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LIBRARY 
 
 OF THE 
 
 University of California. 
 
 GIFT OF 
 
 Class 
 
Digitized by the Internet Archive 
 
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 Microsoft Corporation 
 
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ELEMENTAEY AND 
 CCXNSTEUCTIONAL GEOMETEY 
 
ELEMENTARY 
 
 AND 
 
 Constructional Geometry 
 
 BY 
 
 EDGAE H. NICHOLS, A.B. 
 
 OF THE BROWNE AND NICHOLS SCHOOL, CAMBRIDGE, MASS. 
 I 
 
 
 OF THE ^ 
 
 UNIVERSITY 
 
 OF 
 
 ȣiUFORN\L 
 
 NEW YOKE 
 LONGMANS, GEEEN, AND CO. 
 
 LONDON AND BOMBAY 
 1896 
 
/_T3 / i> ^ 
 
 Copyright. 1896 
 
 BY 
 
 LONGMANS. GREEN. AND CO. 
 
 TWOW DIMCTOHY 
 
 miMTINO AND BOOKBINDING COMPANY 
 
 NEW YORK 
 
PREFACE 
 
 This book is designed for pupils beginning Geometry at 
 the age of twelve, or even younger. It is based upon the 
 author's class-room experience with young boys during the 
 last twelve years. It is hoped that it may prove an aid to 
 those teachers who do not want a text-book except as a 
 guide to the pupil, to enable him to review the principles 
 developed in the class by answering for himself a set of 
 questions similar to those asked in the class-room, and by 
 working out practical problems illustrative of the princi- 
 ples studied. It cannot be used as a text-book from which 
 lessons are assigned in advance to the class. Each prin- 
 ciple should be developed by the teacher and the class, 
 working together — the principle coming at the end, not at 
 the beginning, of the study — and then a lesson from the 
 book, covering essentially the same ground, should be as- 
 signed. The book has not been divided into lessons, be- 
 cause sometimes a single paragraph will afford material for 
 several lessons, and, again, several pages can be mastered in 
 one lesson. Each teacher should be free to develop his 
 subject in his own way without artificial lines of division. 
 
 The main object of the first year's study is to make the 
 pupils perfectly familiar with the use of their tools, so that 
 in the study of Theoretical Geometry the construction of 
 the figures will present no difficulty. In the pursuit of 
 this object a great many useful working principles are 
 learned, which greatly simplify the more advanced study 
 of the subject. 
 
 The proof of principles is introduced very gradually, 
 
 183979 
 
VI PREFACE 
 
 and the use made of even the few proofs given must de- 
 pend upon the age and the ability of the particular class 
 concerned. 
 
 From the. beginning it is insisted that the pupil .shall 
 have a clear idea of the exact direction of any line named 
 with two letters. No slovenliness of thought should be 
 permitted in this matter. For greater accuracy of expres- 
 sion two words have been coined — symparalhl and anti- 
 parallel. Parallel lines that have the same direction 
 are called symparallel lines, and those that have exactly 
 opposite directions are called antiparallel lines. Thus, 
 antiparallel lines form an angle of 180° with each other, 
 but symparallel lines form no angle. It is found that this 
 distinction prevents all confusion in the study of parallels, 
 and helps to make the pupils think clearly and express 
 themselves with accuracy. Simple and natural signs are 
 used for these words. 
 
 Attention is called to the treatment of areas. The au- 
 thor has found that the subject lends itself especially to 
 the object aimed at in the first year's study. An indefinite 
 field for the ingenuity of the pupils is opened, and often 
 excellent results have been obtained. In his manipulation 
 of areas the ordinary problems of construction (often 
 reserved as the very last subject of Geometry) become a 
 mere matter of course to the pupil, and his familiar ac- 
 quaintance with valuable principles is assured. A similar 
 treatment of volumes opens a still larger field, if the time 
 allotted to the subject allows it. 
 
 A proper use of the Summary will add greatly to the in- 
 terest and to the value of the subject. In the first place, 
 each teacher can word his definitions to suit his own con- 
 victions, and can decide what principles he will take up, 
 and how those principles shall be expressed. In the sec- 
 ond place, the pupil feels an added interest in the princi- 
 
PREFACE vii 
 
 pies, because he has helped to decide upon the best word- 
 ing of them. An important feature of the work, there- 
 fore, is the training that it gives in accuracy of expression. 
 The teacher ought first to make the pupils express the prin- 
 ciples in their own words, and then by interpreting their 
 English literally, to bring out any inaccuracies of expres- 
 sion. 
 
 A pupil who has acquired a familiar knowledge of the 
 principles developed in this book should be able to take up 
 the study of Theoretical Geometry, both plane and solid, 
 in a text-book where no complete proofs, but suggestions 
 only, are given to aid in the solution of the more difficult 
 problems and theorems. He should be encouraged to con- 
 sult, after his own solution, the text-books that contain 
 proofs (of which there should be several kinds in the school 
 library), in order to develop his critical power and to add 
 life to the subject. 
 
 The author has in mind the preparation of a Geometry 
 to supplement this book, in which the principles of Plane 
 and Solid Geometry will be developed according to the sug- 
 gestions just made. He will be grateful to any who will 
 call his attention to typographical errors and ambiguous 
 expressions in this book that may have crept in, in spite of 
 the greatest care on the part of himself and of friends, who 
 have kindly read and re-read the proofs. He is indebted 
 to one of his former pupils for the best of the diagrams. 
 
 Edgar H. Nichols. 
 
 December, 1895. 
 
CONTENTS 
 
 Preface 
 
 
 
 EXERCISES 
 
 PAGE 
 V 
 
 Section I. 
 
 Introductory . 
 
 
 1-23 
 
 1 
 
 Section II. 
 
 Introductory . 
 
 
 24-58 
 
 5 
 
 Section III. 
 
 Lines .... 
 
 
 59-74 
 
 10 
 
 Section IV. 
 
 Angles 
 
 
 75-112 
 
 14 
 
 Section V. 
 
 Parallels . 
 
 
 113-121 
 
 22 
 
 Section VI. 
 
 Review 
 
 
 . 122-139 
 
 26 
 
 Section VII. 
 
 Parallels (Continued) 
 
 
 140-171 
 
 28 
 
 Section VIII. 
 
 Triangles . 
 
 
 172-195 
 
 32 
 
 Section IX. 
 
 Review 
 
 
 196-209 
 
 39 
 
 Section X. 
 
 Isosceles Triangles 
 
 
 . 210-246 
 
 41 
 
 Section XI. 
 
 Triangles (Continued) 
 
 
 
 
 Section XII. 
 
 Quadrilaterals 
 
 
 281-323 
 
 55 
 
 Section XIII. 
 
 Division of Lines 
 
 
 324-366 
 
 65 
 
 Section XIV. 
 
 Multiplication of Lines ; Area 
 
 i 367-398 
 
 72 
 
 Section XV. 
 
 Moulding Areas ; Equivalen- 
 
 p 
 
 
 
 Figures .... 
 
 399-436 
 
 79 
 
 Section XVI. 
 
 Peculiarities of Squares 
 
 
 87 
 
 Section XVII. 
 
 Areas of Similar Figures 
 
 . 533-565 
 
 104 
 
 Section XVIII 
 
 Circles and Inscribed Angles 
 
 } 566-625 
 
 111 
 
 Section XIX. 
 
 Volumes .... 
 
 . 626-678 
 
 123 
 
 Index . 
 
 
 
 . 137 
 
 Summary of Facts, Definitions, and Prd 
 
 (CIPLl 
 
 2S 
 
 . 139 
 
ELEMENT AEY 
 
 AND 
 
 CONSTRUCTIONAL GEOMETRY 
 
 SECTION I. 
 
 1. In the group of drawings on the third page are out- 
 lines of several familiar bodies or solids. Learn their 
 names, and find as many examples as you can of each kind 
 of solid in the objects about you. Can you name any 
 other solids not represented there ? 
 
 2. One of the aims that we shall keep before us in our 
 first study of Geometry will be to discover the characteris- 
 tics of these and other common solids. We shall study at 
 first with special care the nature of their bounding sur- 
 faces and of their bounding lines ; we shall learn how to 
 make accurate drawings and measurements of them ; and 
 we shall study the peculiarities of form in the lines, sur- 
 faces, and solids. 
 
 3. In looking at the physical solids about us, we think 
 not only of their form, but also of the material of which 
 they are made, of their color, and of their fineness of fin- 
 ish ; but in dealing with geometrical solids, we think of 
 the form only. If, for example, you move a cube of glass 
 from its place, and then imagine the space which it first 
 occupied to be bounded by surfaces and lines, it will help 
 
2 ELEMENTARY AND 
 
 you to form an idea of a geometrical solid. You must, 
 however, be on your guard against one danger. Your 
 glass cube, however skilfully made, will have imperfec- 
 tions in it ; the lines will not be true, or the surfaces will 
 not be perfectly smooth : but the geometrical cube of your 
 imagination is absolutely perfect ; every line is true, every 
 surface perfectly smooth. It is a great aid to the imagina- 
 tion to use physical solids for illustration ; but you must 
 never forget that the geometrical solids which they repre- 
 sent, and which you are to study, have none of their im- 
 perfections. In all your drawings, too, you must aim to 
 have your work as near perfection as is possible ; to this 
 end you must provide yourselves with well-sharpened hard 
 pencils, a good ruler, and a strong pair of compasses. It is 
 better to have one pencil set aside for drawing only, and to 
 have that sharpened with a chisel edge for use with a ruler. 
 But you must never imagine that the solids have any of the 
 faults of your drawings. 
 
 4. The geometrical solid, then, is a portion of space 
 bounded by imaginary surfaces. It is limited in three di- 
 rections at right angles to each other ; or, in other words, 
 it is said to have three dimensions, length, breadth, and 
 thickness. Sometimes height or depth is used in place of 
 thickness. Can you think of any practical examples where 
 height or depth would be a better word than thickness ? 
 Do you know of any word to take the place of breadth ? 
 
 5. Examine a cube (Fig. 1). Which dimension do you 
 call its length, which its breadth, and which its thickness ? 
 Is there any difference in the magnitude of these dimen- 
 sions ? Can a cube have unequal dimensions ? 
 
 6. Can you draw a cone or a cylinder with equal dimen- 
 sions ? 
 
 7. What do you consider to be the height of a pyramid ? 
 
 8. Can you suggest any practical way of finding the 
 height of a (physical) pyramid ? 
 
 9. Which of the solids represented must have equal di- 
 mensions from their very nature ? 
 
CONSTRUCTIONAL GEOMETRY 3 
 
 Fig. 1. 
 
 Fig. 2. 
 
 Fig. 3. 
 
 Fig. 4. 
 
 Fig. 5. Fig. 6. 
 
 Fro. 7. 
 
 Fig. 8. 
 
4 ELEMENTARY AND 
 
 10. Is there any solid in the group which cannot have 
 equal dimensions without changing its nature ? 
 
 11. How would you proceed to measure the dimensions 
 of a cone of wood ? 
 
 12. The boundaries of a solid are called surfaces, and the 
 intersections of the surfaces are called lines. The bound- 
 ing surfaces form the faces of the solid ; the intersections 
 of the faces form the edges of the solid. 
 
 13. How many faces are needed to bound a cube ? How 
 many edges ? How many faces and edges are needed for a 
 triangular prism ? for a cone ? for a cylinder ? for a sphere ? 
 
 14. How many and what dimensions has a surface ? Is 
 the leaf of your book a surface or a solid ? Why ? Is one 
 surface any thinner than another in Geometry ? Is a coat 
 of paint a surface P Why ? 
 
 15. Can you, by placing a great many surfaces together, 
 one upon another, form a solid ? Is it right to say that a 
 surface is a very thin piece of a solid ? 
 
 16. How many and what dimensions has a geometrical 
 line ? You speak of drawing a line. Is what you draw a 
 geometrical line ? What is it ? Why ? What is the proper 
 name in Geometry for a "telegraph line" ? for a pencil 
 mark ? When you speak of a "fine line" or of a "coarse 
 line," what are you really describing ? 
 
 17. The ends of a line are points. What dimensions can 
 a point have ? Can you make a point with a pencil ? 
 When you indicate the position of a point by a pencil mark 
 or dot, what is the proper geometrical name for the dot ? 
 
 18. How many points are indicated in Fig. 1 ? (page 3) 
 in Fig. 3 ? in Fig. 4 ? 
 
 10. By means of what yon have already learned, complete 
 the sentences that follow, and copy them neatly on the 
 blank leaf at the end of the book headed " Summary of 
 Facts, Definitions, and Principles, " or copy them on some 
 special page of your note-book as yon may be directed. 
 
 20. Solids have dimensions, and are bounded 
 
 by 
 
CONSTRUCTIONAL GEOMETRY 5 
 
 21. Surfaces have dimensions, and are bounded 
 
 by 
 
 22. Lines have dimensions, and are bounded 
 
 by ••• 
 
 23. Points have dimensions ; they show position 
 
 only. 
 
 SECTION II. 
 
 24. If it is not possible to make points, or to draw lines, 
 are we not blocked at the very beginning of our course ? 
 Not so, if we remember what our drawings really amount 
 to ; they are merely aids to the imagination ; the dot, which 
 stands for a point, enables us to be sure that all are think- 
 ing of the same point or position ; the line A B is an aid to 
 the eye, and makes it easier to remember that a line is 
 imagined between the points A and B ; but the imagina- 
 tion should always go behind the poor imitation of a line 
 to the absolutely perfect line represented by it. If the line 
 is not needed to aid the imagination, it is better not to 
 draw it ; for instance, in the figures of Section I. a great 
 many lines are left for the imagination to supply, because 
 the mind does not need them to form a true picture. Give 
 examples of omitted lines in those figures. With the ex- 
 planation just given we can 
 speak of drawing lines and 
 of making points without 
 causing any confusion. 
 
 25. Imagine a point A 
 to move from its position to 
 the position B. It is clear 
 that you can imagine it 
 to move along several 
 paths. How many ? How 
 many dimensions will any 
 one of these paths have ? Why? What, then, is the proper 
 name for the imaginary path ? See 20-23. 
 
6 ELEMENTARY AND 
 
 26. We can, therefore, describe a line in another way, 
 which is better for our purposes upon the whole, as fol- 
 lows : A line is the path traced by a point as it moves 
 from one position to another. Record this definition in 
 the u Summary." 
 
 27. There are four paths or lines between A and B pict- 
 ured in the figure ; the upper path is called a curved line ; 
 the next, A C D B, is called a broken line; the next, A B, 
 a straight line ; and the bottom one a mixed line. 
 
 28. Write out a definition of each kind in your own 
 language, aiming to deeoribe the characteristics of each so 
 clearly that there will be no confusion among the kinds. 
 Bring your results into the class-room, and after deciding 
 upon the best wording, learn the definitions which ere 
 selected as the best. After this, if a line between two 
 points is mentioned, a straight line is always to be imagined 
 unless otherwise gpecifled. 
 
 '.".». Select examples of the different kinds of lines from 
 the figures 1 to 8 on page :*. 
 
 30. Can you, without removing your pencil from the 
 pftper, and without going over the same path twice, trace 
 the outline of Figs. 1, 2, 3, 5, and 
 
 31. Can you trace all four paths in Fig. 9 with a con- 
 tinuous movement ? 
 
 :;.'. With your compasses 
 trace a circumference. Write 
 as good a definition as you can 
 of a circumference viewed as 
 the path of a moving point. 
 
 Draw three concentric cir- 
 cumferences. (Look up the 
 meaning of the word concen- 
 tric, if necessary.) 
 
 33. Construct Fig. 10 with 
 your compasses, beginning Pio. 10. 
 
 with the central circumference and using one radius for 
 all the circumference-. What is meant by the radius of a 
 
CONSTRUCTIONAL GEOMETRY 7 
 
 circumference ? Take great pains with this figure, and do 
 not give up until you have an exact and neat copy of it. 
 
 34. Construct Fig. 11, beginning with three points one 
 inch from each other. 
 
 Fig. 11. 
 
 Fig. 12. 
 
 35. Construct Fig. 12, putting the centres at the corners 
 of a square an inch each way. 
 
 36. Construct a spiral (Fig. 13) by using as the centre for 
 the first semi- circumference 
 A C B and every alternate 
 one, and by using A as the 
 centre for the second semi- 
 circumference B D E and 
 every alternate one. 
 
 37. Construct Fig. 14. 
 
 38. Construct Fig. 15, by 
 first placing two drawing-pins 
 or small tacks about 2J in. 
 apart at M and N ; then, 
 fastening a piece of fine 
 thread 3 in. long to the two pins, with your pencil pressed 
 firmly against the thread, trace the curve. This curve is 
 called an ellipse; Try the experiment of putting the pins 
 nearer together without altering the length of the string. 
 
 Fig. la 
 
 OF THE 
 
 UNIVERSITY 
 
8 
 
 KLKM E\TMIY A XI) 
 
 Put the pins close together ; what kind of a curve do yon 
 now trace ? The ellipse, in addition to being a beautiful 
 curve, has a peculiar interest to us because it is the path 
 
 Fio. 15. 
 
 Fio. 14. 
 
 which the earth traces as it moves about the sun. The sun 
 is at one of the places represented by the pins. 
 
 39. Imagine a point to move from A to D (Fig. 16), 
 tracing the line A D, and then imagine the line A D to 
 move, without tipping, to a 
 new position, B C. Can the 
 line take more than one path f 
 How many ? Make illustra- 
 tions for your answers. 
 
 40. How many dimensions 
 has any one of these paths ? 
 Why ? What should the path 
 be called? See 20-23. 
 
 41. Can you move a line in 
 such a way that it will not generate a surface ? 
 
 42. Can you trace the lateral, or side, surfaces of Fig. 
 1 by the movement of a line as described in 40 ? of Fig. 2 ? 
 of Fig. 4 ? of Fig. 5 ? 
 
 43. How would you move a line to generate the lateral 
 surface of Fig. 3 ? 
 
 Fio. 16. 
 
CONSTRUCTIONAL GEOMETRY 
 
 9 
 
 A 
 
 Fig. 17. 
 
 44. What kind of line would you move to generate the 
 surface of Fig. 6, and in what way would you move it ? 
 
 45. Can you generate the lateral surface of Fig. 5 by 
 the movement of a curved line ? 
 
 46. Draw a curved line ABGD (Fig. 17). Move a line 
 A P, without tipping it, so that 
 
 A shall remain on the curve. 
 Will a surface or a solid be gen- 
 erated ? Why ? 
 
 47. Compare the surfaces 
 which have been generated in 
 exercises 39-42, and tell which 
 are plane and which are curved 
 surfaces. Write a good defini- 
 tion of a plane surface that will 
 distinguish it from a curved 
 surface. Select the best definition from those handed in 
 by the class, and record it in the " Summary." 
 
 Suggestion : Think how a carpenter tests with his straight- 
 edge the evenness of a floor or table. 
 
 48. Imagine the point A to move to D (Fig. 18), generat- 
 ing the line A D ; the line A D 
 to move to B C, generating the 
 surface A B C D ; and the sur- 
 face A B C D to move to the 
 position E F G H. Could the 
 surface take more than one path 
 in moving to its new position ? 
 Illustrate your answer. How 
 many dimensions in any one of 
 the paths ? 
 
 49. Could you generate by 
 the motion of a surface the solids of Figs. 1, 5, 6, and 8 ? 
 
 50. In what way could you move a surface to generate 
 the solids of Figs. 3, 6, and 7 ? 
 
 51. In what two ways could you generate the cylinder, 
 Fig. 5 ? 
 
10 ELEMENTARY AND 
 
 52. Can you move a surface so that it will not generate 
 a solid P 
 
 53. Review from 24, and complete the sentences that fol- 
 low, following the directions given in 19 : 
 
 54. A line is the path traced 
 
 55. A straight line il the path traced 
 
 56. A surface is the path traced 
 
 57. A solid is generated when 
 
 58. A solid of revolution is generated when 
 
 SECTION III. LINES. 
 
 59. Since a straight line is merely the path traced by a 
 point as it moves, without change of direction, from one 
 position to another, the line can have only two character- 
 istics, which are determined by the direction in which the 
 point moves and by the distance which it moves. It is 
 v. iv important in naming a line to indicate, if possible, the 
 direction of it ; it is, therefore, important to put first the 
 point from which the moving point is imagined to start; 
 
 \ample, the line A B . . Q 
 
 ( Kg. 19) is the path traced ' 
 
 by a point moving, without 
 
 change of direct ion. from A P 
 
 to B, not from B to A. N 
 
 The direction of the line u 
 A B is, therefore, from A to 
 B : the length of A B is the *»• u 
 
 distance between A and B. Describe the lines M N, M A, 
 \ M. B M (Fig. 19) in a similar way. It ought not to be 
 necessary for you to represent the lines in this case to get a 
 clear idea of them. The points that limit the lines should 
 enable you to form a distinct picture of the lines in your 
 mind. 
 
 60. In what wavs does the line B A differ from I» M ': 
 
CONSTRUCTIONAL GEOMETRY 11 
 
 How does M N differ from MNP? (Fig. 19). How does 
 M N differ from N M ? 
 
 61. How many different straight lines can you draw from 
 M/oN ? (Fig. 19). Can you give a reason for your answer ? 
 
 An axiom is a truth that is assumed as self-evident, so 
 that it needs no reason. Can you state the axiom that you 
 have made use of in your answer to 61 ? 
 
 62. Can you draw through M and N more than one 
 straight line ? (The straight lines that you draw must 
 differ in some way. ) Can you draw between M and N as 
 the limiting points more than one straight line ? Compare 
 line MNP with M N, and N M with M N before giving 
 your answer, and be sure that the axiom of 61 is worded 
 exactly. 
 
 63. Sometimes, when the direction of a line is of no con- 
 sequence, a line may be named by a single letter; for 
 example, A B (Fig. 19) might be called " I" (Small letters 
 are used when it is desired to name a line in this way.) It 
 is clear, however, that the name " I '* gives no suggestion as 
 to whether the line points from A to B, or from B/oA; 
 that is, there is doubt about one of the two characteristics 
 that are possible for a line. Therefore, unless the direc- 
 tion is of no consequence, or unless the figure in itself makes 
 doubt impossible, it is unwise to use a single letter in naming 
 a line. Of course a line should never be named B A, if A B 
 is the line in one's mind. Beginners cannot be too careful 
 about this point. 
 
 64. On one indefinite straight line mark two lines that 
 shall differ in length, but not in direction ; also on one 
 straight line mark two lines that shall differ in direction, 
 but not in length. 
 
 65. On one straight line mark two lines that differ both 
 in length and in direction. 
 
 66. In the exercises that follow pay no attention to the 
 direction of the lines, but try to estimate accurately the 
 lengths of the lines. Estimate the shorter lines in inches 
 or centimeters, the longer lines in feet or decimeters. 
 
12 ELEMENTARY AND 
 
 Make your estimate first, and record it ; then measure the 
 line, and record the result in a second column ; in i third 
 column put the error made. 
 
 Make five lines, a, b, c, d, e, of different lengths and in 
 different positions ; estimate their lengths in inches, and 
 record results as explained above. 
 
 G7. Make a line, and try to make five others, in various 
 positions, of the same length ; record your errors in milli- 
 meters. 
 
 68. Make a six-sided figure with sides varying consider- 
 ably in length. Estimate the lengths in centimeters, and 
 record estimated lengths, true lengths, and errors in three 
 columns as before. 
 
 Let one of the class put ten different lines on the 
 board, keeping to himself the true lengths of the lines ; 
 let the rest of the class hand in to him their estimates in 
 feet, and let him average the errors made by each one. 
 
 7<>. Make a horizontal line ; try to make a vertical line 
 and a slanting or oblique line of the same length as the 
 horizontal line. BeoOfd errors in millimeters. It is best 
 to measure a line with your compasses, and to apply your 
 compasses to the ruler. 
 
 71. Estimate in inches the height of a chair (to the 
 seat), the dimensions of a window-pane, the width of a 
 door. Record results as before. 
 
 Note. — Exercises of a similar nature can be given prof- 
 itably, until each pupil becomes fairly expert in estimating 
 lengths. 
 
 72. In comparing the lengths of two lines, a and b, we 
 say that a is equal to b, a is greater than b, or a is less than 
 b, according to the facts, recording the result in one of the 
 following ways : a = b, a > b or a < b. Note the meaning 
 of the signs, which are used in comparing any two magni- 
 tudes. 
 
 Make two broken lines of three parts each ; measure the 
 total length of eacli line, and put the proper sign of the 
 three signs just explained between the two results. 
 
CONSTRUCTIONAL GEOMETRY 13 
 
 73. Make three different lines a, b, and c ; make one 
 line equal to a + b + c. Make another line equal to a + 
 b — c ; another equal to a — b + c. Put the proper sign 
 between a + b — c and a — b + c. 
 
 74. Making use of the lines a, b, c, d, and e in Fig. 20, 
 find the line equal to a — b + c — d +e; find also b — 2c 
 
 Fig. 20. 
 
 + d — 3d ; a — 3c + b — e. Record the length of the 
 answer in millimeters for each case. Since all of the class 
 are using the same lines, the answers should come out alike. 
 Do they ? If not, can you account satisfactorily for the 
 differences, supposing each one to have taken the greatest 
 possible care ? 
 
 Although measuring a line with compasses and ruler 
 seems a very simple thing in theory, in practice it is a very 
 hard thing to do, when great accuracy is required. This 
 is due partly to the fact that the points, which represent 
 the ends of the lines, are very far from being true points, 
 and partly to the inaccuracy of our tools. Two lines that 
 seem to be of the same length, when measured with an 
 ordinary ruler, might prove to differ distinctly when 
 measured with a more delicate tool such as the micrometer 
 screw. Do you know what a micrometer screw is ? If not, 
 let some of the class find out what it is, for the next lesson, 
 securing one, if possible, for actual use. Do you know of 
 any other device for measuring lengths with great ac- 
 curacy ? 
 
 The result of your work in the exercises of 74 and what 
 has just been said should teach you that it is not safe 
 to rely wholly upon measurement when it is to be de- 
 cided whether two lines are of equal length or not. Lines 
 
14 
 
 ELEMENTARY AND 
 
 that seem to the eye or to the tools to be equal are often 
 unequal ; while lines that seem to be unequal arc often 
 equal. The sides of a true square, for instance, we know 
 must be equal from the nature of the square, but the sides 
 of the figure that we draw to represent the square may be 
 unequal. You must, of course, use your tools to draw 
 figures to aid your imagination ; but you must think more 
 of the imaginary figures than of the actual figure W0D by 
 you, and you must not think it a sufficient reason to prove 
 two lines equal in length to say that you have measured tin- 
 lines and found them to be equal. Some reason, based 
 upon the nature of the figure as in the case of the square 
 above, must be given, if possible. 
 
 SKCTION IV. AXGLKS. 
 
 75. In the last few exercises you have thought only of 
 the length of the lines ; now consider their direction. 
 If two lines do not point exactly alike, they are said to 
 make an angle with each other. Thus in the oblong 
 ABC I) K PG (Fig. W), the 
 edge A B is said to make an 
 angle with the edges AD, B C, 
 B I\ B B, and D G, because it 
 does not point as they du. 
 
 Does A B make an angle 
 with DO? Why? 
 
 Does A B make an angle with 
 G F ? with C B ? Why ? 
 
 You have very likely used the 
 word " angle " in the place of the word 
 in saying that a house was in the angle of two streets, you 
 meant that it was just at the corner in the space between 
 the two streets ; or, perhaps, you have thought of an angle 
 as the Space included between two lines. But in Geometry 
 
 "corner;" thus, 
 
CONSTRUCTIONAL GEOMETRY 
 
 15 
 
 the angle has to do solely with the direction of the lines ; 
 if the lines point alike, they form no angle ; if they differ 
 in direction, they form an angle. To make an angle larger, 
 we do not produce its sides that they may inclose more 
 space between them, but we make the difference in direc- 
 tion greater by turning one of the lines away from the 
 other. Starting with your compasses closed, turn one leg 
 gradually away from the other : the two legs at once 
 begin to point in different 
 directions so that they form 
 an angle with each other ; 
 as you keep on turning, 
 the legs point farther and 
 farther away from each 
 other, going through the 
 various positions shown in 
 Fig. 22, and the angle grows 
 larger and larger. In time Fig. 22. 
 
 the turning leg points exactly opposite to A in the direc- 
 tion F. If we turn the leg still farther to the position of 
 G, there is not so much difference in the way the legs 
 point as there was when the turning leg was at F ; in 
 other words, the angle is growing smaller again ; if we keep 
 on turning, presently the leg will point again as A does, 
 and there will be no angle formed. 
 
