD and DB into which the bisector CD of the angle C divides the base AB ? Ans. AD = 8^% in., DB = 5{i in. I 2. Prove by means of § 147 that the line joining the middle points of two sides of a triangle is parallel to the third side and equal to half of it. (Compare with § 89.) ! 3. State and prove the converse of Theorem III. [Hint. Prolong AC to E, making CE = BC. Draw EB. (Fig.llO.) Prove AC/AD= CE/DB, and hence CD WEB. Then Zx = Zw and Zy = Zz. But Zw=Zz (since CB = CE) . Therefore Zx=Zy.] 4. An ordinary closed rubber band is stretched out and held against a board by means of two pins A and B. One of the halves of the band is stretched by means of a pencil point into the form of an elastic triangle ABC, with the pencil point at C. How must the pencil point move in order that the ratio CA/CB shall remain constant ? Ill, §150] PROPORTIONAL LINE-SEGMENTS 135 150. Theorem IV. If a series of parallels be cut by two lines, the corresponding segments are proportional. — w \^ X • c/ \f \j \E \ Fig. Ill Given the parallels AH^ BI, CJ, etc., cut by the lines AD and HK. To prove that AB/HI= BG/IJ= CD/JK. Proof. Draw AE II HK. Now AC/AF = AB/AO = BC/GF Why ? and AC/AF=CD/FE. Why? Therefore AB/AG = BC/GF = CD/FE. Why ? But AG = HI, GF = IJ, FE = J/iT; Why ? hence ^5/iJ/ = BC/IJ = CD/ JiT. Ax. 9 Note. All work on proportional segments is based on §§ 145, 150. See §§ 149, 151, 153, etc. Compare also §§ 91-94. EXERCISES 1. Prove Theorem IV when the two lines AD, HK (Fig. Ill) intersect at some point between the parallels AH, DK. 2. Prove the fact stated in Cor. 1, § 92 by means of § 150. 3. If the three segments cut out of one transversal by four parallel lines are 2 in., 3 in., 4 in., respectively, what can be inferred concerning the perpendicular distances between the parallels ? If the distance between the first pair is 1 in., what are the distances between the others ? 136 PROPORTION SIMILARITY [III, § 151 151. Problem 1. To divide a given line into parts proportional to any number of given Ihies. 1 * A^.^ \ '. ^ .B ^ ^-^ \ \ \ , b d D" Fig. 112 Given the line AB. Required to divide AB into parts which shall be proportional to the given lengths a, b, c, d. Construction. Draw AC making any angle with AB. Measure off AG = a, GF= b, FE = c,En = d. Join the last point D thus determined with B, and through E, F, and G draw parallels to BD cutting AB in H, I, and J. Then the segments AJ, JI, IH, HB, divide AB as required. Proof. AJ/AG = JI/GF = IH/FE = HB/ED ; § 150 hence AJ/a = Jl/b = IH/c = HB/d. Note. This problem includes as a special case, the division of a given line into any number of equal parts ; for if we take a = 5 = c = c?, we have AG =GF = FE=^ ED. EXERCISES 1. Divide a line 10 in. long into 8 equal parts by Note, § 151. Show how to do the same exercise by § 8. 2. Divide a line 10 in. long into 7 equal parts by § 151. Can this exercise be done by § 8 ? 3. State when § 8 can be used in place of Note, § 151 to divide a given line into a number of equal parts. Ill, § 152] PROPORTIONAL LINE-SEGMENTS 137 4. How could the page of a notebook be ruled (accurately) into three equal columns ? 5. Divide a given line into parts which are to each other as 1:2:3. 6. Draw a line AB and divide it into fifths ; thereby find the point on AB one fifth the distance from A to B. Find a pomt 2/5 of AB from A. y 7. Given an angle ABC and a point P any- where within it. To construct a line through P such that the portion of the line lying within the angle shall be bisected by P. [Hint. Through P draw PD II BC. Then lay off DE=DB.] '^ 8. Through a point P without an angle ABG to draw PE so that PJ9 = Z>JS7. B 9. State and prove a theorem concerning the segments into which the non-parallel sides of a trapezoid are divided by a series of parallels to the bases. 10. Layers of rock in the earth are nearly parallel, but they are not usually level (horizontal). Show that the ratios of the thicknesses of layers of different kinds of rock can be found by measuring the thicknesses of the cuts made by a vertical mine-shaft, or a deep well. 152. Definitions. In the proportion a/& = c/r7, d is called the fourth proportional to a, 6, and c. Thus, 6 is the fourth pro- portional to 1, 2, and 3. If 6 = c, so that a/h = h/d, d is called the thiid proportional to a and b. 138 PROPORTION SIMILARITY [III, § 153 153. Problem 2. To find the fourth proportional to three given lines. a A a C b B , b , -, "^\^^. \ \ c ^^F Fig. 113 Given the three lines a, b, and c. Required to construct their fourth proportional ; that is, to find a line x such that we shall have a/b = c/x. Construction. Draw a line AB of indefinite extent and from A lay off ^(7 equal to a and CB equal to b. Draw a second line AF of indefinite extent, making any con- venient angle at A, and along AF lay off ^6r equal to c. Join CG. Draw BH II CG and cutting AF at //. Then GH is the required line x. [The proof is left to the student. Use § 145.] EXERCISES 1. Find a fourth proportional to three lines 2 in., 4 in., and 5 in. long, respectively. Find a fourth proportional to three lines 2 in., 5 in., and 4 in. long, respectively. 2. Show how to construct a third proportional to two given lines. 3. To enlarge a line segment I in the ratio 2/3 is to find a new line segment x such that 2/3 = l/x. Show how to do this geometrically. 4. Draw a triangle ; then enlarge its sides in the ratio o/5, using geometric constructions only. See § 21, p. 19. Ill, § 154] SIMILAR TRIANGLES 139 PART HI. SIMILAR TRIANGLES AND POLYGONS 154. Definitions. That a figure may be enlarged or reduced in size without essentially changing its appearance otherwise is a part of common knowledge. Thus we drew such enlarged and reduced figures in § 21. Throughout this book rather small figures are printed; these the student almost always enlarges in drawing on paper or at the board. In general, nearly all drawings represent objects of a size different from that of the drawing; thus house plans are usually drawn (§ 21) on the scale of -^ inch to one foot. Enlargements, or reductions, are made by increasing, or de- creasing, all lengths in a figure in the same ratio throughout, so that all lengths in the new drawing are proportional to the corresponding lengths in the original figure. Two triangles which have their corresponding sides propor- tional are called similar. Fig. 114 represents two similar triangles, each side of the one being twice that of the other. A Fig. 114 The fact that two triangles ABC and A'B'C are similar is expressed by the relation ABC ^ A'B'C In general, if one figure is obtained from another by enlarge- ment or reduction, that is, if all lengths that can be drawn in one are proportional to the corresponding lengths in the other, the two figures are said to be similar. Two triangles are said to be mutually equiangular when their corresponding angles are equal. lid PROPORTION Similarity [iii, § 155 1S5. Theoreiti V. If two triangles are mutually equiangular, they are similar. a'^ ^b' Fig. 115 Given the mutually equiangular A ABC, A'B'C To prove that A ABC ~ A A'B'C. Proof. On CA lay off CD = C'A' and on CB lay off CE = CB'. Draw DE. Then, in the A DEC and A'B'C we have CD = CA', CE = CB', Const, and ZC = ZC. Given Therefore A DEC^ A A'B'C. Why? Whence Z Oi)J5; = Z CM'5'. But ZA=Z C'A'B'. Given Hence Z (7i)jE; = Z A. Why ? It follows that DE II ^5 ; Why ? , 1 'CZ> CE CA' CB' „ . .^ hence also, — = — or -^ = -^- § 146 CA' A'B' In like manner it can be shown that CA AB Therefore AABC^AA'B'C. §154 156. Corollary 1. Two triangles are similar if two angles of the one are equal respectively to two angles of the other. ^7157. Corollary 2. Two right triangles are similar if an acute angle of the one is equal to an acute angle of the other. Ill, § 158] SIMILAR TRIANGLES 141 158. Theorem VI. If two triangles are similar, they are mutually equiangular, {Converse of Theorem F.) Fig. 116 Given the two similar A ABC and A'B'C ; that is, two A such that -^ = -^=-:^. C'A' c'b: a'b' To prove that A ABC and ABIC are mutually equiangular. Proof. On CA lay off CD= C'A', and on CB lay off CE = CB'. Join D and E. Then CA/CD=CB/CE. Therefore, DEW AB\ whence, Z CDE= Z .4 and Z Z)^0 = Z 5. Given § 147 Why? It follows that the A CDE and CAB are mutually equi- angular ; whence also therefore But whence and hence Therefore Why? §155 §154 Given Why? A CDE -- A CAB ; AB/DE= CA/CD = CA/C'A' CA/C'A' = AB/A'B'', AB/DE = AB/A'B', DE=A'B'. Th. A, Cor. 1, § 144 A CDE ^ A C'A'B'. Why ? All three of the A CAB, CDE, and C'A'B' are therefore mutually equiangular ; hence, in particular, the A ABC and ABIC are mutually equiangular. 142 . PROPORTION SIMILARITY [III, § 158 EXERCISES 1. Show that all equilateral triangles are similar. 2. Show that all isosceles triangles that have the same ver- tex angle are similar. 3. Prove that if each of two triangles are similar to a third, they are similar to each other. 4. How high is a house whose shadow is 144 ft. long when that of a gate post 5 ft. high is 12 ft. long ? Ans. 60 ft. 5. If a triangle be stretched out so that each side becomes one half longer than at the beginning, will the size of the angles be changed ? Why ? 6. In order to determine the length AB of a lake, set two stakes at convenient points, D and E, such that AD is perpen- dicular to AB, while DE is perpendicular to AD. Now, from stake E sight the point B and take note of the point C where the line of sight crosses AD. Show how to find AB by measuring ED, DC, and CA. 7. Show that if each of the sides of a triangle is reduced or enlarged in the same scale (§ 21), the angles are unchanged. Thus it is said that enlargement does not distort a figure. 8. A triangular tin plate expands in the same ratio along all straight lines when heated. Show that its shape is not changed. 9. Prove that two triangles that have their sides parallel or perpendicular each to each are similar. 10. Prove that the diagonals of any trapezoid divide each other proportionally. 11. Let P and Q be any two points on the sides of an angle AOB. Show that the triangles formed by perpendiculars from P and Q to the sides opposite them, respectively, are similar. Ill, § 159] SIMILAR TRIANGLES 143 159. Theorem VII. Tivo triangles are similar if an angle of the one equals an angle of the other and the in- cluding sides are proportional. Given the A ABC, A'B'C, such that Z C = ZC and CA/C'A' = CB/C'B'. To prove that A ABC ^ A A'B'C. Proof On CA lay off A'' ^B CD = CA\ and on CB lay Fig. 117 off CE = CB' Join D and E. Then A CDE ^ A CA'B'. Also we have CA/CD^GB/CE; and hence DE II AB. Therefore Z CDE = Z. A, and Z DEC = Z J5 ; whence A yl5(7 and Z>^(7 are mutually equiangular; and A ABC and A'B'C are mutually equiangular. Therefore A ^5(7 ~ A'B'C. §35 Given §147 Why? Why? Why? §155 EXERCISES 1. Compare the various conditions under which two triangles are similar with those under which two triangles are congruent. 2. If the angle at the vertex of a triangle is held fixed while the sides which include the same angle are stretched out by half their length, how do the angles in the resulting tri- angle compare with those in the original triangle? Why ? 3. Two rods AB and CD are hinged together at a point 0, the hinge being placed on each rod one o third the length of the rod from one end. ,^^ — .0 Show that AC = BD/2, whatever the a Z DOB may be. Show also that the ratio AD/CBdoes not change when Z Z>OB changes. liAB— CD. 144 PROPORTION SIMILARITY [in, § 160 160. Definition. If in the proportion a/h = c/d we have = 6, then h (or c) is said to be the mam proportional between a and d. Thus, 2 is the mean proportional between 1 and 4, since we have 1/2 = 2/4. 161. Theorem VIII. If, in any right triangle, a perpendic- ular is drawn from the vertex of the right angle to the hypote- nuse, the two right triangles thus formed are similar to each other and to the given triangle. Given the right-angled A ABO having its right angle at C. Given also the perpen- dicular CD drawn from C to the hypotenuse AB. To prove that A ADC, CDB, and ABC are similar. Proof. These A are mutually equiangular ; therefore they are similar. Fig. 118. Why? Why? 162. Corollary 1. In any right triangle the perpendicular from the vertex of the right angle to the hypotenuse is the mean proportional between the segments of the hypotenuse. Proof. The A ADC, CDB being similar, their corresponding sides are proportional. Upon comparing the sides AD and CD, in the one with those which correspond to them in the other, namely, the sides CD and DB, we obtain AD/ CD = CD/DB, which was to be proved. 163. Corollary 2. If, in any right triangle, a perpendicular is drawn from the vertex of the right angle to the hypotenuse, each side of the right triangle is the mean proportional between the hypotenuse and the segment adjacent to that side. Proof. The A ADC and CDB are each similar to A ABC. Hence AB/AC = AC/ AD and AB/CB = CB/DB. HI, § 164] SIMILAR TRIANGLES 145 164. Problem 3. To find the mean proportional he- tween two lines. Given the two lines a and h. To construct their mean propor- tional ; that is, to con- struct a line x such ' that a/x = xfh. ' • Construction. Draw a line of in- definite extent AB^ and from A lay- off AC=a. From the point C lay off CD = h (Fig. 119). On AD as a diameter draw a semicircle. At C erect CR ± AB, meeting the semicircle at E. Then CE is the required line x. Proof Draw AE and DE. Then /. AED is a right angle. Why ? Complete the proof by means of § 162. EXERCISES 1. Find the mean proportional between 9 and 4. Answer the same question for 1 and 3. What essential difference exists between the kinds of answers in the two cases ? (See § 128.) 2. The altitude drawn to the hypotenuse of a certain right- angled triangle divides the hypotenuse into segments which are of length 16 in. and 4 in., respectively. How long is the altitude ? Answer the same question when the segments are each 10 in. long ; also when they are respectively 15 in. and 5 in. 3. If a perpendicular is drawn from any point on a circle to a diameter, what relation exists between the perpendicular and the segments into which it divides the diameter ? (See § 133.) 4. If one of the sides of a right-angled triangle is three times the other, in what ratio does the perpendicular divide the hypotenuse ? Ans. 1 : 9 146 PROPORTION SIMILARITY [III, § 165 165. Definitions. Similar Polygons. Two polygons are said to be similar when each may be decomposed into the same number of triangles similar each to each and similarly placed. Thus, in the figure, the polygons ABCDE and A'B'C'D'E' are similar. For, if we draw the lines DA, DB, DAI, D'B', ABCDE is decom- posed into three triangles that are similar to the three triangles into which A'B'O'D'E' is decomposed. A regular polygon has been defined in § 96 as one which is both equilateral and equiangular. 166. Theorem IX. Regular ^polygons of the same mimber of sides are similar. Given the two poly- gons ABCDE AB'C'D'E'. and To prove that ABCDE ^ A!B'C'D'E'. Proof. Through D draw DA, DB, and through U draw D'A!, D'B', thus dividing each polygon into three triangles. Then the A DEA and D'E'A' are both isosceles, Why? and they have the equal vertex angles at E and E', Given Therefore A DEA - A D'E'A!. (Compare Ex. 2, p. 142.) Now prove that A DAB ~ A D'A'B' by showing that Z DAB = Z D'A'B' and that AD/A'D' = AB/A!B', To do this, first sliow that Z EAD = Z E'AHy, Ill, § 167] SIMILAR POLYGONS 147 167 . Theorem X . The perimeters of two similar poly- gons are to each other in the same ratio as any two cor- responding sides. Given the similar polygons ABCDE,A!B'C'D'E'. See Fig. 120 To prove that perimeter of ABODE ^ AB perimeter of ^'5'0'Zy^' A'B'' Proof. We have AB^BG^^CD^DE^ Given A'B' B'C CD' D'E'' Therefore AB + BC-^-CD + DE ^ AB , rp. jr . . 44 A'B' + B'G'+C'D' + D'E' A'B" ' '^ that is perimeter of ABODE _ AB ' perimeter of A'B' G'D'E' ~ A'B' ' EXERCISES 1. Give the proof of Theorem IX for a regular hexagon ; for a regular octagon. 2 In two similar polygons two corresponding sides are 15 and 20 in., respectively. If the perimeter of the first is 5 feet, what is the perimeter of the second ? Ayis. 6| ft. 3 The perimeters of two similar polygons are to each other as 5 . 8. A side of the first is 1 ft. What is the length of the corresponding side of the other ? 4. Prove that the perimeters of similar polygons are in the same ratio as any two corresponding diagonals of the polygons. 5. The perimeter of an equilateral triangle is 3 ft. Find the side of an equilateral triangle whose altitude is one half the altitude of the first triangle. Generalize your result into a theorem relating to similar triangles. 6. Prove that the perimeters of similar triangles are to each other in the same ratio as any two corresponding medians. 148 PROPORTION SIMILARITY [III, § 168 PART IV. PROPORTIONAL PROPERTIES OF CHORDS, SECANTS, AND TANGENTS 168. Theorem XI. If two chords intersect icitliin a circle, the product of the segments of the one is equal to the product of the segments of the other. Given the chords AC and BD intersect- ing at K. To prove that KA • KC = KB - KD. Proof. Draw AD and BG. In the A AKD and BKC we have ZD = AC, each being measured by AB/2. § 132 Likewise ZA= Z.B. Why ? Therefore AAKD^ABCK; Why? , KA KD ^^"^^ ^ = ^' whence KA • KO = KB • KD. Fig. 122 Why EXERCISES 1. A chord of a circle is divided into two segments of 5 in. and 3 in., respectively, by another chord one of whose segments is 4 in. How long is the second chord ? A^is. 7 j in. 2. The greatest distance to a chord 8 in. long from a point on its intercepted (minor) arc is 2 in. What is the diameter of the circle ? 