ra MI \ Irving 2MOEIAM Stringham GEOMETRY FOR BEGINNERS. BY G. A. HILL, A.M. BOSTON: PUBLISHED BY GINN, HEATH, & CO. 1884. VV Entered according to Act of Congress, in the year 1880, by GEORGE A. HILL, in the office of Jie Librarian of Congress, at Washington. J. S. CUSHING & Co., PKINTERS, 101 PKAKL STREET, BOSTON. PREFACE. ' I **HE formal method of teaching Geometry, as we find it in Euclid JL and in the common text-books, whatever may be its merits, is a very bad method for beginners. It makes the study of the subject unnecessarily dry, tedious, and difficult ; and it ignores, in most cases, the great law of mental development, that clear perceptions and intui- tions must precede the intelligent use of the faculties of comparison and reasoning. The usual consequence is that this method exercises the memory more than the intelligence, and utterly fails to impart any habits of thought that are useful in after-life. In the present work, a method is employed which seems to the author much better suited to promote the natural growth of the mental powers. The method will speak for itself to those who will take the trouble to examine it. A marked feature of it consists in the numerous exercises which are to be worked by the learner. By this kind of work, if the exercises are well chosen and faithfully studied, the inventive faculty is called into play, and the habit of seizing quickly the true relations of things is acquired. And it is just here, in its power to form this habit of mind, that the value of the study of Geometry, as a prepar- ation for the varied duties and labors of life, can hardly be over-esti- mated. Does not the possession or want of this habit constitute, in most instances, the chief intellectual difference between a man who succeeds in the world and a man who does not ? In Germany the true worth of Geometry as an educational means is better recognized than elsewhere; and the simpler parts of the subject, 800542 IV PREFACE. treated much as in this work, have been introduced with the happiest results into the common schools. The author, while residing in Ger- many in 1877 and 1878, made himself familiar with their methods and text-books, and he has freely used the knowledge thus acquired in writing the present work. As regards the subjects treated in the last foui 1 chapters, the author was not satisfied with what he found, even in the best German text- books, and he has planned and written these chapters according to his own ideas. To make the study both more interesting and more useful, much attention has been given to the practical uses of Geometry ; such, for example, as measuring inaccessible distances, and computing lengths, areas, and volumes. In a few cases the method of Limits has been used. By this means rigor of proof has been secured ; and when the method is presented in its simplest form (as on page 230) , the author believes that it is quite within the comprehension of the average boy fifteen years old. If answers to the exercises are generally desired they will soon be published. The author desires to express his obligations to Prof. G. A. WENT- WORTH, who has kindly read the proof-sheets, and furnished him with many valuable suggestions. G. A. HILL. CAMBRIDGE, June i, 1880. CONTENTS. PAGE. CHAPTER I. INTRODUCTION 1 I. Space ( i). II. The -Cube ( 2-5). III. The Cylinder ( 6). IV. Bodies, Surfaces, Lines, and Points ( 7-15). V. Generation of Space Magnitudes ( 16). VI. Straight and Curved Lines ( 17). VII. Plane and Curved Surfaces ( 18). VIII. Cornered and Curved Bodies ($ 19). IX. Geometry ( 20). CHAPTER II. STRAIGHT LINES 17 I. Direction of One Line ($ 21-25). H. Change of Direction. The Circle ( 26-28). III. Directions of Two Lines ( 29, 30). IV. Di- rections of Three Lines ( 31). V. Length of a Line ( 32-36). VI. Ra- tio of Two Lines ( 37, 38). VII. Units of Length ( 39-41). VIII. Measuring a Line ( 42, 43). CHAPTER III. ANGLES 51 I. Definition of an Angle ( 44). II. Magnitude of an Angle ( 45, 46). III. Magnitudes of Particular Angles ( 47-49). IV. Measure of an Angle ( 50-52). V. Angles made by Two Lines ( 53, 54). VI. Angles made by Three Lines ( 55-58). CHAPTER IV. TRIANGLES 73 I. Sides of a Triangle ($$ 59-63). II. Angles of a Triangle ( 64-66). III. Similarity, Equivalence, and Equality ($ 67, 68). IV. Equal Tri- angles ( 69-79). V. Some Consequences of the Equality of Triangles ( 80-93). VI. Applications ( 94-98). CHAPTER V. QUADRILATERALS Ill I. Sides and Angles of a Quadrilateral ( 99, 100). II. Different Kinds of Quadrilaterals ( 101-105). III. Construction of Quadrilaterals ( 106-111). IV. Subdivision of a Line ( 112-114). VI CONTENTS. PAGE. CHAPTER VI. POLYGONS 123 I. Sides and Angles of a Polygon ($ 115-117). II. Regular Polygons ( 118-122). III. Symmetrical Figures ( 123). CHAPTER VII. AREAS 136 I. Units of Area ( 124, 125). II. The Areas of Polygons ( 126-135). III. Practical Exercises and Applications ( 136-141). IV. Theorem of Pythagoras ($ 142, 143). V. Transformation of Figures ( 144-150). VI. Partition of Figures ( 151-155). CHAPTER VIII. SIMILAR FIGURES 171 I. Proportional Lines and Figures ($ 156-158). II. Similar Triangles ( 159-168). III. Problems and Applications ( 169-176). IV. Simi- lar Polygons ( 177-179). CHAPTER IX. THE CIRCLE 199 I. Sectors, Angles at the Centre, Chords, Segments ( 180-184). II. In- scribed Angles ( 185-188). III. Secants and Tangents ( 189-192). IV. Two Circles ($ 193, 194). V. Inscribed and Circumscribed Fig- ures ( 195-207). VI. Length of a Circumference ( 208-211). VII. Area of a Circle ( 212-218). CHAPTER X. THE ELLIPSE 24O I. The Properties of the Ellipse ( 219-221). II. Construction of Ellipses ( 222-224). CHAPTER XL PLANES 248 I. Straight Lines in Space ( 225-227). II. A Plane ( 228-230). III. A Plane and a Straight Line ( 231-235). IV. Two Planes ( 236-239). V. Three Planes ( 240). VI. Solid Angles ( 241). CHAPTER XII. GEOMETRICAL BODIES 263 I. Polyhedrons ( 242, 243). II. The Prism, Cylinder, Pyramid, and Cone ( 244-250). III. The Sphere ( 251, 252). CHAPTER XIII. SURFACES AND VOLUMES OF BODIES . 279 I. Surfaces of Bodies ($ 253-258). II. Volumes of Bodies ( 259- 274). III. Exercises and Applications ( 275-279). GEOMETRY FOR BEGINNERS. CHAPTER I. INTRODUCTION. CONTENTS. I. Space ($ i). II. The Cube ($ 2-5). III. The Cylinder ( 6). IV. Bodies, Surfaces, Lines, and Points ( 7-15). V. Generation of Space Magnitudes ( 16). VI. Straight and Curved Lines ($ 17). VII. Plane and Curved Surfaces ( 18). VIII. Cornered and Curved Bodies ( 19). IX. Geom- etry ( 20). L Space. 1. We cannot define space. We know, however, that all things exist in space. I lay a book on the table : it has now a definite position in space on the table. The table is in the space occupied by the room ; the room is in the space enclosed by the house ; the house rests upon the earth ; and the earth is always moving swiftly through space. Thus all things are in space ; but what space is, where it begins, or where it ends, these are questions which no man can answer. Space contains all things, and extends in all directions, without beginning and without end. II. The Cube. 1 2. The Cube {fig. /) occupies a portion of space which is limited on all sides. A limited portion of space is called a BODY (see 7) . Therefore the cube is a body. 1 A model of the cube is supposed to rest on the table or other support, with one face turned towards the eye of the learner. 2 GEOMETRY FOR BEGINNERS. [is- The cube is extended in space in three chief directions : from right 'to left, from front to ^^/(% from 28 GEOMETRY FOR BEGINNERS. [ 3- and at last meet in a point, (9, called their intersection ( 12). Two edges of a table which meet at a corner are an example of lines having different directions ; the corner is their intersec- tion. Exercises. 1. Which lines on the blackboard are parallel? which not parallel? Prolong the latter until they meet. 2. Can two lines, not parallel, intersect in more than one point? 3. How do you read the expression AB || CD? 4. Hold two pencils (#) parallel; (b~) so that they would intersect; (V) so that they are neither parallel nor would they intersect. 5. Hold two books (a) parallel ; (U} not parallel. 6. Hold a pencil and a book (#) parallel ; (b*} not parallel. 7. Point out on the cube (a) parallel edges ; () intersecting edges : (^ edges neither parallel nor intersecting. 8. Examine, in the same way, the edges of the prism and the pyramid. 9. Which edges of the room are parallel to the floor? 10. Give another instance of parallel lines; of parallel planes ; of a line parallel to a plane. 11. Are two vertical lines parallel to each other? Ans. Strictly speaking, No; because, owing to the round shape of the earth, two vertical lines intersect at a point near the centre of the earth. But the dis- tance to the earth's centre (nearly four thousand miles) is so great compared with the distance apart of two vertical lines, such as the edges of a house, etc., that the latter are always regarded as parallel. 