 76. Enough has now been said about an angle to enable 
 us to say what the angle really is. An angle is the differ- 
 ence in direction of two lines. It indicates how much you 
 must turn one of the lines to make it point like the other. 
 The amount that one of the lines has to be turned to make 
 it point like the other is a very good measure of the size of 
 the angle. The angle is not the corner or the place where 
 two lines meet ; it is not the space between the lines ; it 
 is not anything that you can see : but it is a difference 
 to be determined by the judgment, as any other differ- 
 ence is determined. Perhaps the best illustration from 
 familiar things, to help us to form a clear idea of what 
 
16 ELEMENTARY AND 
 
 an angle really is, is to be found in colors or in sounds. 
 We see two shades of red, for instauce. We see that 
 there is a difference, but we do not see the difference 
 itself ; that is something to be determined by judgment. 
 If we are well trained in distinguishing colors, we can 
 estimate the difference in shade with great accuracy. 
 Again, when we hear two different notes on the piano, we 
 can estimate the difference in pitch exactly, if our ears 
 have been trained to distinguish sounds. In like manner, if 
 we train ourselves to distinguish differences in direction, we 
 can estimate the size of the angle with surprising accuracy. 
 We must, however, have some means of expressing our 
 estimates. In colors ire estimate by shades; in sounds, by 
 tones and semi-tones. What is to take the place of these 
 measures in the case of angles ? 
 
 77. In your experiment in 75 (Fig. 22), as you turned 
 the leg of the dividers, you made H point more and more 
 away from the direction of A, but nearer and nearer to 
 the direction of F, which has the direction of AO exactly 
 opposite to that of OA. There must have been one 
 position of O B whin it diverged from A just as much as 
 it did from OF. In that position (0 D in Fig. 88) it 
 made a right angle with A and also with F, so that 
 the adjacent ingiM POD and AOD were equal. In 
 Geometry the right angle, or the angle formed when one 
 line meets another so as to make the adjacent angles equal. 
 serves to divide angles into two large classes. Any angle 
 less than a right angle is called an acnte angle ; any angle 
 larger than a right angle is called an obtuse angle. It is 
 also common to call all angles that are not right angles, 
 whether acute or obtuse, oblique angles. 
 
 78. Select as many right angles as you can in Fig. 21. 
 In an ordinary room the two walls intersect in a corner line 
 which makes a right angle with the edges of the floor. 
 Imagine several lines on the floor starting from a corner. 
 What angles do these lines make with the corner line of the 
 two walls ? Can you draw on the floor or ceiling a straight 
 
CONSTRUCTIONAL GEOMETRY 17 
 
 line that does not make a right angle with this corner 
 line ? 
 
 The right angle is perhaps the most interesting of all the 
 angles. It is the angle that the carpenter must under- 
 stand thoroughly; if he makes mistakes in his right 
 angles, his work can never look well. How many devices 
 of the carpenter for getting true right angles do you know ? 
 
 79. Another angle that plays an important part in 
 Geometry is the angle formed when one line points in a 
 direction exactly opposite to that of another line. This 
 angle is called a straight angle ; it is the largest angle dealt 
 with in elementary Geometry ; it indicates that one of the 
 lines would have to make a half revolution to point like 
 the other. Select examples of straight angles in the room. 
 
 80. The division in 77 of angles into two large classes is 
 not definite enough for our purposes. For greater definite- 
 ness in estimating angles, a 
 semi - circumference is di- 
 vided into one hundred and 
 eighty equal parts called de- 
 grees. The number of de- 
 grees that A (Fig. 23) 
 would have to pass over in 
 reaching the position of 
 is an exact measure of the size of the angle between A and 
 C. If it is found that A would have to turn past thirty 
 divisions, the angle is said to be an angle of thirty degrees- 
 written 30°. And A would make an angle of 30° with 
 any line pointing exactly like C, no matter where in 
 space it might be. 
 
 81. How many degrees in every right angle ? How 
 many degrees in the angle between A and OB? (Fig. 23). 
 How many degrees must there be in any angle to make it 
 obtuse ? How many degrees in a straight angle ? 
 
 82. What do you mean by saying that the angle between 
 A and D is 70° ? 60° ? 90° ? 180° ? Answer each 
 part in a full, clear sentence. 
 
 2 
 
18 
 
 ELEMENTARY AND 
 
 83. The instrument used for measuring angles is a 
 graded half circle of metal or cardboard called a protractor 
 (Fig. 24). The centre is clearly marked by a notch. The 
 
 Wm M 
 
 inner edge is graded into 180 equal parts from right to left ; 
 the outer circle is graded into 180 equal parts from left to 
 right. 
 
 84. To make an angle of 70° with your protractor first 
 draw a line of any convenient length (M N Fig. 25), then 
 place your protractor with the* cen- 
 tre mark at M, and with one of the 
 zero lines along M N : next, put a 
 dot 1* at the point exactly opposite 
 the line marked 70° on your pro- 
 tractor, remove von r protractor, and 
 join M with P. If you wish the 
 line to be above M N, you should 
 use the 70° mark of the inner circle ; 
 if you wish it to be below M N, use 
 the outer circle. Draw both cases. 
 
 85. With your protractor make angles of 22°, 60°, 
 125°, and 170°. 
 
 8G. Draw two lines, A IS and A C crossing each other at A 
 
 M 
 
 N 
 
 Fig. 25. 
 
 4.V 
 
Fig. 26. 
 
 CONSTRUCTIONAL GEOMETRY 19 
 
 (Fig. 26). Explain fully, step by step, the way in which you 
 would use your protractor to measure the angle between 
 A B and A 0. Measure it 
 twice, placing the zero line 
 of your protractor first on 
 A B and then on A C. 
 
 87. For the sake of brev- 
 ity the following signs are 
 used : / = angle ; / s = 
 angles; /CABor/BAC 
 ±5 angle between A B and 
 A C. In using the last sign 
 be careful to put the letter 
 naming the point at which the two lines meet between the 
 other two. The angle between A B and A C could not be 
 written /ABCor/ACB; A, the meeting-point of the 
 two lines, called the vertex of the angle, must be the middle 
 letter. What is the plural of vertex ? If only one angle is 
 formed at a vertex, the angle may be named by the vertex 
 letter. In explaining the meaning of / B A C, you must 
 remember to read the line from the vertex out ; for in- 
 stance, / B A C is the difference in direction between A B 
 and A ; it shows how much you must turn A B to make 
 it point like A 0. It would not do to substitute B A for 
 A B, or C A for A C. If you think for a moment how 
 much you would have to turn A B to make it point like 
 C A, you will see the need of being careful. 
 
 88. Explain the meaning of / MJN R, /NMS, /PTQ, 
 without drawing the lines that form the angles. 
 
 89. Make any figure with five sides. Estimate the angles 
 of the figure ; record your estimate in one column ; the 
 true value, found with your protractor, in another ; and 
 the error in a third column. 
 
 90. Repeat 89, with a four-sided figure. 
 
 91. How would you measure the angle between A B and 
 C A (Fig. 26) ? 
 
 First estimate it and then measure it. 
 
L>0 
 
 ELEMENTARY AND 
 
 92. Estimate the value of the angles 1, 2, 3, 4, 5, 6, and 
 7 in Fig. 27, recording estimates, true values, and errors 
 as in 89. 
 
 m. 2:. 
 
 90° free 
 Repeat 
 
 93. Make angles of 60°, 20°, 110°, 45°, and 
 hand. Test your accuracy with the protractor. 
 this exercise, drawing lines 
 that do not meet. 
 
 94. Divide an angle into 
 two equal parts with your 
 protractor. 
 
 95. Divide an angle into 
 three equal parts with your 
 protractor. 
 
 96. Make an angle CAB 
 (Fig. 28) with the aid of your 
 protractor, and a line R. 
 
 Make an angle with vertex at equal to / CAB. 
 
 97. Make a five-sided figure or pentagon. Make a sec 
 ond pentagon with the same angles hut different sides. 
 
CONSTRUCTIONAL GEOMETRY 21 
 
 98. Make two angles that to your eye seem to be of the 
 same size. Can you test their equality by using your com- 
 passes without the aid of your protractor ? Explain your 
 steps carefully. 
 
 99. Can you make one angle equal to another without 
 your protractor ? Give three illustrations. 
 
 100. Make a four-sided figure, or quadrilateral, and 
 make another quadrilateral with different sides but the 
 same angles, using your compasses and ruler only. 
 
 101. Make two angles, A B C and D E F ; make an angle 
 that you think is equal to the sum of / A B C and / D E F. 
 Test your accuracy with the help of your compasses. 
 
 102. Can you explain how to double an angle ABC? 
 
 103. Can you add two angles ? three angles ? Make 
 your work clear in all the steps. 
 
 104. Draw two angles ; add them with the aid of com- 
 passes and ruler only; test your result by finding the 
 number of degrees in each angle with your protractor. 
 
 105. Make a figure with three sides, called a triangle ; 
 add the three angles formed at the three corners. How 
 many degrees are there in the angle which you get for your 
 answer ? 
 
 106. Make four triangles of different sizes and shapes. 
 Add together the angles of each one, and note the number 
 of degrees in the resulting angle each time. In adding 
 the angles use ruler and compasses only. In gotting the 
 number of degrees use a protractor, if necessary. Do your 
 experiments seem to point to any principle ? 
 
 107. Cut out a paper triangle ABC; cut off the cor- 
 ners along wavy lines, and place the vertices A, B, and 
 C together. Does this experiment confirm the previous 
 experiments ? 
 
 108. Make a quadrilateral, and add the angles formed. 
 Try two figures differing in shape and size. If one line 
 should turn through all the angles of the quadrilateral, how 
 far would it turn ? Measure each angle with a protractor : 
 find the number of degrees in all four angles of each figure. 
 
22 ELEMENTARY AND 
 
 109. Make a five-sided figure, and add the angles formed. 
 
 110. Can you draw any conclusion about the sum of the 
 angles of a figure of many sides ? 
 
 111. Verify the results of 109 and 110, by cutting off 
 the corners of paper figures of the corresponding number 
 of sides. Be careful to mark the corners before cutting. 
 
 112. Can you describe a way of finding the difference be- 
 tween two angles with compasses and ruler only P Make 
 two angles, and find an angle equal to their difference. 
 
 SECTION V. PARALLELS. 
 
 113. We have seen that, if two lines point in the same 
 direction, they cannot form an angle. Do you remember 
 
 why P We have seen also that, 
 
 if they point in exactly op- A B 
 
 potite directions, A B and 
 1 1 0, for example, in Fig. 29, 
 
 they form a straight angle 
 
 C D 
 
 Whether two lines point the Fl °- w 
 
 same way or exactly opposite ways, they are said to be paral- 
 lel, because they run b\j the side of each other. (The name 
 is taken from two Greek words.) It is a great convenience 
 to be able to distinguish the two cases. When two lines are 
 parallel and point the same why the following sign, f f . will 
 be used, and the lines will be called symparallel lines. For 
 instance, A BffCDistobe read A B points in the same 
 direction as C 1), or, more briefly, A B is symparallel to C D. 
 When the two lines are parallel and point in opposite direc- 
 tions, the sign f 4 will be used, and the lines will be called 
 antiparallel lines. Thus ABfiDC means that A B runs 
 in the opposite direction to 1) C, or A B is antiparallel to 
 D C. If single letters are used, the direction of the lines is 
 not indicated ; so that the sign without the arrow-heads 
 
CONSTRUCTIONAL GEOMETRY 23 
 
 can be used. Thus a || b mean that the line "a" points 
 either in the same direction as " b" or in the opposite di- 
 rection to " b." 
 
 114. Draw a line A B, and choose a point C anywhere 
 
 outside of A B (Fig. 30). Can , 
 
 you draw through two dis- A 6 
 
 tinct lines diverging from each 
 
 other and yet symparallel to 
 
 A B ? It does not take long to C • 
 
 see that the problem is impossi- FlQ# ^ 
 
 ble, but when we try to give a reason for it, we find that it 
 
 is one of the truths that is self-evident or axiomatic. It is 
 
 equally self-evident that one line can be drawn through C 
 
 in the plane of the paper pointing like A B. The axiom 
 
 can be stated as follows : Through any point in space one 
 
 line, and only one line, can be drawn in any desired direction. 
 
 Fig. 31a. 
 
 115. Imagine two lines, A B and C D, to have the same 
 direction (Fig. 31a), and imagine that E F crosses these 
 lines at K and M respectively. At once we see the possi- 
 
24 ELEMENTARY AND 
 
 bility of a great many angles. There are six angles with 
 their vertices at K, and six with their vertices at M. Can 
 you account for them all ? 
 
 Can you see, without the aid of your protractor or of 
 your compasses, that some of the angles must be equal from 
 the nature of the case ? 
 
 Look at the angles EKB and E M D, for instance ; 
 B K B shows you how much you must turn K B to 
 make it point like K K ; / E M D shows you how much 
 you must tuni M D to make it point like M E. Now the 
 lines K B and M I) point in the same way at the start ; 
 they also point in tin* same way when in their final posi- 
 tions, K E and M E. Therefore the amount of turning 
 has been the sun.- in both cases. Or, in other words, 
 
 / E K B = /_ KM 1). Another way of looking at the 
 same thing is this : since the direction of K B is the same 
 as that of M D, and since the direction of K E is the same 
 as that of M K, the ilificrrnce in direction between K Band 
 K B must be the same as the difference in direction between 
 M D and M E ; just as if two shades of white are the MUM 
 and two shades of red are the same, the difference between 
 a Bhade of the white and a shade of the red must be the same 
 in one case as in the other. Our knowledge that these two 
 angles are equal belongs to the kind of knowledge referred 
 to in the last part of 74, knowledge, that is, which is based 
 upon the nature of the case, and which is not affected by 
 defective instruments or drawings. Does it make any differ- 
 ence in the equality of the two angles, if F E is more or less 
 inclined to the parallel lines ? 
 
 Make the following principle true by crossing out the 
 words that need to be crossed out : 
 
 Two angles are or are not necessarily equal, if their 
 sides have the same directions respectively. 
 
 Learn the principle by heart, and also copy it in the 
 " Summary." 
 
 116. Write the same principle almost entirely by signs, 
 which have already been explained. 
 
[ UNIVERSITY 
 
 V OF J 
 
 N^lifoRJJ^ 
 
 ^*^ M= = : ^^VONSTRUCTIONAL GEOMETRY 25 
 
 117. Select other pairs of angles in Fig. 31a to which the 
 principle applies. 
 
 118. What can yon say about the angle A K B ? What 
 about the angle E K M ? What about their relation to 
 each other ? Does your knowledge of the facts depend 
 upon measurements, or upon the nature of the figure ? 
 Can you put what you have learned in this exercise into 
 the form of a principle ? Select for the " Summary " the 
 wording finally agreed by the class to be the best. 
 
 119. / B K M is one part of a straight angle. What is 
 the other part ? Two answers can be given to this question. 
 Can the answers have different values ? Why ? When 
 two angles together make a straight angle, the angles are 
 said to be supplements of each other. Can one angle be 
 the supplement of two unequal angles ? Why ? What is 
 the supplement of 75° ? of 110° ? of 82° ? of 60° ? Draw 
 the supplement of each angle of Fig. 27. 
 
 120. There is an axiom that if equals are taken from 
 equals, the remainders are equals. Do you understand what 
 it means, and do you see that the truth has anything to do 
 with exercise 119 ? with the principle of 115 ? 
 
 121. Select in Fig. 31a as many pairs of supplementary 
 angles as you can. Note specially those which are the 
 supplements of the same angle. What can you say about 
 the direction of their sides ? 
 
 Your study of the last three articles should enable you 
 to make the following statement true : Two angles are or 
 are not necessarily equal if their sides have opposite direc- 
 tions respectively. Record the principle in the "Sum- 
 mary." Select as many pairs of equal angles in Fig. 31a as 
 you can that are equal by application of this new principle. 
 
 If A K M, Fig. 31a, is 25°, find all the other angles in 
 the figure. 
 
 IfLAKM = |LMKB 3 find the number of degrees 
 in all the angles. 
 
 IfLBKM = 4xLCMF, find the number of degrees 
 in all the angles. 
 
26 
 
 ELEMENTARY AND 
 
 Scholium. 1 The angles that are formed when one line 
 crosses two other lines are of 
 such great importance in 
 Geometry that they have re- 
 ceived special names. Those 
 without the two lines, or 1, 2, 
 7, 8 in Fig. 31b, are called 
 exterior angles ; those within 
 the two lines, or 3, 4, 5, 6, are 
 called interior angles ; I and 
 8 or 1 and 7 are called alter- 
 nate-exterior angles ; 3 and 
 5 or 4 and <» an- called alter- 
 nate-interior angles ; while 1 
 and 5, 2 and 6, 3 and 7, or 4 and 8, are called correspond- 
 ing angles. 
 
 Fio "lb. 
 
 SECTION VI. REVIEW. 
 
 122. What is the difference between a geometrical solid 
 and a physical solid ? 
 
 128. What is a plane surface ? How is a plane surface 
 tested practically t 
 
 134, How many kinds of lines can you draw between two 
 points A and B ? Which is the shortest of them all ? How 
 do you know that it is the shortest ? How does the straight 
 line A B point — from B to A ? or from A to B ? 
 
 125. Give definitions of concentric and adjacent, with 
 illustrations of their use in Geometry. 
 
 186. What is an axiom ? Give any axioms that you have 
 learned. 
 
 127. Explain how you use a protractor. 
 
 128. Write in sign language : A B is longer than C D. 
 
 1 A scholium is a remark containing some explanation. 
 
CONSTRUCTIONAL GEOMETRY 27 
 
 The angle A B C is less than the angle II S T. A B points 
 in the same direction as C D. 
 
 129. Write in good English the following : 
 /ABC=/EFG because B A f f F E and B C 
 
 f f F G, or because their sides / f respectively. 
 
 130. What is a pentagon ? a quadrilateral ? a triangle ? 
 What is a straight angle ? an obtuse angle ? 
 
 131. Does a straight line change its direction in any part 
 of its course ? Can you place two points so that you can- 
 not draw a straight line through them ? 
 
 If two straight lines pass through the same two points 
 can they diverge from each other ? See 113. 
 
 132. Can you think of a way of testing the straightness 
 of your ruler's edge by two points on your paper ? 
 
 133. What is an angle ? Can you really draw an angle ? 
 What is understood by ' ' drawing an angle" ? What differ- 
 ence in the size of the angle does the length of the sides 
 make ? 
 
 134. Place five points so that no three shall be on the 
 same straight line. How many straight lines can you 
 draw that shall each contain two of these points for its 
 ends ? Can you place the five points sp that only four 
 lines can be drawn each containing two, and only two, 
 points of the five as the limiting points ? 
 
 135. Make as clear as you can the diiference between the 
 straight angle ABC and the straight line ABC. 
 
 136. What is the vertex of the angle MPN, and what 
 are its sides ? When is / M P N a right angle ? 
 
 137. When are two angles supplements of each other ? 
 Give numerical examples and geometrical examples. 
 
 138. In how many points can two straight lines cross 
 each other ? two circumferences ? In how many points 
 can a straight line cross a circumference ? 
 
 139. In what cases can you say that two angles are equal, 
 without measurement with protractor or compasses ? 
 Write out the principles of 115 and 121, making use of the 
 terms explained in the scholium to 121. 
 
28 ELEMENTARY AM) 
 
 SECTION VII. PARALLELS CONTINUED. 
 
 140. Draw two lines A B and A C (Fig. 32) ; choose any 
 point P on A C. How many lines can there be radiating 
 from I' that will make the same 
 angle with P C that A B does 
 with A Of Answer first with 
 the understanding that the line 
 can be drawn from 1' in any di- 
 rection in space ; then answer 
 the same question with the un- 
 derstand in;: t hat the line must 
 be drawn on the same plane sur- 
 face with A B and A C, for in- PiaW - 
 stance, on the plane of your paper. Actually draw the 
 lines in the second case. 
 
 141. Draw two lines. A B and A C, as in Fig. 32. At the 
 point P. wwh the aid of your protractor, make an angle 
 equal to / C A B, drawing your second line P R so that it 
 is on the same side of A C with A B. Since P B will have 
 to diverge from P C just as much as A B does, and since 
 l'K and A B are on the same side of A C and in the same 
 plane, thev must point in the same direction, or, in sign 
 language, if / (' P B = / C A B, P R f f A B. 
 
 1 1 .\ Could two lines possibly point in the same direc- 
 tion and not be in the same plane with each other ? See 
 111. 
 
 143. Repeat the work of 141, using your compasses instead 
 of your protractor to make equal angles. 
 
 144. The last four exercises lead to the following impor- 
 tant principle : If two lines drawn in the same plane 
 diverge to the same extent from one side of a third line, 
 they point in the same direction, or are symparallel. 
 Make clear why it is necessary to say M in the same plane " 
 and " from one side/' by giving illustrations of lines that 
 
CONSTRUCTIONAL GEOMETRY 29 
 
 are not symparallel because they fail to answer all the con- 
 ditions. Write the principle given above, using the nota- 
 tion explained in 121, scholium. 
 
 145. Make a line A B, and choose any point outside of 
 the line. Draw through C a line symparallel to A B. 
 Suggestion : Draw C A as a help line, and use the principle 
 of 144. 
 
 146. Make four lines symparallel to a line A B drawn 
 at pleasure. Will these lines be symparallel to each 
 other ? 
 
 147. Make a three-sided figure. Through each corner 
 draw a line symparallel to the opposite side. 
 
 148. Make a quadrilateral with its opposite sides parallel. 
 Such a figure is called a parallelogram. 
 
 149. Can you discover any truth about the angles of a 
 parallelogram without actual measurement ? Record the 
 principle in the " Summary. " 
 
 150. Make a triangle ABC, and through the vertex C 
 (Fig 33) draw D E f f A B. Select pairs of equal angles, 
 telling in each case how you 
 know that the angles are 
 equal. What can you say 
 about the sum of the angles 
 with vertices at C ? What 
 can you say about the sum of 
 the angles of the triangle ? 
 Compare the result obtained Fia33. 
 
 here with the result of your experiments in 106. Notice 
 the difference in the methods of reaching the conclusion. 
 In this case the conclusion is reached after studying one 
 triangle, in 106 after experimenting with several triangles ; 
 in this case what we know about the angles depends upon 
 the nature of angles, while in 106 our knowledge is based 
 upon our measurements. We might have some doubt as to 
 whether the truths learned about the few triangles with 
 which we experimented in 106 would apply to all triangles ; 
 but we are absolutely sure that the principle discovered in 
 
30 ELEMENTARY AND 
 
 150 applies to all triangles, because it is based on the nature 
 of angles and lines. In this case the principle is proved ; 
 in 106 it is merely tested by experiment. Record the prin- 
 ciple in the "Summary." 
 
 151. Make a triangle ABC with / B A C = 70°, and 
 I A B C = 43°. How large is / A C B ? Why 1 Through 
 each vertex, A, B, and C, in turn draw a line parallel to 
 the opposite side. Four new triangles will thus be formed 
 Give the value of each angle in degrees without using your 
 protractor. 
 
 152. One angle of a triangle is 35°, another 47° ; find 
 the third angle. If two angles of a triangle are 48° and 
 95° respectively, what is the third angle ? 
 
 153. Can you make a triangle whose angles are 90°, 40°, 
 and 60° ? Why ? Can you make a triangle with two ob- 
 tuse angles in it } Can you make a triangle two of whose 
 angles shall be right angles ? 
 
 1 54 Make any triangle. Can you make a second triangle 
 which shall have two angles like two of the first triangle, 
 but which shall have its third angle unlike the third angle 
 of the first triangle ? 
 
 155. If the angles of Fig. 34 are two angles of a triangle, 
 how can you find the third 
 
 angle without using your pro- 
 tractor ? Can you think of 
 more than one way of solving 
 this problem ? 
 
 156. Two angles of a tri- 
 angle are said to determine the 
 
 third angle. Give illustrations of the meaning of this 
 statement. 
 
 157. If one angle of a parallelogram is equal to / x (Fig. 
 34), how can you find the other three angles without con- 
 structing the parallelogram ? See 149. 
 
 158. One angle of a parallelogram is 72°. What are the 
 other angles ? 
 
 159. Find all the angles of a parallelogram, if one angle 
 
CONSTRUCTIONAL GEOMETRY 31 
 
 is 90° ; if one angle is 120° ; if one angle is / y (Fig. 
 34). 
 
 160. How many angles of a parallelogram are required 
 to determine the remaining angles ? Put your answer in 
 the form of a principle, and record it in the " Sum- 
 mary." 
 
 161. In the parallelograms whose angles you have stud- 
 ied thus far, can you discover a principle that applies 
 always to the adjacent angles ? 
 
 162. Extend side C D of the parallelogram A B C D to 
 R. What can you say about the respective directions of 
 the sides of / R D A and / D A B ? What follows from 
 the directions of the sides of the angles ? 
 
 163. Make an angle ABC. At some point P on A B 
 betiveen A and B draw (on the opposite side of A B from 
 B C) a line, P R, that shall diverge from P B as much as 
 B diverges from B P. What is the geometrical name 
 for angles R P B and P B C (121 Scholium) ? Which 
 of the following sentences is true ? P R ff B C or 
 PR/,/ B C ? Write a translation of the correct sentence. 
 Is P R parallel to B C ? 
 
 164. Make a line A B and choose a point C outside of 
 A B. Draw a line through C parallel to A B, without 
 prolonging your help line A C beyond C. Will the line be 
 antiparallel or symparallel to A B ? 
 
 165. Repeat the problem of 164 three times, placing A, 
 B, and in different positions each time. 
 
 166. The last four exercises lead to the following prin- 
 ciple : If two lines are crossed by a third in such a way 
 that the alternate interior or alternate exterior angles are 
 equal, the two lines are parallel. 
 
 167. Combining the principles of 144 and 166, we can 
 say that two lines are parallel, if a third line crosses them 
 in such a way as to make any pair of corresponding angles 
 equal, or any pair of alternate interior or alternate exterior 
 angles equal. 
 
 168. A practical way of drawing parallels by making the 
 
32 
 
 ELEMENTARY AND 
 
 corresponding angles equal is to slide a triangular piece of 
 wood or stiff paper along a ruler until the edge passes 
 through any desired point. 
 For instance, in Fig. 35 it 
 is desired to draw a line 
 through P parallel to A B. 
 Place your ruler in such a 
 position that one edge, x, of 
 your triangle will have the 
 direction of A B ; then slide 
 the triangle along the ruler 
 until the same edge, x, passes 
 through P, and draw M N. 
 The ruler represents a line 
 crossing M N and A B in 
 such a way as to make the corresponding angles equal. 
 1 ♦'.:♦. Make a parallelogram by the method described in 
 168. 
 
 170. Mark off six equal distances on a line A B ; then, 
 by the method of 108, draw parallels through the points of 
 division. 
 
 171. Make a triangle ; then through each vertex draw a 
 line parallel to the opposite side, using the method of 108. 
 
 vie m 
 
 SECTION VIII. TRIANGLES. 
 
 172. Make a triangle ABC (Fig. 36). The sign for tri- 
 angle is l\, and for triangles ^. What is the difference 
 between /\ABC and /ABC? Give as full an answer 
 as you can to this question, naming all the respects in which 
 they differ. Are they alike in any respect ? 
 
 173. What is the difference between £± A B C and 
 /\BAC? What is the difference between / A B C and 
 / B A C ? In how many ways can you name the /\ABC, 
 
CONSTRUCTIONAL GEOMETRY 33 
 
 and in how many ways can you name the / A B C, us- 
 ing the letters A, B, C each time ? 
 
 174. Write a short sketch of the A A B c > telling how 
 many parts it has, what it is by nature, what you have 
 
 Fig. 36. 
 
 learned about its angles, and what fact about its sides you 
 can infer from 124. Also write all you can about /ABO. 
 In writing these sketches, take great pains with your Eng- 
 lish. 
 
 175. After making A A B C (Fig. 36), mark off, on a 
 long line, M N equal in length to A B ; then make /NMR 
 equal to / B A C ; then mark off on M R a length M P 
 = AC; and, finally, join P N, thus completing /\ M N P. 
 Notice that you made the lines M N and M P and the 
 angle NMR to match the corresponding parts of the 
 /\ A B C, but that, after taking these steps, you had no con- 
 trol over the line P N", because you were obliged to draw it 
 between two fixed points P and N. With your protrac- 
 tor compare / M P N with / A C B, and / M N P with 
 /ABC; and, with your compasses, compare P N with C B, 
 recording your results by means of the sign = or the sign 
 4= (unequal) as the case may be. If you find that P N is 
 not equal to C B, write PN ^ C B, but, if you find that 
 P N is equal to C B, write P N = C B, etc. 
 