3. P is a fixed point within a circle through which (point) a chord passes. As the chord swings round P as a pivot, what can be said of the segments of all new chords thus obtained ? 4. If two circles intersect and through any point in their common chord two other chords be drawn, one to each circle, prove that the product of the segments of one chord is equal to the product of the segments of the other. Til, § 170] CHORDS SECANTS TANGENTS 149 169. Theorem XII. If from a point without a circle a secant and a tangent are drawn, the tangent is the mean proportional between the entire secant and its exr terior segment. Fig. 123 To prove that AC /AD = AD/AB. Outline of proof. Draw BD and DC. Then Z.A= ^A, Hen. and /.C = Z. ADB, since each is measured by ^ DB. Why ? Therefore A ACD ~ A ABD, § 155 and hence AC/ AD = AD/AB. Why ? Note. If in Fig. 123, AC is allowed to swing around A as a pivot, the tangent AD meanwhile remaining fixed, we have throughout the motion, AC • AB = AD . This, in fact, is what Theorem XII says. But, since AD remains fixed, this means that the product AG • AB remians constant throughout the motion. This fact may be stated as follows: 170. Theorem XIII. If from a fixed point without a circle any two secants are drawn, the j^f^oduct of one secant and its external segment is equal to the product of the other secant and its ex- ternal segment. Fig. 124 150 PROPORTION SIMILARITY [III, § 170 EXERCISES 1. A secant is drawn from a point without a circle in such a way that its whole length is 9 in., while the part cut off within the circle is 4 in. What is the length of the tangent to the circle from the same point ? Ans. 3 V5 in. 2. Two secants are drawn from the same point without a circle. One is 11 in. long and its external segment is 4 in. The external segment of the second is 2 in. What is the length of the second secant ? 3. One of two secants meeting without a circle is 18 in. long, while its external segment is 4 in. The other secant is divided into equal parts by the circumference. Eind the length of the second secant. 4. A point P is 12 in. from the center of a circle whose radius is 7 in. A secant is drawn from P. Find the product of the entire secant and its external segment. Pj 5. Prove that tangents drawn to two intersecting circles from a point on their common chord are equal. 6. Given two non-intersecting circles and 0', draw any circle ABCD intersecting both of them. Draw the two common chords AB and CD, and extend them to meet at P. Show that tangents drawn from P to the two circles and 0' are equal. 7. Show that the diameter of a circular porch column can be found by measuring certain lines entirely outside the column. Can this still be done if over half the face of the column is buried in cement ? Ill, § 172] CHORDS SECANTS TANGENTS 151 171. Definition. A line segment is said to be divided in extreme and mean ratio when one of its segments is a mean pro- portional between the whole line and ^ the other segment. In Fig. 125, C is ^' ' '^ so located that AB/AC = AC/CB, so ^^<^- ^^ that AB is divided by G in extreme and mean ratio. Note. The division of a line in extreme and mean ratio has been called the golden section or division in golden mean. It •was observed by the ancients that artistic effects frequently result from its use. Thus, a rectangular picture frame will usually give the best effect if its length and width are in the ratio just described; that is, in the ratio of AC to CB in Fig. 125. A similar remark applies to the height of the back of a chair as compared with the length of its legs. 172. Problem 4. To divide a given line segment in extreme and mean ratio. \ \d Given the line A B. i r -''' > Required to divide AB in extreme \l ^^.-^'^' and mean ratio; that is, to deter- l-V mine a point F such that AB/AF = ^^^"^ \ ^ v^ y AF/FB. A F Construction. At BdTAwBC±AB Fig. 126 and equal to half of AB. With C as center and CB as radius draw a circle. Draw AC meeting the circle at E and D. On AB take AF =: AE. Then F is the point of division required. Proof. AD/AB = AB/AE ; Why ? hence 4P^^^ = ^B-A^ . why? AB AE ^ But AB = ED, and AF=AE', Why ? therefore AF/AB = FB/AF, or AB/AF = AF/FB. Why ? 152 PROPORTION SIMILARITY [HI, § 173 PART V. SIMILAR RIGHT TRIANGLES. TRIGONOMETRIC RATIOS 173. Similar Right Triangles. The importance of a special study of similar right triangles results from the simplicity and usefulness of the result, § 157, that two right triangles are similar if an acute angle of one is equal to an acute angle of the other. This proposition may now be restated as follows : 174. Theorem XIV. If two right triangles have an acute angle of one equal to an acute angle of the other, their corre- sponding sides are in the same ratios. B Fig. 127 Given the rt. A ABC and A'B'C, with ZA = ZA\ To prove that their corresponding sides are in the same ratios. Proof. A ABC ~ A A'B'C^ ; § 157 hence the corresponding sides are in the same ratios ; BC^B^, Aa^A;C^^ BC^B^ J. .... AB A'B" AB A'B" AC A'C' that is : 175. Corollary 1. If an acute angle of a right triangle is known, the ratios of the sides are all determined, 176. Corollary 2. If the ratio of any pair of sides of a right triangle is given, the acute angles are determined. Note. The study of the use of these ratios is called Trigonometry. By their use, any part of a right triangle can be found, if any two parts, not both angles, are given, besides the right angle. This process is called solving the triangle. Ill, § 178] TRIGONOMETRIC RATIOS 153 177. Definitions. Let Z A, Fig. 128, be a known acute angle of a right triangle, in which Z C is a right angle ; and let us denote the sides opposite ZA,ZBf Z C, by a, h, c, respectively. Then the ratios a/b, a/c, b/c are all determined, by § 175. These ratios are named as follows : (1) a/c is called the sine of the angled; Fig. 128 (2) b/c is called the cosine of the angle A ; (3) a/b is called the tangent of the angle A. That is, for an acute angle of a right triangle : (1) the sine of the angle = side opposite it -=- hypotenuse ; (2) the cosine of the angle = side adjacent it -f- hypotenuse ; (3) the tangent of the angle = side opposite -;- side adjacent. These ratios are called the trigonometric ratios. 178. Values of the Ratios. Table. The values of the three ratios mentioned in § 177 can be obtained, approximately, from a good figure, for any acute angle A. These values are tabulated on p. 155, to three places of deci- mals, for every angle from 0° to 90°, at intervals of 1°. The student may check the accuracy of any entry in this table by drawing an accurate figure with a protractor, actually measur- ing the sides of it, and then calculating their ratios. Example. The length of the shadow of a tree CB is 44 ft. when the angle CAB between the shadow CA and the line AB is 31°. Find the height of the tree. [Solution. Since BC/AC is the tangent of 31", by § 177, we look in the table, in the col- umn headed tangent, and opposite 31°. This gives the value .601 ; hence BC/AC= Ml ; but ^O = 44 ft. ; it follows that ^**'^44 ft. BC = AC X .601 = 44 X .601 = 26.444 (ft.).] 154 PROPORTION SIMILARITY [III, § 178 EXERCISES 1. How large, in degrees, is an acute angle whose tangent isl? ^B 2. The shadow of a tree is 26 ft. long when the angle of elevation of the sun (Z CAB in the figure) is 45°. How tall is the tree ? 3. One side of a right triangle is 2 in. ; the adjacent angle is 42° ; determine the re- ^ ^** maining side and the hypotenuse, and check by measurement from an accurate figure. Ans. side = 1.8 in. ; hyp. = 2.69 in. ^'4. One side of a right triangle is 2 in. and the opposite angle is 42°; determine the remaining side and hypotenuse. Check by measurement in a figure. "^5. The hypotenuse of a right triangle is 28 in. ; one angle is 32°. Determine the two perpendicular sides. Check. ^'. To determine the height of a tree OA standing in a level field the distance OB = 100 ft. from the base of the tree to a point B in the field is measured, and the angle OBA is then found to be 37°. Find approximately the height by measure- ment in a reduced figure, and by the table, p. 155. 7. Measure two adjacent edges of your study table. Find the angles that the diagonal makes with the edges, (a) by drawing an accurate figure and measuring the angle with a protractor ; (b) by use of the table (p. 155). 8. The tread of a step on a certain stairway is 10 in. wide ; the step rises 7 in. above the next lower step. Find the angle at which the stairway rises, (a) by a protractor from an accurate figure ; (6) from the table, p. 155. Ans. 35° (nearly). 9. What angle does a rafter make with the plate beam (Fig., Ex. 2, p. 43), if the roof is "half-pitch " ; if the pitch of the roof is 1/3. Ans. 45°; nearly 34°. [The pUch of a roof is the rise divided by the entire span.] Ill, § 178] TRIGONOMETRIC RATIOS 155 [The abbreviation hyp means hypotenuse ; ad) means the aide adjacent to the angle ; opp means the aide oppoitite the angle. See also the larger table, p. ili.] Anglb Sine C08INK Tangent Angle Sink Cosine Tangent (opp/hyp) (adj/hyp) (opp/adj) (opp/hyp) (adj/hyp) (opp/adj) 0° .000 1.000 .000 45° .707 .707 1.000 1° .017 1.000 .017 46° .719 .695 1.036 2° .035 .999 .035 47° .731 .682 1.072 3° .a')2 .999 .052 48° .743 .()()9 1.111 40 .070 .998 .070 49° .755 .65(3 1.150 6° .087 .9^)6 .087 50° .766 .643 1.192 ()° .105 .9^)5 .105 51° .777 .629 1.235 7° .122 .993 .123 52° .788 .616 1.280 8° .139 .990 .141 53° .799 Am 1.327 i>° .\m .988 .158 54° .809 .588 1.37(5 10^ .174 .985 .176 56° .819 .574 1.428 11° .191 .982 .194 56° .829 .559 1483 12"^ .208 .978 .213 57° .839 .545 1.540 13° .225 .974 .231 58° .848 .530 1.(500 14° .242 .970 .249 59° .857 .515 1.(5(54 15° .259 .966 .268 60° .86(5 .500 1.732 16° .276 .961 .287 61° .875 .485 1.804 17° .292 956 .306 62° .883 .469 1.881 18° .309 .951 .325 ()3° .891 .454 l.i)63 19° .326 .946 .^4 64° .899 .438 2.050 20° .342 .940 .3<>4 65° .906 .423 2.145 21° .358 .934 .384 ()6° .914 .407 2.246 22° .375 .927 .404 67° .921 .391 2.356 23° .391 .921 .424 68° .927 .375 2.475 24° .407 .914 .445 69° .9M .358 2.605 26° .423 .906 .466 70° .910 .342 2.747 26° .4:38 .899 .488 71° .94f) .326 2.iX)4 27° .454 .891 .510 72° .951 .309 3.078 28° .469 .883 .532 73° .956 .2«)2 3.271 29° .485 .875 .554 74° .961 .276 3.487 80° .500 .86() .577 75° .9(;<) .259 3.732 31° .515 .857 .601 76° .970 .242 4.011 32° .530 .848 .()25 77° .974 .225 4.331 33° .545 .839 .649 78° .978 .208 4.705 34° .559 .8'J9 .675 79° .982 .191 5.145 35° .574 .819 .700 80° .985 .174 5.671 36° .588 .809 .727 81° .988 .15(5 6.314 37° .602 .7i)9 .754 82° .990 .139 7.115 38° .616 .788 .781 8:i° .«H>3 .122 8.144 39° .629 .777 .810 84° .995 .105 9.514 40° .643 .766 .839 85° .996 .087 11.430 41° .65() .755 .s(;9 8()° .'.>98 .070 i4..'m 42° .6<)9 .743 .900 87° .\m .052 19.081 43° .682 .731 .933 88° .999 .oa^ 28(5:36 44° .695 .719 .966 89^ 1.000 .017 57.290 45° .707 .707 1.000 90° 1.000 .000 156 PROPORTION SIMILARITY [in, § 179 179. Theorem XV. Corresponding altitudes divide any tivo similar triangles into two corresponding pairs of similar right triangles. Fig. 129 Given the two similar triangles ^5(7 and A'B'C', and given the corresponding altitudes BD and B'D'- To prove that A ABD ^ A A'B'D', and A DCB ~ A D'CB'. Outline of proof. Show that A ABD ~ A A'B'D' by showing that Z.A = ZA'. (§ § 158, 174.) 180. Corollary 1. Any two similar polygons may be sub- divided into correspondiyig pairs of similar right triangles. Note. Division of a triangle into right triangles is often useful. Example. In Fig. 129, suppose c = 10 in., /.A = 30°, Z.B = 85°. Find Z C, and sides a and b. Solution : First find Z (7=65°. Then /i=cx cosine of 30°=10x^ = 5. Hence a = h ^ sine of O = 5 -^ .906 = 5.52. Again AD = c x cosine of 30^=8.66 ; and DC=a X cosine of 65°=2.33 ; hence b = AD -^ DC =10.99. EXERCISES 1. The base of a certain isosceles triangle is 10 in., and the angle at the vertex is 40°. Find the size of one of the equal angles ; find the length of one of the equal sides (a) by measure- ment, (b) using the table, p. 155. 2. The diagonal of a certain rectangle is 4 ft. One side is 2 ft. Find the angle the diagonal makes with that side, (a) by measurement, (b) using the table, p. 155. Ill, §180] MISCELLANEOUS EXERCISES 157 MISCELLANEOUS EXERCISES. CHAPTER III [The Exercises that involve the iise of the trigonometric ratios are starred, * ] ^1. A building casts a shadow 64 ft. long. A lamp-post 8 ft. high at one corner of the building casts at the same time a shadow 9 ft. long. How high is the building ? 2. Show how to find three fifths of a given line ; five sevenths. " 3. A circle is inscribed in an isosceles triangle. Prove that the triangle formed by joining the points of contact is also isosceles. 4. If each of two polygons is similar to a third polygon, they are similar to each other. Prove this statement. "^5. Prove that in an inscribed quadrilateral the product of the segments of one diagonal is equal to the product of the seg- ments of the other. 6. In the adjoining figure, AB rep- resents a vertical pole, and CD a ver- tical stake, so that D and B are in the same line of sight from a point E on the level ground ECA. What measurements must be taken to find ^^? 7. If the non-parallel sides of a trapezoid are extended to meet in a point P, the lengths of the extensions are propor- tional to the lengths of the original sides. State a proof. 8. If in any triangle, a line is drawn parallel to the base, any line through the vertex divides the base and the parallel into segments that are in the same ratio. State a proof. 9. Show that any two altitudes h and h' of a triangle are inversely proportional to the sides a and a' to which they are perpendicular ; that is, h/h' = a' /a. 158 PROPORTION SIMILARITY [III, § 180 10. Let AE and BD be the tangents at the ends of a diam- eter AB of a circle 0. Draw any line through A, and suppose it cuts the circle at C and meets BD at D. Let E be the point of intersection of AE and BC pro- duced. Show that three similar right tri- angles are formed; and show that AB is a mean proportional between AE and BD. 11. Construct a fourth proportional to three lines, 3 in., 5 in., and 1 in. long, respectively. Show that the resulting line must be | in. long. 12. In general, construct a fourth proportional to two given line segments a and b and the unit line segment. Show that the resulting line has a length b/a in terms of the unit. [Geo- metric construction for division.'] 13. Show how to construct a circle through two given points and tangent to a given line. D c [Hint. Draw a line through the given points A and 5, meeting the given line at C. Then find a mean proportional CD between AC and BC. Two circles are possible according as CD is laid off from C in one direction or the other along the given line.] 14. Show how to constuct a circle tangent to two given lines and pass- ing through a given point P. [Hint. Draw BD bisecting Z ABC. Draw through the given point P a line PD ± BD and extend it to E^ making DE = PD. Now apply Ex. 13 to draw a circle through P and E and tangent to AB. How many solutions are possible ?] 15. Show how to inscribe in a given circle a triangle similar to a given triangle. 16. Show how to circumscribe about a given circle a triangle similar to a given triangle. Ill, § 180] MISCELLANEOUS EXERCISES 159 17. Construct a fourth proportional to three lines, 1 in., 2.5 in., and 3.5 in. long. Show that the resulting line must be 2.5 X 3.5 in. long. 18. Give a geometric construction for multiplying any two numbers, [Hint. It x — ah, then 1/a = 6/ic.] 19. Give a geometnc construction for enlarging any line segment in the ratio 5 : 7. Use this construction to draw a tri- angle similar to a given triangle, with its sides enlarged 5 : 7. 20. Show how to enlarge (or reduce) any line segment in the ratio of two given numbers (or two given line segments). 21. Show how to construct, on a given line segment as one side, a polygon similar to a given polygon. 22. Show that if two circles are tangent externally, any line through their point of tangency forms chords of the two circles that are proportional to their radii. State and prove the anal- ogous theorem when the circles are tangent internally. 23. In order to find the distance between two islands A and 5 in a lake, what distances and angles must be measured in the ad- joining figure ? Compare Ex. 6, p. 142. Show that the figure can be extended, so as to find AB by measur- ing distances only. ^4. If a line is drawn parallel to the parallel bases of a trapezoid, show that the segment cut off on it between one side and one diagonal is equal to that cut off by the other side and the other diagonal. 25. The bases of a trapezoid are 10 in. and 15 in. long, respectively, and the altitude is 8 in. Find the altitude of the triangle formed on the smaller base by extending the non- parallel sides until they meet. Ans. 16 in. . 160 PROPORTION SIMILARITY [III, § 180 26. Construct a mean proportional to two lines 3 in. and 4 in. long, respectively. Construct a mean proportional to two lines, each of which is 2 in. long. 27. Construct a mean proportional to two lines 1 in. and 5 in. long, respectively. How long is it ? v^8. Show how to construct the square root of any given number n by finding the mean proportional between a line of unit length and a line n units long. \_OeometriG construction for square rootJ] 29. The total length of a certain secant drawn from a point P to a circle is 10 in. ; its external segment is 4 in. Find the length of the tangent drawn from P. 30. Taking the radius B of the earth as 4000 mi., find how far from a lighthouse 150 ft. high the light is visible. A closer value of R is 3963 mi. ; is the answer changed seriously by using this more accurate value of E? w =* 31. If one side of a right triangle is double the other, in what ratio is the hypotenuse divided by the altitude drawn to it ? What are the angles of the triangle ? \/32. If one chord of a circle is bisected by another, show that either segment of the first is a mean proportional between the segments of the other. 