12. Are horizontal lines always parallel to each other? 13. Can two vertical planes intersect each other? 14. Can two horizontal planes intersect each other? 15. Give an instance of parallel lines which are vertical ; horizontal ; inclined. 30. The simplest way to draw lines parallel to each other (especially when several are desired) is illustrated in Fig. 26. Besides a ruler, a piece of wood with three smooth, straight edges called a SQUARE, is needed. Keep the ruler at rest, slide the square along the edge of the ruler, and, in different positions of the square, draw lines along its edge ; these lines will he parallel to each other. To draw a line through the point J/, parallel to the 3'-] CHAFl'ER II. STRAIGHT LINES. 29 line P Q, place the ruler and square as shown in the figure, then slide the square along till its edge just touches M \ the line drawn through M t along the edge of the square, will be the parallel required. NOTE. How parallel lines are drawn with ruler and compasses will be explained later. (See $ 78). Exercises. 1. Drawastraight line ; and then, through a point not on the line, a parallel to it. Pig- 26. 2. Draw a straight line, and, through a point on the line, a parallel. 3. Draw five parallel lines about equally distant from each other. 4. Draw free-hand several parallel lines. Test and correct them with ruler and square. IV. Directions of Three Lines. 31. Three straight lines in the same plane present, as regards direction, four cases, (a) The three lines are all parallel ; (b) Two of the lines are parallel ; (c) No two lines are parallel, and all intersect in one point ; (//) No two lines are parallel, and they do not all intersect in a point. These cases are all illustrated in Fig. 27. When necessary, the \ \ (a) \ Fig. 27. lines must be prolonged until they intersect. In case (//) the lines (prolonged, if necessary) intersect in three points, and com- 30 GEOMETRY FOR BEGINNERS. [ 3 2 - pletely enclose a portion of a plane surface, which is called a TRIANGLE. NOTE. The properties of the triangle will be studied in Chapter IV. Exercises. 1. Draw any three lines, and then prolong them (by dotted lines) till they intersect. 2. In how many points can four lines intersect one another? (a) when all four lines are parallel ? () when three are parallel ? (r) when two are par- allel ? ( equal to the line CD. E F The sign = is called the SIGN OF EQUALITY. Fig. 31. Expression of Inequality. The lines AB and E F {Fig. ji) are unequal lines. AB is greater than E F, or EF is less than- A B. In place of the words in italics, we often employ the signs > and <, and write : AB>EF, and EF; c>a; x a, d< b. 7. Draw any four lines, m, n, o, p ; then test their lengths, and write the results with the proper signs (=, or >, or <). 8. Try to draw, free-hand, two equal lines, one horizontal, the other ver- tical ; then test their equality. 9. Draw two lines, each longer than the greatest opening of the dividers. How can you test whether they are equal or not? 35. In testing, with the dividers, the lengths of two lines, the third line with which we compare each of them is the distance be- tween the points of the dividers. We assume that if the two lines are each equal to this distance, they must be equal to each other. This assumption is so very obvious that it almost escapes notice. 36.] CHAPTER II. STRAIGHT LINES. 33 If two lines or any two magnitudes are each equal to a third, of course they are equal to each other. In fact, tjis is a truth so simple in its nature that it cannot be made clearer by trying to prove it. Truths of this kind that is to say, setf-evitftnttniths are called AXIOMS. Besides the axiom here under consideration, there are four others often useful in Geometry. We shall now state them all, and shall refer to them hereafter as here numbered. Axioms. I. If two magnitudes are each equal to a third, they are equal to each other. II. If equals are added to equals, the sums are equal. III. If equals are subtracted from equals, the remainders are equal. IV. If equals are multiplied by equals, the products are equal. V. If equals are divided by equals, the quotients are equal. 36. The length of a line (like its direction) is susceptible of change. We may suppose it either to increase or to diminish. Prolong the line AB (Fig. 32} to any point C ; then the line A C is equal to the sum of the lines < A B and B C ; here we may employ A *~~ ~~ to advantage the sign + of addition, and write, A C= AB + B C. Conversely, the line A B is the difference between the lines A C and B C ; or we may use the sign of subtraction, and write, AB = AC-BC. Prolong a line A B (Fig. 33} by adding to it repeatedly (by means of the dividers) lines B C, CD, D E, etc., each equal to A B. Then A C is twice as long as A B, A D is three times as long, A E four times as long, etc. That is, we have multiplied the 34 GEOMETRY FOR BEGINNERS. [ 36. line A B by 2, 3, 4, etc. It is usual to omit the sign x of multi- plication, and write the results thus : AC=2AB,AD=$AB, A E = 4 AB, etc. Also Conversely, a line can always be divided into equal parts. In the above line, for example, A B is one-half of A C, one third of A D, one-fourth of A E, etc. ; in other words, 234 3 These results may be summed up in the following general truth : Lines, like numbers, may be added, subtracted, multiplied^ and di- vided. Remark. Observe, however, that we have only multiplied and divided a line by a number, and that the result has always been another line. If, for instance, we multiply the line AB by the number 10, we simply repeat the line a certain number of times, and the result is the line A L ; or, writing the operation in the form of an equation, 10 x AB = AL. In the converse operation of division there are two cases : - (a) We may divide the line ALty the number 10, obtaining as the quotient the line A B. In this case we divide the line into a certain number of equal parts. (b) Or, we may divide the line A L by the .line AB, and now our quotient will be the number 10. In this case we find how many times one line is contained in the other line. In this sense division is the comparison, as regards magnitude, of two things of the same kind, and the quotient is their relative magnitude or ratio. 37-] CHAPTER II. STRAIGHT LINES. 35 Exercises. 1. In Fig. 33, what line is equal to BD + DE? toAE A D ? to three times the sum of A B and B C ? to four times the difference of A Cand CD? NOTE. Perform the following exercises with ruler and dividers, and not with the aid of the divisions of a divided rule. 2. Draw two lines ; then draw lines equal (a) to their sum, (3) to their difference. 3. Drw A = CD, and MNPQ; then draw lines equal to A B + MN, CD + PQ,AB MN, CD PQ. Why are the first "two of these lines equal? Why are the last two also equal? 4. Draw a line three times as long as your pencil. 5. Draw a line, A B, and prolong it to a point, P, so that AP= $A B. 6. Explain how to repeat or multiply a line any number of times. 7. Draw a line equal in length to the sum of the edges which bound one face of this cube. 8. Draw a line ; then divide it into two, four, and eight equal parts. NOTE. This is to be done here by repeated trial with the dividers. 9. Divide a line, free-hand, into five equal parts. Test and correct with the dividers. Repeat this exercise several times. VI. Ratio of Two Lines. 37. Definitions. I. If a line is contained in another line one or more times without a remainder, the first line is called a MEASURE, or ALIQUOT PART, of the second ; and the second line is called a MULTIPLE of the first. II. A COMMON MEASURE of two or more lines is a line which is exactly contained in each of them. III. The RATIO of two numbers is the quotient obtained by di- viding one number by the other ; in other words, it expresses how often one of the numbers will contain the other. IV. The RATIO of frvo lines is the number of times one of the lines will contain the other ; it is equal to the ratio of the numbers which express how often a common measure is contained in each of the lines. V. Of the two terms of a ratio, the first is called the ANTECEDENT, the second is called the CONSEQUENT. 36 GEOMETRY FOR BEGINNERS. [ 38. For example : in Fig. 34, where AB = B C = CD D E E F, the line AB is a measure of AB, of A C, of CF, in short, of every line represented in the figure ; therefore A B is a common measure of all these lines. On the other hand, all the lines in the figure are multiples of A B. Fig- 34- If we compare the two lines A B and A F, we see that their common measure A B is contained once in A B and five times in A F; hence, the lines A B and A F are to each other as the num- bers i and 5 ; that is, they have the ratio 1:5. By expressing this ratio as a fraction, we obtain the equation, AB _ i X A B _ i ~AF ~ 5 x AB ~ 5 ' The reciprocal ratio of AF to AB is that of 5 to i, or simply 5. NOTE. In common language, instead of using the term ratio, we say, A B is one-fifth as long as AF; or, reciprocally, A F"\s five times as long as A B. In like manner, it is easy to see from Fig. 34 that The lines A B and A C have the ratio 1:2; 2:5; AE " " 3:4; etc. In the form of fractions, these ratios would be written, ^J?