 Record your work from the beginning according to the 
 
 model which follows, taking pains to record each step as 
 
 you take it, not waiting to give your description after all 
 
 the work is done. Put your figure at the upper rig Jit cor- 
 
 ? 
 
34 ELEMENTARY AND 
 
 ner of your page, unless it is so large that it requires the 
 whole width of the page. 
 
 I. I make ^ABC. Position of figure. 
 
 II. I make M N = A B. 
 
 III. I make /NMR=/BAC. 
 
 IV. I make M P = A C. 
 
 V. I join P with N, thus completing ^MNP. 
 VI. I find that / M P K + or = / A C B (using the 
 
 proper sign). 
 VII. I find that / M N B * or = / A B C. 
 VIII. I find that P N * or = C B. 
 
 t70. Repeat the experiment of 175 three times, using 
 triangles of new shapes each time, and recording your work 
 according to the above model. 
 
 177. Do your experiments in 175 and 176 seem to point 
 to any truth about triangles ? Take care to express the 
 principle in the simplest and most accurate wording pos- 
 sible. 
 
 178. If your work was neatly and carefully done in 1 ;:> 
 and 170, you probably were able to discover the principle 
 involved. But, at the best, you have merely tested* prin- 
 ciple in a few cases ; you have not proved it by showing 
 that it depends upon the nature of triangles, or upon prin- 
 ciples already proved. Let us try to prove the principle 
 that two triangles must be equal in all respects, if they 
 have two sides and the included angle alike respectively. 
 
 There is a very great difference at the very start between 
 testing and proving a principle. In testing a principle, 
 everything depends upon the accuracy of the mechanical 
 work ; if there is the slightest error in making the angles 
 or the lines, the test is a failure ; but in proving a prin- 
 ciple the drawing is simply to aid the imagination, by 
 making it unnecessary to carry a great many lines and points 
 in the mind at once ; the accuracy of the drawing makes no 
 real difference in the results, although an inaccurate draw- 
 ing often misleads the mind ; the mind should be con- 
 
CONSTRUCTIONAL GEOMETRY 35 
 
 stantly on the imaginary lines and angles represented, not 
 on the lines and angles drawn. 
 
 Imagine, then, two triangles, ABO and M N P, to have 
 two sides, A B and A C, and the included angle, B A C, of 
 the first equal respectively to two sides, M N and M P, and 
 the included angle M N P of the second. You are to try 
 to prove that the remaining parts, B C and the angles ABC 
 and A C B of the first triangle, must be equal to the re- 
 maining parts, N P, and the angles MNP and MPNof 
 the second triangle. Imagine /\MNP placed on/\ABC, 
 and try to prove that the triangles must exactly fit one 
 another. How many points can you place before /\MNP 
 becomes fixed in position ? After placing M on A, are you 
 free to place N wherever you please ? In how many places 
 could you put it ? Which one will you choose ? After 
 placing M and N in two definite positions, in how many 
 positions could you place P ? Remember that P must be 
 somewhere on the same plane surface with /\ABC, since 
 you imagined the two triangles placed together. Have you 
 any reason for thinking that P, if on the same side of A B 
 with C, must be on the line AC? What is your reason ? 
 If you can prove that it must be somewhere on A C, do you 
 know exactly where it must be ? Remember that M is 
 fixed at A. How do you know its exact position on A C ? 
 Why is there now no doubt as to the position of N P ? See 
 61, Axiom. If you can find satisfactory answers to the 
 questions asked above, and if you can explain your answers 
 by reasons that depend upon the nature of lines and angles 
 or upon principles already proved (and you ought to be 
 able to do this with ease), you have proved the principle 
 that two triangles are necessarily equal in all respects, if 
 they have two sides and the included angle alike respec- 
 tively. 
 
 179. By the aid of this principle of 178 we are able to 
 decide without measurement that two lines must be equal 
 under certain circumstances, and we have found a new way 
 for determining the equality of two angles without protract- 
 
36 ELEMENTARY AND 
 
 or or compasses ; for two lines or two angles must be equal, 
 if they are corresponding parts of triangles that are known 
 to be equal. 
 
 180. The principle of 178 is of great practical value in 
 enabling us to find the distance from one point to another 
 that cannot be seen from the first point because it is hid- 
 den by a building, or wall, or some intervening obstacle. 
 Suppose you wish to find the distance from a stake on one 
 side of a house to a stake on the opposite side of the house. 
 Represent the position of the first stake by a point A. and 
 that of the second stake by a point B. Then choose a 
 point from which you can see both A and B, and measure 
 the angle A C B. 
 
 To measure the angle A C B, you need an enlarged pro- 
 tractor mounted on a tripod. If possible, get some sur- 
 veyor to show you his theodolite, and to explain to you 
 how lie uses it. If yon divide a semicircle having a radius 
 of four or five inches into degrees, and mount it on a tripod 
 such as you would use for a camera, you can measure angles 
 with sufficient accuracy for your purposes. A stick about 
 a foot long, provided with sights and pivoted at the centre 
 of t he protractor, will take the place of the surveyor's tele- 
 scope. From the centre of the protractor suspend a plumb- 
 line. Placing the tripod so that the point C is exactly 
 below the centre of the protractor, sight the point A, and 
 note the division at which the line of sight crosses the pro- 
 tractor ; then turn the stick until you can sight point B, 
 and again note the division of the protractor crossed by the 
 line of light : from the two readings of the protractor esti- 
 mate / ACB. By " sighting point A " is meant sighting 
 a staff held in a vertical position at A. Next, measure the 
 distances C A and C B. If, with the measurements that 
 you have made, you construct a second triangle in a level 
 place where there are no obstacles, you will have a triangle 
 exactly like /\ A B C, so that you can measure the line cor- 
 responding to A B. 
 
 181. If at any time you should be without a mounted 
 
CONSTRUCTIONAL GEOMETRY 37 
 
 protractor, can you think of any way in which you could 
 use a piece of string and a stake to measure / A C B in 
 the last problem ? 
 
 182. Describe the measurements that you would take to 
 find the distance across a pond. 
 
 183. In practice it is often inconvenient to find a suita- 
 ble place in which to make a triangle as large as the origi- 
 nal triangle. Therefore, instead of making the lines C A 
 and C B as long as they were found to be, we might make 
 them one-half as long, or one-tenth as long, or we might 
 use any convenient fraction of the actual length. In such 
 cases the line found would be the same fraction of the orig- 
 inal line A B. We might go still further, and use a centi- 
 meter or an inch to represent several feet, and we then 
 could draw a plan of the ground on paper. The process 
 just described is called "drawing to scale." In drawing 
 to scale you cannot be too careful ; a slight error makes a 
 great diiference ; for instance, if one centimeter stands for 
 ten feet, an error of a millimeter in your plan would make 
 an error of one foot on the ground. In choosing your 
 scale it is well to reduce the lines as little as the size of 
 your paper will allow. 
 
 How long a line can you represent on your paper, if one 
 centimeter represents ten feet ? How long a line, if one 
 inch represents ten feet ? 
 
 184. Draw lines on your paper representing lengths of 
 150 ft., 161 ft., 84 ft., 26J ft., using the scale 1 cm. = 10 
 ft. 
 
 185. Draw lines representing lengths of 75 ft., 63 ft., 
 110 ft., 87 ft., using the scale 1 in. = 20 ft. 
 
 186. Draw a triangle, and find the lengths of the sides 
 of the triangle which it represents, if the scale is 1 cm. = 
 10 ft. ; if the scale is 1 in. = 10 ft. 
 
 187. If you are to represent lines between 300 ft. and 
 400 ft., what scale will you choose for your paper ? 
 
 188. When the scale is 1 cm. = 5 ft., what do lines 
 7.5 cm., 24 cm., 63 mm., and 2 m. represent ? 
 
38 ELEMENTARY AND 
 
 189. If the scale used is \ in. = 3 ft., how long are the 
 lines which are represented by 2} in., 3| in., 7J in. ? 
 
 190. Two sides, A C and B C, of a triangular field were 
 found to be 150.75 ft. and 200 ft. respectively, while the 
 included angle was 70°. Draw a plan of the field, using the 
 scale 1 cm. = 15 ft., and calculate the length of A D. 
 
 191. Choose two points, A Mid B, in different, but ad- 
 joining, school-rooms in such a way that one point cannot 
 be seen from the other, but that both can be seen by a per- 
 son standing in the doorway between the rooms. Explain 
 what measurements you would take to find the distance 
 A B ; actually take the measurements, choose a scale suited 
 to the size of your paper, draw a plan, and find the length 
 of the line A B. 
 
 Note. — A great many problems of this kind can be 
 given with profit. It is important, in the first few cases at 
 least, that the actual measurements be taken by the pupil. 
 Later, if it is thought best, measurements can be taken 
 from city plans that will enable the pupils to find the 
 shortest distance between well-known points. In all cases 
 mat and accurate work should be insisted upon. It is well 
 t<> have the pupils select scales adapted to the paper used, 
 and to have the same problem solved by different scales. 
 
 L98. In 178 we found that two triangles would be alike 
 necessarily, if they were c 
 
 knOWTI to have two sides 
 and the in eluded angle 
 alike respectively. Let us 
 see if we cannot leave out 
 the word included and 
 show that two triangles 
 must be alike in all re- 
 spects, if they have two 
 sides and an angle alike 
 
 respectively. Make a triangle, ABC, taking heed to make 
 A C longer than B C (Fig. 37). Make another triangle 
 R S T havinir— 
 
CONSTRUCTIONAL GEOMETRY 39 
 
 1st. RS^AC; 
 
 2d. / S R T = / C A B ; 
 
 3d. S T = C B. 
 
 After taking the first two steps, are you obliged to take 
 the third step, which will complete your triangle, in one 
 way, or do you have a choice of ways ? Make the following 
 statement true by crossing out the words that need to be 
 crossed out : Two triangles are or are not necessarily equal, 
 if they have two sides and the angle opposite the smaller 
 side alike respectively. 
 
 193. Making use of the same triangle (A B C, Fig. 37), 
 make a triangle MNP having — 
 
 1st. N P = B C ; 
 
 2d./PNM=/CBA; 
 
 3d. P M = C A. 
 
 You remember that C A was purposely made longer 
 than B C. After taking the first two steps, can you take 
 the third step in more than one way ? What words do you 
 think ought to be crossed out in the following statement : 
 Two triangles are or are not necessarily equal in all re- 
 spects, if they have two sides and the angle opposite the 
 greater of the two sides alike respectively? If, then, 
 we leave out the word included in the principle of 178, 
 what must we put in its place ? 
 
 194. Make three different triangles. Make three new tri- 
 angles by taking in each triangle of the first set two sides 
 and an angle opposite the greater of the two sides chosen. 
 
 195. Make as many triangles as you can by taking in 
 each triangle of the first three triangles drawn in 194 two 
 sides and the angle opposite the smaller of the two sides 
 chosen. 
 
 SECTION IX. REVIEW. 
 
 196. What are parallel lines ? 
 
 197. If a line on the floor is parallel to a line on the 
 ceiling, could you move one to the position of the other 
 
40 ELEMENTARY AND 
 
 without changing its direction ? What would the moving 
 line generate ? 
 
 198. Can you draw two parallel lines so that a plane 
 surface cannot be found which will contain both lines ? 
 Upon what axiom does your answer depend ? 
 
 199. If a line on the floor is not parallel to a line on the 
 ceiling, can you move one line to the position of the other 
 without changing its direction ? What would the moving 
 line generate in reaching the position of the second line ? 
 
 200. Draw several pairs of lines that would not meet 
 within the limits of your paper, if extended. Estimate 
 the angles between them, pair by pair, and record your es- 
 timate ; then, by the method of 168, draw a parallel to 
 one line of each pair that shall cross the second line, meas- 
 ure the angle, and record the true value and the error. In 
 doing this exercise take care to have some of the lines 
 make obtuse angles and some acute angles. 
 
 201. What is an oblique angle ? 
 
 202. What is a geometrical proof ? 
 
 203. One angle of a parallelogram is 50° ; what are the 
 other angles ? 
 
 204. One angle of a parallelogram is twice another ; what 
 are the angles ? 
 
 205. Extend side A B of the triangle A B C to R. Try 
 to show that /CBR is equal to the sum of two angles 
 of the triangle, / B A C and / A C B. 
 
 206. Repeat the problem of 205, extending a line at each 
 vertex of the triangle, and showing what angles of the 
 triangle are needed to make each exterior angle formed. 
 
 207. One acute angle of a right triangle is 70° ; what is 
 the other ? 
 
 208. Make a triangle ABC, and make another like it 
 in as many ways as you can, by principles already estab- 
 lished. 
 
 209. Can two triangles have two sides and an angle re- 
 spectively alike and still be unequal ? 
 
CONSTRUCTIONAL GEOMETRY 
 
 41 
 
 SECTION X. ISOSCELES TRIANGLES. 
 
 210. Make a triangle ABC with A C = B C. Such a 
 triangle is said to have equal legs, and for that reason is 
 called an isosceles triangle, a 
 name that in Greek means 
 " with equal legs." The third 
 side, in this case A B, is called 
 the base of the triangle ; the 
 isosceles triangle is generally 
 thought of as standing on this 
 base, the legs meeting in a point 
 which is called the vertex. In 
 triangles that are not isosceles 
 any side may be thought of as the base, and no one vertex 
 is singled out as the vertex. In an isosceles triangle M P N, 
 if P is the vertex, what is the base, and what are the legs ? 
 
 211. Triangle R S T has R T=S T (Fig. 39). To the 
 eye / R seems like / S. Recall the tests for equal angles 
 thus far discovered, and decide whether any one of them 
 will prove the angles equal. Do the sides point the same 
 
 way respectively ? Do the sides point in opposite direc- 
 tions respectively ? Are they supplements of the same 
 
42 ELEMENTARY AND 
 
 angle ? Study the directions of the sides of the angles 
 carefully, and be able to give a reason for your answer, 
 whether yes or no. If all these tests fail, there is still one 
 more chance for proving the angles equal, becausejn ITU it 
 was found that two angles must be alike, if they are cor- 
 n-ponding angles of triangles that can be proved equal. 
 At first thought this test seems of little use, because only 
 one triangle, K S T, is mentioned here ; but the triangle can 
 be divided into two triangles in a great many ways by lines 
 running from T to different points in the base. Choose 
 from all the possible lines the one that divides the angle 
 B T S in halves. You will be obliged to use your protractor 
 to find this line, although the imagination will supply it 
 easily. It is represented by T 1) E in the figure. Two 
 triangles, /\ R T E and /\ S T E, are now formed ; to aid 
 the imagination it may be well to draw them apart from each 
 other (Fig. 89), or to make a paper isosceles triangle and 
 to cut it along the line corresponding to T D R, 
 
 In comparing ^ R T E and E T 8 we know — 
 
 1st, that T R = T S from the nature of isosceles trian- 
 gles ; 
 
 \M. that T B - 'I' B from the nature of straight lines ; 
 
 3d, that / R T E = / E T S because each is one-half of 
 B T S : so that our knowledge of these two triangles is based 
 upon the nature of the figures and not upon our measure- 
 ments. 
 
 Are the three things that we know about the triangles 
 enough to prove them equal ? Upon what principle of equal 
 triangles do you rely ? Write out the remaining parts 
 of the two triangles that must be equal respectively from 
 the principle stated in 179. You know that there must be 
 in /\ E T S an angle equal to / R of /\ E T K, because 
 the two triangles are alike ; but how do 3*011 know whether 
 / S or / E is the one equal to / R ? In this particular 
 case there can be no doubt, because / E evidently is 
 much larger than / R ; but it is not safe to rely upon the 
 eye in such matters. A method of selecting the corre- 
 
CONSTRUCTIONAL GEOMETRY 43 
 
 sponding angles and lines that can never fail is based upon 
 the fact that the corresponding angles are opposite the 
 sides that have been proved equal, and the correspond- 
 ing sides are opposite the angles that have been proved 
 equal, fn this case E T has been proved equal to E T : 
 opposite ETin/\ETRis/R; opposite E T in /\ E T S 
 is / S ; / R and / S, therefore, are corresponding angles 
 of the two triangles, and are equal by 179. Select the other 
 corresponding parts by the same kind of test. You cannot 
 acquire too much skill in selecting corresponding parts of 
 figures. 
 
 212. The knowledge that we have gained in 211 about 
 isosceles triangles gives us a fifth test for detecting equal 
 angles without protractor or compasses. We now know 
 that the base angles of an isosceles triangle must be 
 equal. 
 
 213. In showing that / R = / S (Fig. 39) we discovered 
 a fact that is of even greater importance, namely, that 
 /RET=/TES. What kind of angle must each 
 one be ? Why ? State the fact as a principle relating to 
 the bisector of the vertical angle of an isosceles triangle. A 
 principle that is discovered while we are trying to prove 
 another principle is called a corollary. 
 
 214. There is a second corollary resulting from the proof 
 in 211 about the point E, in which the bisector of the verti- 
 cal angle strikes the base. State it, and record the state- 
 ment decided by the class to be the best. 
 
 215. Can you without your protractor tell the number of 
 degrees in any angle of /\ E T R ? 
 
 216. If / R of A R T S (Fig. 39) is 70°, what are the 
 angles at S and T ? Why ? 
 
 217. What are the base angles of an isosceles triangle, if 
 the vertical angle is 70° ? if it is 60° ? if it is 90° ? 
 
 218. If the base angle of an isosceles triangle is twice the 
 vertical angle, what is the value of each angle of the tri- 
 angle ? 
 
 219. Make a triangle with all three sides, or legs, equal. 
 
44 
 
 ELEMENTARY AND 
 
 1*0' 
 
 -X. 
 
 B 
 
 Ki.i ML 
 
 What can yon say about its angles ? Why ? Sncli a tri- 
 angle is called an equilateral triangle. Why ? 
 
 220. The vertical angle of an isosceles triangle is twice 
 the base angle ; can you tell the degrees in each angle ? 
 
 221. In Fig. 40, ^ ABC is 
 an isosceles triangle ; /CBD 
 formed by extending the base 
 A B is 140°. What are the 
 angles of /\ A B C ? 
 
 222. In Fig. 41/\MNS 
 is an isosceles triangle whose 
 base angle, M, is 60°. M S is extended so that S P = MS. 
 Can yon name in degrees all the 
 angles of the figure P What is 
 / PM Iff 
 
 223. What would be the value of 
 
 1 V N M (Fig. 41) if we should take 
 
 2 M = 50° ? What, if / II = 70° ? 
 if / M = .v. Can you discover any 
 principle about / V S M ': 
 
 224. When two lines form a right 
 angle thej are said to be perpendicu- 
 lar to each other. The sign for per- 
 pendicular is J_. 
 
 What examples of perpendicular 
 lines have we had in Section X. ? 
 (;i\c examples of perpendicular lines 
 in the room about you. 
 
 225. Can you use the principle of 223 to draw a perpen- 
 dicular to a line A B at B ? Suggestion : Choose a point 
 C corresponding to S in Fig. 41, and outline the necessary 
 isosceles triangles. 
 
 '.'-. Draw six different lines, and draw perpendiculars to 
 them, using the method of 225. Reduce the construction 
 lines as much as you can in the last two or three cases. If 
 you thoroughly understand the principle, you can so reduce 
 the help lines that they will hardly be noticed. 
 
 Flo. 4L 
 
CONSTRUCTIONAL GEOMETRY 
 
 45 
 
 227. In proving the fact about the base angles of an 
 isosceles triangle (211), it was necessary for you to imag- 
 ine the vertical angle bisected, or else to use the protrac- 
 tor to bisect it. For our proof the actual bisecting line 
 was not needed (why ?) ; but, as a matter of practical con- 
 struction, it is important to be able to bisect an angle, 
 and a construction problem is not looked upon as solved 
 in Geometry, unless the ruler and compasses are the only 
 tools used. Can you bisect an angle with ruler and com- 
 passes only ? Can you prove that your method is a sound 
 one ? 
 
 If unable to bisect an angle for yourselves, study the fig- 
 ure given here (Fig. 42), and try to give reasons for the 
 statements made. 
 
 1st. With your compasses make C A = C B, using C as 
 centre. 
 
 2d. With A and B as 
 centres, make A K = B K, 
 putting K below A B. 
 
 3d. Join C K and A B, 
 crossing each other, at D. 
 
 l./CAB=/CBA. 
 Why? 
 
 2. /DAK= £ DBK. 
 Why? 
 
 3. £ CAK= /CBK. 
 Why? 
 
 4. ACAK = ACBK. 
 Why? 
 
 5./DCA=/DCB. 
 Why? 
 
 If the five statements made can be shown to be necessar- 
 ily true, there can be no doubt that the line C D K bisects 
 the angle A C B. 
 
 228. Make six different angles, and bisect them. In the 
 first four cases go through the steps carefully, giving all 
 the reasons which prove in the end that the angle is 
 
46 
 
 ELEMENTARY AND 
 
 bisected. In the last two cases reduce the help lines to 
 those which arc absolutely necessary. 
 
 229. Draw a triangle and bisect each angle. 
 
 230. Draw a triangle, ABC. Extend A B, B C, and 
 C A. Bisect the external angles formed. 
 
 231. Could K have been between C and the line A B in 
 227 ? Study Fig. 43, where C A = C B, and D A == D B, 
 before answering. Are enough 
 facts known about the figure to 
 prove that A A D = AC B D? 
 If so, what follows about an- 
 gles A C D and B C D ? 
 
 232. Now that we know how 
 to bisect an angle without a pro- 
 tractor, we know also how to 
 draw a perpendicular to a line 
 (see 213) in a new way, ami 
 how to bisect a line at the same 
 time (see 214). Give an illustra- 
 tion of bisecting a line by a per- 
 pendicular line, explaining I 
 step taken. 
 
 >. Make six different lines 
 bisect them with 
 method described 
 
 Wm m 
 
 in various positions ami 
 perpendicular lines according to the 
 in 232. Leave to the imagination all 
 lines and points not absolutely necessary for your work, but 
 be able to tell what lines need to be imagined. 
 
 234. Make a triangle and find the middle point of each 
 side. Do you notice anything ]>eculiar about the bisecting 
 perpendiculars when they are prolonged ? Experiment 
 with four different triangles, making at least one of the 
 triangles an obtuse-angled triangle, and note whether or not 
 there is the same peculiarity always. 
 
 235. Make an equilateral triangle, and draw semi-circum- 
 ferences without the triangle on each side as a diameter. 
 
 236. Draw a circumference about each side of a square, 
 A B C D, as a diameter. 
 
CONSTRUCTIONAL GEOMETRY 47 
 
 237. Can you, by the method of 232, draw a perpendicu- 
 lar to the line A B at any point of A B except the midde 
 point ? Give three illustrations. 
 
 238. Draw a line, A B, choose a point P (not the centre) 
 on A B, and draw a perpendicular to A B at P, first by 
 the method used in 226, and then by the method just ex- 
 plained. 
 
 239. Can you use the method of 232 to draw a perpen- 
 dicular to A B from a point P outside of A B ? Give three 
 illustrations. 
 
 240. Draw a line, A B, and choose a point, P, above and 
 beyond the end of A B. How will you proceed to draw a 
 perpendicular to A B from P in this case ? 
 
 241. Draw a triangle, ABC. From each vertex draw a 
 perpendicular to the opposite side. Do you notice any 
 peculiarity of the three perpendiculars ? Repeat the ex- 
 periment three or four times, taking an obtuse-angled 
 triangle twice. 
 
 242. The lines drawn in 241 from the vertices of a tri- 
 angle perpendicular to the opposite sides are called the 
 altitudes of the triangle, because they represent the heights 
 of the triangle as it stands on its different bases. Does the 
 altitude of a triangle ever join the vertex with the middle 
 of the base ? Is the altitude ever as long as a side of the 
 triangle ? 
 
 243. With the aid of the isosceles triangle, wholly or 
 partly drawn, lines and angles can be bisected, perpendicu- 
 lars to lines can be drawn, and equal angles may be dis- 
 covered. You cannot, therefore, become too familiar with 
 what has been shown about the isosceles triangle. Try to 
 answer the following test questions without reference to 
 your note-books. If you are obliged to look up the answers, 
 you will know that you have not sufficiently mastered the 
 subject, and that you have not been studying in the right 
 way. 
 
 1st. If two isosceles triangles have the same base, what 
 can you say about the line that joins their vertices ? 
 
48 ELEMENTARY AND 
 
 2d. How many truths can you give about the line that 
 bisects the vertical angle of an isosceles triangl. 
 
 3d. If you should draw three different isosceles trial Lgi« 
 on the same base, could you say anything about the positions 
 of their vertices ? 
 
 4th. If a base angle of an isosceles triangle is 20°, what 
 are the remaining angles ? If the vertical angle is 20°, 
 what are the remaining angles ? 
 
 ■'II. Make a straight angle. How does this differ from 
 making a straight line ? Bisect and also quarter the 
 straight angle. 
 
 845. Make an angle of G0° without your protractor; 
 then divide it into four equal angles. Divide a semi-cir- 
 cumference into twelve equal parts. 
 
 2 1 ;. Divide a line, A B, into four equal parts ; into eight 
 equal parts. Can you divide it into three equal parts with 
 tin- aid of isosceles triangl i 
 
 SECTION XL TRIANGLES— Coxti mi i». 
 
 c p 
 
 247. Make a triangle, ABC (Fig. 44). Make a second 
 triangle, M N P, as follows : 
 1st. Make M N = A B. 
 2d. Make / M = / A. 
 3d. Make / N = / B, thus completing A M X P. 
 
CONSTRUCTIONAL GEOMETRY 49 
 
 Over what parts of the triangle M N P have you had 
 control, and over what parts have you had no control ? 
 
 Can you think of any reason why / P must be equal to 
 / C ? Review 178, and then show that /\MP, if 
 imagined to be placed upon /\ABC, would exactly coin- 
 cide with /\ABC. Describe with great care the placing 
 of /\ M N P, and show, by giving reasons, what must be the 
 position of the rest of the triangle after you have placed 
 two of its points. 
 
 248. Can you complete /\ M N P by making, 
 1st. M N = A B ; 
 
 2d. I M = I A ; 
 
 3d. I P = / C ? 
 
 You will at first be puzzled to decide where to put the 
 vertex of the angle like / C ; but a little ingenuity will 
 enable you to find the right position without measuring 
 A C, B C, or / B. See 155, if you need a hint. 
 
 Will this triangle necessarily be equal to the triangle 
 ABC? Show, by placing /\MNP upon A A B C, wheth- 
 er or not it will exactly coincide with it. 
 
 249. As a result of 247 and 248 make the following 
 statement true, and record the corrected statement in the 
 summary : 
 
 Two triangles are or are not necessarily equal in all re- 
 spects, if two angles and a side of one are equal respectively to 
 two angles and a corresponding side of the other. 
 
 250. Can you construct a triangle that shall have a side 
 A B, and its angles equal to those of /\ A B C, and yet be 
 unequal to A A B C, because the side equal to A B is not in 
 a corresponding position ? How many such triangles 
 can you construct ? 
 
 251. Make any two angles whose sum will not amount 
 to a straight angle ; call these angles x and y ; also make 
 any line a. Construct as many triangles as you can that 
 shall have the angles x and y and the line a. Can you 
 make two unequal triangles that shall have the side a be- 
 tween the angles x and y ? 
 
 4 
 
50 ELEMENTARY AM> 
 
 252. A person wishing to find the distance from his 
 house on the sea-shore to an island, sighted from his 
 house, represented by point A in Fig. 45, a prominent ob- 
 ject, B, on the island, and also a place, C, on land, which 
 
 could be easily reached, and from which B could be seen ; 
 after noting the angle BAG, he measured A C, and 
 sighted B and A from C, noting the angle A C B. With 
 these measurements before him, he constructed on paper 
 the triangle MNP, having / M = / A, and / N = / C. 
 How did he estimate the distance A B ? 
 
 253. Let the class calculate the distance from the school- 
 house to a tree which can be seen from two windows of the 
 room, making the proper measurements with as few sug- 
 gestions as possible from the teacher. The measurements 
 may be taken in the class, and the calculations may be 
 made later, if the time seems too short. 
 
 254. What additional measurement would be needed to 
 determine the height of the tree ? 
 
 255. Describe and take the measurements that would be 
 necessary to determine the length of a ladder that would 
 reach to a particular branch on the same tree, if the ladder 
 rests on the ground five feet from the tree. 
 