33. The greatest distance from a chord 10 in. long to its intercepted arc is 3 in. Find the radius of the circle. 34. A curved pane of glass to fit a window in a round tower is bent in the arc of a circle. If the width of the frame is 30 in. and if the greatest distance from the glass to a horizontal line joining its edges is 1.5 in., find the radius of the arc. * 35. The radius of a circle is 7 ft. What angle will a chord of the circle 11 ft. long subtend at the center? Check by measurement in a reduced figure. 36. A railroad curve is to have a radius of 250 ft. What is the greatest distance from the track to a chord 100 ft. long ? Ill, § 180] MISCELLANEOUS EXERCISES 161 37. A railroad curve turns in the arc of a circle ; the greatest distance from the track to a chord 100 ft. long is 7 ft. ; find the radius of the arc. Aiis. 182+ ft- * 38. The width of the gable of a house is 34 ft. The height of the house above the eaves is 15 ft. ; find the length of the rafters and the angle of inclination of the roof. Find the pitch of the roof. See Ex. 2, p. 43, and Ex. 9, p. 164. * 39. To find the distance across a lake between two points A and B, a surveyor measured off 80 ft. on a line AC perpen- dicular to AB ; he then found Z ACB=46°. Find AB. * 40. A kite string is 250 ft. long and makes an angle of 40° with the level ground. Find (approximately) the height of the kite above the ground, neglecting the sag in the string. * 41. The shadow of a vertical 10-foot pole is 14 ft. long. What is the angle of elevation of the sun ? Ans. About 35.5°. * 42. A chord of a circle is 21.5 ft. long, and the angle which it subtends at the center is 41°. Find the radius of the circle. V * 43. The base of an isosceles triangle is 324 ft., the angle at the vertex is 64° 40'. Find the equal sides and the altitude. * 44. The base of an isosceles triangle is 245.5 and each of the base angles is 68° 22'. Find the equal sides and the altitude. V* 45. The altitude of an isosceles triangle is 32.2 and each of the base angles is 32° 42' ; find the sides of the triangle. f 46. Find the length of a side of an equilateral triangle cir- cumscribed about a circle of radius 15 in. * 47. Show that the sine of the angle A in Fig. 129 is h -^ c, and that the sine of C is ^ -f- a. Hence show that the sines of the angles A and C are proportional to the sides a and c opposite them. [Sine Law.] CHAPTER lY AREAS OF POLYGONS. PYTHAGOREAN THEOREM 181. Area of a Rectangle. The fundamental principle, men- tioned in the Introduction (§ 25), that the area of a rectangle is equal to the product of its base by its height, will be presupposed in what follows in the present chapter. The principle states that if the base and the height of a rectangle are, respectively, a and b, then its area (in terms of the correspond- ing square unit) is a • b. ' The following corollaries result from this principle : 182. Corollary 1. The area of a square is equal to the square of its side. 183. Corollary 2. The areas of two rectangles are to each other as the products of their bases and altitudes. 184. Corollary 3. Two rectangles that have equal altitudes are to each other as their bases; two rectangles that have equal bases are to each other as their altitudes. 185. Definition. Whenever two geometric figures have the same area they are said to be equal in area, or equivalent. The equality in area of two figures is denoted by the symbol =. Thus, the equation AABC = AA'B'C means that the two triangles have equal areas. EXERCISES 1. How many tiles each 8 in. square will it take to tile a floor 30 ft. long and 18 ft. wide ? 2. On a sheet of squared paper ruled in tenths of an inch (see p. 23), how many small squares are there in one square inch? 162 IV, § 185] RECTANGLES 163 3. Compare the areas of two rectangles whose altitudes are equal but whose bases are respectively 10 in. and 7 in. 4. The area of a rectangle is 400 sq. ft. and its altitude is 20 ft. What is the alti- tude of another rectangle having the same base but whose area is 300 sq. ft ? 5. The accompanying figure represents the cross section of a steel beam. The dimen- 192, ^ = 8. Find sions in millimeters are : 6=96, ^(;=12, h the area of the cross section. 6. Show that the adjoining figure il- lustrates the algebraic identity a (b + c) = ah + ac. t 7. Show that the following figures illustrate the algebraic identities (1) (a-\-b)(a-h) = a'-b'', (2) (a -i- by = a^ -{- 2 ab -{- b^ • (3) (a-by = a^-2ab-^b\ F E D o 1 A < K C < 'B^ ^-' L_.h J^a.b» I h^ -ab (a-b)' m (1) (2) (3) 8. Draw a figure to illustrate each of the identities (1) a{})-&)=^ab-ac\ (2) {x-\-a){x ^^^x" ■{-{a-\-b)x-^ab. 9. Draw a rectangle 5 in. long by 6 in. wide. Show that either diagonal divides the rectangle into two congruent right triangles. Find the area of each of these triangles. Ana. 15 sq. in. 10. Find the area of a right triangle whose two sides are 4 in. and 7 in. long, respectively. 164 AREAS [IV, § 186 186. Theorem I. The area of a parallelogram is equal to the product of its base by its altitude. Given the parallelogram ABCD with h its base and a its altitude. To prove that the area of the parallelogram ABCD is equal to a 'h. Proof. Draw AF 1. CD prolonged. Then ABGF is a rec- tangle having the base h and the altitude a. In the right A AFD and BGC we have AF = BG and AD = BC. Why? Therefore A AFD ^ A BGC. Why ? Taking away A AFD from the figure ABCF. the parallel- ogram ABCD remains. Taking away A BGC from the same figure, the rectangle ABGF remains. Therefore O ABCD = rectangle ABGF. But rectangle ABGF=ab, § 181 whence, O ABCD = ah. 1^187. Corollary 1. (a) Two parallelograms are to each other as the prodticts of their bases and altitudes. (b) Two parallellograms that have equal bases and equal alti- tudes are equal in area. I 188. Corollary 2. Two parallelograms that have equal alti- tudes are to each other as their bases; two parallelograms that have equal bases are to each other as their altitudes. IV, § 188J PARALLELOGRAMS 165 EXERCISES 1. In a certain parallelogram the acute angle included be- tween the sides is 30°. If the base and altitude are respec- tively 14 in. and 10 in., what is the area ? Answer the same question in case the included angle is 60°. 2. In the accompanying figure are a number of par- allelograms each hfeving the same base and altitude. Compare their areas. Does the area of the parallelo- gram depend upon the angles included by its sides ? 3. Prove that the lines joining the middle points of the opposite sides of a parallelogram divide it into four parallel- ograms that are equal in area. 4. Construct a parallelogram double a given parallelogram and equiangular to it. How many solutions are possible ? 5. Construct a parallelogram double a given parallelogram and having one of its angles equal to a given angle. 6. What is the locus of the intersection of the diagonals of a parallelogram whose base is fixed and whose area is constant ? 7. ABCD is a jointed parallelogram frame, that is, it con- sists of four pieces of stiff material hinged at each of the points A, B, C, and X), and such that side AD = side 5(7 and side AB = side DC. If the base AB is held fixed while DC is raised and lowered into vari- ous positions, will the areas of the various parallelograms be changing ? If so, what will be the greatest area obtainable and what the least, provided that AB = 6 in. and BC = 4: in. ? 166 AREAS [rV, §189 189. Theorem II. The. area of a triangle is equal to one half the product of its base and its altitude. Given the triangle ABC, having the base b and the altitude h. To prove that the area of the triangle ABC is equal to ^ hb. Proof. Construct the O ABCD. Then LJABCD = hb. §186 But A ABC = \n ABCD. § 82 Tl^refore AABG = ^hb. "iSO. Corollary 1 . (a) Tn-o triangles are to each other as tJie •products of their bases and altitudes. (b) Tico triangles that have equal bases are to each other as their altitudes. (c) Two triangles that have equal altitudes are to each other as their bases. (d) Tvjo triangles that have equal bases and, equal altitudes are equal in area. EXERCISES 1. What is the area of a right triangle whose sides are respectively 3 in. and 5 in. ? 2. Draw several triangles having equal bases and equal altitudes, as an illustration of § 190 (d). Are such triangles necessarily congruent ? 3. Compare two triangles with equal altitudes if the base of the tirst is two thirds that of the second IV, § 190] TRIANGLES 167 ^A. Show that any median of a triangle divides it into two equal triangles. Is the same true of any altitude ? Give reason. *^. What is the locus of the vertices of all triangles having a common base and the same area ? 6. An ordinary elastic rubber band is stretched out and placed around two pins A and B which are stuck into a board. The pins are placed at the extrem- ities of a diameter of a circle, as indicated in the figure. One of the halves of the band is now thrust aside by means of a pencil point so that the band becomes stretched out ^ into the form of an elastic triangle having one of its vertices on the circle at C. If the pencil be now moved about on the circumference, the band meanwhile slipping over the point, va- rious triangles are formed, such as those indicated by dotted lines in the figure. Do these triangles all have the same area ? If not, what is the greatest area obtainable for any triangle, and what the least, provided that the distance between the pins is 6 in. ? n. Show that the diagonals of a parallelogram divide it into four equal triangles. 8. If a line is drawn from the vertex of a triangle to any point P in the base, show that the areas of the two triangles formed are to each other as the segments of the base made by P. 9. Prove that if the middle points of two sides of a tri- angle are joined, a triangle is formed whose area is one fourth the area of the given triangle. 10. Show that the area of a rhombus is equal to one half the product of its diagonals. 11. Prove that the area of an isosceles right triangle is equal to one fourth of the area of the square erected upon its hypotenuse. 168 AREAS [IV, § 191 191. Theorem III. The area of a trapezoid is equal to the product of its altitude and one half the sum of its bases. D Fig. 132 Outline of Proof. Draw the diagonal BD. Then A ABD = ah/2 and A BCD = bh/2 ; hence ABCD = ^ + M = //^ ^ 2 2 I 2 Why 192. Corollary 1. TTie area of a trapezoid is equal to the prod- uct of its altitude and the line joining the mid-points of the non- parallel sides. D .C Fig. 133 [Hint. To prove area of ABCD = h • EF. It may be easily shown that EF = (AB + DC)/2. See § 89.] Note. The median of a trapezoid is a straight line that joins the middle points of the non-parallel sides. Thus, in Fig. 133, EF is the median of the trapezoid ABCD. The corollary of § 192 is frequently stated in the following form : The area of a trapezoid is equal to the product of its alti- tude and its median. IV, § 193] TRAPEZOIDS 169 — r 28-: EXERCISES 1. Find the area of the trapezoid whose bases are 6 in. and 4 in. respectively and whose altitude is 3 in. Ans. 15 sq. in. 2. Find the area of a trapezoid whose median is 8 in. and whose altitude is 6 in. i30' 3. An excavation for a railway track is 28 ft. deep, 130 ft. wide at the top, and 90 ft. wide at the so' bottom. What is the area of its cross section ? 193. Theorem IV. Tivo triangles that have an acute angle of the one equal to an acute angle of the other are to each other as the products of the sides including the equal angles. Given the A ABC and A'B'C having the Z C common. A ABC AC'BC To prove that Proof. Drawls', the same altitude /i, A A'B'C A'CB'G Then, since the triangles ABC, ^B'Ohave and likewise Multiplying, we obtain A ABC AAB'C' AABfC A A'B'C A ABC BG BCf AC A'C ^ AC'BC A A'B'C A'C'BG § 190, (c) 170 AREAS [IV, § 19d 194. Theorem V. Similar triangles are to each other as the squares of any two corresponding sides. Fig. 135 Given the similar A ABC and A'B'C. To prove that A ABC AB^ A A'B'C A'W Proof. AA=Z.A. Therefore A ABC AC-AB A A'B'C A'C'A'B' But AC AB A'C A'B' Therefore A ABC AB'AB AW A A'B' C A'B' . A'B' a'B'^ Why? § 193 Why? 195. Corollary 1. The areas of two similar polygons are to each other as the squares of any tivo corresponding sides. [Hint Divide the polygons up into the same number of similar trian- gles, as in Fig. 120, § 165, then apply Theorem V, together with § 144, Theorem H.] Note. Since any two corresponding lines in two similar figures are proportional to two corresponding sides, it follows that the areas of any tivo similar polygons are to each other as the squares of any two correspoyiding lines. Compare, in par ticular, Exs. 4 and 6, p. 171. IV, § 195] SIMILAR TRIANGLES 171 EXERCISES 1. Compare the areas of two similar triangles whose corre- sponding sides are in the ratio 3 : 4. Ans. 9 : 16. 2. Draw on heavy cardboard two triangles whose sides are in the ratio 1 : 2. Cut these triangles out and weigh each of them. Show that their weights should be in the ratio 1 : 4. 3. The areas of two similar triangles are 100 square feet and 64 square feet respectively. Compare the lengths of their cor- responding sides. Answer the same question when the given areas are respectively 31 square feet and 17 square feet. What distinction is to be made between the two cases ? 4. Prove that the areas of two similar triangles are to each other as the squares of any two corresponding altitudes. 5. One side of a polygon measures 8 feet and its area is 120 square feet. The corresponding side of a certain similar polygon measures 20 feet. What is the area of the second polygon ? / 6. Prove that the areas of any two similar polygons are to each other as the squares of any two corresponding diagonals. 7. If one squai-e is double another, what is the ratio of their sides? ^ / 8. In the figure, BE is per- pendicular to OD while BF is perpendicular to AD. Prove that AB 'BE = A1)' BF. 9. In what ratio must the altitude of a triangle be divided by a line drawn parallel to the base in order that the area of the triangle may be divided into two equal parts ? A71S. 1 : ( V2 - 1). [Hint. Call h llie altitude of the given triangle and let xbe the altitude of the small triangle cut off by the parallel to the base. Form an equation between h and x, solve for x and then form x f {h — x).] 172 AREAS fIV, § 196 196. Theorem VI. The Pythagorean Theorem. The square on the hypotenuse of a right triangle is equivalent to the sum of the squares on the two sides. Given the rt. A ABC having AB as its hypotenuse. To prove that AB' = AC + BC\ First Proof. Draw CPl. AB and prolong it to meet FHat G. Draw CF and BE. Since A ACD and ACB are rt. A, DGB is a straight line. Similarly, AC J is a straight line. In A ABE and AFC, AF = AB and AE = AC. Why ? Also Z BAE = Z CAF, since each consists of a right angle plus the angle BAC. Therefore A ABE ^ A AFC. Why ? But the rectangle AFGP = 2 • A ^FC, since both have the same base AF and the same altitude FO (or AP). Likewise, the square ACDE = 2 • A ABE, since each has the same base AE and a common altitude AC. Therefore rectangle AFGP == square ACDE. Ax. 9 Show similarly that rectangle BPGH = square CBIJ. Therefore, AFGP + BPGH = ACDE + CBIJ. Ax. 1 That is. AB' =A& + BCf. IV, § 197] PYTHAGOREAN THEOREM 173 Second Proof. By § 163, Aff = AB ' AP; and CB" = AB-PB; hence AC^ + Cl? = AB{AP + PB) = A&. Note 1. This second proof is the same in principle as the first, for AC — square ACDE, and AB • AP = rectangle AFGP and AB- PB = rectangle PGHB. Eachjroof consists essentially in showing that Off = AB • PB and AG^ = AB - AP. The second proof might have been given in connection with §163. Note 2. It should be observed that Theorem VI, though re- lating specifically to areasy furnishes at once a rule for finding the length of the hypotenuse of any right triangle when the lengths of its sides are known. Thus, if a and b are the sides, the hypotenuse h will be determined by the formula h= ^a?-\-h'^. Similarly, the theorem furnishes a rule for finding either side of a right triangle when the hypotenuse and the other side are known, the formula then being a = VA^ — b\ These two formu- las are of great value in mathematics. Note 3. Aside from its scientific value. Proposition VI is of great interest historically. Though its origin is not known exactly, it is supposed to have been first proved by the Greek mathematician Pythagoras, who died about 500 B.C. Pythag- oras in his later life settled in Italy and was identified with a group of mathematicians and philosophers known as the Pythagorean School. The school itself was disrupted after about 200 years, owing to political disturbances ; but its influ- ence continued a strong factor in the study apd development of mathematics. 197. Corollary 1. TTie square on either side of a right triangle is equivalent to the square on the hypotenuse diminished by the square on the other side. 174 AREAS [IV, § 197 EXERCISES 1. A baseball diamond is a square 90 ft. on a side. What is the distance from home plate to second base? (Extract square root correct to two decimal places.) Ans. 127.27 ft. ^2. A ladder 30 ft. long is placed against a wall with its foot 8 ft. from the wall. How far is the top of the ladder from the ground ? 