_i- ^_*_ BD ! - 2 ^_1 etc A C~ 2' Z^~7~ ' AF~ $ ' AE~ 4' Exercise. What is the ratio of A D and A B (Fig. 34) ? AD and A C? A D and AD? A D and A E ? AD and A F? A Cand A E? A E and . / F? Write the values both with dots (: ) and as fractions. In these ratios, which term is the antecedent? which the consequent? 38. In order to find the ratio of two lines, apply the shorter line to the longer line as often as possible ; if the shorter line is a ' 3 8.] CHAPTER II. STRAIGHT LINES. 37 measure of the longer, if, for example, it is contained in the longer exactly five times, then i : 5 is the ratio of the shorter line to the longer. If, however, the shorter line is not a measure of the longer, the problem becomes more difficult. After we have applied the shorter line to the other line as many times as possible, there is a remainder less than the shorter line. For example : in the case of the lines MNand PQ (Fig. 35), when MNhas been applied twice from P towards Q, there remains the portion R Q of the longer line. To meet this difficulty, apply R Q to M ' N as often as possible ; then apply the new remainder SjVto R Q, which contains it ex- actly twice. This series of operations is precisely like the process of finding the greatest common divisor of two numbers ; it is, in fact, the method of finding the greatest common measure of two lines. P R Fig- 35- From what precedes we see that JR Q = 2 SN; MN= 3 R Q + SN= 6 SN + SN= 7 P Q = 2MN+ RQ = I46W + 2 SN hence P Q - l6 SN - l6 and MN ^ SN ---> [= S N is the greatest common measure of the two lines ; the longer contains it sixteen times, the other seven times ; the ratio of the lines, therefore, is that of the numbers 16 and 7, either 16 : 7 or 7 : 16. 38 GEOMETRY FOR BEGINNERS. [ 38. It may happen that we cannot find a common measure of two lines by this process, for the reason that the lines have no common measure ; such lines are called INCOMMENSURABLE LINES. Remark. The exact ratio of two incommensurable lines can- not be expressed by numbers ; but by dividing one of the lines into a great number of equal parts, and applying one of these parts to the other line as often as possible, and neglecting the remainder (which will be very small compared with either of the lines) we are able to approximate very nearly to the true value of the ratio. If, for instance, we divide one of two lines into 1000 equal parts, and find that the other line contains more than 600, but less than 60 1, of these parts, we know that the true value of the ratio lies between $$#, and $$j\j. For nearly all purposes we should assume that the ratio was r 6 o(yo> or I \ m other words, we should assume that one line was three-fifths as long as the other. Exercises. 1. In Fig. 36, of what lines is MN a common measure? How many times is it contained in each of them? Of what lines is PQ (=MN) a common measure? How many times is it contained in each? 2. Express the values of the other lines in terms of P Q. D Fig. 36. 3. How many times is i P Q contained in each line in Fig. 36 ? Is it a common measure of all of them? If we divide P Q into three, four, five, etc., equal parts, will one of the parts be a common measure of all the lines? 4. Of what lines is 2 PQ a common measure? 2 MN? 5. What is the greatest common measure of MN and A B? of A B and CD? of MN, ABvn&CD? of MN, A B, CD and 39-] CHAPTER II. STRAIGHT LINES. 39 6. Find the ratio of P Q to each of the other lines in Fig. 36, and write the results both fractionally and with dots. 7. Find the ratio of M ' N to all the other lines, and write the results both fractionally and with dots. 8. Find the ratio of A B to the other lines, and write out the results. 9. What is the ratio of CD to E F? What is the reciprocal ratio? 10. Draw any two lines, and then find their ratio. VII. Units of Length. 39. In comparing the lengths of lines with one another a great amount of time and trouble is saved by employing units of length. Definitions. LA UNIT of length is a known length which men agree to employ as a common term of comparison, or common measure for all lines. Units of length receive individual names, as the foot, the yard, the meter, the mile, etc. 11. To MEASURE a line is to find how often it will contain a unit of length. In other words, to measure a line is to find its ratio to the unit of length. III. The number of times a line contains a unit of length is the NUMERICAL VALUE of the line with reference to that unit; joined to the name of the unit, the whole expression is called the LENGTH of the line. Examples. 6 feet, 9 meters, 12 miles. NOTE. In olden times, units of length were usually taken from some part of the human body. The word foot indicates its origin. The cubit was the length of the forearm, from the elbow to the end of the middle finger. The first attempt to es- tablish a standard yard in England was in 1120, in the reign of King Henry I., who ordered that the yard, the ancient ell (Latin ulna, elbow, arm), should be of the exact length of his own arm. At the present day, horses are described as so many " hands high," and a place is sometimes said to be so many "paces distant." The span, the palm, and \htfathom all derive their origin from the human body; the fathom was the distance between the extreme points of the outstretched arms. But the units of length now recognized by law among civilized nations are defined by the State with the utmost precision, by reference to a standard made of plati- num or other hard metal, which suffers no sensible change from age to age. This standard is kept at the capital of the nation, and carefully made copies of it ar? preserved in public buildings in the chief cities where the unit is employed. 40 GEOMETRY FOR BEGINNERS. [ 40 40. The units of length established by law, and used for mosl purposes in the United States and in Great Britain, are the INCH, the FOOT, the YARD, the ROD, the MILE. 12 inches (in.) = i foot (ft.). 5^ yards = i rod. 3 feet = i yard (yd.). 320 rods = i mile. NOTE. The geographical or sea mile is one-sixtieth of a degree of a meridian of the earth and is equal to 1.15 common or statute miles, very nearly. Exercise. How many inches in one yard ? in one rod ? in one mile ? 41. The units of length now employed by the people of most civilized nations, and by the men of science of all nations, are called METRIC units, from the name of the fundamental unit, the meter. 1 They form, taken together, a part of the Metric System of Weights and Measures, first adopted by the French just after the great revolution of 1792. Definition. The METER the fundamental unit from which all other units in the metric system are derived is defined by law as the length of a certain platinum bar kept at Paris. NOTES. i. The nations which have adopted the metric system take from the Paris meter the most perfect copies which can be made, and in the capital of each nation are preserved a set of metric measures and weights with which those used by the people must agree. 2. The founders of the metric system defined the meter as one ten-millionth part of the distance from the Equator to the North Pole. Extraordinary efforts were made to construct a platinum bar which should have this exact value. An arc of a meridian passing through France and Spain was measured with extreme care, and by combining the results with other known facts the distance from the Equator to the Pole was calculated. The platinum bar based on these results was supposed at the time to be exactly what it was denned to be. But not long afterwards errors in the measurements were discovered, and it is now known that the Paris meter (the platinum bar) is less than its intended value by an amount not exceeding ^ of the whole. The other units of length are multiples and divisions of the meter as follows : Ten meters are called one DEKAMETER. One hundred meters are called one HEKTOMETER. 1 The word meter comes from a Greek word meaning measure, and is pro- nounced me6-ter. The French write the word mttre. 41.] CHAPTER II. STRAIGHT LINES. 41 One thousand meters are called one KILOMETER. One tenth of a meter is called one DECIMETER. One hundredth of a meter is called a CENTIMETER. One thousandth of a meter is called a MILLIMETER. NOTES. i. Ten thousand meters is also called a MYRIAMETER.. 