 256. A ship captain sights a light-house at an angle of 
 70° from his course. After sailing three miles on the 
 same course, he sights it airain at an angle of 100° from 
 his course. How far was he from the light-house each 
 
CONSTRUCTIONAL GEOMETRY 51 
 
 time ? How far, when he passed nearest to the light- 
 house ? 
 
 257. Observe two points on a house or tree, one of which 
 is directly over the other. Take the measurements that 
 will enable you to calculate the distance between the two 
 points without leaving the room. 
 
 258. Can you think of any reason why you can not use 
 the method used in 256 to find the distance to a star ? 
 
 259. A man, standing on the bank of a stream at a point 
 A, sees at an elevation of 60° the top of a tree, which stands 
 at a point B on the opposite shore ; he walks away from the 
 shore on the line B A for 80 ft. ; he then sees the top of 
 the same tree at an elevation of 30°. Draw a plan to scale, 
 and find the height of the tree and the width of the river. 
 
 260. Construct a right triangle with one leg 3 in. and 
 the adjacent acute angle 50°. 
 
 261. Construct a triangle having one angle 30°, another 
 65°, and the side joining the vertices of these angles 3J in. 
 
 262. Construct a triangle, ABC, with / A = 70°, / B 
 = 60°, and side B C = 2 in. 
 
 263. How would you find the distance between two 
 islands without leaving the main land ? 
 
 264/ In measuring a triangular field, A B C, a man found 
 the side A B to be 300 yds. ; he found the angle A B C to 
 be 95° ; when he reached C, he was not satisfied with his 
 measurement of B C. Instead of retracing his steps to meas- 
 ure B C, he sighted A from C, finding the angle A C B to be 
 40°. How did this help him 
 to find B C ? What was the £ 
 
 length of B C ? of A C ? /\ \ 
 
 265. Construct a triangle, yS N. 
 
 A B C, by making A B any yS ^^ 
 
 desired length, and by mak- £ jjj ^ 
 
 ing / B = / A. Will this ^ 
 
 triangle have equal legs, A C 
 
 and B C ? If you determine your answer by the appear- 
 ance of the figure, or by testing the lengths with your com- 
 
52 ELEMENTARY AND 
 
 passes, you will doubtless answer that A C and B C are 
 equal. But is there any reason in the nature of things why 
 they must be equal ? Let us investigate. Imagine / A C B 
 bisected by a line, C D, and, to aid the imagination, draw 
 the bisector C D by the method of 227. Examine carefully 
 ^ADC and BDC. Point out what parts of the first 
 have their equals in the second. Are there enough equal 
 parts to make the triangles necessarily equal in all respects ? 
 What principle of equal triangles do you rely upon ? After 
 satisfying yourselves that the triangles are equal, arrange 
 your work as follows : 
 
 In A A D and A B D c & is known, 
 
 1st, that / A = / B (by construction) ; 
 
 2d, that /ACD=/BCD (each / = J of / A C B) ; 
 
 3d, that C D = C D (same line). 
 
 .-. AABC = ABCD. 
 
 (Quote principle that you rely upon here.) 
 
 .*. A C = B C (give reason here). 
 
 . •. A A C B is isosceles. 
 
 .'. A triangle must be isosceles, if its base angles are 
 equal. 
 
 266. Why could we not have said at once in 265 that AD 
 = B D, and that /CDA=/CDB, thus enabling ofl to 
 use a different principle for testing the equality of the tri- 
 angles ? 
 
 267. Make a triangle with its base angles 60°. What 
 must the third angle be ? What kind of triangle is it ? 
 
 268. From a ship that was sailing in a straight course, 
 A B, a light was sighted at an angle of 40° with the course 
 A B ; after the ship had gone four miles, the same light 
 was sighted at an angle of 70° with B A. How far was the 
 ship from the light at the first observation ? 
 
 269. Make a right triangle, ABC, with right angle at 
 C and with acute angle A equal to 60°. How many degrees 
 in angle B ? Draw a line, C K, in such a way that / K C B 
 = / K B C. Study the figure thus formed, and select all 
 the angles whose value in degrees you know without the aid 
 
CONSTRUCTIONAL GEOMETRY 53 
 
 of a protractor ; also select any lines that you can show to 
 be alike without measurement. Do you discover any truth 
 relating to the longest and shortest lines of a right triangle 
 one of whose angles is 60° ? Note the distance of K from 
 A, B, and C. Compare what you note in this case with 
 what you learned in 225. K is on A B. 
 
 270. Repeat the experiment made in 269, this time 
 making / A = 70°. Make /KCB^/KBCas before. 
 Give the value of all the angles formed. Can you say the 
 same thing about K that you did in 269 ? 
 
 271. Make a right triangle, ABC, without knowing the 
 exact value of acute angle A. Draw C K as before so that 
 /KCB= /KBC. Calling /_ A = / x, can you name 
 all the angles formed by means of / x 9 Can you show 
 that what you learned about K in 269 and 270 still holds 
 true ? 
 
 272. If with the middle point of the longest side or 
 hypotenuse of a right triangle as a centre, you describe a 
 circumference, will it pass through the vertices of the 
 right triangle ? Why ? The principle involved in this 
 problem plays an important part in later problems, so that 
 it is well to record it, and to make special effort to remem- 
 ber it. It may be stated thus : The middle point of the 
 hypotenuse of a right triangle is the same distance from the 
 three vertices of the right triangle. 
 
 273. If, when walking on a straight level road, A B, you 
 sight from A a tree across a field at an angle of 70° with 
 A B, at what angle from B A will you have to sight the 
 tree to know that you have walked a distance equal to that 
 from A to the tree ? 
 
 274. Wishing to cut down a tall tree, a man found that 
 it was of great importance to know how far the tree would 
 reach when felled. Having no instruments with him, he 
 fastened two straight sticks together at an angle of 45°. 
 How could he get an angle of 45° degrees with only a piece 
 of string and a stake to work with ? Can you describe 
 how he used the simple instrument to find the height of 
 
54 ELEMENTARY AND 
 
 the tree and the distance wanted ? Why did he choose 
 the angle 45° ? 
 
 275. How many ways of recognizing equal angles without 
 measurement have you learned thus far ? Give them all. 
 
 276. How many ways of recognizing equal lines without 
 measurement do you know ? 
 
 277. Give the three tests for recognizing equal tri- 
 angles. 
 
 278. Make a triangle (A B C, Fig. 47). Make M N = 
 
 B M 
 
 Fio. 47. 
 
 A B. With M as centre and with A C as radius describe 
 an arc. With N as centre and B C as radius describe an 
 arc cutting the first arc at P and Q. Complete the tri- 
 angles M P 1ST and MQN. Clearly these triangles have 
 the same three sides that triangle ABC has. Is there 
 any reason for thinking that /\ M N P has the same angles 
 as/\MNQ? If only one angle is the same in both, the tri- 
 angles must be equal in all respects. Why ? Let us aim 
 to show that /MPN=/MQN. Join P Q. What 
 kind of triangle is^PMQ? Why ? 
 
 .-. / MPQ= /_ Why? 
 
 Likewise /NPQ=/ Why ? 
 
 .-. /MPN=/ Why? 
 
 .-. A Mp N=AMQN. Why? 
 Should you place /\ A B C in the plane of the paper with 
 
CONSTRUCTIONAL GEOMETRY 55 
 
 A B upon its equal M N, where would C have to fall ? 
 Why? Hence/\ABC = A MNP = A M Q N - Hence 
 a new principle for recognizing equal triangles : Two tri- 
 angles are equal in all respects, if the three sides of one are 
 equal to the three sides of the other respectively. 
 
 279. In Fig. 47 can you tell the number of degrees in 
 the angle between M N and PQ? Give reasons. 
 
 280. If, in A A B C, A B is 7 in - lon & and A C is 5 in. 
 long, how short can B C be ? How long can it be ? What 
 reasons can you give for your answers ? 
 
 SECTION XII. QUADRILATERALS. 
 
 281. Construct quadrilaterals with your compasses and 
 ruler as follows : 
 
 1st. A quadrilateral no two of whose sides are parallel. 
 *. Such a quadrilateral is called a trapezium. 
 
 2d. A quadrilateral only two of whose sides are parallel. 
 This figure is a trapezoid. 
 
 3d. A quadrilateral whose opposite sides are parallel. 
 What is this figure called ? 
 
 282. If one line crosses two others, what angles must be 
 alike to make the lines parallel ? Is there more than one 
 pair of angles whose equality will make the lines parallel ? 
 
 283. How many degrees must there be in the sum of the 
 four angles of a quadrilateral ? See 106-110. 
 
 284. Can you make a trapezium with two of its sides 
 alike ? with three of its sides alike ? with all four of 
 its sides alike ? In the last case, draw a line joining two 
 opposite corners, dividing the quadrilateral into two tri- 
 angles, and, from your study of these triangles, decide 
 whether the figure may be a trapezium or must be another 
 kind of quadrilateral. 
 
 285. Can you make a trapezium with two angles alike ? 
 
56 ELEMENTARY AND 
 
 with three angles alike ? with four angles alike ? Can 
 you make a trapezium with three of its angles 60° each ? 
 Why ? Can there be three angles of 70° in a trapezium ? 
 
 286. Can you make a trapezoid with two sides alike ? 
 with three sides alike ? with two angles alike ? with three 
 angles alike ? 
 
 287. Draw a trapezoid, A B C D, making A B and C D 
 antiparallel. It is possible to fix the value of / D from 
 the value of / A. Do you see how ? If / A = 70°, 
 what is / D ? If / B = 25°, what is / C ? Does / A 
 help you to find / C ? 
 
 . 288. If, in making a trapezoid A B C D, you first make 
 / A = 70° and / B = 40°, can you make fe C and D 
 of any size that you please, or are they fixed in value ? 
 Give a reason for your answer. 
 
 289. How many angles of a parallelogram can you make 
 before all of the angles are fixed in value ? 
 
 290. The most interesting of the quadrilaterals are the 
 parallelograms, of which there are several varieties. Those 
 with their sides all alike and with their angles 90° each are 
 called squares. 
 
 Those that have their sides alike but their angles ob- 
 lique are called rhombuses. Those that have right angles 
 but adjacent sides unequal are called rectangles. Those 
 that have oblique angles and adjacent sides unequal are 
 called rhomboids. 
 
 291. Construct models of each kind of parallelogram, 
 describing the steps in the order in which you take them 
 and at the time at which you take them. 
 
 292. Select examples of the different kinds of parallelo- 
 gram from the school-room. What kind of parallelogram 
 is a sheet of writing-paper ? a diamond ? a window-pane ? 
 
 293. Can you make a rhomboid with three of its angles 
 alike ? Why ? 
 
 294. Is any proof needed to show that the opposite sides 
 of a parallelogram are parallel, pair by pair ? Why ? 
 
 295. Can you draw a parallelogram whose opposite sides 
 
CONSTRUCTIONAL GEOMETRY 57 
 
 are not equal ? Does your answer to this need proof ? 
 Why? 
 
 296. Can you see any reason why a parallelogram should 
 not be defined in the first place as a quadrilateral with its 
 opposite sides parallel and equal f 
 
 Note. — It is important at this point to make clear, if 
 possible, the danger of imposing too many conditions upon 
 lines and angles, and also, in the same line of thought, to 
 impress upon the pupil the importance of drawing the 
 figures exactly according to the conditions, except in rare 
 cases. If the condition imposed is that the lines be made 
 parallel, they should be made parallel, and not equal, al- 
 though it may be possible to prove that they are parallel 
 when made equal. The subject should not be left until, 
 by varied illustration, some impression is produced upon 
 the pupil. It will generally be found necessary to return 
 to the subject at short intervals. Abundant opportunity 
 for testing the pupil's understanding of the subject may 
 be found in questions about the line bisecting the vertical 
 angle of an isosceles triangle. 
 
 297. If a parallelogram really cannot be drawn with its 
 opposite sides unequal, there must be some reason for the 
 fact based upon the nature of lines and angles, or upon 
 principles already established. Draw an angle, DAB; 
 complete a parallelogram by making D C f f A B and 
 B G f f A J), taking pains to make A B and A D un- 
 equal. Draw a line, D B, joining two opposite vertices 
 (the line is called a diagonal), and study the triangles thus 
 formed. It should be a simple matter for you to prove 
 /\ABD=/\BCD. Arrange your work in a vertical 
 line, putting no more than one fact and its reason on a 
 line. Follow this model : 
 
 I. I make D C f f A B. 
 
 II. I make B C f f AD, completing a parallelogram. 
 To prove A B = D C and A D = B C. 
 III.. Draw diagonal D B and compare ^A B D and D B C. 
 IV. / = / (Insert reason.) 
 
58 ELEMENTARY AND 
 
 V. / = / (Reason.) 
 
 VI. Side = Side (Reason.) 
 
 VII. .-. A •••• = A •••• (Reason.) 
 VIII. .\, etc. 
 
 298. It may help you to understand 297 to cut a paral- 
 lelogram something like A B C D from stiff paper, and to 
 cut the parallelogram into two parts along the diagonal 
 D B. Can you place the two triangles formed by cutting 
 the paper parallelogram so that they will not form a paral- 
 lelogram ? Is a quadrilateral whose diagonal divides it 
 into two equal triangles necessarily a parallelogram ? 
 
 299. Can you place the two paper triangles of 298 in 
 more than one position so that they will form a parallelo- 
 gram ? 
 
 300. Does it make any difference which diagonal you 
 draw ? Will the parallelogram be divided into two equal 
 triangles in each case ? 
 
 301. How does a rhomboid differ from a rhombus ? 
 
 302. How does a rhombus differ from a square ? 
 
 303. How does a rectangle differ from a square ? 
 
 304. Notice that, although squares, rhombuses, and rec- 
 tangles are not rhomboids, they differ from the rhomboid 
 merely in having special characteristics in addition to 
 those of the rhomboid. Hence any general principle that 
 is proved about the rhomboid is applicable to the rhombus, 
 the square, and the rectangle; but it is dangerous to as- 
 sume that a principle proved true of a square, a rhombus, 
 or a rectangle can be applied to a rhomboid, because the 
 principle may depend upon the special characteristics of the 
 former figures. Therefore, when trying to prove a prin- 
 ciple that will apply to all .parallelograms, be careful to 
 draw a rhomboid for illustration, or to have a rhomboid in 
 your imagination. 
 
 305. Into what kind of triangles does a diagonal divide a 
 square ? a rectangle ? a rhombus ? 
 
 306. Can you construct a parallelogram with one angle 
 90° and another 80° ? Why ? 
 
CONSTRUCTIONAL GEOMETRY 59 
 
 307. If the adjacent sides of a parallelogram are equal, 
 is there any doubt as to what kind of parallelogram, the 
 figure is ? 
 
 308. Can you make a rhomboid with one side 3 in., an- 
 other 2 in., and a diagonal 5J in. ? Why ? 
 
 309. Make a rhomboid, A B D (Fig. 48). What pre- 
 cautions must you take to be sure that the figure is a 
 rhomboid ? Draw both 
 
 diagonals, A C and B D, 
 crossing at M. 
 
 1st. Make a record in a 
 column as in 297 of all 
 the pairs of equal angles 
 that you can find in the 
 figure, giving a reason in 
 
 fe , ' ° ° Fig. 48. 
 
 each case. 
 
 2d. Select pairs of lines that you can declare to be equal, 
 because of principles already established, giving the prin- 
 ciple in each case. 
 
 3d. Select four pairs of equal triangles in the figure, stat- 
 ing the proper principle for each case. 
 
 4th. From the equal triangles show that certain lines in 
 the figure must be equal to each other that were not known 
 before to be equal. What can you say of the point M ? 
 How do the diagonals divide each other ? Write and re- 
 cord a principle stating the fact that you have just learned 
 about the diagonals of a parallelogram. 
 
 310. Draw a square, a rhombus, and a rectangle, and 
 notice how their diagonals divide each other. Do the di- 
 agonals ever cross at right angles ? Can you account for 
 their crossing at right angles by the principles that enabled 
 you to draw a perpendicular to a line ? 
 
 311. Do the diagonals ever cross at a point which is the 
 same distance from all four corners ? You have already 
 learned a principle that should enable you to answer this 
 question ; can you recall it ? 
 
 312. What kinds of parallelogram can you inscribe in a 
 
60 ELEMENTARY AND 
 
 circumference ? A figure is inscribed in a circumference 
 when the circumference passes through all the vertices of 
 the figure. 
 
 312. On smooth, stiff card-board draw two squares with 
 sides If in. long. In the first square draw one diagonal. In 
 the second square draw the lines represented in the second 
 figure of the plate on the next page. Cut the squares into 
 seven pieces, numbered as in the diagram. 
 
 Select all the pairs of equal lines in the second square ; 
 also give the number of degrees in every angle formed. 
 
 313. With the seven pieces formed by cutting the squares 
 of 312 a great variety of interesting figures can be formed. 
 For instance, by placing them as in the fourth figure of 
 the plate, an old-fashioned chair like the third figure of the 
 plate is formed. 
 
 314. Try to form figures like those given in the next 
 three plates. Each time you must use all seven pieces. I 
 have seen over three hundred figures formed with these 
 seven pieces, many of them bearing quaint resemblances to 
 common objects. With squares of wood, celluloid, or 
 ivory, divided in the manner outlined, you may get much 
 amusement by attempting new forms. 
 
 315. Arrange the seven pieces in such a way as to form 
 a rectangle ; a rhomboid ; a square. Notice the side of the 
 square formed. Which of the original lines forms its side ? 
 How much larger is the new square than the old one ? Is a 
 method of finding a square twice as large as a given square 
 suggested ? 
 
 316. If you put pieces 4, 5, and 7 together so that they 
 form a square, how will that square be related to piece 3 ? 
 What part of the original square will it be ? 
 
 317. Write, for review, short sketches of the square, 
 rhombus, and rectangle, following as a model the following 
 sketch of the rhomboid : The rhomboid is a quadrilateral 
 which has two pairs of parallel sides, its adjacent sides un- 
 equal, and no one of its angles equal to 90°. Each diagonal 
 divides the rhomboid into two equal triangles; the two 
 
CONSTRUCTIONAL GEOMETRY Gl 
 
 □ 
 
ELEMENTARY AND 
 
 y 
 
 LT^ 
 
 ^_7 
 
 \ 
 
CONSTRUCTIONAL GEOMETRY 
 
 63 
 
 /\ 
 
 
 N/ 
 
 X 
 
 7 
 
 W 
 
 \ 
 
 7 
 
Fi<;. 49. 
 
 64 ELEMENTARY AND 
 
 diagonals bisect each other, and divide the rhomboid into 
 four triangles equal pair by pair. One angle of a rhomboid 
 determines all the angles, because the opposite angles are 
 equal, and the adjacent angles are supplementary. 
 
 318. Draw a quadrilateral, ABCD (Fig. 49), by making 
 DCffABand DC = AB. When A D and B C are 
 joined, will the figure be 
 a parallelogram ? Draw 
 B D, and see if you can- 
 not select enough equal 
 parts to prove /\ABD 
 = /\ B D C. Arrange 
 your work neatly, as al- 
 ready explained to you 
 in 297. Draw as many 
 conclusions as you can ; and, as a result of your investiga- 
 tion, make the following statement correct : A quadrilat- 
 eral is or is not necessarily a parallelogram, if it has two sides 
 both equal and parallel. 
 
 319. Place two unequallmes, M Pand N Q, with their mid- 
 dle points together ; make the lines cross at an oblique angle. 
 Can you show what kind of quadrilateral MNPQ must 
 be ? Record the principle, when you have discovered it. 
 
 320. Could you so place the lines M P and N Q of 319 
 that the figure MNPQ would be a rhombus ? 
 
 321. If you wish to place two sticks so as to make a 
 rectangular kite, what conditions must the sticks fulfil ? 
 Would the same conditions enable you to place the sticks 
 so as to make a square kite ? 
 
 322. Make a quadrilateral, A B C D, by making A B = 
 D C and A D = B C. Will the quadrilateral have any pecul- 
 iar form ? Can you prove it by drawing, as a help line, the 
 diagonal B D ? As a result of your investigation complete 
 
 the following sentence : A quadrilateral be a if 
 
 it has its opposite sides equal pair by pair 
 
 323. Can you draw a quadrilateral with two pairs of 
 equal sides that does not come under the principle of 322 ? 
 
CONSTRUCTIONAL GEOMETRY 65 
 
 SECTION XIII. DIVISION OF LINES. 
 
 324. Draw a scalene triangle, ABC 
 scalene triangle is one that has 
 its sides unequal. A scalene 
 triangle bears the same relation 
 to triangles that a rhomboid does 
 to parallelograms. It is the 
 kind of triangle that should be 
 drawn for illustration when no 
 particular kind is specified. 
 
 Find M, the middle of A C, 
 and draw M N symparallel with 
 A B. (Let your construction 
 lines be light but visible.) 
 
 Draw a help line N K f f MA. (Note that help lines 
 are always dotted lines.) 
 
 Study the figure thus formed, and write answers to the 
 following questions, giving reasons in each case : 
 
 How many and what angles are equal to / A ? 
 
 What angle is equal to / C ? to / B ? 
 
 What kind of quadrilateral is A K N M ? 
 
 What lines can be inferred to be equal from the nature 
 of the quadrilateral A K N M ? 
 
 How many angles of A M N C can be found mated in 
 AKBN? 
 
 Is any side of A M N C known to be equal to a side of 
 A K B N ? 
 
 Is A M N C equal to A K B N ? 
 
 If you have been able to answer these questions, and to 
 support your answers with good reasons, you have proved 
 that C N = N B and M N = K B ; so that N is the middle 
 point of B C, and M N is one-half of the base line A B. 
 
 325. The principle involved in 324 is one of the most 
 important in Geometry. Stated in general language it is 
 as follows : 
 
 5 
 
B 
 
 66 ELEMENTARY AND 
 
 A line drawn from the middle point of one side of a triangle 
 parallel to the base bisects the other side, and is itself one- 
 half of the base. 
 
 326. This principle suggests a new and simple method of 
 bisecting lines by using common ruled paper, where the lines 
 are drawn equal distances apart. A sheet of theme-paper 
 is excellent for this purpose, because there is a red line at 
 right angles to the blue ruled lines, and because the paper 
 is broad. Along the red line number successive lines, 0, 1, 
 2, 3, etc. (Fig. 51). Place the line to be bisected on line 
 2 with one end at point 2 ; con- 
 nect the other end, A, with point 
 by your ruler's edge ; 1 B is 
 clearly £ of 2 A. The line A 
 need not be drawn ; the same 
 piece of paper, therefore, may be 
 used for bisecting an indefinite 
 number of lines. 
 
 327. On what other lines be- 
 tween and 12 could you place 
 the line to be bisected, and on 
 
 what lines respectively would the half-line appear ? Bisect 
 a line by placing it on three different lines, and compare 
 your results. Can you see any principle to guide your 
 choice of a line on which to place the line to be bisects I 'i 
 
 328. Make a triangle, and then make another with its 
 sides one-half as long, using the new method for getting 
 the half-lines. 
 
 329. Make a quadrilateral — not a parallelogram — and 
 find the middle points of its sides by the method of 
 Join the middle points of the sides in order by straight lines. 
 What kind of quadrilateral will be formed ? Can you ac- 
 count for its shape ? 
 
 Suggestion : Draw one diagonal of the original quadri- 
 lateral, and apply what you learned in 325 and 318. The 
 principle of 325 does not apply immediately, but it leads 
 very easily to the right principle. 
 
CONSTRUCTIONAL GEOMETRY 67 
 
 330. Make a rhomboid, and repeat the experiment of 
 329. Can you account for the shape of the new parallelo- 
 gram formed ? Can it be anything but a rhomboid ? 
 
 331. Repeat the experiment of 329, using a rectangle. 
 What kind of parallelogram will be formed by joining the 
 middle points of the sides ? 
 
 332. Try the same experiment, using a rhombus at first. 
 Account for the shape of the second parallelogram formed. 
 
 333. Draw a trapezoid, A B C D (Fig. 52). Find M, 
 the middle point of AD, 
 draw M N f f A B, and study 
 the figure formed, using as a 
 
 help line the diagonal D B. m/- -^ 
 
 What reason have you for 
 thinking K the middle of 
 DB? A ^ 
 
 Compare M K with A B. FlG - 52 - 
 
 Have you any reason for thinking N" to be the middle point 
 of B C ? Suggestion : Study the ^ D B C. Compare K N 
 with D C. Compare M N with A B and D C combined. The 
 line MNisa famous line in the trapezoid. It is called the 
 median of a trapezoid because it passes though the middle 
 points of both legs. Notice carefully how the median was 
 drawn. It was not drawn from middle point to middle point, 
 but from one middle point parallel to the two bases, and was 
 proved to pass through the middle point of the second leg, a 
 characteristic from which it gets its name. Perhaps a more 
 important characteristic is that it is half of the sum of the 
 bases, or the average of the two bases. It is very important 
 that you should become familiar with both characteristics 
 and with the following principle : The line drawn from the 
 middle point of one leg of a trapezoid parallel to the bases bi- 
 sects the other leg, and is itself one-half the sum of the bases. 
 
 334. Make two lines, a and b, and find their average by 
 the principle of 333, and with the aid of the ruled sheet of 
 paper used in 326 ; also add the two lines a and b, and find 
 one-half of the resulting line by 326. Compare the results. 
 
 OF THE 
 
 ( UNIVERSITY 
 
68 ELEMENTARY AND 
 
 335. What is the average width of a board which is 12 
 in. at one end and 9 in. at the other ? Do this both by 
 Geometry and by Arithmetic, and compare the results. 
 
 336. Find the average diameter of an ordinary wooden 
 pail. Can you find the average diameter of a barrel ? of 
 a base-ball bat ? 
 
 337. In a trapezoid, ABCD, find the median, if A B = 
 11| in. and C D = 5 J in. 
 
 338. Find the base A B of a trapezoid, A B C D, if C D 
 = 8 in., and the median, M N, =12 in.; if C D = 11 in. 
 and the median, MN, = 13 \ in.; if C D = 7£ cm., and 
 M N = 4| cm. 
 
 339. On a line, A D (Fig. 53), mark off five equal 
 lengths with your compasses. Draw A B making an 
 oblique angle with A D, and through the points of divis- 
 ion on A D draw parallels to A B terminating in a line 
 B C. Measure the upper two parallels and calculate the 
 rest. 
 
 340. Could you calculate all the lines from measurements 
 of the first and third lines, 
 D C and E L? of the first and 
 fourth lines, D C and G M ? 
 
 341. On a sheet of ruled 
 paper draw two non-parallel 
 lines in such a way that they 
 shall intercept \ in. on the 
 top line and 1 in. on the line 
 below the top. Calculate the 
 lengths intercepted on all the 
 lines. " B 
 
 342. On a piece of theme- 
 paper mark the points at which the red line crosses the blue 
 lines, 0, 1, 2, 3, 4, 5, etc., in succession ; draw from the 
 point a line, Y, crossing obliquely the blue lines at 
 points A, B, C, D, E, etc., respectively. If you rule the 
 paper yourselves, take pains to have the lines drawn at 
 equal distances and strictly parallel. Study the triangle 
 
CONSTRUCTIONAL GEOMETRY 69 
 
 2 B. What can you say about the line 1 A ? the line A ? 
 If you give the name x to the line 1 A, what name will you 
 give to the line 2 B ? 
 
 In the next place study the trapezoid 1 A C 3. The line 
 
 2 B is manifestly the median of the trapezoid, and, there- 
 fore, the average of the two bases. It follows that the line 
 
 3 C must be equal to 3x, for 2x is the average between x 
 and 3x. By studying the different trapezoids in order, 
 you will see that 4 D is four times as long as 1 A, or \x ; 
 that 5 E is five times as long as 1 A, or 5x, etc. What 
 would be the distance cut off on line number 10 by Y ? 
 on line number 16 ? Wherever you can, confirm your re- 
 sults by the middle-point principle of 326 and 327. Thus 
 by studying /\04 D you see that 4 D ought to be twice 
 as long as 2 B, or 4x, the result already obtained. 
 
 343. Does it make any difference in the relative value of 
 the lines cut off whether Y diverges more or less from 
 the red vertical line ? 
 
 344. If you should place a line one inch long on line 
 number 10 — placing one end of the inch line on the red 
 line and joining the other end with — what part of an 
 inch would be found on line No. 1 ? on line No. 3 ? Have 
 you ever seen a ruler for getting tenths of an inch arranged 
 on this principle ? Cannot some one of the class obtain 
 such a ruler for illustration in the class-room ? 
 