3. A tree is broken 20 ft. from the ground. The top strikes the ground 20 ft. from the foot, while the other end of the broken part remains attached to the trunk. How high was the tree ? *^4. Two columns 60 ft. and 40 ft. high respectively are 30 tt. apart. What is the distance between their summits ? 5. Show that it is possible to have a right triangle whose hypotenuse and two sides have the respective values 5, 4, 3. In general, can a right triangle be found whose hypotenuse and two sides have any three given values as h, a, b ? If not, when is it possible ? 6. Find the altitude, and then the area, of an equilateral triangle having a side equal to 6 in. 7. Find the area of an equilateral triangle whose altitude is 6 in. 8. Prove the theorem stated in Ex. 11, p. 167, by means § 196. 9. Show that the square of the diameter of a circle is equal to the sum of the squares of any two chords drawn from a point on the circle to the ends of the diameter. (§ 133.) 10. A kite (see Ex. 21, p. 123) is inscribed in a circle whose diameter is 24 ft. If the length of one of the two longer sides of the kite is 18 ft., how long is one of the shorter sides ? 11. Show that the sum of the squares of the four sides of any kite (Ex. 30, p. 87) is equal to twice the sum of the squares of the four segments of the cross formed by its diagonals. IV, § 198] PYTHAGOREAN THEOREM 175 12. Show that the square of the length of the tangent AP to a circle from a point P plus the square of the radius of the circle is equal to the square of the distance OP from P to the center of the circle. 13. In the figure of Ex. 12, let OP=p, OA = r, Show by means of P^x. 12 that AP^t Jf — 7 .2_ (p-r)(p + r). Since p + r \s the length of the whole secant from P through to the opposite side of the circle, and since p — r is the external segment of this secant, show that the preceding equation also results from § 169. 14. In any right triangle ABC (see Fig. 126, p. 151), draw a circle with radius CB about C as center. Using the lettering of Fig. 126, reprove Th. VI by means of § 169, by showing that AB' = AE.AD=(AO- EC)(AO + CD) = (AC-CB)(AC+CB) ^AC^-CB", or AC''=AB^-]-GB\ 198. Definition. The projection of a line AB upon another ^— iB .8 ! ! A/ ! M , / i M N M N M N 1/ N j i ° A 1 ^^^ Fio. 137 line CD is the portion of CD cut off between the perpendicu- lars drawn from the extremities of AB to CD. Thus, in the figure, MN is in each instance the projection of AB upon CD. 176 AREAS [IV, § 199 199. Theorem VII. In any triangle the square on the side opposite the acute angle is equal to the sum of the squares on the other tioo sides diminished hy twice the product of one of those sides and the projection of the other upon it A K — P— ^ Fig. 138 Given the A ABO in which C is an acute angle. Let a, b, c be the sides opposite the angles A, B, C respectively and let p represent the projection of b upon a. To prove that c^ = a^ -{- b^ — 2 ap. Proof. Draw the altitude h upon the base a. Then, in Fig. 138, c2 = 7i2 + (a-p)2. 52_^2_|_^2_2a^_^p2 Why? Why? Why? The re- But hence or c^ = a2 + 62 _ 2 ap. Likewise, in Fig. 139, we have e'^ = h^ -{-{p — ay maining details in this case are left to the student. EXERCISES 1. Show that if b = c in Fig. 138, a^ = 2ap, or a = 2 p. Compare § 43. 2. Show that if c = a in Fig. 138, b^ = 2 ap. 3. Show, from Ex. 2, that the base of any isosceles triangle is a mean proportional between one of the equal sides and twice the projection of the base upon it. IV,/§200] PYTHAGOREAN THEOREM 177 200. Theorem VIII. In any obtuse triangle the square on the side opposite the obtuse angle is equal to the sum of the squares on the other two sides increased hy twice the product of one of those sides and the projection of the other upon it. Given the obtuse A ABC in which C is the obtuse angle. Let a, b, c be the sides opposite the angles A, B, C respectively and let p represent the proiec- tion of upon a. To prove that c^ = a^ + 6^ 4. o ap. [The proof, being similar to that of § 199, is left to the student.] EXERCISES , [The symbol * means that the exercise requires the use of trigono- metric ratios.] 1. How does the square on any side of a triangle opposite an acute angle compare with the sum of the squares on the other two sides ? Answer the same question for the side opposite the obtuse angle in an obtuse angled triangle. 2. Show that either Theorem VII or VIII when applied to a right triangle gives Theorem VI. 3. Find the area of an isosceles triangle whose side is 11 in. and whose base is 8 in. \A. Prove that the sum of the squares on the diagonals of a parallelogram is equal to the sum of the squares on the four sides. * 5. Show that the projection of any line AB on another line CD (Fig. 137) is equal to the product of that line and the cosine of the angle between AB and a parallel to CD through A. * 6. By means of Ex. 5, show that Theorem VII gives (^ = a}-\-W — 2ahx (cosine of C). [The Cosine Law.] 178 AREAS [IV, § 201 201. Problem 1. To construct a square ivhose area shall be equal to the sum of the areas of tivo given squares. Given the squares li and S having respectively the sides r and s. Required to construct a square T whose area shall be the sum of the areas of B and S. Construction. Draw a right A having r and s as sides. Upon the hypotenuse h of this triangle construct a square T. This is the square desired. The proof follows immediately from Theorem VI and is therefore left to the student. EXERCISES 1. Show how to construct a square equal in area to the sum of three given squares. Generalize your answer to the case of any number of given squares. 2. To construct a square equal in area to the difference of two given squares. 3. Show that a square erected on the diagonal of a given square is equal to twice the given square. 4. Show that the square erected on half the diagonal of a square is equal to half the given square. 5. Construct a square equal to a given triangle. [Hint. The side of the square is the mean proportional between the base and half the altitude of the triangle. Why ?] IV, § 202] CONSTRUCTIONS 179 202. Problem 2. To construct a triangle whose area shall be equal to that of a given polygon. Given the polygon P^ ABODE. y^^T^^^ Required to construct // ' \ ^^N^ a triangle equivalent to ^/ / l \ \J\^ ABODE. f\/ j \ \^5 Construction. Draw //\ / ^^ .y^\^ DA and through E draw __y_ V 1*/^ /N» EF II DA and meeting FA B G BA prolonged at F. Fig. 142 Draw DF. Then the polygon FBOD has one side less than the polygon ABODE, but is equivalent to it. (See proof below.) This process may be continued until the last polygon reached is the triangle desired. Proof. A AED = A AFD, since each has the same base AD and their altitudes are equal. §«i.90, d Adding polygon ABOD to both members of this equation, we obtain Polygon FBOD = Polygon ABODE. Similarly we have Polygon FBOD = A FGD. Therefore A FOD is the triangle required. EXERCISES 1. Construct a triangle equivalent to a square whose side is 2 in. Is your triangle the only such triangle ? 2. Construct a square equivalent to a given parallelogram. [Hint. The side of the square is the mean proportional between the base and altitude of the parallelogram. Why ?] 3. How could a square be constructed that would be equiva- lent to a given pentagon ? [Hint. Construct a triangle equal to the given pentagon ; then pro- ceed as in Ex. 5, p. 178.] 180 AREAS [IV, § 202 MISCELLANEOUS EXERCISES. CHAPTER IV [The symbol * means that the exercise involves Trigonometric ratios.] 1. Find the area of a square whose diagonal is 30 ft. long. 2. Find the difference in boundary between a rectangle whose base is 16 ft. and a square equal to it whose side is 12 ft. 3. Prove that the diagonal of a trapezoid divides it into two triangles whose areas are proportional to the bases. 4. Show, by Ex. 3, that the perpendiculars let fall upon one diagonal of a trapezoid from the ends of the other diagonal are proportional to the parallel bases. 5. Show how to find, from proper measurements, the area of a vacant lot bounded by four streets, only one pair of which are parallel. 6. A field of the form of a right triangle containing 9 acres is represented on a map by a right triangle whose sides are 17 in. and 25 in. On what scale is the plan drawn ? [1 acre = 4840 sq. yd!] 7. What is the locus of all points such that the sum of the squares of the distances of any one of them from two fixed points is equal to the square of the distance between those two points ? 8. Obtain the formula for the area of an isosceles right triangle whose hypotenuse is h. 9. Obtain the formula for the area of an isosceles triangle whose base is b and whose side is a. 10. From Ex. 9 derive a formula for the area of an equi- lateral triangle. 11. If one side of a right triangle is three times the other, how long is the hypotenuse as compared with the shorter side ? 12. In Ex. 11, what are the lengths of the segments of the hypotenuse made by a perpendicular from the vertex of the right angle, in terms of the smaller side of the triangle ? IV, § 202] MISCELLANEOUS EXERCISES 181 13. A channel for water has a cross section which is of the form of a trapezoid ABCD. If AD and BC are known and the total height H of the channel is known, find the area of the cross section. If the depth of the water is h, find the area of the cross section of the water in the channel ; (1) when h = H/2 ; (2) when h = H/i ; when 7i = 2 H/S. Find the length of CD in terms of H if the angle ADC is 45°. If the water level is EF, find CF in terms of h, if the angle ADC is 45°. 14. Do the areas mentioned in Ex. 13 change if BC, H, and AD remain fixed, but the angles at B and C are changed as represented in the figure ? Explain your answer. 15. The area of a field may be found in the following way : Run (stake out) a line (base line) AB. From certain points in the boundary of the field the distances to this line are measured, as a, b, c, d, e, /, g, and the distances, j, k, I, m, n, 0, are also meas- ured. Complete the de- scription of how to proceed. 16. To lay off a right angle, carpenters frequently use three sticks 3 ft., 4 ft., and 5 ft. long, respectively. Show that these sides form a right triangle. 17. Show that any multiples of 3, 4, 6, may be used in place of 3, 4, 5 of Ex. 16. 18. Can you discover any three integral numbers a, b, c, not multiples of 3, 4, 5, for which a^ + b^^c^? 182 AREAS [IV, § 202 \l9. Prove that if the middle point of one of the non-parallel sides of a trapezoid is joined to the extremities of the other non-parallel side, the area of the triangle thus formed is equal to half that of the trapezoid. 20. TF is a wall having a round corner upon which a gutter is to be placed, and it is desired to find the radius of the circle of which ABC is a quadrant. If the line AC measures 24 ft., show that the desired radius will be about 17 ft. 21. Find the diagonals of a rhombus whose side is 6 ft. 1 in. and whose area is 9 sq. ft. 22. Find a formula for the altitude h of an equilateral tri- angle in terms of one side a, and find the ratio of this altitude to (1) the whole side, (2) one half of one side. * 23. From Ex. 22 show that the sine of 60° is equal to V3/2. Show also that the cosine of 60° is equal to i; and that the tangent of 60° is equal to V3. (See § 177.) Check by means of the table, p. 155. * 24. By means of an isosceles right triangle, show that the sine of 45° and the cosine of 45° are each equal to V2/2. Check by the table, p. 155. * 25. Turning the figure of Ex. 23 into the new position shown below, find the sine, the cosine, and the tangent of 30°. Check by the table, p. 155. * 26. Ex. 11 may be restated as fol- lows ; if the tangent (§ 177) of an angle is 3, what is the cosine of that angle? the cosine of the angle is 1/VlO. Show, by Ex. 11, that Check by the table, p. 155. IV, § 202] MISCELLANEOUS EXERCISES 183 O 27. How much leather is required to cover a window seat, in the shape of half of a regular hexagon, if the length of each of its three sides is 5 ft. ? 28. Show how to construct a line parallel to the bases of a given parallelogram or triangle that shall divide the figure intotwo parts that are equal in area. ^9. Prove the Pythagorean theorem by means of some one of the adjoining figures. C B G ^G K E H J r L (a) Note. Figure (a) is said to be the one used by Pythagoras, who discovered the theorem. Figure (6) is due to Bhaskara, a native of India, about 1150 A.D. The figure used in § 196 is that used by Euclid, who wrote a famous treatise on Geometry about 250-300 b.c. 30. Prove the fact stated in Ex. 9, p. 167, by means of § 189. 31. A room is 18 ft. long, 14 ft. wide, and 10 ft. high. Find the length of one diagonal of the floor. Then find the length of a string stretched through the center of the room from one corner at the floor to the farthest opposite corner at the ceiling. Ans. 24.9 ft. 32. If n is any integer, show that 2 w, n^— 1, and n -|- 1 are integers that are proportional to the three sides of a certain right triangle. Find these three integers when w = 2; when w = 4 184 AREAS [IV, § 202 33. Any polygon described on the hypotenuse of any right triangle as one side is equal to the sum of the two similar polygons drawn on the sides, respectively, with the corre- sponding side as one side of the polygon. Prove this statement. [Hint. Suppose the similar polygons are triangles, as in the figure. Then P/^ = a7c2, B/q = byc^ ; hence Q = P -^ B.] ^34. Show how to construct the side of an equilateral triangle whose area is the sum of two given equilateral triangles. 35. In any triangle ABC let us write s = (a 4- & + c)/2, where a, b, c, are the three sides. Show^ that Area A ABC == V s(.s — a){s— b){s — c). Outline of Proof. Let A be an acute angle, and let h be the altitude from (7; then we have h' = a'-AD\ b' = a'-{-c'-2c'AD, by §199; but hence h' 2 r«' + C2 - b^' 2c by But 4c2 = [2 ac-\-(a^-\-c' - b')'][2 ac-(a' + c' - b')-] -^ ^ c". 2ac + (a'-\-c^- b') = (a^ + 2 ac + c^) - b' = (a + cf-b^={a-{-c-{-b){a + c-b) . /a -j- & + c\ /a 4- & + c , \ . / , n = %— 2— "j (2 b^isis- b). Likewise 2ac-(a^-^c'-b')=b'-(a-cy = 4.(s-c)(s-a), Hence /i^ = 4 s (s — a) (s — b)(s — c) -h c^, and Area A ABC = h • c/2 = Vs(s — a)(s — b){s — c), IV, §202] MISCELLANEOUS EXERCISES 185 36. Find the area of a triangular field whose sides are, re- spectively, 80 rd., 220 rd., 200 rd. [1 acre = 160 sq. rd.] 37. Find the area of a parallelogram whose sides are, re- spectively, 8 in. and 10 in. long, and one of whose diagonals is 15 in. long. Ans. 74.0 sq. in. 38. Find the area of a triangle, given two sides 15 ft. and 25 ft., and their included angle 30°. * 39. Find the area of a triangle, given two sides 15 ft. and 25 ft., and their included angle 40°. Ans. 120.5 ft. 40. Show that the area of any triangle is s • ?•, where s = (a-{-b+c)/2, as in Ex. 35, and r is the radius of the in- scribed circle. (See § 124.) 41. Show that the area of any polygon circumscribed about a circle is half the sum of its sides times the radius of the circle. 42. Comparing the two results of Exs. 35 and 40, show that the radius of a circle inscribed in any triangle is r = V(s — a)(s — b){s — c)/s. 43. Show that in any triangle one of whose angles is 120°, the square on the side opposite the larger angle equals the sum of the squares on the other two sides plus the product of those sides. 44. Show that in any triangle one of whose angles is 60° the square on the side opposite that angle is equal to the sum of the squares on the other two sides diminished by the product of these two sides. * 45. Show, in Figs. 138 and 139, that h = bx (sine of Z C): hence show that the area of the triangle ABC is Area A ABC=\ ah X (sine of Z C). * 46. Prove Theorem IV, § 193, by means of Ex. 45. CHAPTER V REGULAR POLYGONS AND CIRCLES 203. Regular Polygon. A polygon that is both equiangular and equilateral is called a regular polygon. (See § 96.) EXERCISES 1. Construct a quadrilateral that is equilateral but not equiangular. What are such quadrilaterals called ? 2. Construct a quadrilateral that is equiangular but not equilateral. What are such quadrilaterals called ? 3. Draw a quadrilateral that is neither equiangular nor equilateral. 4. Construct a regular polygon of four sides. What are such quadrilaterals called ? 5. Is an equilateral triangle a regular polygon ? Why ? 6. What is the size of each angle of a regular polygon of seven sides ? (See § 97.) 7. If three equal rods are hinged together at their ends in the form of an equilateral triangle, is the framework rigid ? 8. If four equal rods are joined together at their ends by hinges, is the framework formed necessarily a square ? Can the angles be changed without changing the lengths of the sides ? 9. If five equal rods are hinged together at their ends in the form of a regular pentagon, is the framework rigid ? 10. Will the framework of Ex. 8 be rigid if a stiff rod is in- serted along one diagonal ? Along how many diagonals must rods be inserted in the framework of Ex. 9 to make it rigid ? 