2. The names of the multiples de&ameter, hectometer, kilometer, myn'ameter are formed by prefixing to the word meter the Greek numerals deka (io) P hecto (100), kilo (1000), and myria (10000) ; for the divisions of the metre, the Latin numep als, deci, centi, and milli are used. Abbreviations. Meter = m ; dekameter = dkm ; hecto- meter = hm ; kilometer = km ; decimeter = dm ; centimeter =* cm ; millimeter mm. TABULAR VIEW. km. hm. dkm. m. din. cm. mm. I = 10 = IOO = IOOO = IOOOO =IOOOOO = IOOOOOO I = 10 = IOO = IOOO = IOOOO = IOOOOO I = IO = IOO = IOOO = IOOOO I = 10 = IOO = IOOO I ' = IO = IOO I IO This table shows that a given number of units of one value are reduced to units of the next lower value by multiplying by 10, and to units of the next higher value by dividing by 10. Examples. 1. Reduce 32 meters to decimeters, and also to dekameters. Ans. 32O dm ; 3.2 dkm - 2. Reduce 0.764 of a kilometer to meters. Solution. Since we wish to reduce kilometers to the third lower unit, we must multiply by 10 three times. As in decimal fractions, this is done by moving the decimal point three places to the right. Ans. 764- 3. Reduce 376 millimeters to kilometers. Solution. Here we have to reduce millimeters to the sixth higher unit; there- fore, we must divide by 10 six times. That is, move the decimal point six places to the left. Ans. 0.000376^", Remarks. i. The units most used are, the kilometer (for long distances), the meter, and the centimeter and millimeter (for lengths less than one meter) . 42 GEOMETRY FOR BEGINNERS. [41. 2. In practice, only one unit is employed to express a length. We say 76" or 76o mm (not y dm 6 cm ) ; 332 (not 3 hm 3 dkm 2 m , and not 33 dkm 2 m ), etc. 3. In reading infraction of a unit, the word signifying the num- ber of decimal places may be omitted ; thus, 4.5 may be read : four meters five; 4.5 2 m may be read : four meters fifty-two, etc. This is analogous to the common way of reading money; $2.75, for instance, is read : two dollars seventy-five (or, yet more sim- ply : two seventy-five) . In the case of money, the omitted word is cents ; with metric units it is tenths, hundredths, thousandths, etc., according to the number of decimal places. 4. In adding, subtracting, etc., metric units, they must first all be reduced to a common unit, as, for example, the meter. Thus, 6.4 m -f- 27 cm = 6.4 m f 0.27 = 6.67 m - COMPARISON OF METRIC AND ENGLISH UNITS. kilometer = 0.62137 miles (about |- of a mile), meter = 1.0936 yards (about 39-3- inches), centimeter = 0.3937 inch (about f- of an inch), millimeter = 0.0394 inch (about ^V of an inch). mile =1609 meters (about if- kilometers), yard = 0.9144 meter (about -JT meter), foot = 0.3048 meter (about 30 centimeters). inch = 2.54 centimeters (about 2^ centimeters). NOTE. It might be well for the pupil to learn by heart the approximate values (not the others) just given in parentheses; but the best way to become familiar with the values of the meter and its divisions is by comparing a meterstick with a yardstick, by comparing the divisions on a divided rule, and by solving exercises like the following. Exercises. 1. Give the metric units of lengths thus: 10 millimeters make I centimeter, 10 centimeters make I decimeter, etc. 2. Write a tabular view of the metric units of length. 3. Write with abreviations : 8 meters; 72 kilometers; 324.6 dekameters; $7.34 hectometers; 9.27 centimeters; 4.2 decimeters; 13 millimeters. 4I-] CHAPTER II. STRAIGHT LINES. 43 4. Read the following : 37.5 ; 947 ; 846.7 cm ; l8 dm ; 9.O4 km ; 45.2o6 hm ; 77-96 dkm . 5. Reduce SS44 mm to meters and to kitometers. 6. Reduce 3-27 km to meters and to millimeters. 7. Reduce I2.4 hm to all the lower units. 8. Reduce i8o cm to all the higher units. 9. What part of a kilometer is i m ? i cm ? i mm ? 10. 25 cm are what part of a meter? also what part of a kilometer? 11. 24.32 m + 6 dm 4- i8 dm -f- 26o cm = how many meters? 12. 4 dkm 4- 4 m 4- 4 dm 4" 4 cm 4- 4 mm = how many meters? 13. 6 km 4- 6 hm 4- 6 dkm 4- 6 m 4- 6 dm = how many kilometers? 14. A shopkeeper sells ribbon as follows : to the first customer 8 m , to the second 7O Cm , to the third 4.5, to the fourth 2 dm , to the fifth 8o cm ; how many meters does he sell in all? 15. What is- the difference in length of two nails, one of which is 4 cm , the other 28 mm long? 16. From io m take io mm . 17. From ioo km take ioo mm . 18. 87 cm X 36 = how many meters? 19. 42.64 m X 80 = how many kilometers? 20. If the post-office is 900 from my house, and I go to it twice a day for letters, how many kilometers do I walk in a week? 21. I receive an order for 20 rulers, each to be 3O cm long; what length of wood is required? 22. How many rails each 7.5 m long are required to build a track 3oo km long? 23. How long would it take an express train going at the average rate of 4O km an hour to travel from Boston to San Francisco, a distance of 55OO km ? 24. If i m of silk costs $3.00, what costs i dm ? i cm ? i mm ? i dkm ? i hm ? i km ? 25. Reduce the following values to English equivalents : velocity of an ex- press train = 6o km an hour, height of Mt. Blanc = 481 5, usual height of a barometer = j6 cm . 26. Reduce the following values to metric equivalents : velocity of light = 190,000 miles a second; distance from Boston to New York = 236 miles; average height of a man = 5 feet 8 inches. 27. Which is greater, 3 km or 2 miles? 3 dm or I foot? 28. What unit would you use to express the distance from New York to Chicago? the height of a mountain? the length of a room? the dimensions of a sheet of paper? the thickness of a soap-bubble? 44 GEOMETRY FOR BEGINNERS. [42. 29. Why is the Metric system also called a decimal system? 30. Which do you consider preferable, metric units or the foot, yard, etc.? VIII. Measuring a Line. 42. In order to measure a line various instruments are em- ployed. Among them we may mention the Divided Rule, the Yardstick (or Meterstick), the Tape-Rule or Roulette, and the Surveyor's Chain. Lines on paper are generally measured by placing beside them a divided rule. A very useful kind of divided rule is a rule one foot long, divided on one side into inches and parts of an inch, on the other into decimeters, centimeters, and millimeters. On the blackboard a yardstick or meterstick is usually employed. Longer lines, such as the length of a room or of a garden, can be more rapidly measured by the help of a tape-rule or roulttte. These are made of cloth or flexible steel, wound around an axis, and enclosed in a small case suitable for the pocket. They are made of various lengths up to one hundred feet. In measuring lines on the ground surveyors employ a chain sixty-six feet long, containing one hundred links, or a steel tape one hundred feet long. French surveyors use a chain one dekameter long, containing fifty links. If a chain or steel tape is not at hand, a rope divided by knots into yards or meters may be employed. In the absence of any better means of measuring a line, pacing must be resorted to. The average value of a pace must be found by counting how many paces are made in walking a certain known distance, as ten meters. Pacing is not a very exact way of meas- uring a line, but it is a rapid way, and is always at our disposal. In many cases a line is not measured directly, but its length is computed from lines of known length by methods which will be described hereafter. Indeed, it often happens that the direct measurement of a line is impossible ; for example, the breadth of a river, the height of a mountain, the diameter of the earth, and, in 43 -J CHAPTER II. STRAIGHT LINES. 45 general, the distance of an inaccessible object (as the sun or the moon) . Exercises. 1. Make a line 3 dm long, and divide it into centimeters. 2. Draw lines with lengths as follows : i6 cm ; o.6 dm ; I75 mm . 3. Draw lines with lengths as follows : j6 cm ; I4 dm ; i.36 m . 4. Draw, free-hand, a line 3 dm long. Measure it with the divided rule, and find what error you have made. Repeat the exercise three times. 5. Measure, in metric units, with the divided rule, the following lines, and write out the results: () Length, breadth, and thickness of a book; (<) length and breadth of your desk; ( a distance of 48 meters by a line -^ m = 9.6 cm long ; a length of 66.5 meters by a line |^(f m = 13.3'" long, etc. And, in general, the reduced length would be found by dividing the real length by 500. Conversely, the real length of a line would be found from the reduced length by multiplying the reduced length by 500. Thus, & line 8 centimeters long on the paper, would represent a line which was in reality 8 cm X 500 = 4ooo cm 40 long. Scales are named by expressing the ratio of reduction, either fractionally, thus, T ^ TU5F T5&4'S> 3T68tf> F3&60- 23. Which scale reduces lines the most : a scale of I inch to i mile, or a scale of I centimeter to I kilometer? 24. If the scale to which a map is made is not known, how can it be found? 440 CHAPTER III. ANGLES. 51 CHAPTER III. ANGLES. CONTENTS. I. .Definition of an Angle (44). II. Magnitude of an Angle (45,46). III. Magnitudes of Particular Angles ( 47-49). IV. Measure of an Angle ($ 50-52). V. Angles made by Two Lines ( 53, 54). VI. Angles made by Three Lines ( 55-58). L Definition of an Angle. 