 345. How would you use these same ten lines to obtain 
 jfo of an inch ? T f 7 of an inch ? 
 
 346. Compare the line 3 C, on line No. 3, with the line 5 E 
 on line No. 5 ; also compare line 4 D with line 7 G. Then 
 explain how you would get § of a line, $ of a line, f of a 
 line, T V of a line by means of a sheet of ruled theme-paper. 
 
 347. Place any line, P Q, on the tenth line, with P at point 
 10. What part of P Q will be found between your ruler 
 and point 7, if you place your ruler's edge from to Q ? 
 
 348. Can you get J of a line by using the same sheet of 
 paper ? Explain your method. Draw a line, P Q, and get 
 | of it ; % of it ; 2 J of it. 
 
70 
 
 ELEMENTARY AND 
 
 349. Make a trefoil (Fig. 11), with the centres about one 
 inch apart. Then make a figure just like it in shape 
 —or similar to it— with all the lines $ as long. Then 
 enlarge the original figure by making the lines } as long. 
 
 350. Make a rose window (Fig. 54) by putting the cen- 
 tres of the arches at points on 
 a £ in. circumference which are 
 f of an inch apart. Reduce the 
 figure by making the lines ^ as 
 long ; also enlarge the figure by 
 making the lines £ as long. 
 
 351. With only ten ruled lines 
 can you find -fc of a line? & of 
 aline ? T J T of aline ? 
 
 352. Can you find $ of a line 
 too long to put on your paper ? r, ° M - 
 
 353. Find a line -^ as long as your desk. 
 
 354. Can you find f T of a line with a sheet of paper con- 
 taining only ten parallel lines ? 
 
 355. It will be convenient for the future to refer to your 
 ruled sheet of theme-paper as your dividing tool; for with 
 it you can get almost any fraction of a line that you wish. 
 
 356. If, instead of putting point on the top line, we had 
 put it on line 5, say, and about half-way between the ed| 
 and if we had numbered the lines above this zero line, 
 1, 2, 3, 4, 5, and the numbers below the zero line, 1, 2, 
 3, 4 5, 6, 7, etc.; and if we had drawn two oblique lines 
 meeting at across all the lines, we should have had a 
 picture of an instrument called proportional dividers, a 
 very useful instrument in reducing or enlarging a figure. 
 The principle is the same as that of your dividing tool, but 
 the tool is more convenient than yours in actual practice. 
 Any line between the oblique lines on line 5 above the zero 
 line is the same in length as the corresponding line on line 
 5 below the zero line, and this latter line we know to be f 
 as long as the length intercepted on line 7 by the oblique 
 lines. Make a diagram to illustrate what has been said, 
 
CONSTRUCTIONAL GEOMETRY 71 
 
 and see if you can follow it. Proportional dividers are di- 
 viders pivoted, not at one end as your dividers are, but some- 
 where between the ends at a point, corresponding to 0, that 
 can be changed at pleasure by a sliding screw. If it is 
 wished to reduce the lines of a figure to $ of their actual 
 size, the dividers are pivoted so that the legs on one side of 
 the pivot are f as long as those on the other side. Now, 
 when the longer legs are opened any distance, the shorter 
 legs are opened only f as far ; so that all the reduced lines 
 can be found with great rapidity without changing the pivot 
 screw. The figures, pp. 61-63, were reduced from actual 
 blocks by proportional dividers. If possible, secure a pair 
 of proportional dividers to illustrate the important principle 
 involved. 
 
 REVIEW. 
 
 357. Review from 281, and come into the class prepared 
 to write a short sketch of any of the quadrilaterals thus 
 far studied. 
 
 358. Do the diagonals of a trapezoid or a trapezium ever 
 bisect each other ? 
 
 359. What can be said of the diagonals of a rhombus that 
 cannot be said of the diagonals of a rectangle, and vice versa f 
 
 360. A quadrilateral when cut along one diagonal, was 
 found to be divided into two equal pieces ; can you say 
 what the shape of the quadrilateral was ? Why ? 
 
 361. Two sides of a quadrilateral were found to be equal 
 and to be symparallel ; what possible shapes could the fig- 
 ure have had ? 
 
 362. In studying a certain quadrilateral a pupil noticed 
 that the point where the diagonals crossed was the same dis- 
 tance from all the corners ; do you know what quadrilateral 
 he was studying ? 
 
 363. If you have two lines of unequal length, how will 
 your dividing tool enable you to tell whether the shorter is 
 more or less than T \ of the longer ? 
 
72 ELEMENTARY AND 
 
 364. What principle of the trapezoid leads to the con- 
 struction of your dividing tool ? 
 
 365. With a pair of proportional dividers how would you 
 reduce the figures on the plates following 314 so that the 
 lines would be § as long ? 
 
 366. If on side A B of /\ A B C a point, P, is chosen 
 I of the distance from C to A, and if from Pa line, P Q, is 
 drawn symparallel to the base A B, what part of A B will 
 P Q be ? What part of C B will C Q be ? 
 
 SECTION XIV. MULTIPLICATION OF LINES. 
 AREAS. 
 
 367. We have already learned how to add lines and to 
 subtract lines ; in certain cases, with the aid of the dividing 
 tool, we can divide one line by another ; for the line inter- 
 cepted on the ninth parallel by Y of the dividing tool will 
 contain the line intercepted on the fourth parallel by Y 2^ 
 times. The question naturally arises whether or not two 
 lines can be multiplied together. 
 
 Draw a line, M N (Fig. 55), and with your dividing tool 
 find M Q = { M N. Place the line M Q at right angles to 
 M N, and complete the rectangle M N S Q. Now divide 
 M N into nine equal parts with your dividing tool. How 
 many of these parts will M Q contain ? Why ? Through 
 the points of division in M N draw parallels to M Q, and 
 through the points of division in M Q draw parallels to 
 M N. The small figures formed will be squares. Why ? 
 How many of them ? 
 
 How many squares would there have been, had you made 
 M Q = £ of MN? How many squares in the figure 
 MTRQ? If each one of the divisions of M N is called " b," 
 each one of the little squares is called a " square b " or " b 
 square " (written b 2 ). It is easy to see from experiments 
 
CONSTRUCTIONAL GEOMETRY 
 
 73 
 
 with different rectangles in Fig. 55 that any rectangle con- 
 tains as many " square b's " as there are units in the product 
 obtained by multiplying the number of b's in the base by the 
 number of b's in the side. Thus the rectangle TNSR, 
 with two b's in the base and five b's in the side, contains 
 two lines five or ten " square b's," written 106 2 . 
 
 In describing a floor we say that it contains 300 square 
 feet — meaning by that that it covers as much surface as 
 
 h 
 
 Mhr 
 
 N 
 
 Fig. 55. 
 
 300 squares each 1 ft. long and 1 ft. wide would cover ; and 
 we find out how many square feet it covers by measuring 
 the length and width in feet, and by multiplying the two 
 results together. In the same way we measure M N and 
 M Q in b's, and multiply the number of b's in M N by the 
 number of b's in M Q, thus obtaining the number of 
 u square b's " covered by the rectangle MNSQ. 
 
 368. It is, therefore, a simple step to say that the product 
 of two lines represents a surface, bounded by a rectangle of 
 which the two lines are adjacent sides. To multiply two 
 lines, then, according to this understanding of the terms, 
 we simply complete a rectangle with the two lines as ad- 
 jacent sides. It is not necessary to form each time a pict- 
 
74 ELEMENTARY AND 
 
 ure of the little squares that correspond to b 2 in Fig. 55. 
 On the contrary, it is better to think of the rectangular sur- 
 face as a whole in thinking what the product of two lines 
 means. 
 
 369. Draw three pairs of lines, and multiply the lines of 
 each pair together. It will be well to review the method of 
 drawing a perpendicular at the end of a line given in 8 ■.'.".. 
 
 370. Draw three rectangles, and tell what lines must be 
 multiplied together to produce them. 
 
 371. In problems relating to the surface, or area, of rec- 
 tangles, the upright side is called the altitude of the rectan- 
 gle, and the line on which the rectangle is supposed to 
 stand is called the base. The principle stated in the pic- 
 ceding articles of this section may be expressed as follows : 
 The product of the base and the altitude of a rectangle is the 
 surface of the rectangle. 
 
 372. Although the simplest product of two lines is a rec- 
 tangular surface, fortunately we are not limited to this sur- 
 face. 
 
 Draw a rhomboid A B C D (Fig. 56 (a)), and draw 
 B H 1 A B. Cut the /\ B II C from the right of Fig. 56 (a) 
 
 (a.) FlQ 66. (6.) 
 
 and place it to the left of A D, as in Fig. 56 (b). How 
 do you know that B C will fit on A D, and that D K will 
 be in a straight line with II D ? How does the surface of 
 the rectangle ABHK compare with the surface of the 
 rhomboid A B C D ? What product represents the rectan- 
 gle ? What product represents the rhomboid therefore ? 
 Illustrate this article by cutting a rhomboid out of paper. 
 
CONSTRUCTIONAL GEOMETRY 
 
 75 
 
 373. The surface, or area, of the rhomboid is the same 
 as the surface of the rectangle ; but the rhomboid is not 
 equal to the rectangle because one figure would not fit the 
 other exactly. When two areas are the same the figures 
 are said to be equivalent ; a new sign, =o, is very com- 
 monly used to express this condition, although it is not a 
 sign universally agreed upon. The sentence / J ABC 
 D O 1 | A B H K is translated thus : The rhomboid 
 A B D is equivalent to the rectangle ABHK; and 
 the meaning is that the figures have the same area. 
 
 374. What is the altitude of the rhomboid A B C D, 
 Fig. 56 (a) ? Translate this sentence : / 7 A B C D =o= 
 A B x B H. 
 
 375. Make four rhomboids of different shapes, and draw 
 the lines whose product gives the area in each case, record- 
 ing the result according to the geometrical sentence in 374. 
 Is there more than one answer in each case ? 
 
 376. Multiply two lines so that the product shall be a 
 rhomboid. Can you get more than one rhomboid for an 
 answer ? How many ? 
 
 377. Can you multiply two lines and obtain the result- 
 ing rhomboid at once, without first drawing the rectangle 
 of the two lines ? 
 
 378. Make two parallel lines, A F and K P (Fig. 57). 
 Make AB = OD = EF, and complete the rectangle 
 A B H K and the two rhomboids D M N and E F P 0. 
 Which of the three parallelograms has the largest surface ? 
 Why ? Make clear what product represents each surface. 
 
76 
 
 ELEMENTARY AND 
 
 379. The sum of the bounding lines of a figure is OoDed 
 its perimeter. Do the three parallelograms of Fig. 57 
 have equal perimeters ? 
 
 380. If a man has an acre of ground, will it make any 
 difference to him, in the cost of fencing it, whether the 
 field is in the shape of a rectangle or of a rhomboid ? 
 Which is the more economical shape ? 
 
 381. Can you arrange a square foot of paper in the form 
 of a rhomboid with a perimeter of fifty feet ? Can you 
 make the perimeter as long as you please without changing 
 the surface of the paper ? as short as you please ? 
 
 382. In building a cistern of a fixed height to hold a cer- 
 tain amount of water, does it make any difference in the 
 number of bricks used whether you make the bottom of 
 the cistern in the shape of a rhomboid or of a rectangle ? 
 
 383. Which lets in the more light, a diamond-shaped 
 window pane or a rectangular window pane, if the perim- 
 eters of the two panes are the same ? 
 
 Make two lines, a and b. Make a rhomboid equivalent 
 to a x b which shall have a perimeter three times the sum 
 of a and b. 
 
 384. If a farmer wants to have as little fencing as possi- 
 ble to build, and if he is obliged to have his field in the 
 shape of some parallelogram, 
 what parallelogram ought he 
 to choose ? 
 
 385. Under what conditions 
 are two parallelograms with 
 the same base equivalent ? Il- 
 lustrate your answer, and re- 
 cord the principle involved. 
 
 386. Can the product of two 
 lines ever be a triangular sur- 
 face ? This is merely another 
 way of asking whether a tri- 
 angular surface can be changed into an equivalent rectan- 
 gular surface. Draw a scalene triangle, ABC (Fig. 58). 
 
CONSTRUCTIONAL GEOMETRY 77 
 
 Find the middle point, M, of A C, and draw M N parallel 
 to A B. Draw from C a line, C H 1 A B. (Show your 
 construction lines. ) What is C H called ? What part of 
 C His CD? Why? 
 
 Erect perpendiculars to A B at A and B. Extend D M 
 and D N until they meet these lines at K and S respec- 
 tively. Show that A D N = A S K B and that AC M D = 
 AMKA. 
 
 IsAABC=o=[^]ABSK? Why? 
 
 By what line must you multiply A B to produce 
 A ABC? 
 
 Follow the steps of this article with a paper triangle 
 which you cut along the lines indicated. 
 
 387. Draw five different triangles, and draw side by side 
 the lines whose product in each case represents the area of 
 the triangle. In two of the cases make a base angle obtuse. 
 
 388. Make three pairs of unequal lines. Form three tri- 
 angles whose surfaces will be represented by the products of 
 the lines respectively taken pair by pair. 
 
 389. Is there only one triangle to be found equivalent 
 to the product of two lines, a and b ? If more than one, 
 how many can be found ? Do their perimeters differ ? 
 
 390. Make two parallel lines ; on the lower line, A D, 
 
 Fig. 59. 
 
 take three equal lengths, A B, E F, and C D ; on the up- 
 per line take three points, P, R, and Q, at random ; com- 
 plete the triangles A B P, E F R, and C D Q. Which 
 has the largest area ? Which has the longest perimeter ? 
 If you wish to find the triangle which has the shortest 
 
78 ELEMENTARY AND 
 
 perimeter possible, provided that the base equals A B and 
 that the vertex is on P Q, what shape will you give the tri- 
 angle ? 
 
 391. Pin the ends of a short piece of elastic to a line A B 
 at the points A and B ; then form a triangle A B P with 
 the elastic without stretching the elastic ; draw through 
 P a line, P Q, parallel to A B, and pull the vertex P of the 
 elastic triangle along P Q. In this way you can mould the 
 triangle A B P into a triangle having almost any desired 
 shape. Does the area change in the process ? When is 
 the perimeter shortest ? If the elastic could be stretched 
 indefinitely, would there be any limit to the perimeter ? 
 
 392. Can you mould the elastic triangle of 391 into a 
 right triangle ? an isosceles triangle ? an equilateral tri- 
 angle ? 
 
 393. Learn by heart the following principle : 
 
 If the vertex of a triangle is moved parallel to the base, the 
 area of the triangle remains unchanged as long as the base 
 remains unchanged. 
 
 394. Make a triangle, ABC, and mould it into an equiv- 
 alent triangle with one side twice as long as A C without 
 changing the base A B. 
 
 395. Make a triangle, ABC, and mould it into an equiv- 
 alent isosceles triangle ; mould it also into a triangle with 
 base angle 70°. 
 
 396. Make two lines, a and b, and find the triangle 
 represented by their product, without constructing an aux- 
 iliary rectangle. Repeat the problem with three different 
 pairs of lines. 
 
 Note. — The problem should be repeated until it is as easy 
 to form the " triangle of two lines " as it is to form the 
 "rectangle of two lines." 
 
 397. Make a pyramid with a square base. Draw the 
 lines, five pairs, the sum of whose products, pair by pair, 
 represents the total surface of the pyramid. 
 
 398. If three triangular tin boxes of equal height, with 
 bottoms and tops shaped like the three triangles of Fig. 59, 
 
CONSTRUCTIONAL GEOMETRY 
 
 79 
 
 were to be made, which box would hold the most sand ? 
 Would one box require more tin than another ? Why ? 
 Can you draw pictures of the boxes ? What is the geomet- 
 rical name for the sides of the boxes ? When boxes are made 
 with triangular bottoms, what shape is usually chosen ? 
 Do you understand why ? 
 
 SECTION XV. EQUIVALENT FIGURES. 
 MOULDING OF AREAS. 
 
 399. The principle of 393 enables you to mould triangles 
 and all plane figures of geometry almost as easily as if the 
 figures were made of elastic or of clay. For instance, sup- 
 pose that two triangles, A A B C and A B D E of Fig. 60, 
 are of the same height, and are standing, side by side, on 
 the same base line as in Fig. 60. They can be moulded 
 
 Fig. 60. 
 
 into one triangle, A C D, with perfect ease. Do you see 
 why A B C D is equivalent to A B E D ? 
 
 Note. — In this figure and in future figures, when figures 
 are moulded, the lines along which vertices are moved 
 will be indicated by light dotted lines, as all help lines 
 should be indicated ; the resulting lines of the remodelled 
 figure will be indicated by double dotted lines. In your 
 
80 ELEMENTARY AND 
 
 drawings follow the same practice, unless you are able to 
 use different colors for the different kinds of lines. 
 
 400. Add together two triangles which have the same 
 altitude, but different bases. Repeat with three pairs of 
 triangles. 
 
 401. Can you add three triangles that have the same 
 altitude ? Can you do this problem with only one double 
 dotted line ? 
 
 402. Find an isosceles triangle equivalent to three tri- 
 angles that have the same altitude. 
 
 403. Mould three triangles having the same altitude into 
 a right triangle. 
 
 404. Make six equilateral triangles from one point, O, 
 forming the six-sided figure, or hexagon, represented in 
 Fig. 61. With a sharp 
 
 knife cut the figure along 
 the lines O A, O B, etc., 
 beginning at O and not cut- 
 ting quite up to points A, 
 B, C, etc., except at point 
 F. Place the figure thus 
 formed so that A, B, C, D, 
 E, F will form a straight 
 line. Can you form one 
 triangle as large as the 
 hexagon ? IIow long will 
 its base be ? How tall will *io. « 
 
 it be? 
 
 405. Could you mould a square into a triangle by a pro- 
 cess similar to that of 404 ? a parallelogram ? 
 
 406. Make a quadrilateral, A B C D, Fig. 62. The 
 problem is to mould this quadrilateral into an equivalent 
 triangle. The principle of 393 suggests a method. Di\ Mta 
 A B C D into two triangles by the diagonal B D. C D B 
 can now be looked at as a triangle standing on the base D B 
 with vertex at C. C can be moved parallel to D B as far 
 as you wish to move it, and D C and B C will follow the 
 
CONSTRUCTIONAL GEOMETRY 81 
 
 moving C just as the legs of the elastic triangle in 391 
 followed the moving vertex. When C reaches R, in the 
 line of A B, /\ C D B has been moulded into /\RDB, and 
 the quadrilateral A B C D has been moulded into /^ARD. 
 
 Fio. 62. 
 
 This process can be illustrated very prettily on a moulding- 
 tray with clay or sand in the shape of a quadrilateral, and 
 should be so illustrated if possible. 
 
 407. If you had moved C in a line antiparallel to D B, 
 where would you have stopped the movement ? Illustrate 
 with a figure, using the same quadrilateral as in 406. 
 
 408. Draw the diagonal C A, Fig. 62, and mould the 
 quadrilateral into a triangle in two ways, first moving D 
 symparallel with C A, and in a second figure moving D 
 antiparallel to C A. 
 
 409. Make a quadrilateral, and mould it into an isosceles 
 triangle. 
 
 410. Make a quadrilateral, and mould it into a right tri- 
 angle. 
 
 411. Can you reverse the process of 406, and mould a tri- 
 angle ADR (Fig. 62) into a quadrilateral A B C D ? Sug- 
 gestion : Choose any point between A and R for B, and draw 
 D B as a guide line, to which you must make R C parallel. 
 If all the lines and points of Fig. 62 were blotted out ex- 
 cept /\ARD, what would be your chances of reproducing 
 quadrilateral A B C D ? 
 
 6 
 
82 ELEMENTARY AND 
 
 412. Make a^ABC and mould it into a trapezoid. 
 What difficulty can you see in moulding a triangle into a 
 parallelogram by this process ? 
 
 413. In moulding a triangle, ABC, into a quadrilateral 
 (Fig. 63), if, after choosing the point P between A and B for 
 one vertex of your quadrilateral, you should move B parallel 
 to P C, not stopping until you 
 should reach A C extended, 
 the figure formed would be a 
 new triangle, A P Q. The 
 base of ^ A B C has been 
 shortened and the altitude 
 lengthened. Hence it is pos- 
 sible to shorten the base of a 
 triangle to any desired length. 
 Illustrate this article with a 
 clay or sand triangle, pushing 
 
 in the base to any desired point and noticing the increase 
 in altitude. 
 
 414. In 408 you changed one quadrilateral, A B C D, 
 Fig. 62, into four triangles by using different guide lines, 
 D B and C A, and by moving C and D first in one direc- 
 tion and then in the opposite direction. Place these four 
 triangles on one line, ride by side, and shorten all the 
 bases to one length, thive-fourths of the line A B. h 
 your work is accurately done, where should the vertices 
 of the raised triangles bo ? Why ? Take pains to have 
 this construction an almost perfect specimen of workman- 
 ship. 
 
 415. In 413 you shortened the base line just as much or 
 as little as you pleased, but the vertex rose to a point that 
 was out of your control. You, of course, knew that the 
 more you shortened the base the higher the vertex rose, 
 but you could not fix the level to which the vertex would 
 rise, at least without further construction or arithmetical 
 calculation. Can you raise a triangle to any desired alti- 
 tude ? Try to solve the problem by yourselves, but, if you 
 
CONSTRUCTIONAL GEOMETRY 
 
 83 
 
 is 
 
 find it too hard, study Fig. 64, where /\ A B C is the orig- 
 inal triangle and H P is the desired altitude. M B is the 
 guide line, parallel to. which 
 you draw C Q, to determine 
 the length A Q of the short- 
 ened base. This problem is 
 just the reverse of 413, 
 and you can easily see that 
 your construction is right by 
 shortening the base line to 
 A Q, and by noticing that 
 you would move B along a 
 line symparallel with Q to 
 
 M. If Q C is the guide to B M in one construction, it 
 not hard to see that M B becomes the guide to C Q in the 
 reverse construction. 
 
 416. Make three different triangles and raise the vertex 
 to a level fixed upon beforehand. 
 
 417. In 415 we shortened the base by cutting off some- 
 thing at the right end, moving B in as far as Q. Try to 
 solve the same problem by cutting off a length at the left 
 end, thus pushing A in toward B. Suggestion : Extend 
 B C until it crosses P M at N, Fig. 64, and draw a new 
 guide line, N A. Compare the 
 
 resulting base line with A Q of 
 Fig. 64. 
 
 418. Make three triangles, 
 and raise the vertices to any- 
 desired level by pushing the 
 left end of the base toward the 
 right end. 
 
 Practise changing triangles 
 to triangles of greater altitude 
 until it is a matter of indiffer- 
 ence to you on which side of 
 the triangle you draw your construction lines. 
 
 419. Can you now take the necessary steps to lower the 
 
84: ELEMENTARY AND 
 
 vertex of a triangle without changing the area ? Make a 
 triangle, ABC, and lower the vertex to the level of P, Fig. 
 65. Do this twice : first, so that the. position of A shall not 
 be changed ; second, so that the position of B shall not be 
 changed. Compare the two bases obtained. 
 
 4901 Place two triangles of different altitudes side by 
 side. First, raise the lower one to the level of the other, 
 and add the two triangles ; second, lower the taller one to 
 the level of the shorter, and add the two. Compare the 
 two triangles resulting from addition in the two cases by 
 changing one to the level of the other. 
 
 421. Place two triangles, one of which is a great deal 
 taller than the other, side by side. Change them to a com- 
 mon level by raising one and lowering the other ; and then 
 add the results. In changing levels work from the left 
 side of the left-hand triangle, and from the right side of the 
 right-hand triangle, that the vertex where the triangles join 
 may not be changed in the process. 
 
 422. Place three triangles of different heights side by 
 side as in Fig. G6. 
 
 1 1.. «« 
 
 Change ^ A B C and D F G to the level of A B D E 
 without changing the positions of B and D. Add the re- 
 sulting triangles. Aim to complete the problem in as few 
 steps as possible, indicating your steps by the kind of line 
 
 drawn. 
 
CONSTRUCTIONAL GEOMETRY 85 
 
 423. Make a triangle and a quadrilateral that seem to 
 you to have the same area. Change the quadrilateral to an 
 equivalent triangle, and change this triangle to the level of 
 the first triangle, and draw a triangle representing your 
 error in estimation, if there be any error. 
 
 424. Make a quadrilateral free hand that seems equivalent 
 to the sum of a triangle and a quadrilateral that you have 
 drawn. Test your accuracy as in 423. 
 
 425. Change a pentagon into an equivalent triangle. 
 Suggestion : First mould the pentagon into a quadrilateral, 
 and then mould the quadrilateral into a triangle. The 
 drawing is much less confused, if you work on different 
 sides of the figure in taking the two steps. 
 
 426. Find a triangle equivalent to the sum of a pentagon 
 and a quadrilateral. 
 
 427. Can you change a triangle into a pentagon ? 
 
 428. Figures having six, seven, eight, or any number of 
 sides can be changed to triangles by reducing the number 
 of sides one at a time, just as the pentagon in 425 was 
 changed to a quadrilateral, and the resulting quadrilateral 
 was changed to a triangle. After two steps the lines be- 
 come confused, unless a new figure is drawn. 
 
 All plane figures that are bounded by straight lines are 
 called polygons. Special names have already been given to 
 polygons of three, four, five, and six sides. What are they ? 
 
 429. Can you give a rule for changing any polygon into 
 an equivalent triangle, that will hold good no matter how 
 many sides the polygon has ? 
 
 430. Draw a circumference and divide it into twenty- 
 four equal parts. Can you do this without a protractor ? 
 Form a polygon of twenty-four sides by joining the points 
 of division. The polygon is inscribed in the circumference. 
 What is meant by this statement ? This polygon can be 
 moulded into a triangle by a process much simpler than the 
 one which you have described in 429 ; for, by joining the 
 vertices with the centre of the circumference, you form 
 twenty- four equal triangles. See 404, where a hexagon was 
 
86 ELEMENTARY AND 
 
 divided in this way. Since the twenty-four triangles have 
 the same altitude, they can be changed into one triangle 
 very easily. Describe the method. How long a base 
 will the resulting triangle have ? How tall will it I 
 
 431. Describe how you would find a polygon of forty- 
 eight sides inscribed in a circumference from the inscribed 
 polygon of twenty-four sides of 430. How could you change 
 this polygon into a triangle? How long would the I 
 be-' 
 
 432. In the same way that you got an inscribed polygon of 
 48 sides from one of 24 sides, you could get one of % sides 
 from one of 48 sides, and then you could get polygons of 
 r*.\ 384, 70s sides in succession, continuing the process 
 until the polygons seemed to the eye to coincide with the 
 circle. Each time the polygon could be moulded into a tri- 
 angle having a base equal to tlie perimeter of the polygon 
 and an altitude equal t<» the distance from the centre of the 
 circumference to one of the sides of the polygon. 
 
 438. It is not difficult to see from 482 that fche circle 
 itself could be moulded into a triangle having a base 
 equal to the oiroumferenoe and an altitude equal to the 
 radios. The difficulty if to find a straight line equal to 
 the circumference. Make the following experiments and 
 bring your results into the class : Roll a cylinder of wood 
 about two inches in diameter in a straight line, making it 
 turn ten times. By marking the lowest point of the cylin- 
 der at the start and the |>oint on the table just below this 
 point, it is | simple matter to find the distance rolled and 
 the number of turns made. Compare the distance rolled 
 with the diameter. Push a 28-inch bicycle until the wheel 
 has made ten revolutions Measure the distance covered 
 and compare it with the diameter. Perform each experi- 
 ment several times and record the results in a neat table. 
 
 How many times as long as the diameter do you find the 
 circumference to be on the average ? 
 
 434. Using the results of 433, change five different 
 circles into essentially equivalent triangles. 
 
CONSTRUCTIONAL GEOMETRY 
 
 87 
 
 435. Can you tell why it is a much more difficult prob- 
 lem to change a triangle into a circle ? Suggestion : 
 Think how much longer than the altitude the base of the 
 triangle ought to be, to enable you to call the altitude of 
 the triangle the radius of the equivalent circle. 
 
 436. Draw a particular triangle that you can turn into 
 a circle with as much accuracy as the results of 433 allow. 
 
 SECTION XVI. PECULIAEITIES OF SQUAEES. 
 
 437. Make an isosceles right triangle. The problem is 
 to change it into an equiv- 
 alent right triangle with a 
 leg equal to any chosen 
 length, say P Q, longer 
 tha^n A B. One way of 
 solving this problem is to 
 raise the vertex by the 
 method of 415, until the 
 height A R is equal to 
 P Q. Take the necessary 
 steps and find the equiva- 
 lent triangle A S R. 
 