186 V, §204] MUTUAL RELATIONS 187 204. Theorem I. If a circle is divided into a number of equal arcs : (a) the chords joining the points of division form a regular inscribed polygon ; (b) tangents draivn at the points of division form, a regular circumscribed polygon. s Given the circle divided into the equal arcs AB, BC, CD, DE, and EA by the chords AB, BC, CD, DE, and EA ; and given the tangents PQ, QR, RS, ST, and TP, touching the circumference at the points B, C, D, E, and A, respectively. To prove that (a) ABCDE is a regular polygon ; (h) PQRST is a regular polygon. Proof of (a). Since AB=BC= CD=DE=EI), Given Arcs ABC, BCD, CDE, DEA, and EAB are equal. Why ? Hence Z ABC = Z BCD = Z CDE, etc. Why ? Also AB = BC=CD= DE, etc. Why ? It follows that ABCDE is a regular polygon. § 203 Proof of (6). Show that A APB ^ A BQC by showing that AB = BC, Z PAB = Z QBC (§ 137), and Z PBA = Z QCB. Likewise A APB ^ A CRD, etc. Then show that PQ = QR = etc. ; and Z P =Z Q = etc. ; and thus prove the polygon PQRST a regular polygon. 188 REGULAR POLYGONS AND CIRCLES [V, § 205 205. Theorem II. (a) A circle may he circumscribed about any regular polygon ; (b) a circle may also be inscribed in it. Given the regular polygon ABCDE. To prove that (a) a circle may be circumscribed about it ; (p) Si circle may be inscribed in it. Proof of (a). Pass a circle through the points A, Bj and C. §120 Join the center with the vertices of the polygon. Then Moreover and Therefore so that CD. §§ 103, 203 § 203 §40 Ax. 2 §35 Why? OB = OC, and AB Z CBA = Z DCB, Z CBO = Z OCB. Z OB A = Z OCD, AOAB^A OCD', hence OA = OD, and the circle passes through D. In like manner show that the circle passes through E. Hence the circle is circumscribed about the polygon. § 114 Proof of (&). The sides of the regular polygon are equal chords of the circumscribed circle. Therefore they are equally distant from the center. § 109 The circle with as a center and the perpendicular from the center to any side as radius is inscribed in the polygon. § 114 V, §207] MUTUAL RELATIONS 189 206. Definitions. The common center of the circumscribed and inscribed circles is called the center of the regular polygon. The radius R of the circumscribed circle is called the radius of the regular polygon. ^ ^D The radius r of the inscribed circle is called the apothem of the regular polygon. The perimeter of a polygon is the sum of the lengths of its sides. 207. Theorem III. The area of a regular polygon is equal to half the product of its apothem and perimeter. Given the regular polygon ABODE ••♦, Fig. 145. Let its perimeter be denoted by p and its apothem by r. To prove that ABCD •-. =pr/2. Proof. Draw the radii OA, OB, 0(7, etc., Fig. 145, thus making as many triangles as the polygon has sides. The altitude of each triangle is the apothem r, and each base is a side of the polygon ; hence, the area of each triangle is half the product of r and one side of the polygon. § 189 Since the sum of all the bases is the perimeter p of the poly- gon, and the sum of the areas of the triangles is the area of the polygon, it follows that Area of ABCD ••• =pr/2. Compare this result with Ex. 41, p. 185. 190 REGULAR POLYGONS AND CIRCLES [V, § 207 EXERCISES 1. If n represents the number of sides of a regular polygon, show that the angle at the center subtended by one side is 4/ii right angles. 2. Find the angle at the center subtended by one side of an equilateral triangle ; of a square ; of a regular pentagon ; of a regular hexagon. Draw these figures by means of a protractor. 3. Prove that any angle of a regular polygon is supple- mentary to the angle at the center subtended by one side. 4. Eind the number of degrees in the angle at the center subtended by one side of a regular octagon. Eind the size of one angle of the octagon by means of Ex. 3 ; and then check your answer by § 97. Draw the figure. 5. How many sides has a regular polygon if the angle at the center subtended by one side is 40° ? 6. If one angle of a regular polygon is 140°, what is the number of its sides ? 7. Eind the area of a regular hexagon whose side is 6 in. 8. Two squares have for their sides 8 and 11 in., respec- tively. Eind by § 207 the ratio of their areas. 9. Two squares have areas of 144 sq. in. and 225 sq. in., respectively. What is the ratio of their perimeters ? ^10. Eind the ratio of the perimeters of squares inscribed in and circumscribed about the same circle. Ans. 1 : V2. 11. The perimeter of a regular inscribed hexagon is 30 in. Find the perimeter of a regular hexagon circumscribed about a circle of twice the diameter. Ans. 69.28 in. 12. Eind the area of an equilateral triangle circumscribed about a circle whose radius is 12 in. Ans. 748.25 sq. in. 13. Find the apothem of an equilateral triangle of which one side is 6 in. (Use Th. XXXIII, § 102.) Hence find its area by Theorem III. Compare with the result of Ex. 6, p. 174. V, § 210] MUTUAL RELATIONS 191 208. Areas of Inscribed and Circumscribed Regular Poly- gons. The area of a circle is evidently greater than the area of any regular polygon inscribed in it. (17, §31.) By doubling the number of sides, we obtain regular inscribed poly- gons whose areas are more nearly equal to that of the circle, as in- dicated in the figure. Thus the area of a regular in- scribed octagon, for example, is evi- dently more nearly equal to the area of the circle than is the area of an inscribed square. The area of a circle is also evidently less than the area of any circumscribed polygon. Perimeters. The perimeter of any inscribed polygon is evidently less than the length of the circumference (Post. 3, § 28) ; we shall also assume that the length of the circumfer- ence is less than that of any circumscribed polygon. 210. Areas and Lengths of Circles. Since the regular inscribed and regular circumscribed polygons approach, both in length and in area, nearer to one another and to the circle, as the number of sides is doubled, we may say : If the number of sides of the regular inscribed and regular circumscribed polygons is repeatedly doubled^ (a) their areas approach the area of the circle as a common limit; (b) their perimeters approach the length of the circumference of the circle as a common limit. It is also obvious that the apothem of the inscribed regular polygon will approach the radius of the circle, under the same circumstances. 192 REGULAR POLYGONS AND CIRCLES [V, §211 211. Theorem IV. The circumferences of two circles are to each other as their radii. Fig. 147 Given two circles and 0' whose radii are r and r'. To prove that the length of their circumferences c and c' are to each other as their radii ; that is, c/c^ = T/r\ Proof. Draw inscribed regular polygons P and P\ of the same number of sides in each circle, let the perimeters of these polygons be p and p', and let the corresponding sides be s and s'. Then P^P', §166 and p/p' = s/s'. § 167 But s/s' = r/r', since the triangle two of whose sides are s and r is similar to the triangle two of whose sides are s' and r'. Hence p/p' = r/r'. Ax. 9 Since p and p' approach c and c' if the number of sides is repeatedly doubled, the difference p/p' — c/c' can be made as small as we please by doubling the number of sides repeatedly; hence also r/r' — c/c' can be made as small as we please. But r/r' and c/c' do not change at all as we increase the number of sides of the polygons ; hence r/r' = c/c', for if r/r' and c/c' differed from each other, their difference would be fixed, and therefore not as small as we might please. V, §214] LENGTH OF CIRCUMFERENCE 193 212. Corollary 1. The ratio of a circumference to its diameter is the same for all circles. For, since c/c' = r/r' (§ 211), we have also, if d and d' are the diameters, d/d' = 2 r/2 r' = r/r' = c/c'. Hence by alternation (Theorem D, § 144), c/d = c'/d'. 213. The Number ir. The number obtained by dividing the circumference of any circle by its diameter, which, by § 212, is the same for all circles, is denoted by the Greek letter tt (pronounced pi). This number, which is about 3^ (or, more accurately still, 3.1416), will be computed approximately later in this chapter. 214. Corollary 2. In any circle c = ird, or c = 2 ttt, ivhere r is the radius, d the diameter, and c the length of the circumference. For, since tt = - (§ 213), c = ird] or since d=2 r, c = 2 Trr. d EXERCISES 1. Find the length of the circumference of a circle of radius 2 ft. [Take tt =^ 3f ] Ans. 4 n ft., or 12.57 ft. 2. If the radius B of the earth is 4000 mi., what is its circumference at the equator? [Take tt = 3}.] How much is this result affected by taking the more accurate values E = 3963 mi., tt = 3.1416. 3. Find the diameter of a circle whose circumference is 40 in. 4. Measure the circumference of some round object, such as a porch column, by stretching a string around it tightl}^ and then measuring the string. ' Now compute the diameter. 5. How wide must a piece of tin be cut in order to be made into a stovepipe 8 in. in diameter ? Ans. 8 n in., or 25f in. 6. Show how to find quickly the approximate length of wire in a coil by measuring the diameter of the coil, and count- ing the number of strands. 7. What is the length of an arc that subtends an angle of 60° at the center of a circle whose radius is 5 ft. ? 194 REGULAR POLYGONS AND CIRCLES [V,§21o 215. Theorem V. The area of a circle is equal to one half the product of its radius and its circumference. Fig. 148 Given the circle O, with radius r, circumference c, and area A. To prove that A = rc/2. Proof. Circumscribe a regular polygon about the circle. Let A denote its area andy its perimeter. Then A' = rp' /2. § 207 As the number of sides of the regular circumscribed polygon is increased, p' approaches c as its limit. § 210 Hence rp' /2 approaches rc/2 as a limit. Also A' approaches ^ as a limit. § 210 Therefore the difference between A and rc/2 must be as small as we please, as in § 211. It follows, as in § 211, that A = rc/2. The student should state carefully all of the remaining steps in the argument. 216. Corollary 1. The area of a circle is equal to tt times the square of its radius, that is, A = irr^. For, by § 215, A = rc/2 ; but c = 2 Trr ; hence A = ttt^. 217. Corollary 2. The areas of two circles are to each other as the squares of their radii. V, § 219] AREA OF CIRCLE 195 Note. The very famous problem of " squaring the circle," that is, of constructing the side of a square whose area equals that of a given circle, depends on determining the value of tt. We now know that this construction is impossible with ruler and compasses ; but the ancient Greeks and the Schoolmen of the Middle Ages spent much time attempting to do it. 218. Sectors. The area of a sector bears the same ratio to the area of a circle as the angle of the sector bears to 360°. For example, the area of a sector whose arc is 36° is 1/10 the area of the circle. 219. Circle divided into Sectors. A circle may be cut into equal sectors and arranged as in the following figure. Fig. 149. Circle divided into Sectors Each sector approximates the shape of a triangle. The area of each sector is equal to one half the product of its arc (base) by its radius (altitude). Their sum equals one half the whole circumference times the radius. EXERCISES [Exercises that involve the number v should be worked through first in terms of the symbol v ; then the value 2>\ should be substituted for it. Thus the area of ^ of a circle whose radius is 6 inches is ^ • 6^ . tt sq. in. = 9 TT sq. in. ; substituting t = 3f, we find the result 28f sq. in.] 1. Find the area of a circle whose diameter is 14 in. 2. Find the area of a square circumscribed about a circle whose radius is 2 ft. What is the ratio of the area of the circle to the area of this square ? Ans. 16 sq. ft. ; 7r/4, or 11/14. 196 REGULAR POLYGONS AND CIRCLES [V,§219 3. The circumference of a circle is 14 ft. Find its radius and its area. 4. The area of a circle is 24 sq. in. Find its radius. 5. What is the change in the area of a circle, if its radius is multiplied by 2 ? by 5 ? loj n ? 6. The effect of an explosion in a large powder plant was felt for a distance of 10 mi. in every direction. How large an area was affected ? 7. A tinner is to cut the largest possible square from a cir- cular piece of tin. What proportion of the tin will be left ? 8. Find the area of a sector in a circle of radius 3 ft. if the angle of the sector is 45° ; 60° ; 30° ; 80°. 9. If a tinner cuts from the same sheet of tin two circular pieces, one of which is twice as wide as the other, how much heavier is the larger one ? 10. If it is desired to cut two circular weights out of a flat piece of metal, how much wider must the larger be to weigh twice as much, the thickness being the same ? 11. A steel rod whose cross-section is 1 in. square weighs 3.4 lb. per foot of length. Find the weight per foot of length of a round rod 2 in. thick of the same material. 12. Find the weight per foot of length of a waterpipe whose outer diameter is 3 in., and whose inner diameter is 2.75 in., made of the material mentioned in Ex. 11. 13. Show that the area of a circular ring /^^^^^^>\ contained between two concentric circles of '^W t^-"^^^ diameters D and d is ^^ ^'^^^ B-d D + d '\^^^^W\ (UP- -d'\ A=7ri : 1, Or^ = 7r r<-t-*k- — d— >t*i-»; 14. How much water per hour (in cubic feet) will flow through a pipe whose inner diameter is 3 in.. if the water is flowing 3 ft. per second ? V, § 221] AREA OF CIRCLE 197 220. Problem 1. Given the side and radius of a regular in- scribed polygon, to find the side of a regular inscribed polygon of double the number of sides. Given AB, the side of a regular inscribed polygon of radius r. Required to find AC, a. side of the regular inscribed polygon of double the number of sides. Solution. Draw the diameter CEj and the radius AO. Also draw AE. Now OD ± AB at its middle point. Hence or and Also that is, Olf^r'-iAB", OD = Vr^ J J^, CD = r- vy - \ aW. AC^=CE'CD = 2r'CD; AC=^2r(r--Vr^-iAF = Vr(2 r - V4 r^ - AW). Why? Why? 163 221. Corollary 1. If r = 1, and s = the side of the inscribed polygon, 198 REGULAR POLYGONS AND CIRCLES [V, §222 222. Problem 2. To compute approximately the value of tt. The perimeter of a regular hexagon inscribed in a circle of unit radius is 6 units. (Why ?) By using the formula in § 221 and computing successively the perimeters for polygons of 12, 24, 52, . • •, and 768 sides we get the following results : NiTMBEK OF Sides Length op one Side Length of Pekimeteb 12 .51763809 6.21165708 24 .26105238 6.26525722 48 .13080626 6.27870041 96 .06643817 6.28206396 192 .03272346 6.28290510 384 .01636228 6.28311544 768 .00818126 6.28316941 By continuing this process it is found that the first five figures in the decimal remain unchanged. Hence 6.28317 is a close approximation to the circumference of a circle whose radius is 1. Since the diameter is 2, the ratio tt of the circum- ference to the diameter is, approximately, IT = 5:?^ = 3.14159 (usually written 3.1416, or 3|). Note. A still more accurate value can be computed by con- tinuing the preceding process. It has been proved that the number tt cannot be expressed precisely by any finite decimal. The fact that tt cannot be expressed precisely is equivalent to the statement that the diameter and the circumference of a circle are incommensurable to each other (§ 128). But we can obtain, by the preceding process, as great accuracy as we please. The value is known to over 700 decimal places. To ten places it is TT = 3.1415926536, but such accuracy is never neces- sary in any ordinary affairs. As a curiosity, we quote the value : TT = 3.1415926535897932384626433832795028841971693993751. V, § 224] AREA OF CIRCLE 199 223. Problem 3. To inscribe a square in a given circle. Given the circle 0. To inscribe a square in circle 0. ^ Construction. Draw two diameters AC and DB perpendicular to each other. Draw AB, BC, CD, and DA. Then ABCD is the square desired. Proof. [The proof is left for the student.] 224. Problem 4. To inscribe a regular hexagon in a given circle. I g Given the circle 0. To inscribe a regular hexagon in circle 0. Construction. Draw the radius OA and with ^ as a center and radius OA draw an arc cutting the circle in B. Then AB is the side of the hexagon desired. Outline of Proof central angle 60°. EXERCISES 1. Show how to inscribe an equilateral triangle. 2. Show how to inscribe a regular polygon of twelve sides. 3. Since one side of an inscribed square subtends a central angle 90°, and one side of a regular inscribed hexagon subtends a central angle of 60°, show that if one vertex of the square coincides with one vertex of the hexagon, the next vertices are at the extremities of an arc of 30°. 4. Show how to inscribe a regular polygon of eight sides. ^6. Show how to inscribe a regular polygon of twenty-four sides directly from an inscribed regular octagon and a regular inscribed hexagon. Fig. 152 Draw OB and show that AB subtends a 200 REGULAR POLYGONS AND CIRCLES [V, §225 225. Problem 5. To inscribe a regular decagon. Given the circle 0. Required to inscribe a regular decagon in the given circle. Fig. 153 Construction. Draw the radius OA and divide it in extreme and mean ratio at O, having the larger segment next to the center. § 172 Then OA:OG = OG: GA, and OG is the side of the decagon required. By applying OG ten times to the circle as a chord, the desired regular decagon is formed. Proof. Draw OB and BG. Since OA/OG = OG/GA and OG = AB, we have OA/AB = AB/ GA ; hence A OAB ~ A ABG. Therefore ZO = Z ABG and Z AGB = Z ABO. But Z OAB= ZABO', hence Z AGB = Z GAB, and AB = BG = GO. Therefore Z = Z GBO. But we have shown Z = Z ABG ; hence, addin'g, 2 ZO=Z OBA = Z OAB. Now ZO-\-Z OBA + Z OAB = 2 rt. A, whence 5 Z = 2 rt. z^, or Z0 = 1/5 of 2 rt. A, or 1/10 of 4 rt. A. Therefore the arc AB is 1/10 of the circumference, and the chord AB, equal to OG, is a side of a regular inscribed decagon. Const. Why? Why? Why? Why? Why? Why? Why? Why? V, § 225] AREA OF CIRCLE 201 EXERCISES 1. Show how to inscribe a regular pentagon in a circle. 2 Show how to inscribe regular polygons of 20, 40, etc. sides in a given circle. 