44. When we open the legs of the compasses (Fig. J/) they point in different directions : they are now said to" make an ANGLE with each other. Replace, in thought, the legs by geometrical lines A B and A C, and we have an angle as understood in Geome- try. Definition. An ANGLE is the difference in direction of two lines. The two lines are called the SIDES of the angle, and the point where they meet is called the VERTEX.! If the lines do not actually meet (see Fig. 25, p. 27), the point where they would meet if prolonged, is the vertex of the angle. An angle is named in one of three ways : (1) By a small letter placed just inside the vertex; (2) By a large letter placed just outside the vertex ; (3) By three letters, one at the vertex, the others on the sides. 1 Plural, vertices. 52 GEOMETRY FOR BEGINNERS. [45- In the last case, in reading or writing the angle, the letter at the vertex must stand between the others. For example : the angle of the lines A B and A C (Fig. 37) is called either the angle d, or the angle A, or the angle B A C or CAB. In general, the word angle need not be written ; the let- ters d, or A, or A C, or CAB are sufficient. / Exercises. 1. Name the angles drawn on the black- / board. 2. Draw four angles ; name them in different ways. 3* If (as in Fig. j 20 > 3> 40. 5 minutes? 9. How long does it take the minute-hand to describe a right angle? the hour-hand? 10. What kind of an angle does the hour-hand describe in I hour? in 3 hours? in 5 hours? in 6 hours? in 8 hours? in 9 hours? in 12 hours? 11. What kind of an angle does a vertical line make with a horizontal line? 12. Draw a right angle with one side vertical ; with one side horizontal; with one side inclined. 48. Definitions. I. A straight line is said to be PERPEN DICULAR to another straight line when it makes a right angle with z>, II. Two lines not perpendicular to each other are said to be INCLINED to each other. Examples. The lines AE and C G (Fig. 41}, the sign -f in arithmetic, a common cross, the slats of a window (Fig. 42) , are examples of perpen- dicular lines ; the lines A B and A C (Fig. 39) are inclined lines. The sign 1 is sometimes used for the word perpendicular; thus : A EL CG is read : the line A E is perpendicular to the line C G. Exercises. 1. Give examples of perpendicular lines ; of inclined lines. 2. What is the difference be- tween a perpendicular line and a Fig. 42. ' vertical line ? 49-] CHAPTER III. ANGLES. 57 3. If two lines are perpendicular to each other, and one of them is ver- tical, must the other be horizontal? 4. If two lines are perpendicular to each other, and one of them is hori- zontal, must the other be vertical ? 5. Give an example of two perpendiculars, one of which is vertical, the other horizontal. 49. Among the most common problems which the surveyor, carpenter, draughtsman, etc., have to solve, are (/.) To erect a. perpendicular at a given point of a given line. (//.) To let fall a perpendicular from a given point to a given line. On the ground, surveyors use various instruments, the simplest of which is the Surveyors' Cross {Fig. 43). It consists of a block of wood having two saw-cuts made very precisely at right angles to each other. This block is fixed on a pointed staff on which it can turn freely. To erect a perpendicular, set the cross at the point of a line where a perpendicular is wanted, then turn the block till on looking through one saw- cut you see the ends of the line ; then the other saw-cut will point out the direction of the perpen- dicular. To let fall a perpendicular to a line from some object (as the corner of a field, a tree, etc.), set the cross at a point of the line which seems to i &' 43 ~ the eye about right, then note how far from the object the perpen- dicular at this point strikes, and move the cross that distance ; re- peat this operation till the correct point is found. On paper, or the blackboard, both of the above problems are readily solved by means of the Ruler and the Square. Explain how. NOTE. The method of solving these problems on paper with the compasses will be shown in Chapter IV. Exercises. 1. Draw a straight line, and then, free-hand, (ci) erect a perpendicular at a point on the line, () let fall a perpendicular from a point not on the line. Test and correct the results by the aid of the ruler and the square. 58 GEOMETRY FOR BEGINNERS. [ 50. 2. How many perpendiculars can "be erected at a given point of a given line? How many can be let fall from a given point to a given line? 3. Draw a horizontal line, and, at five points on it, erect perpendiculars. What kind of lines are they? 4. Draw a vertical line, and, at five points on it, erect perpendiculars. What kind of lines are they? 5. Draw an inclined line, and, at five points on it, erect perpendiculars. What kind of lines are they? 6. Draw a horizontal line, erect a perpendicular, and at a point of this perpendicular erect also a perpendicular. What direction has the second per- pendicular? 7. Erect a perpendicular at a point of a line on the ground. 8. Choose a point in the yard (playground or field), and let fall a perpen- dicular to the nearest side of the yard. IV. Measure of an Angle. 50. Angles (like lines) are measured by choosing a unit, and then comparing other angles with this unit. The unit for measur- ing angles is obtained by dividing the right angle (which would be too large a unit) into ninety equal parts, and calling one of these parts a degree. The degree is subdivided into minutes and seconds. One ninetieth part of a right angle is called a DEGREE. One sixtieth part of a degree is called a MINUTE. One sixtieth part of a minute is called a SECOND. Abbreviations. Degree = ; minute = ' ; second = ". Thus, 24 1 7' 8" is read : 24 degrees 1 7 minutes 8 seconds. The values in degrees of the angles defined in 47 are as fol- lows : A right angle = 90. An acute angle < 90. A straight angle =180. An obtuse angle > 90. Three right angles 2 70. A concave angle < 1 80. Four right angles = 360. A convex angle > 180. Exercises. 1. How many seconds in i? in io? in 45? in 360? 2. Reduce 48 54' 36" to seconds. 3. Reduce 120000" to degrees. 51-] CHAPTER III. ANGLES. 59 4. Add 37 48' 35", 28 39', and 78 9' 55". 5. From 128 15' 31" take 69 42' 18". 6. Multiply 1 8 35' by 2, 3, 4, and 5. 7. Divide 72 27' by 2, 3, 4, and 5. 8. Reduce to degrees 5 R (5 right angles); 6 R ; 8 A 1 / 12 A 5 ; % R ; R; % R; & A'/ i AV i R; % R; \ R; T \> R; T V A. 9. How many right angles in 540? 900? 225? 30? 5? 10. Find the ratio between 10 and three right angles. 51. In measuring angles we take advantage of a simple rela- tion between angles and the arcs intercepted between their sides, and described with the same radius from their vertices as centres. In Fig. 44 compare the angles A O B, A O C, A O D, with the arcs A B, AC, AD, intercepted by their sides. We see that the greater the angle the greater the corresponding arc, or arc intercepted by its sides. This conclusion is general : in the same circle, the greater the angle at the centre the greater the corresponding arc, and the less the angle the less the arc. Again, if the two angles A OB and FOG are equal, the corresponding arcs A B and FG are also equal. For the angles, being equal, may be placed one on the other so that they will coincide ; and then the arcs A B and F G must also coincide, since all the points of one arc are at the same distance from the centre O as the points of the other arc. In like manner it can be shown that if the arcs A B and F G are equal, the corresponding angles at the centre, AOB and FOG, will also be equal. The same reasoning may be applied with the same result to any angles and their corresponding arcs ; hence, in general : Fig. 44. 60 GEOMETRY FOR BEGINNERS. [ 51. I. Eqtial angles at the centre of a circle intercept on the cir- cumference equal arcs. II. Equal arcs on the circumference correspond to equal angles at the centre. Accordingly, the circumference of a circle is divided into 360 equal arcs, each corresponding to an angle of i at the centre. These arcs are also called degrees, and are subdivided like the angu- lar degree into minutes and seconds. And, in practice, an angle is measured by finding how many degrees, minutes, and seconds there are in the corresponding arc, it being obvious that the num- ber of angular degrees, minutes, and seconds in an angle is the same as the number of arc degrees, minutes, and seconds in the arc described with any radius from the vertex of the angle as a centre, and intercepted between its sides. We have just said that the arc employed to measure an angle may be described with any ra- dius. With a longer radius we have, it is true, a longer arc (Fig. 45), but each division of the arc is also correspondingly increased in length, so that the number of divisions (degrees, etc.) in the entire arc remains the same as before. We are now prepared to an- swer the question : How can the equality of two angles be tested without placing one of them upon the other ? We reply : two an- gles are equal, if, when their corresponding arcs are measured (as shown in the next section), these arcs are found to contain the same number of degrees, minutes, and seconds. NOTE. How an arc can be divided into degrees, etc., is a question which must be deferred until we know more about the properties of the circle. 