 438. A much more 
 roundabout way of solv- 
 ing the problem of 437 — - 
 but for the present pur- f 
 poses a better way — is to 
 proceed as follows : move C, Fig. 68, along a line parallel 
 to B A, until it reaches a point E that is at a distance P Q 
 from A, thus moulding /\ABC into /\ABE. 
 
 Now imagine /\ABE to be pivoted at A, and to be 
 turned through a quarter of a revolution into the position of 
 /\ACR. A A ^oAABC. Why? Now move C 
 
 Fig. 67. 
 
88 
 
 ELEMENTARY AND 
 
 parallel to A R to S, thus moulding /\ACR into ^ASR. 
 /\ACRoA ASR - Wh y ? HenceAASRoABC; 
 since /\ASR lias the desired leg A R = A E = P Q, the 
 problem is solved. Compare the result with the result of 
 the last article. 
 
 439. Using the method of 438, turn isosceles right tri- 
 angles into right triangles with one leg of any chosen length 
 
 \ *x. // 
 
 X V 
 
 X. x 
 
 \ "*•». ' 
 
 \ \ 
 
 \ -V- 
 
 X \ 
 
 X \ 
 
 \-' .' "* 
 
 x * 
 
 *)x/ 
 
 ^ \ \ 
 
 
 ^. \ * 
 
 / /N 
 
 
 I 
 
 //' 
 
 Flo. 68. 
 
 longer than the leg of the isosceles triangle, until you be- 
 come perfectly familiar with the process. 
 
 440. What is the angle formed by S C and S A, Fig. 68 ? 
 Why ? What kind of triangle is A A C E ? 
 
 441. Can you turn an isosceles right triangle ABC into 
 a right triangle ASR, Fig. 68, without actually drawing 
 lines B E and C R ? Try three cases of this kind. The 
 lines should be imagined, though not drawn. 
 
 442. The isosceles right triangle A B C, Fig. 68, is one 
 half of a square, and the right triangle A S R is one half of 
 
CONSTRUCTIONAL GEOMETRY 
 
 89 
 
 Fig. 69. 
 
 a rectangle. Draw a figure showing the square and rectangle 
 meant. What is true of their areas ? Why ? 
 
 The process described in 438 gives you the means of 
 moulding a square into a rectangle having one side any de- 
 sired length longer than the side of 
 the square. Do you see how ? 
 
 443. Turn four squares into rectan- 
 gles. 
 
 444. Turn one square into four dif- 
 ferent rectangles. 
 
 445. In 437 and 438 we have seen 
 how to turn an isosceles right triangle 
 into a right triangle with unequal legs, 
 in two different ways. Can a right 
 triangle with unequal legs be changed 
 to an isosceles right triangle ? Try 
 first to reverse the method of 437, and 
 to change a right triangle A S E, Fig. 69, to an isosceles 
 right triangle A B C by lowering the vertex E to the proper 
 level C, so that the extended base A B 
 shall be equal to the reduced height 
 A C. How will you take the first step ? 
 Why is it difficult to draw a guide line, 
 parallel to which E may be moved ? 
 
 446. In the next place, try to reverse 
 the method of 438. To help the imag- 
 ination, place triangle A S E (Fig. 70) 
 in a position something like that of tri- 
 angle A S E in Fig. 68. 
 
 The problem still is to find C, to which 
 to move the vertex S. We know that C 
 is somewhere on a line drawn parallel to 
 E A from S, but we do not know exactly 
 where. It is through the triangle corre- 
 sponding to A A C E (Fig. 68) that we 
 shall be able to find C. In 440 you learned that /\ACE 
 is a right triangle. How can you find the point corre- 
 
 Fig. 70. 
 
90 ELEMENTARY AND 
 
 sponding to E ? Review 272, and, with the aid of the 
 principle therein stated, find point C. A C will be the side 
 of the isosceles triangle desired. Draw A 13 at right angles 
 to A C and equal to it. Complete triangle ABC. 
 
 447. Change six different right triangles into isosceles 
 right triangles by the method of 44G. Vary the position 
 of the triangles, that you may not depend upon having your 
 figure placed as in Fig. 08. 
 
 44*. Change a rectangle into a square, following the 
 method suggested by 44M. Hepeat three times. 
 
 449. Change a triangle into a square. Suggestion : First 
 change the triangle into a rectangle. 
 
 450. Change a pentagon into a square. Make the steps 
 in their proper order perfectly clear. 
 
 •l"'l. Add a triangle and a quadrilateral, and change the 
 resulting figure into a square. 
 
 452. Put three equal squares side by side, so as to form a 
 rectangle. Mould the rectangle thus formed into a square 
 form. 
 
 Find a square three times as large as a given square. 
 Find a square five times as large as a given square. 
 
 454. Cut a square into five equal rectangular strips, find- 
 ing J of one side with your dividing tool. Change one of 
 these strips into a square form. How is this new square re- 
 lated to the original square ? 
 
 455. Find i Bqnare \ of a given square. See 454. 
 466. Find ■ square ^ of a given square. See 454. 
 
 457. Make a pentagon. Can you find a square that shall 
 be j as large as the pentagon ? 
 
 458. In changing a square into a rectangle, 438, yon 
 found that a gnat many rectangles could be formed equiva- 
 lent to the square. The shape of the rectangle depends 
 upon the distance which the vertex of the square is moved 
 along C E (see Fig. 68). In changing a rectangle into a 
 square, how many results can you get ? 
 
 459. Make a square, A B C D, Fig. 71. Change the 
 square into a rectangle eight times by moving each vertex 
 
CONSTRUCTIONAL GEOMETRY 
 
 91 
 
 in the two directions indicated by the arrows at the vertex 
 (Pig. 71). 
 
 Move each vertex the same distance, and compare the re- 
 sulting rectangles. 
 
 Note. — The object of this ex- 
 ercise is to give you practice in 
 changing squares into rectangles 
 with the squares in different 
 positions. If you find any one 
 position hard for you, practise 
 that case until it is as simple as 
 any of the others. 
 
 460. Draw a square ABCD 
 (Fig. 72), and change it, by 
 438, to rectangle K H A F. The 
 fact that the square is equivalent to the rectangle may be 
 
 " OAF x AH. 
 x ES 
 
 JK * 
 
 V- 
 
 D C 
 A R 
 
 
 c l +' 
 
 Fig. 71. 
 
 ] S K F E. 
 
 expressed as follows : A D 
 
 Show that ED 2 =G=EP 
 
 Suggestion : Imagine that D 
 has been moved to A, parallel 
 toNE. 
 
 461. What kind of quadri- 
 lateral is S H A E, which is 
 made up of the two rectangles 
 S K F E and K H A F ? 
 
 Show that AE 2 O AD 2 + 
 D~E 2 - 
 
 462. Does the truth stated in 
 461 apply to all right trian- 
 gles ? 
 
 463. The principle of 460 and 
 461 stated in words is as fol- 
 lows : The square formed on the hypotenuse of a right tri- 
 angle is equivalent to the sum of the squares formed on 
 its sides. 
 
 This principle is called the Pythagorean principle, be- 
 
92 ELEMENTARY AND 
 
 cause the discovery of it is attributed to Pythagoras, a famous 
 Greek mathematician, who lived about 600 B.C., and who 
 did much to encourage the study of Geometry. How 
 Pythagoras made his discovery is not known. The 
 method of showing its truth here given is essentially that 
 devised by Euclid, who lived three hundred years later than 
 Pythagoras, and who was the most famous of the ancient 
 Geometers. 
 
 404. The Pythagorean principle makes the addition of 
 squares even simpler than the addition of triangles ; for it 
 is necessary merely to place the squares to be added on the 
 legs of an imaginary right triangle, as in Fig. 72. The 
 square on the hypotenuse is the square desired. 
 
 465. Add two squares. Repeat the process four times, 
 using different squares. 
 
 466. Can you form two unequal squares on one line, A B, 
 as a side ? If A B is a given line, can you form AB 2 with- 
 out any further description ? 
 
 What is meant by the statement that a square is deter- 
 mined by one of its sides ? 
 
 467 . Can v on find the side of a square equivalent to 
 MN* + §T, where M N and S T are given lines, without 
 constructing any square ? Describe your process carefully. 
 It is better, in the majority of cases, not to draw the squares 
 actually, but to use the sides of the squares. 
 
 408. Given two lines, a and b, find «* + Ir in n square form. 
 
 Find a square equivalent to a* + a 7 or 2 a 1 ; to «* + 2 «* 
 or 8 n 2 . Find a square equivalent to 3 «* by 453, and com- 
 pare the result with that just obtained. Compare your 
 method <>f getting the side of 2a* here with the method 
 suggested in 315. 
 
 469. Given three lines a, b, c. Can you find a square 
 equivalent to a* + #* + c*? Find a* + a* + a* by this 
 method. Find 3a 8 by the methods of 453 and 468, and 
 compare the results with that obtained here. 
 
 470. Change two triangles into squares, and add the re- 
 sulting squares. Add the two triangles without changing 
 
CONSTRUCTIONAL GEOMETRY 93 
 
 them to squares ; change the resulting triangle to a square, 
 and compare the results obtained by the two methods. 
 
 471. Make a square on a given line, a. Make a square 
 on a line equal to 2a. Compare the two squares, a 2 and 
 (2a) 2 . How much larger than the first square is the 
 second ? 
 
 472. Make a square on the line da as a base, and com- 
 pare the resulting square with a 2 . 
 
 473. What is the side of a square equivalent to 16 a 2 ? to 
 25 a 2 ? to 100 a 2 ? 
 
 474. Make a square on the line a. Draw a line M N" 
 equal to 3a and, at right angles to M N, draw N P equal 
 to a. Join M P, and call it I. How many times as big 
 as a 2 is b 2 ? 
 
 475. Can you devise a simple method of finding 5 a 2 from 
 the experiment in 474 ? Find a square five times as large 
 as a 2 by the method of 453, and compare the result with 
 that obtained by your new method. 
 
 476. Can you draw a right triangle by first drawing the 
 hypotenuse ? Eecall what you know about the middle 
 point of the hypotenuse. 
 
 477. Make a line A B, and draw a semi-circumference, 
 of which A B is the diameter. Join any point, C, of this 
 semi-circumference with A and B. Can you tell the size 
 of / B C A ? Give a reason for your answer. 
 
 478. Make a square, a 2 , of the same size as that used in 
 474 and 475. With a line A B 
 equal to 3a as a diameter, de- 
 scribe a semi-circumference. See 
 Fig. 73. Make a line AC, 
 called a chord of the circum- 
 ference, equal to a. Draw the 
 chord C B. Fig. T3. 
 
 Draw squares on A B, A C, and B C. 
 
 How much larger than A C 2 is A B 2 ? 
 
 How much larger than AC 2 is CTB 2 ? Why ? 
 
 479. Draw a new semi-circumference of the same size as 
 
94 ELEMENT WHY AND 
 
 that of 478, and this time draw a chord A D equal to 2a. 
 Join D B, draw the squares as before, and compare all the 
 squares with a 2 . 
 
 480. Draw two squares, a 2 and #*. Can you find a square 
 equivalent to the difference between a 2 and #* ? Describe 
 the steps carefully. 
 
 481. Cut from a square piece of paper, at any place, a 
 smaller square. If the remaining piece of paper could be 
 moulded into a square form, how long would the sides of 
 the square be ? Show, by a neat construction, how you 
 would obtain the desired length. 
 
 482. Make three pairs of squares and subtract the smaller 
 square of each pair from the larger, obtaining your result 
 in the form of a square. 
 
 483. It is a simple matter, as you have seen, to find a 
 square equal to 4a 2 , 9a 2 , 10a 2 , 25a 2 , etc., when a 2 isght n. 
 Can you, by simple addition or subtraction of these easily 
 obtained squares, find squares equal to 7a 2 , 13a 2 , 12a 2 , 20a 2 , 
 3a 2 ? 
 
 Whenever you see two combinations that will give the 
 same result, it is well to try both and to compare the an- 
 swers. After the first two or three problems, it is not 
 necessary to draw the actual squares. Y<>u will find that 
 you need to deal with the sides only ; for a square is easily 
 formed in imagination, if its side is clearly seen. 
 
 484. Not receiving sufficient Light from a certain square 
 window, a man deeided to make an opening twice as large 
 without changing its shape; he, accordingly, made his 
 window twice as long and twice as wide as before. How 
 much more light was let in ? What should he have done 
 to find the proper length for his window ? 
 
 485. Two farmers have square fields to fence ; if the field 
 of the first farmer is four times as large as that of the 
 second, how much more fencing will he need ? Draw 
 plans of the fields before making a final answer ? Can you 
 account for the difference between the relative cost of fenc- 
 ing and the relative size of the fields ? 
 
CONSTRUCTIONAL GEOMETRY 95 
 
 486. On a piece of cardboard draw a diagram, showing 
 how you would cut the cardboard to make it fold into a 
 hollow cube with one-inch edges. If you were to make a 
 hollow cube with two-inch edges, how much more surface 
 would your diagram cover ? How much more sand would 
 your second cube hold than the first would hold ? How 
 much cardboard would you save in making one large cube 
 like the second one, in place of making enough small 
 cubes like the first to hold an equal amount with the 
 second ? 
 
 487. If you want to make a cubical box that will have 
 an outside surface just twice as large as that of the first 
 cubical box of 486, how will you proceed to draw the dia- 
 gram by which to cut out the box ? 
 
 488. Can you draw a diagram that will fold into a cube 
 with a surface three times as large as that of the first cube 
 of 486 ? 
 
 489. You cannot become too familiar with the peculi- 
 arities of squares that have been dwelt upon in this sec- 
 tion. You will find later that a great many other figures 
 behave like squares ; but this will do you little good, un- 
 less you know well the peculiar characteristics of squares, 
 which enable you to add them, to subtract them, to show 
 what effect the change in a side will produce upon the sur- 
 face, and to show how a change in the size of the square 
 affects the side. You ought also to be expert in moulding 
 any polygon into a square form. 
 
 490. Cut from a piece of paper a square 3 in. by 3 in. 
 Can you, by adding strips to two sides only, keep the 
 paper in the form of a square ? 
 
 If you add strips an inch wide, how long must the strips 
 be to keep the square perfect ? How many square inches 
 in the strips added ? 
 
 491. Cut a strip J in. wide of the right length to make 
 a perfect square when added to two sides of the 3 in. 
 square of 490. How many square inches in the strip ? 
 
 If the strip is 2 in. wide, how long should it be ? 
 
96 ELEMENTARY AND 
 
 492. Imagine a square 7 ft. by 7 ft. How long a strip 
 will be needed to enlarge the square to a square 9 ft. by 9 
 ft., if the strip is added on two sides ? How many square 
 feet in the strip ? 
 
 493. Imagine a square 80 ft. by 80 ft. How long a strip 
 will be needed to enlarge this to a square 87 ft. by 87 ft. ? 
 How many square feet in the strip ? 
 
 494. Calculate the lengths and areas of the strips needed 
 in each of the following cases : — square 70 ft. by 70 ft., 
 strip 9 ft. wide ; square 130 ft. by 130 ft., strip 5 ft. wide ; 
 square 90 ft. by 90 ft., strip 6 ft. wide ; square 700 ft. by 
 700 ft., strip 20 ft. wide. 
 
 495. Can you state the law by which you can find the 
 length of the strip from the length of the square and the 
 width of the strip ? 
 
 496. Make a square 3 in. by 3 in. If you have sixteen 
 square inches of paper which can be cut into rectangular 
 strips, how wide will you make your strips, that you may 
 have a perfect square when you shall have added the strips 
 to the 3 in. square ? The strips in this and the follow- 
 ing problems are to be added on only two sides of the 
 square. 
 
 Solve the problem of 496, if the original square is 
 9 ft. by 9 ft. and the material for addition has a surface of 
 63 sq. ft. 
 
 498. By adding a strip that contained 889 sq. ft. to a 
 square 60 ft. by 60 ft, a square was obtained ; how wi<l< 
 and how long was the strip ? 
 
 499. Calculate the lengtli and width of the strips in the 
 following cases : — square 80 ft. by 80 ft., area of strip 656 
 sq. ft. ; square 50 ft. by 50 ft., area of strip 981 sq. ft. ; 
 square 170 ft. by 170 ft., area of strip 2076 sq. ft. 
 
 500. Can you make a rule for finding the length and 
 width of a strip needed to make a perfect square, when 
 added to a known square, from the side of the known 
 square and the area of the strip ? 
 
 501. Arrange 784 sq. ft. in the form of a perfect square 
 
CONS TE UGTIONAL GEOMETR Y 
 
 97 
 
 by first making as large a square as you can, using only 
 round numbers like 10, 20, 50, 100, 300, etc., for the side 
 of the square, and by then adding the material left in the 
 form of a strip according to the examples given in articles 
 496-499, and according to the rule in 500. 
 
 The following model for the arrangement of the work is 
 suggested. In the model the problem is to arrange 4489 
 sq. ft. in a perfect square. 
 
 Dimensions 
 of strip. 
 
 
 127 
 
 7 
 
 Material. 
 4489 
 3600 
 
 889 
 
 Side of square. 
 60 (1st guess) 
 
 7 (width of strip) 
 
 67 answer. 
 
 Explanation. — A square 70 
 ft. by 70 ft. would require 4900 
 sq. ft. of surface ; therefore, in 
 round numbers, a square 60 ft. Fig. u. 
 
 by 60 ft. is the largest possible 
 
 square. The material left after forming this square (Fig. 
 74), is 889 sq. ft. ; the length of the strip to be added must 
 be at least 2 x 60 ft. or 120 ft. ; the width is roughly 889 
 -T- 120 or 7 ; since the strip must be as much longer than 
 120 ft. as it is wide, 7 ft. is added to 120 ft. to obtain the 
 true length of the strip. The surface covered by the strip 
 is 127 ft. x 7 ft. or 889 sq. ft. All the material is thus 
 used, and a perfect square is formed. In writing the first 
 estimate, 120 ft., of the length of the strip, the cipher was 
 written a little above the line, to leave room for the addi- 
 tion of the width to the length as soon as the width could 
 be guessed. It is well to draw a figure like Fig. 74 to aid 
 the imagination, until the process is perfectly familiar to 
 you, so that no aid is needed. 
 
 502. Arrange the following surfaces in square forms, fol- 
 lowing the model given in 501 : 1156 sq. ft. ; 6889 sq. ft. ; 
 7 
 
98 
 
 ELEMENTARY AND 
 
 5184 sq. ft. ; 2401 sq. in. ; 1G9 sq. yds. ; 729 sq. ft. : 
 sq. in. In doing the last problem follow this model : 
 
 Ditnetiftions 
 of strip. 
 
 Material . Side of sq uare 
 
 418609 600 (1st guess) 
 B0OOOO 
 
 width of 1st 
 strip 
 
 640i 8lde of M 
 \ square 
 
 width of 2d 
 strip 
 
 answer. 
 
 7ft foJi-Jht 
 
 uo* 
 
 
 Log* 
 
 «M 
 
 Fio. 75. 
 
 In getting the width of the 
 first strip, do not try to guess anything except round num- 
 bers like 30, 40, 50, etc. ; and do not try to add all of the 
 material in one strip. The explanation is the same as that 
 given in the previous article ; two steps instead of one are 
 necessary. 
 
 503. Arrange the following surfaces in a square form, ob- 
 taining the length of the side of the square in each case: 
 <j. ft. ; 146161 sq. ft. ; 98686 sq. ft. ; 00516 sq. ft. ; 
 120409 sq. ft. 
 
 I n arranging the following surfaces in a square form, 
 you will find that some material will be leftover. Calcu- 
 late the side of the square and the amount left over in each 
 case :— 536 sq. ft. : 4376 sq. ft. ; 59785 sq. ft. 
 
 505. If, after adding one or two strips to a square, there 
 is not enough material left to make a strip 1 ft. wide or 1 
 in. wide, according to the unit used, it is possible to add 
 strips that are so many tenths of a foot, or tenths of an inch 
 in width : and then to add other strips whose width is 
 measured in hundredths of a foot or hundredths of an inch, 
 and so on until the material left is not worth noticing. See 
 the following example, where 281 sq. ft. are to be arranged 
 in the form of a square : 
 
CONSTRUCTIONAL GEOMETRY 
 
 99 
 
 Dimensionsof strip. 
 
 
 
 26 
 
 Material. 
 
 281. 
 100. 
 
 10 
 6 
 
 Side of square, 
 first guess 
 
 181. 
 156. 
 
 width of 1st strip 
 
 6 
 
 16. 
 
 .7 
 
 side of 2d square 
 
 33.7 
 
 .7 
 
 25.00 
 22.89 
 
 width of 2d strip 
 
 33.46 
 
 .06 
 
 2.1100 
 2.0076 
 
 16.7 
 .06 
 
 side of 3d square 
 width of 3d strip 
 
 33.523 
 
 .003 
 
 .102400 
 .100569 
 
 16.76 
 .003 
 
 side of 4th square 
 width of 4th strip 
 
 
 .001831 
 
 16.763 
 
 answer. 
 
 281 sq. ft., therefore, when moulded into the form of a 
 square, will make a square whose side is 16.763 ft. ; the 
 part unused, a little under two-thousandths of a square 
 foot, is not worthy of notice in ordinary cases. Multiply 
 16.763 by 16.763, and see how much the product lacks of 
 amounting to 281. 
 
 506. Arrange in the form of a square the following sur- 
 faces ; calculate the side to the thousandth of your unit : — 
 69 sq. ft. ; 351 sq. ft. ; 48 sq. ft. ; 10 sq. in. ; 5 sq. in. ; 3 
 sq. in. ; 2 sq. in. 
 
 507. Make a square decimeter. Draw the diagonal and 
 measure it carefully. Arrange in the form of a square two 
 square decimeters, calculating the length of the side as in 
 506. Compare your two answers. 
 
 508. The side or base of a square is sometimes called the 
 root of the square. The square root of a surface is one 
 of the two equal lines that, multiplied together, will pro- 
 duce the surface ; in other words, it is the side of the square 
 into which the surface may be moulded. Thus the square 
 root of 4489 sq. ft. in 501 was found to be 67 ft. This 
 fact would be written in sign language as follows : — 
 V 4489 sq. ft. = 67 ft. Notice the sign for square root. 
 You have already found the square root of 1156 sq. ft., 
 
100 ELEMENTARY AND 
 
 2401 sq. in., 1G9 sq. yds., in 502 ; write the results ob- 
 tained, making use of the proper signs. 
 
 In arithmetic the square root of a number is one of the 
 two equal factors that, multiplied together, will prodm-c 
 the number. The process of getting the square root of a 
 number is the same as that of getting the square root of 
 a surface explained in articles 501-505. 
 
 509. In 507 you drew a line equal to V 2 sq. dm. 
 without any arithmetical work. Can you think of a 
 way to draw a line equal to V 5 sq. in. ? Compare your 
 answer with the answer found in 50G by the for mer method. 
 
 510. Draw lines equal to V 2 sq. in., V 10 sq. in., 
 
 V 3 sq. in. ; explain your steps ; measure the resulting 
 lines ; and compare the answers with those obtained in 500. 
 
 511. Can you draw a line that shall represent the 
 V|sq. in.; V | sq. in. ; V \ sq. in. (See 454.) In actu- 
 ally doing this work, it is better to draw to scale, taking 
 a lino much longer than an inch to represent the side of 
 the square inch. 
 
 512. On a line mark <»1T lengths equal to \ L tq. dm., 
 
 V 2 sq. dm., V 3 sq. dm., etc., up to V 10 sq. dm. 
 
 513. On a lin e mark off lengths equal to V J sq. dm., 
 
 V I sq. dm., V } sq. dm., etc., up to V <fo sq. dm. 
 
 514. With the scales you have just made in 512 and 513, 
 ami with the aid of what you know about adding and sub- 
 tracting squares, you can find the square root of a great 
 many surfaces with great rapidity. Use the scales in find- 
 in g the followin g square roots : v 17 sq. dm.; V 12 sq. dm.; 
 
 V 24 sq. dm. ; V 5} sq. dm. ; V 2J sq. dm.; V T \ sq. dm. 
 
 515. Iu articles 501-606 you saw how to find the side of 
 a square from its surface. The ability to do this adds 
 greatly to the practical value of the Pythagorean principle 
 that the square on the hypotenuse of a right triangle con- 
 tains as much surface as the squares on the two legs. 
 If the legs of a right triangle are 7 in. and 3 in., the 
 square on the hypotenuse must contain 49 sq. in. + 9 sq. 
 
CONSTRUCTIONAL GEOMETRY 101 
 
 in. or 58 sq. in. Show that the length of the hypotenuse 
 is 7.61 in. 
 
 516. Measure the length and width of the bottom of a 
 chalk box, and calculate the length of the diagonal across 
 the bottom. 
 
 517. Measure the length and width of the school-room, 
 and calculate how far apart the diagonally opposite corners 
 are. 
 
 518. A table is 8 dm. long and 6.6 dm. wide ; how long 
 a straight stick could be placed on the table so that no point 
 of the stick would be beyond the edge of the table ? 
 
 519. A drawer is 8 dm. long, 3.5 dm. wide, and 1 dm. 
 deep ; what is the longest straight stick that can be placed 
 wholly within the drawer ? 
 
 520. A ladder 13 ft. long was placed against a tree, with 
 its foot 5 ft. from the base of the tree ; how far up the tree 
 did the ladder reach ? 
 
 521. A boy walked round two sides of a rectangular field, 
 11 yds. by 60 yds. ; how much would he have saved, if he 
 had gone diagonally across the field ? 
 
 522. In raising a pole 65 dm. long a man rested the up- 
 per end against a house ; he found the lower end to be 33 
 dm. from the house ; the next time that he rested the pole 
 against the house he found that its end was 16 dm. from 
 the house ; how many decimeters had he raised the upper 
 end the second time ? Make an estimate of the amount ; 
 then calculate it ; and compare the results. 
 
 523. An isosceles triangle has a base 24 in., and legs 37 
 in. long ; how long is the altitude of the triangle ? How 
 many square inches in its surface ? 
 
 524. An equilateral triangle has sides 12 in. long. Can 
 you calculate its altitude and its area ? 
 
 525. Make a right angled triangle with one leg 3 in. and 
 the other leg 4 in. Calculate the hypotenuse. 
 
 526. The triangle that you have just made is a most use- 
 ful one in practical work. If you wish to nail two straight 
 sticks together so that their edges shall be at right angles 
 
102 ELEMENTARY AND 
 
 to each other, you can mark the points 3 in. or 3 ft. along 
 one stick and 4 in. or 4 ft. along the other, and then nail 
 a stick across the two so that there shall he a length of 5 in. 
 or 5 ft. between the points marked. See whether the edges 
 of a box are really at right angles to each other by this test. 
 Show how you could set a corner post at right angles to a 
 platform by the aid of the 3, 4, 5, right triangle. 
 
 527. A pyramid is formed by four equilateral triangles 
 resting on t he sides of a square base G in. by in. Calcu- 
 late the entire surface of the pyramid. 
 
 528. A rhombus has sides 13 in. long and one diagonal 
 24 in. long ; how long is the other diagonal, and what is 
 the area of the rhombus ? 
 
 529. Divide a circumference whose radius is 6 in. into 
 arcs of 60 ° each ; join the ends of these arcs, forming a 
 regular hexagon. Divide the hexagon into six triangles 
 from the centre, and calculate the area of the hexagon. 
 
 530. The end of a house is 20 ft. by 20 ft. to the eaves; 
 the ridge pole is 30 ft. above theground. Find the length on 
 the roof from ridge pole to eaves, and the number of square 
 feet in the gable end. Can you tell the piteh of the rouf } 
 
 581. A diamond-shaped window pane has one edge 8 in. 
 long and one angle 60°. Can you calculate the Dumber 
 of square inches of glass in the pane ? Would more or less 
 light come in, if the pane were in the shape of a square with 
 one side 8 in. } 
 
 An equilateral triangle has sides 18 in. long. Draw two 
 of its altitudes, ami calculate how far from the vertices of 
 the triangle they cross each other. 
 
 Divide a circumference whose radius is 10 in. into arcs 
 of 90° each : join the ends of the arcs, forming a square. 
 Calculate the sides and the area of this square. 
 
 A circumference is drawn through the corners of a 
 square whose sides are 8 in. long. Can you calculate the 
 radius of the circuinferen- 
 
 The diagonals of a rhombus are IS ft. and 20 ft. respec- 
 tively. Calculate the sides of the rhombus. 
 