3. Show that if one vertex of a regular inscribed pentagon coincides with one of a regular hexagon inscribed in the same circle, the uext vertices are extremities of an arc of 12°. Hence show how to inscribe a regular polygon of 30 sides. 4. Show how to inscribe a regular polygon of 15 sides directly by using the regular inscribed pentagon and the in- scribed equilateral triangle. MISCELLANEOUS EXERCISES. CHAPTER V 1. The radius of a circle is 2 in. Find the length of the circumference ; . the area, w^ 2. The area of a circle is 98 sq. ft. Find the diameter. 3. The diameters of two circles are 4 ft. and 9 ft., respec- tively. Find the ratio of their areas. 4. How many people can be seated at a round table 54 in. in diameter when it is extended 4 ft., allowing 2 ft. to a person ? 5. Find the perimeter of a regular hexagon inscribed in a circle whose radius is 1 ft. ; 3 f t. ; a ft. - 6. The perimeter of a regular inscribed hexagon is 48 ft. What is the diameter of the circle ? 7. Find the perimeter of a regular circumscribed hexagon, if the radius of the circle is 1 ft. ; 3 ft. ; r ft. 8. If the radius of a circle is r, find the area of the in- scribed equilateral triangle; of the circumscribed equilateral triangle. > 9. Show that the area of the inscribed equilateral triangle equals one fourth the area of the circumscribed equilateral triangle. 202 REGULAR POLYGONS AND CIRCLES [V, §225 10. A cow is tethered at the end of a 50 ft. rope, which is fastened to the corner of a barn. The barn is 25 ft. wide and 60 ft. long. Over how much area may the cow graze ? 11. How many revolutions per mile does a 28-in. bicycle wheel make ? Ans. 720. 12. The boiler of an engine has 200 tubes, each 3 in. in diameter, for conducting the heat through the water. Find their total cross sectional area. 13. Construct a regular inscribed pentagon, and draw all the diagonals. Show that the sum of all the angles in the vertices of the resulting five-pointed star equals two right angles. 14. A circular piece of brass has a radius of 10 in. and it is desired to cut a hole through it equal in area to one half the disk. What should be the radius of the hole ? 15. The central angle whose arc is equal to the radius of the circle is called a radian. It is often used as a unit of measure of angles. Show that 1 radian = 180° -r-7r= 57.3° approximately. ^16. A circle is circumscribed about a right triangle and two others are described with the sides as diameters. Prove that the large circle equals the sum of the two small ones. r ^mmSmm^)^ 17. Semicircles are constructed on the three sides of a right triangle as in the adjacent figure. Show that the sum of the semicircles AO'CD and BO"CE is equal to the semicircle AOBF. ^'^mMB^Q 18. Construct a circle equal to the area of two given circles. v^ 19. If the limit of safety for the surface speed of an emery stone is 5500 ft. per minute, what is the diameter of the largest wheel that can safely make 1500 revolutions per minute? Ans. 11/(3 tt) ft., or 14 in. V, §225] MISCELLANEOUS EXERCISES 203 20. Can a piece of paper 6 in. wide be used to wrap up a circular mailing roll whose radius is 1 in. ? 21. How many laps around a circular running track whose diameter is 125 yd. are necessary to make up a distance of 8 mi.? 22. Find the length of the curved portion of a railroad track that connects two straight portions at right angles to each other, given that the curved portion is an arc of a circle of radius 200 ft., tangent at its extremities to the straight portions of the track. Find the length of the curved portion if the angle between the straight portions is 60°. 23. What is the total pressure on the piston of an engine if the cylinder is 20 in. in diameter and the gauge shows 76 lb. per square inch ? 24. Four pumps each with a diameter of 5 in. are used in a mine. If one pump were used to remove the same amount of water in the same time, what would be its diameter, every- thing else being the same? Ans. 10 in. 25. When the gauge shows a steam pressure of 100 lb. per square inch, what is the total pressure tending to blow the cylinder head out, if it is 18 in. in inside diameter ? 26. If the water in a 3-in. (inside diameter) water main is flowing at the rate of 5 ft. per second, how much water is passing a given point per minute? 1 gal. = 231 cu. in. 27. A water main of 6-in. diameter is continued beyond a certain point by a pipe of 4-in. diameter. If the water in the 6-in. pipe is running at the rate of 2 ft. per second, how fast is the water in the 4-in. pipe running? 204 REGULAR POLYGONS AND CIRCLES [V, § 225 28. A conduit for carrying water is circular in form and is 10 ft. in diameter. Find the area of the total cross section. If the water level is at EF, find the area of the cross section of the water if the angle EOF is 90° ; if EOF is 60°. ^_ water 29. Find the amount of water flow- ing through the conduit of Ex. 28 per minute if Z.EOF= 60° and the speed is 2 ft. per second. 30. Find the length EGHF of the portion of the circular outline, Ex. 28, which is wet when the water reaches EF if EOF is 45°; if EOF is 60°; if EOF is equal to 90°. (This so-called " wetted perimeter " is of the greatest importance in determining friction and therefore the resistance of the pipe to the water flow.) * 31. Find the area of the cross section of the water in the conduit of Ex. 28 when Z EOF= 80°. Ans. 73.41 sq. ft. * 32. Find the area of the cross section of the water in the conduit of Ex. 28 when the distance from to EF is 4 ft. 33. How fast is the point on the rim of a wheel moving, in feet per second, if the wheel is 2 ft. in diameter and is rotating at an angular speed of four revolutions per minute? 34. How far does a carriage move when one of its wheels revolves (without slipping) through five complete revolutions, if the diameter of the wheel is 4 ft. ? 35. How many times will the wheel of a bicycle revolve, if it is 28 in. in diameter, in going 3 mi. ? If the bicyclist goes 3 mi. in 20 min., how many revolutions does the wheel make per minute? 36. Find the angular speed of a car wheel that is 20 in. in diameter, when the train is going 40 mi. per hour. V, § 225] MISCELLANEOUS EXERCISES 205 37. The radius of the earth is approximately 4000 mi., and the earth makes one revolution per day. What is the speed, in miles per hour, of a point on the equator? of a point whose latitude is 30° ? 38. Compare the speed of a point on the equator of the earth, in miles per hour, with the speed of an express train going 60 mi. per hour. Compare it with the speed of a point on the rim of a flywheel 2 ft. in diameter that is making 100 revolutions per second. 39. If a hollow pipe has an inside diameter d and an outside diameter D, the thickness of its walls t is equal to (D — d)/2. Show that the area of the cross section of the metal is A D-{-d , Show that D-\-d = 2(2) - t) ; hence show that A = 'ir{Dt- t^). 40. Show that the area bounded by two concentric circles of radii r and R and two radii of the larger one, is equal to half the product of its altitude (R — r) and the sum of its two circular sides. (Compare § 191.) 41. Show how to find the area of a city lot bounded by two circular streets that have a common center, and two of their radii, if the lengths of the circular arcs are 120 ft. and 160 ft., and the straight line boundaries are 75 ft. long. APPENDIX TO PLANE GEOMETRY MAXIMA AND MINIMA 226. Definitions. Let P be a fixed point upon the circum- ference of a given circle and let a series of chords be drawn through this point. The longest one of all these chords is the diameter PQ (Ex. 2, p. 91). This fact may be briefly stated by saying : Of all cliords through P, the diameter PQ is the maximum (greatest). Again, of all regular polygons that can be inscribed in a given circle, the one whose area is least is the inscribed equilateral tri- angle, a fact which may be stated by saying : Of all regular Fig. 154 Fig. 155 polygons inscribed in a circle, the equilateral triangle has the minimum (least) area, or simply, is the minimum. Likewise, of all the straight lines that can be drawn from a fixed point to a given line, the perpendicular is the minimum (§ 77). These and all other similar considerations constitute the subject of maxima and minima in Geometry. The maximum of several quantities is the greatest among them ; the minimum is the least among them. We shall now state and prove a num- ber of theorems related to this subject. 206 §227] MAXIMA AND MINIMA 207 227. Theorem I. Of aJl triangles that have the same two given sides, that in which these sides in- clude a right angle is the maximum. Given the right A ABC and any other A EBC con- structed upon the side BC and having its side EB AB. To prove Proof. Draw Then But Therefore Whence also A ABC > A EBC. ED ± BC. EB>ED. EB = AB. AB>ED, A ABC > A EBC. §75 Given Why? (6), § 190 EXERCISES 1. What is the maximum line that can be drawn within a rectangle and terminated by the sides ? What is the minimum line through a given interior point ? 2. Draw a circle and take any point P within it. Construct the maximum line and also the minimum line from Pto the circumference. Repeat, using a point P outside the circle. 3. In the figure, C is the cylinder of a steam engine, P the piston, PE the piston rod, RL the connecting rod, and IF the driving wheel. As the engine works, describe the position of the connect- ing rod when the area of the triangle PLO is a maximum. (0 represents the center of the wheel W.) Answer the same question for a minimum triangle RLO. 208 APPENDIX [§ 228 228. Definition. Fij^-ures having equal perimeters are called isoperimetric. 229. Theorem II. Of all isoperimetric triangles having the same base, the isosceles is the maximum. Given the isosceles A ABC and any other A DBO constructed upon EC and isoperimetric to ABC. To prove A ABC > A DBC. Fig. 157 Proof. Prolong BA to E, making AE = BA, and draw EC. A circle having A as center and BE as diameter can now be drawn through the points B, C, and E. Why ? Therefore Z BCE is a right angle. Why ? Now draw AG and DF II BC also draw DH = DC, and join B and H. Then AB + AG=DB + DC= DB-\- DH=BE. Given But DB + DH > BH. Why ? Therefore BE > BH. Whence also CE > CIT. Why ? But CG = \CE and (7i<^= i C^. Why ? Therefore CG^ > CF-, Why? and hence A ABC > A Z)J5(7. Why ? 230. Corollary 1. Of all isoperimetric triangles, the equilateral is the maximum. §231] MAXIMA AND MINIMA 209 231. Theorem III. Of all isoperimetric polygons having the same number of sides, the maximum is the one that is equilateral. F A Given the maximum polygon ABODE of all those that can be drawn isoperimetric to each other, having the same number of sides. To prove that ABODE is equilateral. Proof. If ABODE is not equilateral, at least two of its sides, as AB and AE, must be unequal ; and we may construct on the diagonal BE an isosceles A BFE which is isoperimetric with A ABE. Then A BFE > A ABE. § 229 Therefore BODEF > ABODE. But this is contrary to hypothesis. Therefore, AB = AE and ABODE is equilateral. EXERCISES 1. Show that of all triangles inscribed in a semicircle, that is greatest which has the diameter as base and the radius, r^ as altitude. Its area is equal to r\ 2. Compare the area of the triangle of Ex. 1 with that of the semicircle ; with the area of a square inscribed in the circle. 3. Show that of all isoperimetric quadrilaterals, the maxi- mum is a rhombus ; show furthermore that it is a square. 4. Prove that of all parallelograms having given sides, the rectangle is the maximum. What does this theorem become when stated with reference to jointed frames ? (See Ex. 7, p. 165.) 210 APPENDIX [§232 232. Theorem IV. Of all x>olygons with sides all given but one, the maximum (in area) can be inscribed in a semicircle having the undetermined side for its diameter. C Fig. 159 Given the lengths of all sides of the polygon ABCDE, except EA ; and given that ABCDE is the maximum polygon (in area) that has the given sides and any other side EA. To prove that ABCDE can be inscribed in a semicircle whose diameter is the remaining side, EA. Proof. From any vertex, as C, draw CA and CE. The A ACE must be the maximum of all A having the given sides CA, CE; otherwise, by increasing or diminishing the /.ACE, meanwhile keeping the lengths of CA, CE un- changed, as also the form of the figures, ABC, CDE, but allow- ing A and E to slide along MN while C is lowered or raised, we can increase the A ACE, while the rest of the polygon remains unchanged in area. Hence, unless A ACE is the maximum, we can by such processes increase the area of ABCDE. To increase in this way the area of ABCDE is, however, to deny the hypothesis that ABCDE is the maximum polygon. Hence, the A ACE is the maximum that can be drawn, having the sides AC, CE. Therefore, A ACE is a right triangle. § 227 Hence, also, (7 lies on the semicircumference of which AE is diameter. Why ? Likewise, every vertex can be shown to lie on the semicir- cumference whose diameter is AE-, that is, the maximum polygon can be inscribed as stated in the theorem. §234] MAXIMA AND MINIMA 211 233. Theorem V. Of all polygons ivith the same given sides, that ivhich can be inscnbed in a circle is the maximum. B^^_ ^^^^ B' ^C' Fig. 160 Given a polygon ABODE inscribed in a circle and mutually equilateral to another given polygon, A'B'O'jyE', which can- not be inscribed. To prove ABODE > A'BfCUE', Proof. From A draw a diameter AF and join F to the adjacent vertices, D and O. On 0'D\=OD) construct AO'F'D'^ACDF and draw AF'. Then, AEDF > AE'DF' and ABOF > A'B'O'F'. § 232 Adding these two inequalities, ABOFDE > A'B'O'F'D'E'. Take away from the two figures the equal A OFD, CF' U , and we have, ABODE > A'B'CD'E'. 234. Corollary 1 . Of all isoperimetric polygons of a given number of sides, the maximum is regular. EXERCISE 1. If in the figure of § 232 we regard the sides AB, BO, etc., as stiff rods and assume that the rod AB is attached to the rod BO by means of a hinge at B, with a similar arrange- ment at each of the joints, we have what is known as a jointed frame. Considering all the different forms which this jointed frame ABODE can assume, what can be said of that one whose area is the maximum ? 212 APPENDIX [§235 235. Theorem VI. Of tivo isoperimetric regular polygojis, the one having the greater number of sides has the greater area. Fig. 161 Given a regular polygon of three sides (equilateral triangle) ABO and a regular polygon of four sides (square) Q, and let ABO and Q be isoperimetric. To prove that Q > ABO. Proof. Draw OD from O to any point in AB, and upon OD construct a triangle ^Z> (7 which shall be congruent to ADO'^ that is, such that ED = AO and EO^AD. The figure DBOE thus formed is an irregular polygon of four sides which by construction has the same perimeter as ABO, and hence the same as Q. Also, it has by construction the same area as ABO. But Q > the irregular polygon DBOE. § 231 Whence Q > ABO. In like manner it can be shown that a regular polygon of five sides is greater than an isoperimetric square, and so on. 236. Corollary 1. The circle is the maximum of all isoperi- metric plane closed figures. EXERCISES 1. Give the proof that a regular polygon of five sides is greater than a corresponding isoperimetric square. 2. Show that a round can will hold more than a square can of the same perimeter, the two cans being of equal height. §238] MAXIMA AND MINIMA 213 237. Theorem VII. Of all regular polygons of the same area, that which has the greatest number of sides has the minimum perimeter. Q q" Fig. 1G2 Given the equivalent regular polygons Q and Q', of which Q' has the greater number of sides. To prove that the perimeter of Q' > the perimeter of Q. Proof. Construct a regular polygon ^' having the same perimeter as ^ and the same number of sides as Q. Then Q' > Q". Therefore Q > Q'\ Whence the perimeter of Q> the perimeter of Q". But the perimeter of Q' = the perimeter of Q". Therefore the perimeter of Q > perimeter of (^. § 235 §182 Cons. 238. Corollary 1. Of all plane closed figures that are equal in arettf the circle has the minimum perimeter. MISCELLANEOUS EXERCISES. MAXIMA AND MINIMA 1. From two given points on the circumference of a circle to draw two lines meeting on a tangent to the circle and making a maximum angle with each other. 2. To inscribe the maximum rectangle in a circle. 3. To inscribe the maximum rectangle in a quadrant. 4. Prove that of all triangles having the same base and area, the isosceles triangle has the minimum perimeter. [Hint. Prolong AB to E, making AE — AC Draw EB and AB. Prove that t\ACB^t\ AEB so that BG - BE. Now use 3, § 28.] B' ./' TABLES TABLE I Ratios of the Sides of Right Triangles and Chords and Arcs of a Unit Circle TABLE II Squares and Square Roots of Numbers Cubes and Cube Roots of Numbers TABLE III Values of Important Numbers including Units of Measurement TABLE I RATIOS OF THE SIDES OF RIGHT TRIANGLES AND LENGTHS OF CHORDS AND ARCS OF A UNIT CIRCLE EXPLANATION OF TABLE I 1. Ratios of the Sides of Right Triangles. If an angle given in the Angle Column is one acute angle of a right triangle: The Sine Column gives the ratio of the side opposite the angle to the hypotenuse ; The Tangent Column gives the ratio of the side opposite the angle to the side adjacent to the angle. To find the Cosine of any angle, take the sine of the comple- ment of that angle. 