52.] CHAPTER ill. ANGLES. 61 Exercises. 1. Find the angle described by the hour-hand of a watch in I hour; 2 hours; 5 hours; 12 hours. 2. Find the angle described by the minute-hand of a watch in i, 5, 10, 15, 20, 30, 45 minutes. 3. In how many minutes does the minute-hand describe an angle of i? 60 ? 225? 300? 4. What angle do the hands of a watch make with each other at I o'clock? at 2, 3, etc., up to 12 o'clock? 5. The earth turns on its axis in 24 hours ; what angle does a point on the surface describe in I hour? in 6, 12, 15 hours? 52 hours? 52. On paper or the blackboard, when great accuracy is not required, angles are measured with an instrument called a PRO- TRACTOR (Fig. 46). It is a semicircle, made of paper, horn, Fig. 46. brass, or silver, the circular edge of which is divided into 180 equal parts or degrees. To measure an angle with the protractor, place the centre of the instrument on the vertex of the angle, and its zero line on one side ; then read off on the edge of the protractor the division through which the second side of the angle passes. On the ground, surveyors and engineers employ for measuring 62 GEOMETRY FOR BEGINNERS. [52. angles, costly instruments called THEODOLITES. A cheap substitute for a theodolite is shown in Fig. 4.7. It consists of two pieces of wood shaped like rulers mounted on a vertical axis, by a pin driven Fig. 47. through their exact centres. The vertical needles inserted near the end of the rulers are used for sighting. In place of the needle nearest the eye, it is better to employ a thin strip of wood, A t having a fine vertical slit ; and in place of the other needle, a ver- tical wire fixed in a light frame, B. By the help of this instrument, arid a protractor, one can mea*. ure with considerable accuracy an angle on the ground for in- stance, the angle M ON (Fig. 48). 53-] CHAPTER III. ANGLES. 63 Exercises. 1. On the blackboard are several angles. Measure each with the protractor, and write the result between its sides. 2. Draw five different angles. Estimate the value of the first in degrees; then measure it with a protractor. Proceed in like manner with each of the other angles. Give the results in 3 columns: in column I, the estimated val- ues ; in column 2, the measured values ; in column 3, the differences between the estimated and the measured values. 3. From a point on a straight line draw two lines, both on the same side of the given line ; measure each of the three angles thus formed, andadd the results. What is the sum? What ought the sum to be? 4. Through a point draw three lines ; measure the six angles thus formed about the point ; add the results. What is the sum? What ought it to be? 5. Open the legs of the compasses so that they make angles of 90; 60; 45; 30. 6. Imagine two lines drawn from your eye, one to the right-hand upper corner of the room, the other to the left-hand upper corner ; what angle would the lines make? 7 . With the aid of the protractor make an angle equal to a given angle ? 8. Draw angles equal to 20; 30; 40; 60; 90; 100; 150; 180; 15; 45; 79; 81; 142. 9. Explain how the instrument in Fig. 4.7 is constructed, and how you would proceed to measure by means of it the angle M O N'\n Fig. 48. V. Angles made by Two Lines. 53. The angles A O Cand C OB (Fig. 49) are called adjacent angles. Definition I. Two angles are ADJACENT when they have the vertex and one side common, and the other sides are opposite parts of the same straight line. _ We see from the figure that Since this equation would hold true in whatever direction the line O C was drawn, we conclude that in all cases The sum of two adjacent angles is two right angles, or 180. 64 GEOMETRY FOR BEGINNERS. [ 54. Definition II. If the sum of two angles is 180, each is called the SUPPLEMENT of the other. Adjacent angles are always supplements of each other. Exercises. 1. If one of two adjacent angles is a right angle, what is the other? 2. If one of two adjacent angles is acute, what is the other? 3. If one of two adjacent angles is obtuse, what is the other? 4. If one of two adjacent angles is known, how is the other found? 5. How is the supplement of a given angle found? 6. Find the adjacent angles of the following angles: 10; 30; 45; 75; 99; 100; 179; 15 48'; 79 13' 52". What is the supplement of each of these angles? 54. Whenever two lines cross one another there are formed about the intersection of the lines four angles, a, b, c, d (Fig. jo). Of these angles the pairs a and b, b and c, c and d> d and a are adjacent angles ; the pairs a and c 9 b and d are called vertical angles. Definition. Two angles are called Fig. co. VERTICAL angles if they have a common vertex, and the sides of one angle have opposite directions to the sides of the other angle. If we measure two vertical angles, as a and c (Fig. 50) , either with a protractor or by cutting them out on a piece of cardboard and laying them one upon the other, we shall find that they are equal. But without measuring them we can prove that they must be equal, if we bear in mind what has been said about adjacent angles in the last section. Since b is an adjacent angle to both a and c, therefore, and 55-] CHAPTER III. ANGLES. 65 Hence it follows from Axiom I. (state the axiom) that, a+b=b+c subtract b = b There remains a = c For, if equals are taken from equals, the remainders are equal (Axiom III.). Since this reasoning holds good, however the intersecting lines cut each other, we come to the general conclusion, that Two vertical angles are always equal to each other. Remark. In the above reasoning we have made use of the general truth shown in the last section ; namely, that the sum of two adjacent angles is 180 ; and also of Axioms I. and III. By reasoning upon these truths we have proved or demonstrated a new truth ; namely, that two vertical angles are equal. Geometrical truths which are capable of proof by reasoning from known truths are called THEOREMS. Exercises. 1. Draw adjacent and vertical angles and name them. 2. What is the sum of the angles a and b (Fig. 50)! Why? 3. What is the sum of the angles a, b, c, and d (Fig. jo~) ? Why? 4. One of the angles formed by two intersecting lines is 36; find the others. 5. One of the angles formed by two intersecting lines is 90 ; find the others. 6. Draw any two intersecting lines, and find the values of the angles formed at the intersection. ? If one of the angles formed by two intersecting lines is known, how are the others found? 8. Each of two vertical angles is 45; find the value of each angle in the other pair of vertical angles. 9. Prove that b = d (Fig. 50) by reasoning like that used above to show that a=c. VI. Angles made by Three Lines. 55. Thus far the angles considered have had a common ver- tex j let us now consider angles formed about two different ver- 66 GEOMETRY FOR BEGINNERS. [55- tices. Such angles are formed when two straight lines are cut by a third line. The most important case, and the only one which we shall exam- ine, is that in which two parallel lines, AB and CD (Fig. jz), are cut by a -% third line, E F. There are formed about the two points of intersection _ eight angles. These angles receive D special names. The four angles, c, d, n> m, which lie between the parallel lines, are called INTERNAL ANGLES ; the other four, a, b, 0, /, are called EXTERNAL ANGLES. An external angle and an internal angle, as a and m, having dif- ferent vertices, and lying on the same side of the intersecting line, are called CORRESPONDING ANGLES. 1 An external angle and an internal angle, as a and , having dif- ferent vertices, and lying on different sides of the intersecting line, are called CONJUGATE ANGLES. Two external angles, as a and p, or two internal angles, as d and m, having different vertices, and lying on the same side of the in- tersecting line, are called OPPOSITE ANGLES. Two external angles, as a and o, or two internal angles, as d and n, having different vertices, and lying on different sides of the in- tersecting line, are called ALTERNATE ANGLES. Exercises. 1. Name the four pairs of corresponding angles in Fig. 51 ; the four pairs of conjugate angles ; the four pairs of opposite angles ; the four pairs of alternate angles. 2. Which are external, which internal opposite angles? which are ex- ternal, which internal alternate angles? 8 Which is the angle corresponding to o? conjugate to p? opposite to m ? alternate to n ? _ 1 _ . __ . 1 They are also called external-internal angles. 56.] CHAPTER III. ANGLES. 