CONSTRUCTIONAL GEOMETRY 
 
 103 
 
 532. Models of solids having four, eight, or twenty faces 
 in the shape of equilateral triangles, may be made from card- 
 board by drawing the patterns of Fig. 76, and by cutting 
 the cardboard half through along the lines that separate 
 the triangles. Bend the cardboard until the edges come 
 together, and glue the edges. Calculate the total surface 
 of each solid, if one side of the equilateral triangle is 1 in. 
 
 Fig. 76. 
 
104 ELEMENTARY AND 
 
 SECTION XVII. AREAS OF SIMILAR FIGUBE& 
 
 Pio. T7. 
 
 533. With your dividing tool, divide one side A C of a 
 scalene triangle, ABC (Fig. 77), into seven equal parte, 
 and draw lines parallel to the base through the points of 
 division. Draw D S parallel to C B, and draw E S the 
 diagonal of the parallelogram D E II S. Show that the 
 trapezoid V II K I) is made up of three triangles equal to 
 A 1> B (\ What part of A C F II is [\ D C E ? Bow 
 much larger is A ^ H C than A D E C ? How much 
 longer is F 1 1 than D E ? F C than DC? II C than EC? 
 Into how many triangles like DEC can you divide the 
 noond trapezoid K IS II FF Give the number of triangles 
 likeDE in the third trapezoid MNRK; in the fourth 
 trapezoid UVNM: in the fifth trapezoid PQVU; in 
 the sixth trapezoid A B Q P. How much bigger than 
 ADECisAABC? How much longer are the sides of 
 A A B C than the corresponding sides of A D E C ? Com- 
 pare the sides of A C K R with the corresponding si 
 A C D E ; also compare the surfaces of the two triangles. 
 
CONSTRUCTIONAL GEOMETRY 105 
 
 Compare the sides of /\ C M N with the corresponding sides 
 of /\ C D E ; also compare the surfaces of the two triangles. 
 Compare the surfaces and the corresponding lines of 
 ^ C U Vand C F H ; of ^ C P Qand C M N ; of ^ C A B 
 and CUV. 
 
 534. Triangles like those studied in 533, where the cor- 
 responding angles are equal, and where the corresponding 
 sides have the same relation to each other, are called simi- 
 lar triangles. The investigations suggested in 533 point to 
 the law that similar triangles have the same peculiarities 
 as squares in the matter of their surfaces. Thus you have 
 learned that, to quadruple a given square, you make a 
 square with its sides twice as long as those of the given 
 square. In the same way, to quadruple a given triangle, 
 you make another triangle similar to it in shape, with the 
 sides respectively double those of the given triangle. Com- 
 pare the triangle in Fig. 77, that is nine times /\CDE 
 with /\CDE, and see if its sides have the relation to 
 those of /\ C D E that you would expect from your knowl- 
 edge of squares. Do the same with the triangle that is 
 twenty-five times as large as /\ C D E ; with the triangle 
 that is thirty-six times as large as /\ C D E. 
 
 535. In Fig. 77, one of the triangles is two and one- 
 quarter times as large as /\ C F H. Find this triangle 
 and record the result in this way : — /\ .... o fA^FH. 
 Do the sides of the triangle found bear the relation to 
 those of /\ C F H that you would expect from your knowl- 
 edge of squares ? 
 
 536. The sides of /\ C U V are f of the corresponding 
 sides of /\ C F H. If similar triangles resemble squares in 
 the matter of areas, how much larger than /\CFH should 
 /\ C U V be ? Test your answer by counting the number 
 of triangles like /\ C D E in each of the larger triangles 
 concerned. 
 
 537. All of the experiments made from 533 to 536 con- 
 firm the statement that similar triangles are like squares 
 in the way in which the areas change as the sides 
 
106 ELEMENTARY AND 
 
 change. To give a complete proof of this statement 
 would carry us too deeply into the principles of Qeontt 
 for our present purposes. The principle may be Accepted 
 as a good working principle until it can be shown to fail, or 
 until it can be proved to be true in all cases. 
 
 538. A little further study of Fig. 77 would show that 
 the altitudes of the .various similar triangles are related to 
 each other just as are the sides of the corresponding tri- 
 angles ; thus the sides of /\ C F II are f of the OOTrefpond- 
 ing sides of /\ C U V, and the altitude of /\ 1 II ifl also 
 { of the altitude of i\ C U V, because the parallels D E, 
 F H, and K R, etc., cut all lines between and A B into 
 seven equal parts. Draw a figure lik ',',. ami draw 
 the Altitude from C to A B, and show that it is divided into 
 seven equal parts by the parallels. This fact gives additional 
 reason for believing the principle of 537 to be true ; for the 
 essential peculiarity of squares, so far as their area is DOH- 
 oernedj il that no change can take place in the base of a 
 square, unless it also takes place in the side or altitude of 
 the square; if the base is made | as large, the altitude 
 must also be made | as large, or the figure ceases to be a 
 square : the area of the square, which is nothing but the 
 product of the base and altitude, is affected bjf both changes, 
 and therefore becomes J of f as large, or y as large. Now 
 this same peculiarity is shared by similar triangles ; if the 
 base is made | as large, the altitude must also be made f as 
 large, or else the new triangle ceases to be similar to the 
 old ; the area, which is merely the product of the base and 
 the half-altitude, is affected by both changes, and becomes 
 | of | as large or */ as large. 
 
 539. If similar triangles are like squares, they can be 
 added like squares. Suppose you wish to add /\CK B 
 to A C M X (Kg. 77) ; place C K at right angles to C M 
 (Fig. 78), as if you were to add K* to M 2 ; K If will bo 
 the side of the new triangle corresponding to C K and C M 
 of the given triangles. Make a triangle with the angles of 
 /\ C K R, starting with K M, Fig. 78, in the position of 
 
CONSTRUCTIONAL GEOMETRY 
 
 107 
 
 C K, Fig. 77. The triangle thus formed will be like A 
 OKR and C M N in shape, and will be equivalent to their 
 sum. 
 
 In Fig. 77 there is a triangle 
 that is equivalent to /\ C K R + 
 A C M N. Find this triangle, 
 and compare it with the triangle 
 that you have just constructed. 
 
 540. Make three pairs of simi- 
 lar triangles, and add the trian- 
 gles of each pair by the method 
 described in 539. 
 
 541. Can you subtract one tri- 
 angle from a larger triangle similar 
 to it ? Give two illustrations. 
 
 542. Can you find a triangle twice as large as another 
 triangle and similar to it ? Suggestion : Think how you 
 double a square. 
 
 543. Make a triangle ABC; can you find a similar tri- 
 angle equivalent to3x/\ABC? to 4xAABC?to 
 
 5 x A AB C? 
 
 544. Given a triangle, ABC, can you find a similar trian- 
 gle equivalent to J of A A B C ? to J of A A B C ? to f of 
 AABCFtofofAABC? 
 
 545. Polygons are said to be similar to each other when 
 their corresponding angles are equal, and when their cor- 
 responding sides have the same relation to each other. 
 
 Make a pentagon, and then, with the aid of your divid- 
 ing tool and protractor, make another pentagon similar to 
 it and with sides f as long as those of the first pentagon. 
 
 546. A convenient method of making one polygon simi- 
 lar to another is illustrated in Fig. 79. 
 
 The problem is to make a pentagon similar to A B C D E 
 with sides twice as long respectively. 
 
 Divide A B C D E into three triangles by the diagonals 
 AC, AD. Extend A B to H, making A H = 2 A B. 
 
 Draw H K 
 
 B meeting A C extended in K. 
 
 In the same 
 
108 ELEMENTARY AND 
 
 way draw K M | C I) and M N | D E, completing the pen- 
 tagon AHKM N. The corresponding angles of tin- two 
 polygons are plainly the same. Why ? Can you, with the 
 help of the principle of 325, show that II K, K M, M N, 
 
 
 and N A are respectively twice as long as the corresponding 
 .,f A BODBf 
 How much larger is each triangle of A II K M N than 
 the corresponding triangle of A B C D K !' How mu.h 
 larger is A II K M N than ABOD E? 
 
 547. Make a hexagon, and then make a similar hexagon 
 with sides \ as long as those of the first, using the method 
 described in 
 
 Compare the areas of the corresponding triangles and of 
 the two hexagons. 
 
 548. Do similar polygons resemble squares in respect to 
 their surfaces? 
 
 549. Oan yon make a parallelogram twice as large as a 
 given parallelogram, and similar to it? 
 
 550. (an you make a pentagon similar to those in Fig. 
 .ml equivalent to their sum P 
 
 551. (an you make a pentagon similar to those in Fig. 
 79, and equivalent to their difference ? 
 
CONSTRUCTIONAL GEOMETRY 
 
 109 
 
 552. Can you make a hexagon $ of a given hexagon ? f 
 of a given hexagon ? \ of a given hexagon ? In each 
 case the hexagon must be similar to the given hexagon. 
 
 553. Can you make a circle equivalent to the sum of two 
 given circles ? 
 
 554. Can you subtract one circle from another, giving the 
 difference in the form of a circle ? 
 
 555. Can you change a circular ring, or band, into a 
 circle ? 
 
 556. Can you make a circle twice as large as another cir- 
 cle ? three times as large ? four times as large ? 
 
 557. One circular hole was found to have § of the diam- 
 eter of another ; how much smaller than the second hole 
 was the first hole ? 
 
 558. If you wish to double the amount of light coming 
 through a circular window, should you double the diameter 
 of the window ? 
 
 559. Draw a circumference, Fig. 80, and with your di- 
 viding-tool divide the diameter into five equal parts ; draw 
 semicircumferences on the di- 
 ameters AB, AG, AD, and 
 A E above the diameter A F, 
 and semicircumferences on the 
 diameters B F, C F, D F, and 
 E F below the diameter A F. 
 
 What part of the semicircle 
 A H F is A K B ? 
 
 What part of the whole circle 
 is semicircle A K B ? 
 
 What part of the whole circle 
 is semicircle B M F ? 
 
 The horn - shaped figure 
 A K B F T A is made up of 
 the semicircles A K B + A T F 
 the whole circle is the horn ? 
 
 If the corresponding lines of similar figures have the 
 same relation to each other, what part of the semicircum- 
 
 B M F. What part of 
 
110 ELEMENTARY AND 
 
 ference A II F should the semicircumference A K B 
 What part of the semicircumference A II F should the 
 semicircumference B M F be ? 
 
 Which is the longer path from A to F, the path by the 
 semicircumference A T F, or that by the two semicircum- 
 ferences A K B and B M F ? 
 
 560. By adding and subtracting the right semicircles in 
 Fig. 80, the surface of the strip A L C N F M B K A < an 
 be found. Can you select the right semicircles, and can 
 you estimate what part of the whole circle each semicircle 
 is, and what part of the whole circle the strip is ? Can you 
 estimate the perimeter of the strip ? 
 
 561. In 559 you saw thai the horn AKBFTA is 
 equivalent to the semicircles A K B + A T P — B M P : can 
 you find one semicircle that shall be equivalent to A K B 
 +ATF-BMF? Bee 568 and 654. 
 
 Make a circle ami divide the diameter into eight 
 parti ; draw semicircumferences, as in Fig. 80, cutting the 
 circle into eight strips. Try to show that the circle is thus 
 divided into eight equivalent strips. 
 
 563. Can you find a circle one-half of a given circle ? 
 one-third of a given circle ? two-fifths of a given ein 1. \ 
 
 ."••'.I. Review carefully Bectioni XV., XVI.. and XVII. 
 (in several lessons if necessary). 
 
 565. Write a sketch of the square, describing how squares 
 can be added and subtracted ; how they can be enlarged any 
 number of times without losing their shape; and how they 
 can be divided into as many small squares as may be de- 
 sired. Tell, also, what other figures can be added, sub- 
 tracted, enlarged, or diminished by the methods used for 
 squares. 
 
CONSTRUCTIONAL GEOMETRY 111 
 
 SECTION XVIII. CIRCLES AND INSCRIBED 
 ANGLES. 
 
 566. A circle is a figure bounded by a line, called the 
 circumference, every point of which is equally distant 
 from a point within, called the centre. The sign for circle 
 is 0, plural © ; and that for circumference is O, plural 
 ©. You have already learned that you can measure an 
 angle by placing its vertex at the centre of a circle whose 
 circumference has been divided into three hundred and 
 sixty equal parts, and by noting the number of these parts, 
 or degrees, between the sides of the angle. Any portion of 
 a circumference is called an arc, and the angle formed 
 by the radii through the ends of an arc is said to sub- 
 tend the arc. Thus in Fig. 81 
 
 /BOA subtends the arc B A, 
 written BA. The number of de- 
 grees in the arc B A is the same 
 as the number of degrees in the 
 angle B A at the centre ; if 
 BA = 60°, / B A = 60° ; if 
 BA = 50°, I B A = 50° ; if 
 BA = x°, /_ B A = x° ; so 
 that an angle at the centre of a 
 circle has the same name as the 
 arc which it subtends. This prin- 
 ciple is simply a restatement of the principle of the pro- 
 tractor. 
 
 567. In using a protractor you have always placed the cen- 
 tre of the instrument upon the vertex of the angle. The pe- 
 culiarities of circles enable us often to measure angles with- 
 out a protractor, and also to measure angles whose vertices 
 are not at the centre of the circle. In Fig. 82 A E is a 
 diameter, and AB = 60°. Can you, without a protractor, 
 give the value in degrees of all of the angles in the figure ? 
 
112 
 
 ELEMENTARY AND 
 
 Can you see any simple relation between / BE A ami 
 
 BA P 
 
 5G8. Imagine, or draw, an arc A C = 70°, and the diam- 
 eter A E. Can you tell how many 
 degrees there are in the angle 
 C E A ? Try arcs of <K)°, 75°, 
 30°, 120°, and 150°, from A; 
 and, after joiuing the end of 
 each arc with B, find the value 
 of each angle formed at K with 
 the line E A. Draw a figure 
 like Fig. 82 for each case, mak- 
 ing the arc A B the proper 
 length each time. 
 
 An angle, placed as an- 
 gle B B A is place* 1 in Fig. 
 
 B8, 1 ith its vertex on the ei re u inference, and with its sides 
 running to other points of the circumference, forming 
 chords, is called an inscribed angle. The Experiment! of 
 and 5G8 suggest the principle that inscribed angles, 
 which have a diameter for one tide, subtend arcs of twice their 
 own number of degrees. Try to 
 prove this principle by showing 
 that if / E = *°, All must be 
 equal to &b° Fig. 83. Draw II 
 O ami oompare j_ II A with 
 / x. How is /_ H A related 
 to II \ I 
 
 570. What is the largest in- 
 scribed angle that you can draw 
 with one side a diameter? 
 
 Prom one end of a diameter 
 draw chords that shall make 
 angles of 60°, 45°, 30°, 22J°, 
 
 and 75° with the diameter, using your compasses and ruler 
 only. 
 
 Warn, ^ 
 
CONSTRUCTIONAL GEOMETRY 
 
 113 
 
 Fig. 84. 
 
 571. Place the inscribed angle x, so that the centre of 
 the circle shall come between the 
 
 two chords which form its sides, 
 as in Fig. 84. Will AH still be 
 equal to 2x° ? Draw the diame- 
 ter EOK, dividing / x into 
 two pieces, each of which has a 
 diameter for one side. If you 
 give to piece A E K the name y, 
 the piece KEH will be x — y. 
 AK = 2y° ; KH = 2x° - 2y° ; 
 . • . HA = 2^° + 2x° - 2y° = 
 2x°, so that HA = 2x° as before. 
 
 If / x be inscribed, as in Fig. 85, in such a way that 
 both chords come on the same 
 side of the centre, will the arc 
 A H still contain twice as many 
 degrees as the angle x ? Draw 
 EOK, the diameter, and call 
 / A E K, / y, as before ; 
 
 KA = 2if, and HK = 2x° + 
 2y°. . • . HA = HK - KA = 
 2x° + 2ij° — 2y° = 2x° as be- 
 fore. We therefore can say that 
 any inscribed angle subtends an 
 arc of twice its own number of degrees. 
 
 572. How large an arc must an inscribed angle of 90° 
 subtend ? an angle of 22J° ? an angle of 150° ? 
 
 573. Several arcs of a circumference, when measured, 
 were found to be respectively, 70°, 83°, 90°, 340°, and 
 180° ; how large was each inscribed angle that subtended 
 these arcs ? 
 
 574. An inscribed angle is said to stand on the arc which 
 it subtends. Can you say why all inscribed angles in one 
 circle which stand on the same arc must be equal ? Make 
 
 8 
 
 Fig. 85. 
 
114 
 
 ELEMENTARY AND 
 
 a circle, and inscribe any angle A E II. Make six other 
 angles in the same circle equal to / A E H, using your 
 ruler only. 
 
 575. Make a circle, and make six right angles by means 
 of this circle, using no tool except your ruler to make the 
 angles. Compare the results here with what you have 
 already learned about the middle point of the hypotenuse 
 of a right triangle. 
 
 576. Can you inscribe a regular hexagon in a circle ? 
 Can you determine the number of degrees in each angle of 
 the hexagon by the arc subtends 1 P 
 
 577. Can you inscribe a regular octagon and find the 
 number of degrees in each angle ? 
 
 578. Inscribe an angle of 75° (/ A E II, Fig. 86). 
 Choose any point V in the arc A II and draw the chords 
 II V and A V completing the 
 quadrilateral A V II E. Can 
 you tell the number of degrees 
 in the angle at V P How is 
 / II V A related to / E ? Can 
 you give the value of the angles 
 at II an.l A ? Can you tell 
 the number of degrees in their 
 sum ? 
 
 579. Can yon show what 
 must always Ik* the relation be- 
 tween the opposite angles of 
 an inscribed qaadri late ral ? Is there any parallelogram 
 whose opposite angles have this required relation P 
 
 580. Can you make a cinu inference pass through the 
 four comers of a rhomboid ? of a rhombus ? of a rectangle ? 
 of a square ? 
 
 581. Draw a circle, Fig. 87, and then draw two paral- 
 lel chords, A E and II K. By the aid of A K can you 
 show that the are A II is equal to the arc K E, and thus 
 establish the principle that parallel chords intercept equal 
 arcs? 
 
CONSTRUCTIONAL GEOMETRY 
 
 115 
 
 Fig. 87. 
 
 582. Any chord divides a circle into two segments ; for 
 example, A E, Fig. 87, divides the circle into the two seg- 
 ments AHKE, and AVE. 
 Segments are named by the num- 
 ber of degrees in their arcs ; thus 
 a segment of 200° is one whose 
 arc is 200° ; a semicircle is a seg- 
 ment of 180°. If one of the 
 segments into which a chord di- 
 vides a circle is 120°, what is 
 the other segment ? An angle 
 is inscribed in a segment when 
 its vertex is in the arc of the 
 segment, and when the sides of 
 the angle pass through the ends of the chord that makes 
 the segment. Thus the angle A F E would be an angle 
 inscribed in the segment AVE, Fig. 87. 
 
 583. An angle A T E is inscribed in a segment of 100° ; 
 what is the value of the arc A E on which the angle stands ? 
 of the arc A T E ? of the angle ATE? 
 
 584. How large a segment is required that an angle of 
 120° may be inscribed in it ? 
 
 585. What can you say of all 
 angles inscribed in the same seg- 
 ment ? Why ? 
 
 586. If an angle is inscribed in 
 a semicircle, what must it be ? 
 
 587. Of two inscribed angles, 
 which is inscribed in the larger 
 segment, the larger angle or the 
 smaller one ? 
 
 588. If a line merely touches 
 a circle at one point, so that, 
 when extended in either direc- 
 tion, it does not pass within, or out, the circle, it is called 
 a tangent to the circle ; the point at which it touches the 
 circle is called the point of contact, In Fig. 88, T A is a 
 
116 ELEMENTARY AND 
 
 tangent to the circle with its point of contact at T. Can 
 you explain why the name tangent was selected for this 
 line ? 
 
 589. Draw any radius, as T, Fig. 88, and, at its ex- 
 tremity T, draw a line T A X T. Do this without ex- 
 tending T, if you can. Can you show that T A can- 
 not touch the circle in any point except T, so that it must 
 be a tangent ? Suggestion : Let A stand for any point 
 of the line except T, and show what kind of triangle 
 OTA would necessarily be, if A were on the circumference, 
 and also what would be the value of the / O A T in such 
 a case. Is such a triangle as ^ T A would become, if 
 A were on the circumference, a possible triangle P Why P 
 
 590. Choose several points on the circumference of a cir- 
 cle, and draw tangents touching the circle at these points 
 respectively. 
 
 591. Thi angle A T II, Fig. 89, made by a tangent T H 
 and a chord T A through the point of contact is not called 
 
 an inscribed angle, but like an in- 
 
 scribed angle it contains only yr N. 
 
 one half as many degrees as the / \ 
 
 arc T E A subtended by the chord / \ 
 
 T A. Study ,\ T V formed / -— A| 
 
 by drawing the radius OT and I Q^ 7\ 
 
 F the bisector of the angle \ [ \ / J 
 T O A. What kind of triangle \ rVs / 
 is A T O A P Select as many \ ! / y<£ 
 pairs of equal angles as yon can ^^^ v^^ 
 from your knowledge of the nat- T FT 
 nre of ^ T A and from the *»• »• 
 method of drawing F. How many degrees in / FTO + 
 
 1 T F ? How many degrees in/FTO+/FTH? 
 Why must / V THor / A T n = / T F ? Now 
 / TOF has as many degrees as 'ffe, or one half as many 
 degrees as TA. Hence / AT II has one half as many de- 
 -iv.- ;i< '1' \. or an ant^le formed by a tangent and a chord 
 
CONSTRUCTIONAL GEOMETRY 117 
 
 through the point of contact contains one half as many de- 
 grees as the intercepted arc. 
 
 592. How is any angle inscribed in the larger segment 
 formed by T A, Fig. 89, related to /ATH? How is any 
 angle inscribed in the smaller segment A E T related to 
 / A T H ? 
 
 593. A chord A B intercepts an arc of 80°. How many 
 degrees are there in the angle between A B and the tan- 
 gent at A ? How many degrees in the angle between B A 
 and the tangent at B ? 
 
 594. Draw the tangent to the circle of Fig. 89 at A as 
 well as at T. Can you give any reason for thinking that 
 the angle between the tangent at T and T A is the same as 
 the angle between the tangent at A and AT? 
 
 595. What kind of triangle will be formed by a chord 
 and the two tangents at its extremities ? 
 
 596. Draw a circle with centre 0, Fig. 90, and choose a 
 point, P, outside the circle. It is required to draw a line 
 from P that shall be tangent to the circle. 
 
 If A P is a right angle, A P will be tangent (589). 
 Study Fig. 90, and try to explain what principles were 
 used to determine the points A and B that /sOAP and 
 
118 
 
 ELEMENTARY AND 
 
 OBP might be 90°. Can you see what help line was 
 drawn first ? 
 
 597. Make four different circles, and draw tangents to 
 them from points selected outside of the circles. 
 
 598. If, in Fig. 90, A is 5 in., and P is 13 in., how 
 long is A P ? What principle enables you to answer this 
 question ? 
 
 A ball, whose radius is 7 in., rests in a hollow cone 
 so that its points of contact with the cone are 24 in. 
 from the apex of the cone ; how far is the centre of the ball 
 from the apex ? 
 
 600. If the radius of the earth is 4000 miles, how far 
 out to sea does the horizon of a person extend, if he is C ft. 
 tall, and if he is standing at the water's edge ? 
 
 801. How far can the light of a light-house 105 ft. high, 
 be seen from the surface of the water ? 
 
 H«>w far can the same light be seen from a point in the 
 rigging, thirty feet above the surface of the water ? Con- 
 sider the radius of the earth 4<>oo miles in both cases. 
 
 I >aw a circle with centre at ; then take P at a 
 distance from <> equal to the diameter of the circle. Draw 
 the tangents PA and 
 P B, and the chord 
 
 of contact. A I'.. 
 Can you give the 
 number of degrees 
 in all the angles 
 formed, with rea- 
 sons in each case ? 
 How many degrees 
 in the arc A Q B ? 
 Why ? In the arc 
 AOH? Why ? If 
 the radius of the first circle is 8 in., can you find the lengths 
 of all the lines in the figure, curved and straight t 
 
 603. What part of the circumference is \Q\\. Fig, '.'1 P 
 What part of the earth's surface would a mau see were he to 
 
 A^-- 
 
 Flo. 91. 
 
CONSTRUCTIONAL GEOMETRY 119 
 
 ascend 4000 miles above the earth's surface ? How far 
 would he have to go to see one half of the earth's surface ? 
 
 604. In Fig. 91, if you call angle A P B, x°, can you say 
 how many degrees there must be in the arc AQB ? 
 
 Two tangents make an angle of 30° with each other; 
 how long an arc do they inclose ? 
 
 If you want tangents to meet at an angle of 45°, how 
 many degrees apart must you take the points of contact ? 
 
 605. Can you state a principle concerning the relation be- 
 tween the angle which two tangents form and the arc in- 
 cluded between the points of contact ? 
 
 606. In Fig. 91, assuming that the radius is 8 in., can 
 you find the number of square inches inclosed between the 
 tangents and the arc AQB? Find also the area of the 
 oval A O B Q ; of the smaller segment cut off by A B ; and 
 of the triangle A O B. 
 
 607. Three circles of equal size with a radius of 6 in. are 
 placed in contact as in Fig. 92. 
 
 Can you find the area of the space between the circles ? 
 
120 
 
 ELEMENTARY AND 
 
 What is the area of the parts of the circles without the tri- 
 angle of centres P Q ? 
 
 608. Draw a circle, and draw tangents to this circle at 
 points D, F, and L, Fig. 93, which are 120° apart, thug 
 forming /\ABC. Draw 
 ADFL, and also A H P E, 
 whose vertices are at the mid- 
 dle points of the arcs F L, L D, 
 and 1) F, respectively. Can 
 you account for the shape of 
 the triangles formed P How 
 much smaller than the outside 
 triangle is the smallest tri- 
 angle ? (Recall the principle 
 about the surfaces of similar 
 triangles.) What part of the 
 intermediate triangle is the 
 hexagon ? Find the number 
 
 of square inches in each triangle and in the hexagon, if 
 A B = (» in. Find also the area <»f the circle. 
 
 <'.<>!*. Can yon make a circum- 
 ference pan through two points, 
 A ami B P How many cireuni- 
 terencet P What can you say of 
 the position of the centres of all 
 the circumferences that pass 
 through A and B ? 
 
 610. Can you make a circum- 
 ference pass through any three 
 points, A, B, and II P Bow 
 many circumferences can you 
 draw through A, B, and II P 
 Can you put A, B, and 11 in 
 such positions that a circumference cannot be drawn 
 through them ? 
 
 Gil. Under what circumstances can you draw a circum- 
 ference through four points ? Can you recall any quadri- 
 
 Fiu. M. 
 
CONSTRUCTIONAL GEOMETRY 
 
 121 
 
 lateral through whose corners you could draw a circumfer- 
 ence ? 
 
 612. Make a line, and a point without the line ; can you 
 draw a circumference that shall merely touch the line, and 
 at the same time pass through the point ? How many cir- 
 cumferences fulfilling these conditions can you draw ? 
 
 613. Can you draw a circumference that shall be tangent 
 to the line of 612 at a chosen point, and at the same time 
 pass through the point without the line ? 
 
 614. Can you make a circumference with a given radius, 
 touch a given line, and pass through a given point ? Can 
 the radius be made too long ? Can it be made too short ? 
 
 615. Can you make a circumference pass through the 
 vertices of a scalene triangle ? 
 
 616. Using some round object as a guide, draw a circle ; 
 can you find the centre of the circle ? 
 
 617. Draw free hand a curved line ; can you tell whether 
 or not the curve is a piece of a circumference ? 
 
 Fig. 95. 
 
 618. Can you make a circle touch two intersecting lines ? 
 How many answers can you obtain for one pair of inter- 
 secting lines ? 
 
 619. Can you make a circle touch the three sides of a 
 triangle ? 
 
 620. Can you make a circle touch two lines that do not 
 
122 
 
 ELEMENTARY AND 
 
 meet within the boundaries of the paper, although they are 
 not parallel lines ? 
 
 621. At two points, A and B, of a line (Fig. 95) draw 
 two circles tangent to the line, using different radii of 
 such lengths that the circles will have no point in common. 
 
 Draw the radii, A P and B Q, and join the centres of the 
 circles by the line P Q. 
 
 What kind of quadrilateral is A B Q P ? 
 
 llow many of its angles do you know ? 
 