2. Chords and Arcs of a Unit Circle. If an angle given in the Angle Column is an angle at the center of a circle of unit radius : The Chord Column gives the length of the chord that subtends that angle ; The Arc Column gives the length of the arc that subtends that angle. To find the lengths of chords or arcs of any circle of radius r, multiply the values given in the table by that radius. The table is limited to angles less than 90° ; but to find the chord that subtends an obtuse angle, first take half the angle, find the sine of this half angle, and multiply by 2. This follows from the fact that the chord of any angle is twice the sine oj half that angle. 1] Quantities Determined by a Given Angle iii Angle Sine Tan- gent Chord Arc Angle Sine Tan- gent Chord Arc 0°00' .0000 .0000 .0000 .0000 9° 00' .1564 .1584 .1569 .1.571 10 .0029 .0029 .0029 .0029 10 .1593 .1614 .1598 .1600 20 .0058 .0058 .0058 .0058 20 .1622 .1()44 .1627 .1629 30 ' .0087 .0087 .0087 .0087 :30 .1650 .1673 .1656 .1658 40 .0116 .0116 .0116 .0116 40 .1679 .1703 .1685 .1687 50 .0145 .0145 .0115 .0145 50 .1708 .1733 .1714 .1716 1°00' .0175 .0175 .0175 .0175 10° 00' .1736 .1763 .1743 .1745 10 .0204 .0204 .0204 .0204 10 .1765 .1793 .1772 .1774 20 .02:j;3 .0233 .0233 .0233 20 .1794 .1823 .1801 .1804 30 .0202 .0262 .0262 .0262 30 .1822 .1853 .18.-30 .1833 40 .0291 .0291 .0291 .0291 40 .1851 .1883 .18.59 .18(32 50 .0320 .0320 .0320 .0320 50 .1880 .1914 .1888 .1891 2° 00' .o;m9 .0M9 .0349 .0.349 11°00' .1908 .1944 .1917 .1920 10 .0378 .0378 .0378 .0378 10 .1937 .1974 .1946 .1949 20 .0407 .0407 .0407 .0407 20 .1965 .2004 .1975 .1978 m .04;341 .0640 .0f)40 40 .2193 .2247 .220(i .2211 50 .0669 .0670 .0669 .0069 50 .2221 .2278 .22(35 .2240 4° 00' .0098 .0699 .0698 .0698 13° 00' .2250 .2309 .2264 .2269 10 .0727 .0729 .0727 .0727 10 .2278 .2339 .2293 .2298 20 .0756 .0758 .0756 .0756 20 .2306 .2370 .2322 .2327 30 .0785 .0787 .0785 .0785 30 .2.3;^ .2401 .2351 .23,56 40 .0814 .0816 .0814 .0814 40 .2363 .2432 .2380 .2385 50 .0843 .0846 .0843 .0844 50 .2391 .2462 .2409 .2414 5° 00' .0872 .0875 .0872 .0873 14° 00' .2419 .2493 .2437 .2443 10 .OiWl Mm .0901 .0902 10 .2447 .2524 .2466 .2473 20 .m)29 .0934 .0931 .0931 20 .2476 .2555 .2495 .2502 ■30 .0958 .0963 .0960 .0<)()0 ;io .2504 .2586 .2524 .2531 40 .0987 .091 >2 .0989 .0989 40 .2532 .2617 .2553 .2560 .W .1016 .1022 .1018 .1018 50 .2560 .2648 .2582 .2589 6° 00' .1045 .1051 .1047 .1047 15°00' .2588 .2679 .2611 .2618 10 .1074 .1080 .1076 .1076 10 .2616 .2711 .2639 .2647 20 .1103 .1110 .1105 .1105 20 .2644 .2742 .2668 .2676 30 .1132 .1139 .1134 .1134 30 .2672 .2773 .2697 .2705 40 .1161 .1169 .1163 .1164 40 .2700 .2805 .2726 .2734 50 .1190 .1198 .1192 .1193 50 .2728 .2836 .2755 .27(53 7^ 00' .1219 .1228 .1221 .1222 16° 00' .2756 .2867 .2783 .2793 10 .1248 .1257 .1250 .1251 10 .2784 .2899 .2812 .2822 20 .1276 .1287 .1279 .1280 20 .2812 .2<)31 .2841 .2851 ;% .1305 .1317 .1308 .130<) ;30 .2840 .2962 .2870 .2880 40 .1334 .134^) .1337 .1338 40 .28(Wi .2^)94 .2899 .2i)09 50 .13()3 .1376 .13()0 .1367 50 .2896 .3026 .2927 .29(38 8° 00' .1392 .1405 .1395 .139(> 17° 00' .2924 .3057 .2956 .2967 10 .1421 .1435 .1424 .1425 10 .2952 .3089 .2985 .299(3 20 .1449 .14(;5 .1453 .1454 20 .2979 .3121 .3014 .3025 30 .1478 .1495 .1482 .1484 m .3007 .3153 .3042 ..3054 40 .1507 .1524 .1511 .1513 40 .3035 .3185 .3071 .3083 50 .1536 .1554 .1540 .1542 50 .3062 .3217 .3100 .3113 1 9° 00' Aofyi .1584 .1569 .1571 18° 00' .3090 .3249 .3129 .3142 iv Quantities Determined by . a, Given Angle [I Angle Sine Tan- gent Chord Arc Angle Sine Tan- gent Chord Arc 18°00' .3090 .3249 .3129 .3142 27° 00' .4540 .5095 .4669 .4712 10 .3118 .3281 .3157 .3171 10 .456() .5132 .4697 .4741 20 .3145 .3314 .3186 .3200 20 .4592 .5169 .4725 .4771 30 .3173 .3346 .3215 .3229 30 .4617 .5206 .4754 .4800 40 .3201 .3378 .3244 .3258 40 .4643 .5243 .4782 .4829 50 .3228 .3411 .3272 .3287 50 .4669 .5280 .4810 .4858 19° 00' .3256 .3443 .3301 .3316 28° 00' .4695 .5317 .4838 .4887 10 .3283 .3476 .3330 .3345 10 .4720 .5354 .4867 .4916 20 .3311 .3508 .3358 .3374 20 .4746 .5392 .4895 .4945 30 .3338 .3541 .3387 .3403 30 .4772 .5430 .4923 .4974 40 .3365 .3574 .3416 .3432 40 .4797 .5467 .4951 .5003 50 .3393 .3607 .3444 .3462 50 .4823 .5505 .4979 .5032 20° 00' .3420 .3640 .3473 .3491 29° 00' .4848 .5543 .5008 .5061 10 .3448 .3673 .3502 .3520 10 .4874 .5581 .5036 .5091 20 .3475 .3706 .3530 .3549 20 .4899 ..5619 .5064 .5120 30 .3502 .3739 .3559 .3578 30 .4924 .5658 .5092 .5149 40 .3529 .3772 .3587 .3607 40 .4950 .5696 .5120 .5178 50 .3557 .3805 .3616 .3636 50 .4975 .5735 .5148 .5207 21° 00' .3584 .3839 .3645 .3665 30° 00' ..5000 .5774 .5176 .5236 10 .3611 .3872 .3673 .3694 10 .5025 .5812 .5204 .5265 20 .3638 .3906 .3702 .3723 20 .5050 .5851 .5233 .5294 30 .3665 .3939 .3730 .3752 30 .5075 .5890 .5261 .5323 40 .3692 .3973 .3759 .3782 40 .5100 .5930 .5289 .5352 50 .3719 .4006 .3788 .3811 50 .5125 .5969 .5317 .5381 22° 00' .3746 .4040 .3816 .3840 31°00' .5150 .6009 .5345 .5411 10 .3773 .4074 .3845 .3869 10 .5175 .6048 .5373 .5440 20 .3800 .4108 .3873 .3898 20 .5200 .6088 .5401 .5469 30 .3827 .4142 .3902 .3927 30 .5225 .6128 .5429 .5498 40 .3854 .4176 .3930 .3956 40 .5250 .6168 .5457 .5527 50 .3881 .4210 .3959 .3985 50 .5275 .6208 .5485 .5556 23° 00' .3907 .4245 .3987 .4014 32° 00' .5299 .6249 .5513 .5585 10 .3934 .4279 .4016 .4043 10 .5324 .6289 .5541 .5614 20 .3961 .4314 .4044 .4072 20 .5348 .6330 .5569 ..5643 30 .3987 .4348 .4073 .4102 30 .5373 .6371 .5597 .5672 40 .4014 .4383 .4101 .4131 40 .5398 .6412 .5625 .5701 50 .4041 .4417 .4130 .4160 50 .5422 .6453 .5652 .5730 24° 00' .4067 .4452 .4158 .4189 33° 00' .5446 .6494 .5680 .5760 10 .4094 .4487 .4187 .4218 10 .5471 .6536 .5708 .5789 20 .4120 .4522 .4215 .4247 20 .5495 .6577 .5736 .5818 30 .4-147 .4557 .4244 .4276 30 .5519 .6619 .5764 .5847 40 .4173 .4592 .4272 .4305 40 .5544 .6661 .5792 .5876 50 .4200 .4628 .4300 .4334 50 .5568 .6703 .5820 .5905 25° 00' .4226 .4663 .4329 .4363 34° 00' .5592 .6745 .5847 .5934 10 .4253 .4699 .4357 .4392 10 .5616 .6787 .5875 .5963 20 .4279 .4734 .4386 .4422 20 .5640 .6830 .5903 .5992 30 .4305 .4770 .4414 .4451 30 .5664 .6873 .5931 .6021 40 .4331 .4806 .4442 .4480 40 .5688 .6916 .5959 .6050 50 .4358 .4841 .4471 .4509 50 .5712 .6959 .5986 .6080 26° 00' .4384 .4877 .4499 .4538 35° 00' .5736 .7002 .6014 .6109 10 .4410 .4913 .4527 .4567 10 .5760 .7046 .6042 .6138 20 .4436 .4950 .4556 .4596 20 .5783 .7089 .6070 .6167 30 .4462 .4986 .4584 .4625 30 .5807 .7133 .6097 .6196 40 .4488 .5022 .4612 .4654 40 .5831 .7177 .6125 .6225 50 .4514 .5059 .4641 .4683 50 .5854 .7221 .6153 .6254 27° 00' .4540 .5095 .4669 .4712 36° 00' .5878 .7265 .6180 .6283 1] Quantities Determined by a Giyen Angle V Angle Sine Tan- gent Chord Arc Angle Sine Tan- gent Chord A.C 36° 00' .5878 .7265 .6180 .6283 45° 00' .7071 1.0000 .7654 .7854 10 .5901 .7310 .6208 .6312 10 .7092 1.0058 .7681 .7883 20 .5i)25 .7355 .6236 .6341 20 .7112 1.0117 .7707 .7912 30 .5948 .7400 .6263 .6370 30 .7133 1.0176 .7734 .7941 40 .5972 .7445 .62^)1 .6400 40 .7153 1.0235 .7761 .7970 50 .5995 .7490 .6318 .6429 50 .7173 1.0295 .7788 .7999 37° 00' .6018 .7536 .6M6 .6458 46° 00' .7193 1.0355 .7815 .8029 10 .OOil .7581 .6374 .(J487 10 .7214 1.0416 .7841 .8058 20 .6065 .7627 .6401 .6516 20 .72^4 1.0477 .7868 .8087 30 .6088 .7673 .(>42i) .6545 30 .7254 1.0538 .7895 .8116 40 .6111 .7720 .6456 .6574 40 .7274 1.0599 .7922 .8145 50 .6134 .7766 .6484 .6603 50 .7294 1.0661 .7&48 .8174 38° 00' .6157 .7813 .6511 .6632 47° 00' .7314 1.0724 .7975 .8203 10 .6180 .78(50 .a539 .6661 10 .7333 1.0786 .8002 .8232 20 .6202 .7907 .(5566 .6690 20 .7353 1.0850 .8028 .8261 30 .6225 .7954 .6594 .6720 30 .7373 1.0913 .8055 .824 30 .7490 1.1303 .8214 .8465 40 .6383 .8292 Sim .(5923 40 .750*) 1.1369 .8241 .8494 50 .6406 .im2 .6813 .6952 50 .7528 1.1436 .8267 .8523 40° 00' .6428 .8391 .6840 .(5981 49° 00' .7547 1.1504 .8294 .8552 10 .(M50 .8441 .6868 .7010 10 .75(36 1.1571 .8320 .8581 20 .6472 .8491 .6895 .7039 20 .7585 1.1640 .8347 .8(510 30 .6494 .8541 .6922 .7069 30 .7604 1.1708 .8373 .8(539 40 .6517 .8591 .69.50 .7098 40 .7623 1.1778 .8400 .8668 50 .6539 .8642 .69/7 .7127 50 .7642 1.1847 .8426 .8698 41° 00' .6561 .8693 .7004 .7156 60° 00' .7(560 1.1918 .8452 .8727 10 .65a3 .8744 .7031 .7185 10 .7679 1.1988 .8479 .8756 20 .(5604 .8796 .7059 .7214 20 .7698 1.2059 .8505 .8785 30 .6626 .8847 .7086 .7243 30 .7716 1.2131 .8531 .8814 40 .(i648 .8899 .7113 .7272 40 .7735 1.2203 .8558 .8843 50 .f;670 .8952 .7140 .7301 50 .7753 1.2276 .8584 .8872 42° 00' .6691 .9004 .7167 .7330 51° 00' .7771 1.2349 .8610 .8901 10 .6713 .9057 .7195 .7359 10 .7790 1.2423 .8636 .8930 20 .6734 .9110 .7222 .7389 20 .7808 1.2497 .8663 .8959 30 .6756 .9163 .7249 .7418 30 .7826 1.2572 .8689 .8988 40 .6VVV .9217 .7276 .7447 40 .7844 1.2647 .8715 .9018 50 .6799 .9271 .7303 .7476 50 .7862 1.2723 .8741 .9047 43° 00' M20 .9325 .7330 .7505 62° 00' .7880 1.2799 .8767 .9076 10 .6841 .9380 .7357 .7634 10 .7898 1.2876 .8794 .9105 20 .6862 .9435 .7384 .7563 20 .7916 1.2954 .8820 .9134 30 MM .9490 .7411 .7592 m .79:34 1.3032 .884(> .9163 40 .6905 .9545 .7438 .7621 40 .7951 1.3111 .8872 .9192 50 .6926 .m)i .7465 .76.'>0 50 .79(59 1.3190 .8898 .9221 44° 00' .6947 .9657 .7492 .7(579 53° 00' .7986 1.3270 .8924 .9250 10 .6967 .9713 .7519 .7709 10 .8004 1.3351 .8950 .9279 20 .6988 .9770 .7546 .7738 20 .8021 1.3432 .8976 .9308 30 .7009 .9827 .7573 .7767 30 .80.39 1.3514 .9002 .9338 40 .7030 .9884 .7600 .7796 40 .8056 1.3597 .9028 .9367 50 .7050 .^1942 .7627 .7825 50 .8073 1.3680 .9054 .9396 45° 00' .7071 1.0000 .7654 .7854 54° 00' .80i)0 1.3764 .9080 .9425 vi Quantities Determined by . a Griven Angle [1 Angle Sine Tan- gent Chord Arc Angle Sine Tan- gent Chord Arc 54° 00' .8090 1.3764 .9080 .9425 63° 00' .8910 1.9626 1.0450 1.0996 10 .8107 1.3848 .9106 .9454 10 .8923 1.9768 1.0475 1.1025 20 .8124 1.39M .9132 .9483 20 .8936 1.9912 1.0500 1.1054 30 .8141 1.4019 .9157 .9512 30 .8949 2.0057 1.0524 1.1083 40 .8158 1.4106 .9183 .9541 40 .8962 2.0204 1.0549 1.1112 50 .8175 1.4193 .9209 .9570 50 .8975 2.0353 1.0574 1.1141 65°00' .8192 1.4281 .9235 .9599 64° 00' .8988 2.0503 1.0598 1.1170 10 .8208 1.4370 .9261 .9628 10 .9001 2.0655 1.0623 1.1199 20 .8225 1.4460 .9287 .9657 20 .i)013 2.0809 1.0648 1.1228 30 .8241 1.4550 .9312 .9687 30 .9026 2.0965 1.0672 1.1257 40 .8258 1.4641 .9338 .9716 40 .9038 2.1123 1.0697 1.1286 50 .8274 1.4733 .9364 .9745 50 .9051 2.1283 1.0721 1.1316 56° 00' .8290 1.4826 .9389 .9774 65° 00' .9063 2.1445 1.0746 1.1345 10 .8307 1.4919 .9415 .9803 10 .9075 2.1609 1.0771 1.1374 20 .8323 1.5013 .9441 .9832 20 .9088 2.1775 1.0795 1.1403 30 .8339 1.5108 .9466 .9861 30 .9100 2.1943 1.0820 1.1432 40 .8355 1.5204 .9492 .9890 40 .9112 2.2113 1.0844 1.1461 50 .8371 1.5301 .9518 .9919 50 .9124 2.2286 1.0868 1.1490 57° 00' .8387 1.5399 .9543 .9948 66° 00' .9135 2.2460 1.0893 1.1519 10 .8403 1.5497 .9569 .9977 10 .9147 2.2637 1.0917 1.1548 20 .8418 1.5597 .9594 1.0007 20 .9159 2.2817 1.0942 1.1577 30 .8434 1.5697 .9620 1.0036 30 .9171 2.2998 1.09()6 1.1606 40 .8450 1.5798 .9645 1.0065 40 .9182 2.3183 1.0990 1.1636 50 .8465 1.5900 .9671 1.0094 50 .9194 2.3369 1.1014 1.1665 58° 00' .8480 1.6003 .9696 1.0123 67° 00' .9205 2.3559 1.1039 1.1691 10 .84% 1.6107 .9722 1.0152 10 .9216 2.3750 1.1063 1.1723 20 .8511 1.6212 .9747 1.0181 20 .9228 2.3945 1.1087 1.1752 30 .8526 1.6319 .9772 1.0210 30 .9239 2.4142 1.1111 1.1781 40 .8542 1.6426 .i)798 1.0239 40 .9250 2.4342 1.1136 1.1810 50 .8557 1.6534 .9823 1.0268 50 .9261 2.4545 1.1160 1.1839 59° 00' .8572 1.6643 .9848 1.0297 68° 00' .9272 2.4751 1.1184 1.1808 10 .8587 1.6753 .9874 1.0327 10 .9283 2.4960 1.1208 1.1897 20 .8601 1.6864 .9899 1.0356 20 .9293 2.5172 1.1232 1.1926 30 .8616 1.6977 .9924 1.0385 30 .9304 2.5386 1.1256 1.1956 40 .8631 1.7090 .9950 1.0414 40 .9315 2.5605 1.1280 1.1985 50 .8646 1.7205 .9975 1.0443 50 .9325 2.5826 1.1304 1.2014 60° 00' .8660 1.7321 1.0000 1.0472 69° 00' .9336 2.6051 1.1328 1.2043 10 .8675 1.7437 1.0025 1.0501 10 .9346 2.6279 1.1352 1.2072 20 .8689 1.7556 1.0050 1.0530 20 .9356 2.6511 1.1376 1.2101 30 .8704 1.7675 1.0075 1.0559 30 .9367 2.6746 1.1400 1.2130 40 .8718 1.7796 1.0101 1.0588 40 .9377 2.6985 1.1424 1.2159 50 .8732 1.7917 1.0126 1.0617 50 .9387 2.7228 1.1448 1.2188 61° 00' .8746 1.8040 1.0151 1.0647 70° 00' .9397 2.7475 1.1472 1.2217 10 .8760 1.8165 1.0176 1.0676 10 .9407 2.7725 1.1495 1.2246 20 .8774 1.8291 1.0201 1.0705 20 .9417 2.7980 1.1519 1.2275 30 .8788 1.8418 1.0226 1.0734 30 .9426 2.8239 1.1543 1.2305 40 .8802 1.8546 1.0251 1.0763 40 .9436 2.8502 1.1567 1.2334 50 .8816 1.8676 1.0276 1.0792 .50 .9446 2.8770 1.1590 1.2363 62° 00' .8829 1.8807 1.0301 1.0821 71°00' .9455 2.9042 1.1614 1.2392 10 .8843 1.8940 1.0326 1.0850 10 .9465 2.9319 1.1638 1.2421 20 .8857 1.9074 1.0351 1.0879 20 .9474 2.9(500 1.1661 1.2450 30 .8870 1.9210 1.0375 1.0908 30 .9483 2.9887 1.1685 1.2479 40 .8884 1.9347 1.0400 1.0937 40 .9492 3.0178 1.1709 l.?508 50 .8897 1.9486 1.0425 1.0966 50 .9502 3.0475 1.1732 1.2537 63° 00' .8910 1.9626 1.0450 1.0996 72° 00' .9511 3.0777 1.1756 1.2566 1] Quantities Determined by a Given Anj^le vii Angle Sine Tan- gent Chord Arc Angle Sine Tan- gent Chord Arc 72^00' .9511 3.0777 1.1756 1.2566 81° 00' .9877 6.3138 1.2989 1.4137 10 .9520 3.1084 1.1779 1.2595 10 .9881 6.4348 1.3011 1.4166 20 .9528 3.1397 1.1803 1.262,-. 20 .988(5 6.560(5 1.3033 1.4195 riO .9537 3.1716 1.1826 1.2654 30 .98iK) 6.(5912 1.3055 1.4224 40 .9546 3.2041 1.1850 1.2683 40 .9894 6.82(59 1.3077 1.4254 oO .9555 3.2371 1.1873 1.2712 5(3 .9899 6.9682 1.3m)9 1.4283 73° 00' .9503 3.2709 1.1896 1.2741 82° 00' .9903 7.11.54 1.3121 1.4312 10 .t)572 3.3052 1.1920 1.2770 10 .945 10 .9929 8.3450 1.3274 1.4515 20 .9()28 3. .5656 1.2083 1.2974 20 .9932 8.5555 1.32ii6 1.4.544 30 .di'm 3.()059 1.2106 1.3003 30 .9936 8.7769 1.3:318 1.4573 40 .%44 3.6470 1.2129 1.3032 40 .9939 9.0098 1.3339 1.4603 50 .mo2 3.6891 1.2152 1.3061 50 .9942 9.2553 1.33(51 1.4632 75° 00' .96,59 3.7321 1.2175 1.3090 84° 00' .9945 9.5144 l.a383 1.4661 10 .9667 3.7760 1.2198 1.3119 10 .9948 9.7882 1.3404 1.4(590 20 .9(J74 3.8208 1.2221 1.3148 20 .9951 10.078 l.:3426 1.4719 :^o .9()81 3.86<)7 1.2244 1.3177 30 .9954 10.385 1.3447 1.4748 40 .9689 3.9136 1.2267 1.3206 40 .9^357 10.712 1.34(59 1.4777 r)0 .96i)(j 3.9617 1.2290 1.3235 50 .9959 11.059 IMM 1.4806 76^00' .9703 4.0108 1.2313 1.3265 85° 00' .9962 11.430 1.3512 1.48,35 10 .9710 4.0611 1.2336 1.3294 10 .91>64 11.826 1.3533 1.48(54 20 .9717 4.1126 1.2359 1.3323 20 .9967 12.251 1.3555 1.4893 :iO .9724 4.1653 1.2382 1.3352 30 .99(59 12.706 1.3576 1.4923 40 .97;^X) 4.2193 1.2405 1.3:381 40 .9971 13.197 1.3597 1.49,52 50 .9737 4.2747 1.2428 1.3410 50 .9974 13.727 1.3619 1 .4981 77° 00' .9744 4.3315 1.2450 1.3439 86° 00' .9976 14.301 1.3640 l.,5010 10 .9750 4.3897 1.2473 1.3468 10 Mm 14.924 1.3661 1.5039 20 .9757 4.4494 1.2496 1.3497 20 .9980 15.605 1..3()82 l.,5068 30 .9763 4.5107 1.2518 1.3526 30 .9<)81 16.350 1.3704 l.,5097 40 .9769 4.5736 1.2541 1.3555 40 .9983 17.169 1.3725 1.5126 50 .9775 4.6382 1.2564 1.3584 50 .9985 18.075 1.3746 1.5155 78^00' .9781 4.7046 1.2586 1.3614 87° 00' .9986 19.081 1.3767 1.5184 10 .9787 4.7729 1.2609 l.:3(>43 10 .9988 20.206 1.3788 1.5213 20 .9793 4.8430 1.2632 1.3672 20 .9989 21.470 1..3809 1.5243 30 .9799 4.9152 1.2654 1.3701 30 .9990 22.904 1.3830 1.5272 40 .9805 4.9894 1.2677 1..3730 40 .9992 24.542 1.3851 1.5301 50 .9811 5.0658 1.2699 1.3759 W .99f)3 26.432 1.3872 1.5330 79° 00' .9816 5.144<; 1.2722 1.3788 88° 00' .9994 28.6:36 1..3893 1.5:3.59 10 .9822 5.22r»7 1.2744 l.:i817 10 .090 1.C0161 1.64303 1.68516 1.0a540 1.03850 1.04158 1.044()4 1.04769 1.05072 1.05373 1.05672 1.05970 2.23070 2.23738 2.24402 2.25062 2.25718 2.26370 2.27019 2.27664 2.28305 4.80590 4.82028 4.83459 4.84881 4.86294 4.87700 4.89097 4.90487 4.91868 1.20 1.4400 1.09545 3.40410 1.72800 1.062G6 2.28943 4.93242 1.21 1.22 1.23 1.24 1.25 1.26 1.27 1.28 1.29 1.4641 1.4884 1.5129 1.5376 1.5625 1.5876 1.6129 1.6384 1.6641 1.10000 1.10454 1.10905 1.11355 1.11803 1.12250 1.12694 1.13137 1.1.3578 3.47851 3.49285 3.50714 3.52136 3.53553 3.54965 3.56371 3.57771 3.5916(} 1.77156 1.81585 1.86087 1.90(5(32 1.95312 2.00038 2.04838 2.09715 2.14669 1.06560 1.06853 1.07144 1.07434 1.07722 1.08008 1.08293 1.08577 1.08859 2.29577 2.30208 2.30835 2.31459 2.32079 2.32697 2.33311 2.33921 2.34529 4.94609 4.95968 4.97319 4.986(33 5.00000 5.01330 5.02653 5.03968 5.05277 1.30 1.6900 1.14018 3.60,555 2.19700 1.09139 2.35133 5.06580 1.31 1.32 1.33 1.34 1.35 1.36 1.37 1.38 1.39 1.7161 1.7424 1.7689 1.7956 1.8225 1.8496 1.8769 l.i>044 1.9321 1.14455 1.14891 1.15326 1.15758 1.16190 1.16619 1.17047 1.17473 1.17898 3.61939 3.63318 3.64692 3.66060 3.67423 3.68782 3.70135 3.71484 3.7282V 2.24809 2.299<)7 2.35264 2.40610 2.46038 2.51546 2.57135 2.62807 2.68562 1.09418 1.0969(5 1.09972 1.10247 1.10521 1.10793 1.11064 1.113:M 1.11602 2.35735 2.3111 1.26619 1.2()827 1.27033 1.27240 1.27445 1.27650 1.27854 2.71893 2.72344 2.72792 2.73239 2.73(585 2.74129 2.74572 2.75014 2.75454 5.85777 5.86746 5.87713 5.88677 5.89637 5.90594 5.91548 5.92499 5.93447 2.10 4.4100 1.44914 4.58258 9 26100 1.28058 2.75892 5.94392 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 4.4521 4.4944 4.5369 4.5796 4.6225 4.6656 4.7089 4.7524 4.7961 1.45258 1.45602 1.45945 1.46287 1.46629 1.46969 1.47309 1.47648 1.47986 4.59347 4.60435 4.61519 4.62601 4.63681 4.64758 4.65833 4.66905 4.67974 9.39393 9.52813 9.66360 9.80034 9.93838 10.0777 10.2183 10.3602 10.5035 1.28261 1.28463 1.28665 1.28866 1.2906(5 1.29266 1.29465 1.29664 1.29862 2.76330 2.7(5766 2.77200 2.77633 2.78065 2.78495 2.78924 2.79352 2.79779 5.95334 5.96273 5.97209 5.98142 5.99073 6.00000 6.00925 6.01846 6.02765 2.20 4.8400 1.48324 4.69042 10.6480 1.30059 2.80204 6.03681 2.21 2.22 2.23 2.24 2.2,5 2.26 2.27 2.28 2.29 4.8841 4.9284 4.9729 5.0176 5.0625 5.1076 5.1529 5.1984 5.2441 1.48661 1.48997 1.49332 1.49666 1.50000 1.50333 .1.5066.5 1.50997 1.51327 4.70106 4.71169 4.72229 4.73286 4.74342 4.75395 4.76445 4.77493 4.78539 10.7939 10.9410 11.0896 11.2394 11.3906 11.5432 11.6971 11.8524 12.0090 1.30256 1.30452 1.30648 1.30843 1.31037 1.31231 1.31424 1.31617 1.31809 2.80(528 2.81050 2.81472 2.81892 2.82311 2.82728 2.83145 2.83560 2.815974 6.04594 6.05505 6.06413 6.07318 6.08220 6.09120 6.10017 6.10911 6.11803 2.30 5.2900 1 51658 4.79583 12.1670 1.32001 2.84387 6.12693 2.31 2.32 2.33 2.34 2.35 2.36 2.37 2.38 2.39 5.3361 5.3824' 5.4289 5.475(> 5.5225 5.5696 5.6169 5.