67 4. Draw two parallel lines, and a third line intersecting them. Letter the eight angles e,f,g, h, and/, r, s, t. Then write in six vertical columns (a) the external angles ; (li) the internal angles ; (c) the corresponding angles; (c/) the conjugate angles ; (i) the opposite angles ; (/) the alternate angles. 56. If we conceive the line AB to move along the line E F, remaining always parallel to its first position, then, since its direction does not change, the four angles a, l>, c, d ; which it makes with EF do not change. When A B reaches the par- allel line CD, it will coincide with it in direction, and the four angles a, b, c, d will coincide respectively with the corresponding angles m, ;/, o, p. Moreover, each pair of conjugate angles, as a and ;/, become now adjacent angles ; therefore their sum is 1 80 ( 52) : the same is true of each pair of opposite an- gles, as a and p ; lastly, each pair of alternate angles, as a and o, become now vertical angles, so that they are equal ( 53). Hence, I. 2. 3- 4- a = m. a + n= 2 R. a + p = 2 R. a= o. b = n. b +771=2 R. b + 0=2 R. b = p. C =0. c + / = 2 R. c -f- n 2 R. c = m. d = p. d+ o = 2 R. d-\-m= 2 R. d= n. Putting these results into words, we obtain the following theo- rem : Theorem. If two parallel lines are cut by a third line, /. The corresponding angles are equal ; 2. The sum of two conjugate angles is two right angles ; 3. TJic sum of two opposite angles is two right angles ; 4.. The alternate angles are equal. 68 GEOMETRY FOR BEGINNERS. [' 57. liemark. Theorems often involve consequences which are easily deduced from the theorems : such consequences are called in _ Geometry COROLLARIES. The pres- ent theorem furnishes an example. Corollary. If a = 90 (that is, D ,toowstatw = 90 (that is, that CD-lEF) ; or, to Fig. 53. state the corollary in general terms, If a straight line is perpendicular to one of two parallels, it is also perpendicular to the other. Exercises. 1. Give the proof separately for each of the last three (equations in each of the above four series of equations. Models. b = n, because when A B coincides with CD, b coincides with n. b -J- m = 2 R, because when A B coincides with C D, b becomes adjacent to m, and the sum of two adjacent angles is 180. 2. Find the other seven angles when (7 = 112; when a = 90. 3. Draw two parallel lines, and a line intersecting them; then find the eight angles thus formed, using a protractor as little as possible. 4. If either one of the four parts in the theorem of this section is not true, what can we conclude as to the lines A B and CD? 57. Theorem. If two straight lines are cut by a third line .so that two corresponding angles are equal, then the lines must be parallel. If, for example, a = m (Fig. 52), then is AB \\ CD. For, if we move A B towards C D, keeping it parallel to its first position, during the motion the angle a does not change because the direc- tion of A B does not change. When the intersection of A B and E F coincides with the intersection of CD and F F, AB must coincide in direction with CD, since a = m. Therefore, AB, in its first position, must be parallel to CD. Corollary. By making in this theorem a = m = 90, we ob- tain the following corollary : 58.] CHAPTER III. ANGLES. 69 If two straight lines are both perpendicular to a third line, they are parallel. (Fig. 54.) Remark. All the four properties enumerated in the theorem of the last section are so related that if either one of them is true the other three must also be true, and the two intersected lines must be parallel. Hence arises the importance of corresponding, conjugate, oppo- site, and alternate angles. Before we can assert with absolute cer- tainty that two lines are parallel, we ought to show that the lines, when prolonged farther than even the imagination can reach, would not meet. Such an actual extension of the lines P R is of course impossible ; it is, however, quite un- necessary, for the parallelism of two lines is very simply tested by the angles which are formed when the lines are cut by a third line. Exercises. 1. Prove the corollary, independently of the theorem, with lines drawn and lettered as in &% 54- 2 When more than two lines are perpendicular to a r^ TJ i TV third, are they all parallel to each other? Fig. 54- 58. In Fig. 55, we have drawn AB II CD and EF \\ G H. If we compare the angle a with either of the angles x, y, *, we see that their sides are respec- tively parallel. But there are these differences : (Y.)the sides of x have the same di- rection respectively as those of a ; (it.) the sides of z have opposite directions ; -^ (///.) the sides of y have, one the same direction, the other the opposite direction, ~. . fj s 55* 70 GEOMETRY FOR BEGINNERS. In case (/.), a = b (corresponding angles on the parallels EF and G H}, and x = b (corresponding angles on the parallels C D~ and A B) : therefore a = x (Axiom I.). Theorem I. If two angles have their sides respectively par- allel, and directed the same way from the vertex, the angles are equal. In case (.), x = z (vertical angles). And a = x (Theorem I.). Therefore a = z (Axiom I.). Theorem II. If two angles have their sides respectively par- allel, and directed opposite ways from the vertex, the angles are equal. In case (in.), x + y = 2R (adjacent angles). But a = x (Theorem L). Therefore a -f- y = 2 R. Theorem III. If two angles have their sides respectively par- allel, and directed, the one pair the same way, the other pair opposite ways ', from the vertex, the sum of the angles is two right angles. Exercises. 1. In Fig. 55, if a = 115, what is the value of x? of y ? ofz? 2. Prove Theorem I., using in the proof the angle c in Fig. 55, instead of the angle b. 3. What relation exists between a and s {Fig. 55) ? Why? 4. Prove Theorems I., II., and III., with a new figure, drawn and lettered differently from Fig. 33. REVIEW OF CHAPTER III. QUESTIONS. 1. Define an angle; its sides ; its vertex. 2. How is an angle named? 3. How may an angle be conceived as produced by motion? 4. On what does the magnitude of an angle depend? 5. Define equal angles. 6. Explain (by figures) what is meant by adding, subtracting, multiplying, and dividing angles. CHAPTER III. ANGLES. 71 7 * What is bisecting an angle or a line ? What is the bisector of an angle ? . 8. Define a straight angle ; a concave angle ; a convex angle ; a right angle ; an acute angle ; an obtuse angle. 9. Show that all straight angles are equal ; also, that all right angles are equal. 10. When are two lines perpendicular to each other? when are they inclined? 11. W T hat are two of the commonest problems in surveying and drawing? 12. Describe the surveyor's cross, and how to use it. 13. How are perpendiculars to a given line drawn on paper? 14. Define a degree, a minute, and a second. 15. W 7 hat are the values in degrees of the angles mentioned in Question 8? 16. Explain why an angle can be measured by means of the arc of a circle intercepted between its sides. 17. Why is the measure of an angle the same for any radius of the arc? 18. How is the equality of two angles tested without placing one of them on the other? 19. Describe the protractor, and how to use it. 20. Describe a simple substitute for a theodolite, and how an angle on the ground is measured by means of it. 21. Define adjacent angles ; the supplement of an angle. 22. Show that the sum of two adjacent angles must be 180. 23. Define vertical angles. 24. Two vertical angles are equal. 1 25. What is a theorem ? 26. Define external angles ; internal angles ; corresponding angles ; conju- gate angles ; opposite angles ; alternate angles. 27. If a straight line intersect two parallel lines, the corresponding angles are equal, the alternate angles are equal, and the sum of two conjugate angles, or of two opposite angles, is 180. 28. What corollary follows from this theorem? 29. What is a corollary ? 30. If two lines are cut by a third line so that two corresponding angles are equal, the two lines are parallel. 31. Give a corollary of this theorem. 32. How can you test if two straight lines are parallel? 1 In the review we shall always (as here) simply state the theorem to be proved or problem to be solved The learner is to understand that the proof (or solution) is required. 72 GEOMETRY FOR BEGINNERS. 83. Two angles have their sides respectively parallel; then, (1) If these sides are directed the same way from the vertex, or if they are directed opposite ways, the angles are equal. (2) If the sides are directed, the one pair the same way, the other pair opposite ways, the sum of the angles is 180 EXERCISES. 1. What is the ratio of one right angle to three right angles? of one-third of a right angle to four right angles? 2. How can you test whether two intersecting lines are perpendicular to each other? 3. How can you test whether the two edges of a square are truly at right angles to each other? Suggestion. How was the straight edge of a ruler tested ( 23, Exercise 6)? 