 Yu. OC, 
 
 Draw a line Q II parallel to B A. Describe the two 
 figures into which Q II divides the quadrilateral A B Q P. 
 If the radial of the larger circle, Fig. 95, is 8 in., 
 and that of the smaller is 3 in., and if the centres of the 
 circles are I'.i in. apart, can you find the length of the com- 
 mon tangent AB? of HP? of HQ? 
 
 G23. Can you make a trapezoid, knowing the two bases 
 and one leg, so that the bases will make right angles with 
 the unknown leg ? Suggestion : Try to reproduce Fig. 95 
 from three lines equal in length to P Q, A P, and B Q. 
 
CONSTRUQTIONAL GEOMETRY 123 
 
 624. Can you draw a line that shall touch two given 
 circles"? 
 
 Study Fig. 96, where the two given circles have their 
 centres at and P, and where, by the aid of the circle 
 with P as diameter, the tangent T E was determined. 
 Compare the quadrilateral OTEP with the quadrilateral 
 P A B Q in Fig. 95. 
 
 In Fig. 96 a line, R H, is found, which touches the two 
 circles and passes between them. Can you see how points 
 F and R were determined when this interior tangent was 
 drawn ? 
 
 625. Make two circles and draw an interior tangent to 
 them. Repeat the problem with different circles twice. 
 
 SECTION XIX. VOLUMES. 
 
 626. In 367 and the following exercises you learned that 
 the product of two lines represents a surface ; that the 
 simplest surface that you can form, in multiplying two 
 lines together, is the rectangle ; but that this surface can 
 be moulded into many other forms. 
 
 627. Multiply a line 7 cm. long by one 4 cm. long ; the 
 rectangle, representing their product, contains 28 sq. cm. 
 See rectangle A B C D (Fig. 97), where the 28 sq. cm. are 
 outlined. Draw a line, B F, 5 cm. long, that shall be per- 
 pendicular to both B A and B C, and complete the rectan- 
 gular parallelopiped A B C D E F K. If B S is taken 
 equal to one centimeter, and if the figure is divided as 
 represented, what are the dimensions of the solid S B M. 
 What is its name ? How many solids like it could be made 
 out of the solid H S T D, which forms the top layer? 
 How many layers like HSTD could be cut from the 
 whole solid? How many cubic centimeters (ccm.) could 
 be cut from the whole solid ? How do you obtain the 
 answer to the last question ? 
 
124 
 
 ELEMENTARY AND 
 
 628. How many ccm. of wood are there in a rectangu- 
 lar piece 8 cm. long, 5 cm. wide, and 3 cm. thick. 
 
 How many cubic inches (cu. in.) of ice could be cut from 
 
 2l 
 
 .^z^^S 2 ^ 
 
 H 
 
 f ^Ba 
 
 Flo 9T. 
 
 a rectangular block of ice 24 in. long, 15 in. broad, and 8 
 in. thick ? 
 
 620. Just as you saw that the product of two lines rep- 
 resents a surface, you can now see that the product of three 
 lines represents a solid. The simplest solid that you can 
 form, in multiplying three lines together, is the rectangu- 
 lar oblong, or rectangular parallelopiped formed by using 
 the three lines for the edges which are at right angles to 
 each other. 
 
 The product of two of the lines gives the base of the ob- 
 long ; the third line is the altitude of the oblong. It is not 
 necessary to form a mental picture of the small cubes into 
 wliieh the oblong may be divided eaeli time that you mul- 
 tiply three lines ; in fart it is better to think of the space 
 occupied by the solid as a whole. 
 
 630. Make four sets of three lines each, and draw the 
 rectangular oblong represented by the product of the three 
 lines of each set. 
 
CONSTRUCTIONAL GEOMETRY 
 
 125 
 
 631. Draw the oblong of Fig. 97 again, this time with- 
 out the additional dividing lines. You recall that its dimen- 
 sions are 7 cm. by 4 cm. by 5 cm. Imagine a piece cut from 
 the solid at the left end along the line VTM, and imagine 
 this piece transferred to the right end, with the face DAE 
 
 Fig. 98. 
 
 placed upon its equal, C B F K. Will the new solid 
 V T M E L P H occupy more or less space than the old ob- 
 long ? Will its dimensions be changed ? (It is supposed 
 that the cutting is done without waste.) What shape will 
 the faces of the new solid have ? Will the twelve edges of 
 one solid be any longer than the twelve edges of the other ? 
 Why ? Which of the new faces, if any, cover any larger 
 surface than the corresponding faces of the oblong ? 
 Which solid has the larger upper base ? Which lines must 
 you multiply together to obtain the upper base VTHP? 
 What lines must you multiply together to obtain the solid 
 VTMRLPH? 
 
 632. The solid VTMRLPH (Fig. 98), is called a 
 paralellopiped. Do you understand why ? If possible se- 
 cure two blocks of wood, one a rectangular parallel opiped, 
 and the other a parallelopiped, both of which have been cut 
 from one long rectangular block. The two parallelopipeds 
 
126 BLBMRNTARY AND 
 
 should have one edge (corresponding to edge D C, Fig. 98) 
 the same in both. After coating the blocks with parajfine, 
 that they may not absorb water, sink first one and then the 
 other in a vessel of water, noting carefully how high the water 
 rises in the vessel each time. Does the experiment confirm 
 your answers to the questions of 631 which relate to the vol- 
 ume of the solids ? Try the experiment of weighing the two 
 blocks, and see whether the result agrees with that of your 
 first experiment. 
 
 633. Measure in centimeters the edges of the rectangu- 
 lar block used in 632, estimate the volume of the block in 
 cubic centimeters as in 628, and then verify your estimate 
 by pouring from a vessel graded in cubic centimeters 
 enough water to raise the level of the water in the vessel 
 used in 632 to the same height to which it was raised by 
 sinking the block. 
 
 634. In cutting the parallelepiped from the rectangular 
 block, Fig. 98, do you alter the size of the parallelopiped by 
 changing the directions in which you draw the lines V X 
 and T M to guide your rutting ? Do you change the area 
 of the upper face as you change the direction of V T ? 
 Can you draw V T and T M in such a way that the paral- 
 lelopiped will have four of its faces rectangles ? two of its 
 faces rectangl- 
 
 . Can you multiply three lines together so that the 
 product will be a rectangular parallelepiped? i parallele- 
 piped with four faces rectangles ? with two faces rectangles ? 
 with no faces rectangles ? 
 
 636. Can you give any conditions under which two paral- 
 lelopipeds of different shapes will occupy the same amount 
 of space? Think of the corresponding conditions for 
 equivalent parallelograms. 
 
 I haw any parallelopiped that is not rectangular, and 
 draw the lines whose product represents the volume of the 
 solid. Form the rectangular parallelopiped that is equiva- 
 lent to the first solid. 
 
 G38. Take a block of wood in the form of an oblique- 
 
CONSTRUCTIONAL GEOMETRY 127 
 
 angled parallelopiped ; find the lines whose product repre- 
 sents the volume of the parallelopiped, measure these lines 
 in centimeters, estimate the volume of the solid, and test 
 your answer by the method of 633. 
 
 639. The principle to be derived from Articles 627-638 
 is that a parallelopiped is the product of its three di- 
 mensions, or from another point of view, the volume of a 
 parallelopiped is the product of its base by its altitude. 
 Just as a rectangle can be moulded into any parallelogram 
 which has the same base and altitude as the rectangle, a 
 rectangular parallelopiped can be moulded into any oblique 
 parallelopiped which has the same base and altitude as the 
 rectangular parallelopiped. 
 
 640. Cut a rectangular block of wood into two pieces by 
 sawing along two diagonally opposite edges. The result- 
 ing pieces are called triangular prisms ; the two triangular 
 faces form the bases of the prism, and the three faces in 
 the shape of a parallelogram form the lateral surface. 
 When a rectangular parallelopiped is cut into two triangu- 
 lar prisms, as in this case, the prisms are called right 
 prisms ; for a right prism is one whose lateral edges are at 
 right angles to the two bases. Do the two right prisms 
 formed in the experiment occupy the same amount of 
 space ? Test your answer to the question by weighing the 
 two prisms, or by sinking them in a vessel of water, and by 
 noting the height to which the water rises in each case. 
 
 641. Can you find the three lines whose product repre- 
 sents the volume of one of the prisms of 640 ? Can you 
 draw the rectangular parallelopiped that will occupy the 
 same space as the prism ? 
 
 64^. Can you show that the product of two of the three 
 lines found in 641 represents the base of the prism, and 
 that the third line is the lateral edge, and at the same time 
 the altitude of the prism ? 
 
 In doing this you illustrate an important principle of 
 Geometry : The volume of a right triangular prism is the 
 product of its base by its altitude. 
 
128 
 
 ELEMENTARY AND 
 
 643. Imagine the oblique parallelopiped ( A B C D E I 1 1 . 
 Fig. 99), none of the faces of which are rectangles, to be cut 
 
 Flo. 99. 
 
 into oblique triangular prisms by a plane through the di- 
 agonally opposite edges A B and £ D. Imagine also, that, 
 before the pieces are separated, the parallelopiped is cut 
 along the lines M K and K L, which have been drawn at 
 right angles to E II. Transfer the portion cut off at the left 
 to the right, forming the parallelopiped L V T R M K P. 
 At the same time the triangular prisms B II D 1 
 and B C D E A N will be changed into the prisms VPB 
 M K L. and V T B M L Q. 
 
 If possible, follow the directions just given with a block 
 of wood or a piece of potato in the proper shape. 
 
 Son many faces of the new parallelopiped will be rec- 
 tangles ? If the new parallelopiped is imagined to stand on 
 tin* bast* V T l\ l\ it will be I right parallelopiped, beCftOM 
 its edges, K 1\ M It, etc., are at right angles tc the two 
 faces. Illustrate this by showing that your block, if placed 
 on the base V T R P, or M K L Q, will stand upright, bat 
 that if it is placed on any other base it will lean over. The 
 volume of the right parallelopiped, by 639, will be the 
 product of tin T |{ V bv the altitude K P. The 
 
 triangular prisms V P B M K L. and V T B M L Q, are 
 right prisms. (Why ?) The volume of one of them is, by 
 642, the product of the bow V P B and the altitude K I\ 
 
CONSTRUCTIONAL GEOMETRY 129 
 
 or just one half of the volume of the right parallelopiped. 
 Hence a right parallelopiped, as well as a rectangular paral- 
 lelopiped (640), can be cut into two parts which have the 
 same volume by a plane through two diagonally opposite 
 edges. But the right parallelopiped of Fig. 99 occupies the 
 same space as the original oblique parallelopiped ; and the 
 right triangular prisms occupy the same space as the oblique 
 triangular prisms, so that any parallelopiped— whether rec- 
 tangular, right, or oblique— can be cut into two parts which 
 have the same volume by a plane through two diagonally op- 
 posite edges. 
 
 644. Test the principle just stated in 643 by weighing 
 the two pieces into which parallelopipeds of wood having the 
 required shapes are cut by planes containing two diagonally 
 opposite edges. 
 
 645. The volume of the oblique parallelopiped 
 ABCDEFH (Fig. 99) is, by 639, the product of its base 
 13 C D II by its altitude, the perpendicular distance between 
 its two bases ; the volume of the oblique prism B D II 
 E F A, which is one half of the parallelopiped by 643, is 
 the product of one half of B D II, or 13 D H, by its alti- 
 tude, that is, it is the product of its base B D II by its alti- 
 tude, so that the volume of any triangular prism is the prod- 
 uct of its base by its altitude. 
 
 646. In how many different ways could you cut an ob- 
 lique parallelopiped into triangular prisms by planes con- 
 taining two diagonally opposite edges ? Would there be 
 any difference in size in the various prisms ? in shape ? 
 
 647. Make three lines, a, b, and c ; form a right parallel- 
 opiped equivalent to the product of the lines a x b x c ; 
 also form an oblique parallelopiped equivalent to a x b x c ; 
 form an oblique triangular prism equivalent to a x b x c. 
 
 648. A tin box, in form a right triangular prism, has as 
 its bases two right-angled isosceles triangles with the equal 
 sides 8 cm. long ; the box is 15 cm. tall. How many ccm. 
 of water will the box hold ? How many sq. cm. of tin will 
 be needed for the box and its cover ? 
 
 9 
 
130 ELEMENTARY AND 
 
 649. The inside dimensions of a cistern shaped like a 
 rectangular parallelopiped are 10 ft. x 6 f t. x 8 ft. ; how 
 many gallons of water will the cistern hold, if it takes 7i 
 gallons to fill a cubic foot of space ?• 
 
 650. A right prism 10 in. tall has a regular hexagon for a 
 base ; if each edge of the base is 6 in., how much space 
 does the prism occupy ? 
 
 651. A water-pail in the form of a cylinder is 12 in. tall 
 and 9 in. in diameter; can you tell how many gallons ii 
 will hold if it takes 231 cu. in. to hold a gallon ? (The 
 volume of a cylinder is found on the same principle on 
 which the volume of a prism is found.) 
 
 652. Into a tumbler partly filled with water put a piece 
 of lead or a block of wood of irregular shape, and note 
 carefully how much tin- water rises when the object is com- 
 pletely under water. Estimate the volume of the im- 
 mersed solid after taking the proper measurements. Use 
 a tumbler that has the form of a cylinder. 
 
 . 653. Sink in a tumbler of water a triangular prism, and 
 note the height to which the water rises; next, sink a tri- 
 angular pyramid with a base and an altitude equal to those 
 respectively of the prism ; how much more space does the 
 prism occupy than the pyramid does t Try this experi- 
 ment several times before giving a final answer, and, if pos- 
 sible, use several pairs of prisms and pyramids, making each 
 pair according to the directions above, but varying the 
 sizes of the blocks used. 
 
 »'..". 1. Form two pyramids of the same height on triangu- 
 lar bases that have the same area hut very different forms. 
 Sink these pyramids in water, and note whether one occu- 
 pies more space than the other. 
 
 655. Form several triangular pyramids of the same altitude 
 and with equivalent (not equal) bases, and form one prism 
 with an altitude equal to that of the pyramids, and with a 
 base equivalent to one of the bases of the pyramid. See 
 how many pyramids it takes to displace as much water as 
 the prism displaces. 
 
CONSTRUCTIONAL GEOMETRY 131 
 
 656. Sum up the results of the experiments in 653-655 
 by writing a principle that will enable you to obtain the 
 volume of a triangular pyramid. 
 
 657. Can you extend the principle of 656 so that it will 
 include any pyramid and also a cone ? 
 
 658. In dealing with triangles you saw that you could 
 move the vertex of a triangle along a line parallel to the 
 base without altering the area of the triangle. Do your ex- 
 periments in 653-655 point to any similar principle about 
 triangular pyramids ? What is the principle ? 
 
 In moving the vertex of a triangle without changing the 
 area you were obliged to move it either symparallel with 
 the base or antiparallel to it. But in moving the vertex 
 of a triangular pyramid, you can move it in a countless 
 number of directions without changing the volume of the 
 pyramid ; for a line will be parallel to the triangular base 
 of the pyramid if it is parallel to any one of the countless 
 lines that can be imagined in the triangular base. To get 
 a clear idea of this, place a triangular pyramid on the table ; 
 you will at once see that there are thousands of directions 
 in which you could move the vertex without changing the 
 volume, keeping it in a line parallel to the base or the 
 table. You can therefore 
 mould triangular pyramids 
 with even more freedom than 
 you can mould triangles. 
 
 659. By moulding triangles 
 you can change a polygon 
 into an equivalent triangle. 
 In the same way, by moulding 
 triangular pyramids you can 
 change a pyramid which has a 
 polygon as a base into a trian- 
 gular pyramid. For an exam- 
 ple see Fig. 100, where the ¥m - m 
 pyramid V— A BOD has a quadrilateral as a base. 
 
 The plane V B D cuts the pyramid into two triangular 
 
132 ELEMENTARY AND 
 
 pyramids, V-A B D and V-B C D. The latter may be 
 imagined to stand on the base B D V, with vertex at C. 
 Move C along the line C K, which is parallel to D B, and 
 therefore to V B D. (See 658.) The pyramid C-BDV 
 is thus moulded into the equivalent pyramid K-BOV, 
 which joins with V— A B D to form the triangular pyra- 
 mid V — A K I) ; equivalent to the original pyramid. 
 
 600. Draw three different pyramids with quadrilaterals 
 for bases, and mould each pyramid into its equivalent tri- 
 angular pyramid. 
 
 001. Can you mould two triangular pyramids of the same 
 height, but with different bases, into one triangular pyra- 
 mid ? 
 
 Review the method by which you lowered the ver- 
 tex of a triangle to any desired point without changing the 
 surface of the triangle, and then study the construction which 
 follows, by which the vortex of a pyramid is lowered with- 
 out anv change in the volume of the pyramid. The origi- 
 nal pyramid is V— A B C, Fig. 101, and the point to 
 which the yertex is to be lowered ii P. 
 
 My a plane, P B (a guide plane corresponding to the 
 guide line of Exercise 419), a pyramid, V — P BC, is cut off; 
 the vertex V is then moved to K 
 along V K, which is parallel to 
 P B, and therefore parallel to the 
 base I 1 H ('. which contains P B ; 
 the edges P V, B V, C V, take the 
 respective positions, P K, B K . 
 c K. and the pyramid V-A BC 
 is changed Into the equivalent 
 pyramid P— A K <\ 
 
 668. Draw two triangular pyra- 
 mids with bases on the same plane, 
 but with di tie rent altitudes. Low- 
 er the taller pyramid to the level 
 of the shorter one. Then find one triangular pyramid equiv- 
 alent to their sum. 
 
CONSTRUCTIONAL GEOMETRY 133 
 
 664. Can you raise the vertex of a triangular pyramid 
 to any desired level without changing the volume of the 
 pyramid ? 
 
 665. How much larger is a triangular prism than a tri- 
 angular pyramid which stands on the same base and has 
 the same altitude as the prism ? (See 656.) 
 
 666. If you wish to find a triangular pyramid with a 
 volume equal to that of a given triangular prism, how tall 
 must you make the pyramid, if you make its base equal to 
 the base of the prism ? How large a base must it have, 
 if you keep the height the same as that of the prism ? 
 
 667. A wine-glass shaped like a cone and a tumbler 
 shaped like a cylinder are of the same height and have 
 the same width at the top ; how much more water will the 
 tumbler hold than the wine-glass ? 
 
 668. How much more than the wine-glass would the tum- 
 bler of 667 hold, if it were twice as tall as the wine-glass, 
 other conditions remaining the same ? 
 
 669. In. 432 and the following articles you saw that a 
 circle could be moulded into a triangle with the radius 
 as altitude, if only a line could be found for the base equal 
 in length to the circumference, and by an experiment 
 you found how to get the circumference from the radius 
 with sufficient accuracy for practical purposes. In a simi- 
 lar way a sphere can be moulded into a cone with an altitude 
 equal to the radius, if it is possible to find a base equiva- 
 lent to the surface of the sphere. The following experi- 
 ment will fix in your minds the facts that have been dis- 
 covered about the surface of a sphere by more advanced 
 principles of Geometry than you have yet taken up. Take 
 a ball of wood or of rubber (a baseball would answer very 
 well) and have a tin cylindrical box made, into which the 
 ball will just fit. When the cover is on the box it should 
 touch the ball. Sink first the ball and then the closed box 
 in a small cylindrical jar of water, and notice how much 
 higher the water rises in one case than in the other. You 
 will find that the ball makes the water rise only two-thirds 
 
13± ELEMENTARY AND 
 
 as far as the box does ; so that the ball occupies only two- 
 thirds as much space as the cylinder which just surrounds 
 it. Now imagine the ball moulded into a cone, with the 
 radius of the ball as its altitude, and with the surface 
 of the ball flattened into a circle for its base. The 
 cone will be only one-half as tall as the box, for the box 
 has the diameter of the ball for its altitude, and the cone 
 has the radius of the ball for its altitude. Since the coue 
 is one-half as tall as the box, to occupy as much space as 
 thr box does it would need a base six times as large as the 
 base of the box (see 668) ; but since it occupies only two- 
 thirds as much space as the box does, its base need be only 
 t\\ .-thirds of six times as large as the base of the box, or 
 four times the base of the box. This experiment illustrates 
 the truth, which can be proved by the principles of Geom- 
 etry, that the surface of a sphere is four times the base 
 of a cylinder in which the sphere is inscribed. 
 
 670. If you should cut the bull in halves, and then re- 
 place the hal\<< in the lx>x. the flat side of one half would 
 exactly fit the bottom of the box, and the flat side of the 
 other half would lit the top of the box, while thecur\.«l 
 portions would just touch each other in the centre of the 
 box. What portion <»f the box is empty P 
 
 If you cut an orange in halves, the two circles forming 
 the plane sides of the pieces contain as much surface as the 
 peel (.n OIM Of the pieces. Why P 
 
 671. If yon should cut the box used in o'*.*.» and 670 along 
 a line perpendicular to the two bases, and flatten the lateral 
 surface into a plane surface, you would form a rectangle 
 with the ci renin ference of the original base as abase, and 
 with an altitude equal to twice the radius of the base. If 
 you should also mould one of the bases into a triangle, the 
 triangle would have the circumference of the original base 
 as a base, and an altitude equal to the radios. Since the 
 rectangle has the same base as the triangle, but twice its 
 altitude, the surface of the rectangle is four times that of 
 the triangle. Make this clear by drawing a rectangle and 
 
CONSTRUCTIONAL GEOMETRY 135 
 
 a triangle which fulfil the above conditions, and by chang- 
 ing the triangle into a rectangle. Since the lateral surface 
 of the cylindrical box is four times that of its base, it must 
 be equivalent to that of the ball. Hence another way of 
 describing the surface of a sphere : The surface of a sphere 
 is the same as the lateral surface of a cylinder in which the 
 sphere is inscribed. 
 
 672. Compare the entire surface of a cylinder with that 
 of a sphere which is inscribed in the cylinder. 
 
 673. Can you draw a circle that shall have as much sur- 
 face as a given sphere has ? A sphere is given if its radius 
 is known. 
 
 674. Can you draw a cone that shall have a base whose 
 radius is the radius of a given sphere, and that at the same 
 time shall equal the sphere in volume ? 
 
 675. Find by arithmetic the volume of each of the fol- 
 lowing spheres, the radii of which are 7 in., 6 in., 3 in., 
 and 14 in., respectively. Compare the first answer with 
 the last answer, and the second answer with the third an- 
 swer. 
 
 676. Assuming that the earth is a sphere with a radius of 
 4,000 miles, how many square miles are there on the earth's 
 surface ? 
 
 677. As a result of your study of this section write and 
 record a rule for finding the volume of each of the following 
 solids : a parallelopiped ; a prism ; a pyramid ; a cone ; a 
 cylinder ; and a sphere. 
 
 678. Draw three lines, a, b, and c, and form a parallelo- 
 piped, a prism, a triangular pyramid, and a quadrangular 
 pyramid, equivalent to the product of these lines. Can 
 you draw a cone, a cylinder, and a sphere that shall be 
 equivalent to the product of these lines ? 
 
INDEX 
 
 Altitude, 242. 
 
 Angle, 76; Right, Acute, Obtuse, 
 Oblique, Straight Angles, 77; 
 Exterior, Interior, Alternate 
 Exterior or Interior, Corre- 
 sponding Angles, 121; Inscribed 
 Angle, 569. 
 
 Antiparallel, 113. 
 
 Arc, 566. 
 
 Area, 371. 
 
 Axiom, 61. 
 
 Chord, 569. 
 Circumference, 32. 
 Contact, Point of, 588. 
 Corollary, 213. 
 
 Diagonal, 297. 
 Dimensions, 4. 
 Dividing Tool, 355. 
 Drawing to Scale, 183. 
 
 Edges, 12. 
 Ellipse, 38. 
 Equilateral, 219. 
 Equivalent, 373. 
 
 Faces, 12. 
 
 Geometrical Solids, 4. 
 
 Hexagon, 404. 
 Hypotenuse, 272. 
 
 Inscribe, 312. 
 Inscribed Angle, 569. 
 Interior Tangent, 624. 
 Isosceles, 210. 
 
 Lines, 12, 26, and 27. 
 
 Median of a Trapezoid, 333. 
 
 Parallel, 113. 
 
 Parallelogram, 148. 
 
 Parallelopiped, 627, 632, 643. 
 
 Pentagon, 97 
 
 Perimeter, 379. 
 
 Perpendicular, 224 
 
 Points, 17. 
 
 Polygon, 428. 
 
 Prism, 640. 
 
 Product of Two Lines, 368; of 
 
 Three Lines, 629. 
 Proof, 150. 
 
 Proportional Dividers, 351. 
 Protractor, 83. 
 Pythagorean Principle, 463. 
 
 Quadrilateral, 100. 
 
 Radius, 33. 
 
 Rectangle, 290. 
 
 Rectangular Parallelopiped, 627. 
 
 Rhomboid, 290. . 
 
 Rhombus, 290. 
 
138 
 
 INDEX 
 
 Right Angle, 77; Right Prism, Stood on, 574. 
 
 640; Right Parallelopiped, 643. 
 Root, 508. 
 
 Scalene, 324. 
 Scholium, 121. 
 Segment, 582. 
 Similar, 349. 
 Solid*, 3, 4, 58. 
 Square, 290. 
 Square Root, 508. 
 
 Straight Angle, 79. 
 Supplement, 119. 
 Surface, 12, 47. 
 Symparallel, 113. 
 
 Tangent, 588, 624. 
 Trapezium, 2H1. 
 Trapezoid, 281. 
 Triangles, 105, 210, 324. 
 
 Vebtex, 87. 
 
 8ION8 USED IN ADDITION TO TIIE ORDINARY 8I0N8 OP 
 ARITHMII l« . 
 
 = , it equal to. 
 4 • is not equal to. 
 <• is smaller than. 
 > * is larger than. 
 /_> /k* angle, angles 
 
 A* ^. Mugte, triangles. 
 |. is parallel to 
 
 1 1 . is symparallel with. 
 
 f I . is antiparallel to. 
 _L » is perptuditular to. 
 
 ■O", is equivalent to. 
 . ' .. therefore. 
 1 L rectangle. 
 Ji 7 • rarallelogram, or rhom- 
 boid. 
 
 Oi ®, eireuniferem e. < ire inii- 
 
 ferences. 
 O. €>. eirele, circles. 
 *""\ arc. 
 *J , square root. 
 
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 J£ORH\h. 
 
SUMMAKY OF 
 FACTS, DEFINITIONS, AND PRINCIPLES 
 
SUMMAliV OF FACTS, I >i:iIM TI< >NS, AND PRDfCD 
 
 N-.iK. — Each pupil should enter in thin Summary tin- important 
 principle- and definition*, when he learns them, expressing them in tin- 
 language that MOM most natural to himself, provided that his words 
 accurately express the ideas intended. 
 
 An index should be made, at the time of the entries, on the last Hank 
 
 paf* 
 
 Kadi teacher will hare his own opinion of the principles that he 
 wishes his particular class to become familiar with ; so that the entries 
 Ik -re ought to vary with different teachers, and also with different 
 classes under the same teacher 
 
 To assist those who have no wish to make a selection for them- 
 it is suggested that entries ought to be made, in general, after tli 
 of excr . 76, 77. 7ft, IIS, 
 
 111, 11'., 11H, 120, 121. 144. 14*.», 150, 156, L60, 161, 166, 17-. l.'L'. I '.»;;, 
 106, 111, IIS, 118, 114) Mft, MS, 272, 278, 297, 809, 31H, 319, M 
 ::;:;:. :::i. H ■ "-71. 581, 
 
 658, 669, 671, and 077. 
 
THIS BOOK IS DUE ON THE LAST DATE 
 STAMPED BELOW 
 
 AN INITIAL FINE OF 25 CENTS 
 
 WILL BE ASSESSED FOR FAILURE TO RETURN 
 THIS BOOK ON THE DATE DUE. THE PENALTY 
 WILL INCREASE TO SO CENTS ON THE FOURTH 
 DAY AND TO $1.00 ON THE SEVENTH DAY 
 OVERDUE. 
 
 OCT 11 1937 v ... 
 
 
 
 
 SjMPW 
 
 
 
 AUG 29 1963 
 
 
 mil 
 
 
 
 
 rftc v L-f 
 
 
 0CT2 3'63-9AM 
 
 
 
 
 JUL 231992 
 
 
 3v5! it 
 
 
 utobiSCCIRC OCT ; 
 
 .4'93 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 LD 21-95m 7,37 
 
YB 05206 
 
 III 
 
 U.C. BERKELEY LIBRARIES 
 
 C0312MAQ7E 
 
 
 183979