()(;44 5.7121 1.51987 1.52315 1.52643 1.52971 1.53297 1.53623 1.53948 1.54272 1.54596 4.80625 4.81664 4.82701 4.83735 4.84768 4.85798 4.86826 4.87852 4.88876 12.3264 12.4872 12.6493 12.8129 12.9779 13.1443 13.3121 13.4813 13.6519 1.32192 1.32382 1.32572 1.32761 1.32950 1.33139 1.33326 1.33514 1.33700 2.84798 2.85209 2.85618 2.86026 2.86433 2.86838 2.87243 2.87646 2.88049 6.13579 6.144(53 6.15345 6.16224 6.17101 6.17975 6.18846 6.19715 6.20582 2.40 5.7600 1.54919 4.89898 13.8240 1.3.3887 2.884.50 6.21447 2.41 2.42 2.43 2.44 2.45 2.46 2.47 2.48 2.49 5.8081 5.8564 5.9019 5.9536 6.0025 6.0516 6.1009 6.1504 6.2001 1.55242 1.55563 1.55885 1.56205 l..")(r)25 1.56H44 1.57162 1.57480 1.57797 4.90918 4.91935 4.92950 4.93964 4.<^975 4.95984 4.96991 4.97996 4.98;)<>9 13.9975 14.1725 14.3489 14.5268 14.70(51 14.8869 15.0692 15.2530 15.4382 1.34072 1.34257 1.34442 1.34626 1.34810 1.34993 1.36176 1.35358 1.35540 2.88850 2.89249 2.89647 2.90044 2.90439 2.90834 2.91227 2.91620 2.92011 6.22308 6.231(58 6.24025 6.24880 6.25732 6.26583 6.27431 6.28276 6.29119 Xll Powers and Roots [li n tt2 V^ \/Wn n^ 72i 2.92402 6.29961 2.51 2.52 2.53 2.54 2.55 2.56 2.57 2.58 2.59 6.3001 6.3501 6.4009 6.4516 6.5025 6.5536 6.6049 6.6564 6.7081 1.58430 1.58745 1.59060 1.59374 1.59687 1.60000 1.60312 1.60624 1.60935 5.00999 5.01996 5.02991 5.03984 5.04975 5.05964 5.06952 5.07937 5.08920 15.8133 16.0030 16.1943 16.3871 16.5814 16.7772 16.9746 17.1735 17.3740 l.a5902 1.36082 1.36262 1.36441 1.36620 1.36798 1.36976 1.37153 1.373.30 2.92791 2.93179 2.93567 2.93953 2.94338 2.94723 2.95106 2.95488 2.95869 6.30799 6.31636 6.32470 6.33303 6.34133 6.34960 6.35786 6.36610 6.37431 2.60 6.7600 1.61245 5.09902 17.57<50 1.. 37507 2.9(5250 6.38250 2.61 2.62 2.63 2.64 2.65 2.66 2.67 2.68 2.69 6.8121 6.8(544 6.9169 6.9696 7.0225 7.0756 7.1289 7.1824 7.2361 1.61555 1.61864 1.62173 1.62481 1.62788 1.63095 1.63401 1.63707 1.64012 5.10882 5.11859 5.128,35 5.13809 5.14782 5.15752 5.16720 5.17687 5.18652 17.7796 17.9847 18.1914 18.3997 18.6096 18.8211 19.0342 19.2488 19.4651 1.37683 1.37859 1.38034 1.38208 1.38383 1.38557 1.38730 1.38903 1.3<)076 2.96629 2.97007 2.97385 2.97761 2.98137 2.98511 2.98885 2.99257 2.99629 6.390(58 6.39883 6.40696 6.41507 6.42316 6.43123 6.43928 6.44731 6.45531 2.70 7.2900 1.64317 5.19(515 19.(5830 1 .39248 3.00000 6.46330 2.71 2.72 2.73 2.74 2.75 2.76 2.77 2.78 2.79 7.3441 7.3984 7.4529 7.5076 7.5625 7.6176 7.6729 7.7284 7.7841 1.64621 1.64924 1.65227 1.65529 1.65831 1.66132 1.66433 1.66733 1.67033 5.20577 5.21536 5.22494 5.23450 5.24404 5.25357 5.26308 5.27257 5.28205 19.9025 20.1236 20.3464 20.5708 20.7969 21.0246 21.2539 21.4850 21.7176 1.39419 1.39591 1.39761 1.39932 1.40102 1.40272 1.40441 1.40610 1.40778 3.00370 3.00739 3.01107 3.01474 3.01841 3.02206 .3.02570 3.02934 3.03297 6.47127 6.47922 6.48715 6.49507 6.502944 36.9260 37.2597 37.5954 37.9331 38.2728 38.6145 38.9582 1.49031 1.49181 1.49330 1.49480 1.49(529 1.49777 1.49926 1.50074 1.50222 3.21077 3.21400 3.21722 3.22044 3.22365 3.22686 3.2.3006 3.23325 3.23643 6.91740 6.92436 6.93130 6.93823 6.94515 6.95205 6.95894 6.96582 6.97268 3.40 11.5(500 1.84391 5.83095 39.3040 1.50.369 3.2.31X51 6.97953 3.41 3.42 3.43 3.44 3.45 3.46 3.47 3.48 3.49 11.6281 11.6964 11.7649 11.8336 11.9025 11.9716 12.0409 12.1104 12.1801 1.846<52 1.84932 1.85203 1.85472 1.8.-)742 1.86011 1.86279 1.86548 1.86815 5.83952 5.84808 5.85662 5.86515 5.87367 5.88218 5.89067 5.89915 5.90762 39.(5518 40.0017 40.3536 40.7076 41.0636 41.4217 41.7819 42.1442 42.5085 1.50517 1.50664 1.50810 1.50957 1.51103 1.51249 1.51394 1.51540 1.51685 3.24278 3.24.195 3.24911 3.25227 3.25.142 3.25856 3.26169 3.26482 3.26795 6.98(537 6.99319 7.00000 7.00680 7.013,58 7.02035 7.02711 7.03385 7.04058 XIV Powers and Roots ui n in? V^ VlO/i n^ ^n ^ron 4 3.40851 3.41138 3.41424 3.41710 7.31238 7.318(51 7.32483 7.33104 7.33723 7.34342 7.34960 7.35576 7.36192 H] Powers and Hoots XV n n2 V^ VlOn n^ ifH ^10 n «51 7.64032 7.64603 7.(55172 7.65741 xvi Vi 3wers and Roots cn n n^ Vn VlOw, n^ ^n ^lOn ^100 1* 4.50 20.2500 2.12132 6.70820 91.1250 1.65096 3.55(389 7.66309 4.51 4.52 4.53 4.54 4.55 4.56 4.57 4.58 4.59 20.3401 20.4304 20.5209 20.6116 20.7025 20.7936 20.8849 20.9764 21.0681 2.12368 2.12603 2.12838 2.13073 2.13307 2.13542 2.13776 2.14009 2.14243 6.71.565 6.72309 6.73053 6.73795 6.74537 6.75278 6.76018 6.76757 6.77495 91.73.39 92.34.54 92.9597 93.5767 94.1964 94.8188 95.4440 96.0719 96.7026 1.65219 1.65.341 1.654(32 1.65584 1.65706 1.65827 1.65948 1.66069 1.66190 3.55953 3.56215 3.56478 3.56740 3.57002 3.57263 3.57524 3.57785 3.58045 7.66877 7.67443 7.68009 7.68573 7.69137 7.69700 7.70262 7.70824 7.71384 4.60 21.1600 2.14476 6.78233 97.3360 1.66310 3.58305 7.71944 4.61 4.62 4.63 4.64 4.65 4.66 4.67 4.68 4.69 21.2521 21.3444 21.4369 21.5296 21.6225 21.7156 21.8089 21.9024 21.9961 2.14709 2.14942 2.15174 2.15407 2.15639 2.15870 2.16102 2.16333 2.16564 6.78970 6.79706 6.80441 6.81175 6.81909 6.82(342 6.83374 6.84105 6.848.% 97.9722 98.6111 99.2528 99.8973 100,545 101.195 101.848 102.503 103.162 1.66431 1.66551 1.66671 1.66791 1.66911 1.67030 1.67150 1.072(39 1.67388 3.58564 3.58823 3.59082 3.59,340 3.59598 3.59856 3.60113 3.60370 3.60626 7.72503 7.73061 7.73619 7.74175 7.74731 7.75286 7.75840 7.76394 7.76946 4.70 22.0900 2.16795 6.85565 103.823 1.67507 3.60883 7.77498 4.71 4.72 4.73 4.74 4.75 4.76 4.77 4.78 4.79 22.1841 22.2784 22.3729 22.4676 22.5625 22.6576 22.7529 22.8484 22.9441 2.17025 2.17256 2.17486 2.17715 2.17945 2.18174 2.18403 2.18(;;;2 2.18861 6.86294 6.87023 6.87750 6.88477 6.89202 6.89928 6.90652 6.91375 6.92098 104.487 105.154 105.824 106.496 107.172 107.850 108.531 109.215 109.902 1.67626 1.67744 1.67863 1.67981 1.(38099 1.68217 1.683.34 1.68452 1.68569 3.61138 3.61394 3.61649 3.61903 3.02158 3.62412 3.62665 3.62919 3.63172 7.78049 7.78599 7.79149 7.79697 7.80245 7.80793 7.81339 7.81885 7.82429 4.80 23.0400 2.19089 6.92820 110.592 1.68687 3.63424 7.82974 4.81 4.82 4.83 4.84 4.85 4.86 4.87 4.88 4.89 23.1361 23.2324 23.3289 23.4256 23.5225 23.6196 23.7169 23.8144 23.9121 2.19317 2.19545 2.19773 2.20000 2.20227 2.20454 2.20681 2.20907 2.21133 6.9.3542 6.94262 6.94982 6.9.5701 6.96419 6.97137 6.978.54 6.98570 6.99285 111.285 111.980 112.679 113..380 114.084 114.791 115.501 116.214 116.930 1.68804 1.68920 1.69037 1.69154 1.69270 1.69386 1.69503 1.69619 1.69734 3.63676 3.63928 3.64180 3.64431 3.64682 3.64932 3.65182 3.65432 3.65681 7.83517 7.84059 7.84601 7.85142 7.85683 7.86222 7.86761 7.87299 7.87837 4.90 24.0100 2.21359 7.00000 117.649 1.69850 3.65931 7.88374 4.91 4.92 4.93 4.94 4.95 4.96 4.97 4.98 .4.99 24.1081 24.20(34 24.3049 24.4036 24..5025 24.6016 24.7009 24.8004 24.9001 2.21.585 2.21811 2.22036 2.22261 2.22486 2.22711 2.22935 2.23159 2.23383 7.00714 7.01427 7.02140 7.02851 7.03562 7.04273 7.04982 7.05691 7.0(5399 118.371 119.095 119.823 120.554 121.287 122.024 122.763 123.506 124.251 1.69965 1.70081 1.701()6 1.70311 1.70426 1.70.540 1.70655 1.70769 1.70884 3.66179 3.66428 3.66676 3.66924 3.67171 3.67418 3.67(365 3.67911 3.68157 7.88909 7.89445 7.89979 7.90513 7.91046 7.91578 7.92110 7.92641 7.93171 UJ Powers and Roots XVll n W2 VTi VlOn n^ ^ ^10 w ^100 n 5.00 25.0000 2.23607 7.07107 125.000 1.70993 3.68403 7.93701 5.01 5.02 5.03 5.04 5.05 5.06 5.07 5.08 5.09 25.1001 25.2004 25.3009 25.4016 25.5025 25.6036 25.7049 25.8064 25.9081 2.23830 2.24054 2.24277 2.24499 2.24722 2.24944 2.25167 2.25389 2.25610 7.07814 7.08520 7.09225 7.09930 7.10654 7.11337 7.12039 7.12741 7.13442 125.752 12().50() 127.264 128.024 128.788 129.554 130.324 131.097 131.872 1.71112 1.71225 1.71339 1.71452 1.715(56 1.71679 1.71792 1.71i)05 1.72017 3.68649 3.68894 3.69138 3.69383 3.69627 3.69871 3.70114 3.70:357 3.70600 7.*)4229 7.94757 7.95285 7.95811 7.96337 7.96863 7.97387 7.97911 7.984)34 6.10 26.0100 2.25832 7.14143 132.051 1.72i;^ 3.70843 7.98957 5.11 5.12 5.13 5.14 5.15 5.16 5.17 5.18 5.19 26.1121 26.2144 26.3169 26.4196 26.5225 26.6256 26.7289 26.8324 26.9361 2.26053 2.20274 2.26495 2.26716 2.26936 2.27156 2.27376 2.275<)() 2.27816 7.14843 7.15542 7.16240 7.16938 7.17635 7.18;>3l 7.19027 7.19722 7.20417 133.433 134.218 135.006 135.797 13(5.591 137.388 138.188 138.992 139.798 1.72242 1.72;i55 1.724(57 1.72579 1.72691 1.72802 1.72914 1.73025 1.73137 3.71085 3.71327 3.71569 3.71810 3.72051 3.72292 3.72532 372772 3.73012 7.9^)479 8.00000 8.00520 8.01040 8.01559 8.02078 8.02596 8.03113 8.03629 5.20 27.0400 2.28035 7.21110 140.608 1.73248 3.73251 8.04145 5.21 5.22 5.23 5.24 5.25 5.26 5.27 5.28 5.29 27.1441 27.2484 27.3529 27.4576 27.5(525 27.6676 27.7729 27.8784 27.9841 2.28254 2.28473 2.28692 2.28910 2.29129 2.29a47 2.29565 2.29783 2.30000 7.21803 7.22496 7.23187 7.23878 7.24569 7.25259 7.25<)48 i.mm 7.27324 141.421 142.237 143.056 143.878 144.703 145.532 146.363 147.198 148.036 1.73359 1.73470 1.73580 1.73(591 1.73801 1.73912 1.74022 1.74132 1.74242 3.73490 3.73729 3.73968 3.74206 3.74443 3.74681 3.74918 3.75155 3.75392 8.04(560 8.05175 8.05689 8.06202 8.06714 8.07226 8.07737 8.08248 8.08758 5.30 28.0900 2.30217 7.28011 148.877 1.74351 3.75629 8.0i)267 5.31 5.32 5.33 5.34 5.35 5.36 5.37 5.38 5.39 28.1CJ61 28.:3024 28.4089 28.5156 28.6225 28.729«j 28.8369 28.9444 29.0521 2.30434 2.30651 2.30868 2.31084 2.31301 2.31517 2.31733 2.31948 2.32164 7.28<;97 7.29;i83 7.30068 7.30753 7.31437 7.32120 7.32803 7.33485 7.34166 149.721 loO.odO 151.419 152.273 1.53.130 153.991 154.854 155.721 156.591 1.74461 1.74570 1.74680 1.74789 1.748i)8 1.75007 1.75116 1.75224 1.75333 3.75865 3.76101 3.76336 3.76571 3.76806 3.77041 3.77275 3.77609 3.77743 8.09776 8.10284 8.10791 8.11298 8.11804 8.12310 8.12814 8.13319 8.13822 5.40 29.1600 2.32379 7.34847 157.4<)4 1.75441 3.77976 8.14325 5.41 5.42 5.43 5.44 5.45 5.46 5.47 5.48 5.49 29.2681 29.3764 29.4849 29.5936 29.7025 29.8116 29.9209 ;i0.O304 30.1401 2.325iH 2.3280i) 2.33024 2.33238 2.33452 2.33666 2.33880 2.34094 2.34307 7.35.527 7.36206 7.3688.5 7.37.5(>4 7.38241 7.38918 7.39594 7.40270 7.40945 158.340 159.220 160.103 160.989 161.879 162.771 163 667 164.5()7 165.469 1.75549 1.75657 1.75765 1.75873 1.75981 1.76088 1.76196 1.7(5303 1.76110 3.78209 3.78442 3.78675 3.78907 3.79139 3.79371 3.79603 3.79834 3.80065 8.14828 8.15329 8.15831 8.16331 8.16831 8.17330 8.17829 8.18327 8.18824 xviii Powers and Roots Lli n n2 Vn \/Wn n^ 5 3.92787 3.9,'5003 3.93219 3.9:34:34 8.43901 8.44:3(59 8.44836 8.45:303 8.457(59 8.46235 8.46700 8.47165 8.47(529 6.10 37.2100 2.46982 7.81025 226.981 1.82716 3.93650 8.48093 6.11 6.12 6.13 6.14 6.15 6.16 6.17 6.18 6.19 37.3(?21 37.4544 37.5769 37.69<>6 37.8225 37.i^56 38.0689 38.1924 38.3161 2.47184 2.47386 2.47588 2.47790 2.479<)2 2.48193 2.48395 2.485% 2.48797 7.81665 7.82304 7.82943 7.83582 7.84219 7.84857 7.85493 7.86130 7.86766 228.099 229.221 230.346 231.476 232.()08 233.745 234.885 230.029 237.177 1.82816 1.82915 1.83015 1.83115 1.83214 1.83313 1.83412 1.83511 1.83610 3.93865 3.94079 3.94294 3.94508 3.94722 3.94936 3.95150 3.95363 3.95576 8.4855G 8.49018 8.49481 8.49942 8.50403 8.50864 8.51324 8.51784 8.52243 6.20 38.4400 2.48998 7.87401 238.328 1.83709 3.95789 8.52702 6.21 6.22 6.23 6.24 6.25 6.26 6.27 6.28 6.29 38.5(i41 38.6884 38.8129 38.9376 39.062.5 39.1876 39.3129 39.4384 39.5(^1 2.49199 2.49.399 2.4^)600 2.49800 2.50000 2.50200 2.50400 2.50599 2.50799 7.88036 7.88670 7.89303 7.89937 7.90569 7.91202 7.91833 7.92465 7.93095 239.483 240.642 241.804 242.971 244.141 245.314 246.492 247.673 248.858 1.83808 1.83906 1.84005 1.84103 1.84202 1.84300 1.84398 1.844% 1. 84504 3.9(5002 3.9(5214 3.96427 3.96638 3.96850 3.97062 3.97273 3.97484 3.97695 8.5:3160 8.5:3618 8.54075 8.54532 8.54988 8.55444 8.55899 8..56a54 8.56808 6.30 39.6900 2mm 7.93725 250.047 1.84691 3.97906 8.57262 6.31 6.32 6.33 6.34 6.35 6.36 6.37 6.38 6.39 39.8161 39.9424 40.0689 40.1956 40.3225 40.4496 40.5769 40.7044 40.8321 2.51197 2.51396 2.51595 2.51794 2.5]9<>2 2.52190 2.52389 2.52587 2.52784 7.94355 7.iM984 7.95613 7.96241 7.96869 7.97496 7.98123 7.98749 7.99375 251.240 252.436 253.636 254.840 2rKi.048 257.259 258.475 259.6<)4 2()0.917 1.84789 1.84887 1.84984 1.85082 1.85179 1.85276 1.85373 1.85470 1.85567 3.98116 3.98326 3.98536 3.98746 3.98956 3.99165 3.99374 3.99583 3.99792 8.57715 8.58168 8.58620 8.59072 8.59524 8.59975 8.60425 8.60875 8.61.325 6.40 40.9600 2.52<)82 8.00000 262.144 1.85664 4.00000 8.61774 6.41 6.42 6.43 6.44 6.45 6.46 6.47 6.48 6.49 41.0881 41.2164 41.3449 41.4736 41.(K)25 41.7316 41.8609 41.9904 42.1201 2.53180 2.53377 2.53574 2.53772 2.531^39 2.54165 2.54362 2.54558 2.54755 8.00«)25 8.01249 8.01873 8.02496 8.03119 8.03741 8.04363 8.04984 8.05605 263..375 264.609 265.848 267.090 268.3.36 2(59.586 270.840 272.098 273.359 1.&5760 1.85857 1.85953 1.86050 1.86146 1.86242 1.86338 1.86434 1.86530 4.00208 4.00416 4.00624 4.008,32 4.010:39 4.01246 4.01453 4.01660 4.01866 8.62222 8.(52671 8.63118 8.6.3.566 8.64012 8.64459 8.64904 8.(35350 8.65795 XX Powers and Roots [n n W,2 yJn VlO»i n^ 16 49.7025 49.8436 49.9849 50.1264 50.2681 2.64764 2.(>4963 2.65141 2.65330 2.05518 2.65707 2.66895 2.66083 2.66271 8.37257 8.37854 8.3^451 8.39047 8.39()43 8.40238 8.40833 8.41427 8.42021 344.472 345.948 347.429 348.914 350.403 351.896 353.393 354.895 .3.5<).401 1.91384 1.91475 1.91566 1.916,57 1.91747 1.91838 1.91929 1.92019 l.t)210{) 4.12325 4.12521 4.12716 4.12912 4.13107 4.13303 4.13498 4.13693 4.1.3887 8.88.327 8.88749 8.89171 8.89592 8.90013 8.90434 8.90854 8.91274 8.91(593 7.10 50.4100 2.66458 8.42615 357.911 1.92200 4.14()S2 8.92112 7.11 7.12 7.13 7.14 7.15 7.16 7.17 7.18 7.19 50.5521 50.6944 50.8369 50.9796 51.1225 51.2656 51.4089 51. .5524 51.6^1 2.66646 2.66833 2.67021 2.67208 2.67395 2.67582 2.67769 2.67955 2.68142 8.43208 8.43801 8.44393 8.44985 8.45577 8.46168 8.46759 8.47.'U9 8.47939 359.425 360.944 362.467 363.994 365.526 367.062 368.602 370.146 371.695 1.92290 1.92380 1.92470 1.92560 1.92650 1.92740 1.92829 1.92919 1.9,3008 4.14276 4.1-1470 4.146(54 4.14858 4.15052 4.15245 4.15438 4.15(531 4.15824 8.92531 8.92949 8.93367 8.93784 8.94201 8.94618 8.95034 8.95450 8.95866 7.20 51.8400 2.68328 8.48,528 373.248 1.93098 4.1(5017 8.96281 7.21 7.22 7.23 7.24 7.25 7.26 7.27 7.28 7.29 51.9841 52.1284 52.2729 52.4176 52.5()25 52.7076 62.8529 52.9984 53.1441 2.68514 2.68701 2.68887 2.69072 2.69258 2.69444 2.69629 2.(59815 2.70000 8.49117 8.49706 8.50294 8.50882 8.51469 8.52056 8.52643 8.53229 8.53816 374.806 37(5.367 377.933 379.503 381.078 382.657 384.241 385.828 .387.420 1.93187 1.93277 1.93366 1.93456 1.9:3544 1.93633 1.93722 1.93810 1.93899 4.1(5209 4.10402 4.16594 4.16786 4.16978 4.17169 4.17361 4.17552 4.17743 8.96696 8.97110 8.97524 8.97938 8.98,351 8.98764 8.99176 8.99588 9.00000 7.30 53.2900 2.70185 8.54400 389.017 1.93988 4.17934 9.00411 7.31 7.32 7.33 7.34 7.35 7.36 7.37 7.39 53.43()1 53.5824 53.7289 53.8756 54.0225 64.1696 54..3169 54.4(>44 54.6121 2.70370 2.70.555 2.70740 2.70924 2.71109 2.71293 2.71477 2.71662 2.7184(> 8.54985 8.55.570 8.56154 8.56738 8.57.321 8.57904 8.58487 8.59069 8.59661 390.618 392.223 393.833 395.447 397.0(;5 398.688 400..316 401.947 403.583 1.94076 1.94165 1.94253 1.94341 1.944.30 1.94518 1.94606 1.94694 1.94782 4.18125 4.18315 4.18506 4.18696 4.18886 4.19076 4.19266 4.194.55 4.19644 9.00822 9.01233 9.01643 9.02053 9.02462 9.02871 9.03280 9.0.36H9 9.04097 7.40 54.7600 2.72029 8.602.33 405.224 1.94870 4.198,34 9.04504 7.41 7.42 7.43 7.44 7.45 7.46 7.47 7.48 7.49 64.9081 55.0564 56.2049 55.:«36 55.5025 55.651() 55.8009 5;"). 9504 56.1001 2.72213 2.72397 2.72580 2.72764 2.72947 2.73130 2.7a313 2.73496 2.73679 8.60814 8.61394 8.61974 8.62554 8.63134 8.63713 8.64292 8.64870 8.6544H 406.869 408.618 410.172 411.8,31 413.494 415.161 416.833 418.609 420.190 1.94957 1.95045 1.95132 1.95220 1.95307 1.95.395 1.95482 1.955(59 1.956,56 4.20023 4.20212 4.20400 4.20589 4.20777 4.20966 4.21153 4.21341 4.21529 9.04911 9.05318 9.05725 9.06131 9.06537 9.06942 9.07347 9.07752 9.08156 xxu Powers and Ro ols Cn n n^ V^ VlOw n^ 4 4.530(55 9.76100 9.31 9.32 9.33 9.3i 9.35 9.36 9.37 9.38 9.39 9.40 9.41 9.42 9.43 9.44 9.45 9.46 9.47 9.48 9.49 8(5.6761 8(3.8(524 87.0i89 87.2356 87.4225 87.6006 87.7969 87.9844 88.1721 3.05123 3.05287 3.0.5450 3.05014 3.05778 3.05041 3.00105 3.06208 3.0(5431 9.64883 9.65401 9.65019 9.60437 9.(3(3i)54 9.07471 9.07988 9.(58504 9.(SiK)20 800.054 800.5.-)8 812.166 814.781 817.400 820.026 822.0,57 825.2M 827.0.36 2.10370 2.10445 2.10520 2.10505 2.10671 2.10746 2.10821 2.10806 2.10071 4.53228 4.5:3300 4.53552 4.53714 4.53876 4.54038 4..54109 4.54301 4.54522 9.7(3450 9.70799 9.77148 9.77497 9.77846 9.78105 9.78543 9.78891 9.79239 88.3600 3.0(55i)4 9.605.36 830.584 2.11045 4.54684 9.79586 88.5481 88.7:364 88.9249 89.113(5 89.3025 89.4916 89.6809 80.8704 90.0001 3.0(5757 3.00020 3.07083 3.07246 3.07409 3.07571 3.07734 3.07806 3.08058 9.70052 9.70567 9.71082 9.71597 9.72111 9.72025 9.73139 9.736.53 9.74166 8.33.238 835.807 838.562 841.232 843.000 846.501 840.278 851.W1 854.670 2.11120 2.11105 2.11270 2.11344 2.11410 2.114^ 2.11568 2.11642 2.11717 4.54845 4.55006 4.55167 4.55328 4.55488 4.55649 4.55809 4.55{>70 4.56130 9.79933 9.80280 9.80627 9.80974 9.81320 9.81666 9.82012 9.82:357 9.82703 xxvi Powers a nd Roots [11 n ^2 Vn VlOn n^ ^n