4. Make any angle, and then, by the aid of a protractor, make an angle four times as large ; also, an angle one-fourth as large. 5. Mark three points, join them by lines, then measure the three angles which the lines make with one another. What is their sum? 6. Can you make three angles with two lines? 7. If two lines intersect, and one of the four angles is a right angle, prove that the other three are also right angles. 8. Draw five lines meeting at a point ; how many angles are formed? what is their sum? If the angles are equal, find the value of each in degrees. 9. Ten lines meet at a point so as to form a regular ten-rayed star ; find the value of the angle between any two rays. 10. What is the supplement of i? of 179? of 180? 11. Of two adjacent angles the greater is twice the less ; find the values of the angles in degrees. What is the ratio of each to four right angles? 12. The bisectors of adjacent angles are at right angles to each other. 18. The bisector of one of two vertical angles bisects the other also. 14. If two parallel lines are cut by a third line, then (a) the bisectors of two alternate angles are parallel to each other; (U} the bisectors of two op- posite angles are perpendicular to each other. 15. What is the difference between a theorem and an axiom? between a theorem and a corollary ? 59-] CHAPTER iv. TRIANGLES. 73 CHAPTER IV. TRIANGLES. CONTENTS. I. Sides of a Triangle ( 59-63)- II. Angles of a Triangle ($$ 64-66). III. Similarity, Equivalence, and Equality ( 67, 68). IV. Equal Tri- angles ($ 69-79). V. Some Consequences of the Equality of Triangles ( 80-93). VI. Applications ( 94-98). L Sides of a Triangle. 59. In Fig. 56 are several figures enclosed on all sides by lines. Each is a portion of a plane surface (the surface of the paper) , and is called a Plane Figure. Fig. 56. Definition I. A PLANE FIGURE is a portion of a plane sur- face bounded on all sides by lines. Many surfaces with which we are familiar, for example, the out- side of our houses, the floor and ceiling of a room, the leaf of a book, and, in many cases, gardens, fields, parks, etc., furnish in- 74 GEOMETRY FOR BEGINNERS. [ 60. stances of plane figures bounded by straight lines. Such figures are called Polygons. Definition II. A POLYGON is a plane figure bounded by straight lines. The bounding lines are called the SIDES of the polygon, and their sum is called the PERIMETER of the polygon. Exercises. 1. In Fig, 56 which figures are polygons and which are not polygons? 2. How many sides and how many angles has each polygon in Fig. 56 ? 3. Draw a polygon of five sides ; six sides ; eight sides ; ten sides. How many angles has each? 60. Polygons receive different names according to the num- ber of their sides. At least three straight lines are required to enclose completely a part of a plane surface : the two sides of an angle are not sufficient. Therefore a polygon cannot have less than three sides. Definition. A polygon of three sides is called a TRIANGLE (Fig- 57)- A triangle is usually denoted by three letters standing at the vertices of its angles. Instead of writing the word triangle, the sign A is sometimes employed. Thus the triangle in Fig. 57 is called the triangle A B C, or A A B C. Every triangle has six PARTS, three sides, AB, AC, B C (Fig. 57), and three angles, A, B, and C. Each side, as A B, has two adja- cent angles, A and B, and one oppo- site angle, C ; each angle, as A, is included between two sides, A B and A C, and is opposite to the third side, B C.* Exercises. 1. What angles are adjacent to the side A C (Fig. 57) ? to the side B C ? What angle is opposite to B C? 2. By what sides is the angle B included? the angle C? What side is opposite to B? to C? 6i.] CHAPTER IV. -^TRIANGLES. 61. If we prolong one side of a triangle, the prolongation makes with the adjoining side an an- gle called an EXTERIOR ANGLE of the triangle, and the three angles of the triangle are termed, by way of dis- tinction, INTERIOR ANGLES. In Pig. j8 m is an exterior angle, Fi * 5 8 - b is the adjacent interior angle ; a and c are the opposite interior angles. Exercises. 1. Draw a triangle, and then prolong each side in both direc- tions. How many exterior angles are formed? For each exterior angle point out the adjacent interior angle, and also the opposite interior angles. 2. If an exterior angle of a triangle is 75, find the adjacent interior angle. 62. Theorem. The sum of two sides of a triangle is al- ways greater than the third side. For the straight line AB (Pig. 57) is the shortest line from AtoB ( 32), therefore less than the broken line A C B ; that is, A C + CB> AB. For the same reason it follows that A C + AB>CB, and that A B + B C> A C. Triangles are divided into three classes, according to the lengths of their sides : the EQUILATERAL (Pig. 59, I.), in which the three III. Fig- 59- sides are equal, the ISOSCELES (Pig. 59, II.), in which two of the sides are equal, and the SCALENE (Pig. 59, IH.)> in which the sides are all unequal. GEOMETRY FOR BEGINNERS. [63. Exercises. 1. Can a triangle have for its sides. 2 m , 3, and 6 m ? 8 m , 4, and 2 m ? i m , 2 m , and 3? 3, 5, and y m ? 2. If the perimeter of a triangle is 2 m , what is the greatest value which one side can have? 3. Draw, free-hand (#), an equilateral; (), an isosceles ; (c), a scalene triangle. 4. The perimeter of an equilateral triangle is 1.5; find each side. 5 Can you draw accurately with the dividers an equilateral triangle of given side, say 6o cm ? 63. We may regard a triangle as resting on one of its sides as a base ; this side is then called the BASE of the triangle ; the vertex of the angle opposite to the base is called the VERTEX of the triangle ; and the perpendicular line drawn from the vertex to the base (or base prolonged) is called the ALTITUDE of the tri- angle. If the triangle has a horizontal side, this is usually taken for the base. Thus in the triangle ABC (Fig. 60) we would naturally take the side A B for the base ; then C is the vertex, and CD is the altitude. If A C is taken as the base, B is the vertex, and B E the altitude. If B C is taken as the base, A is the vertex, and A F the altitude. D N M Fig. 60. Notice that the three altitudes intersect in one point. The altitude may be outside the triangle ; this is the case in the triangle MNO, in which MJVis the base, and OP the altitude. The altitude must fall outside if the triangle has an obtuse angle, and one of the sides of this angle is taken as the base. In an isosceles triangle, the side unequal to the other sides is always taken as the base, and the equal sides are usually referred to as the sides of the triangle. 64.] CHAPTER IV. TRIANGLES. 77 Exercises. 1. Draw a triangle having all its angles acute, and then draw the three altitudes. 2. Draw a triangle having an obtuse angle, and then draw the three altiv tudes. How many of the altitudes lie outside the triangle? 3. Draw an isosceles triangle and its altitude. 4. On a given line as base, how many isosceles triangles can be con. structed? How many equilateral triangles? II. Angles of a, Triangle. 64. In order to find the sum of the angles of a triangle ABC (Fig. 61), let us bring them all to a common vertex. For this purpose, suppose a line D E drawn through the vertex C of the tri- angle parallel to the base A B : we thereby make the angles ;;/ and ;z. The angles m and a are equal because they are alternate angles ( 56), and the angles n and b are equal for the same Fig. 61. reason. Hence, a + b-\-c=m-\-n + c (Axiom II.). But m -f- n -h c = 1 80, or a straight angle. Therefore, a + b + c = 1 80 (Axiom I.). This proof applies equally well to any triangle ; therefore, to all triangles ; hence : Theorem. The sum of the angles of a triangle is eqital to two right angles, or 180. Illustration. The truth of this theorem may be shown to the eye as follows : Cut out of paper a triangle, ABC (Fig. 62), and draw its altitude CD. Then fold over the corners on the 78 GEOMETRY FOR BEGINNERS. [65. dotted lines as edges. This will bring the vertices of the three angles to the same point D, and we shall see that the three angles together just make a straight angle. In the same way we might test one triangle after another, find- ing in each case that the sum of the angles was 180 ; after a time we should become convinced that all triangles were subject to this law. But in the proof given above a single case is sufficient ; for a little reflection shows that the same reasoning holds good for all cases. 65. From the preceding important theorem several corolla- ries follow. Corollaries. I. The sum of two angles of a triangle must always be less than 180. 2. A triangle can have only one right, or one obtuse, angle. In other words, two angles of a triangle must be acute ; the third may be either acute, right, or obtuse. If it is acute, the tri- angle is called an ACUTE TRIANGLE (Fig. 6j, I.) ; if it is right, the I. II. Fig. 63. triangle is called a RIGHT TRIANGLE (Fig. 63, II.) ; if it is obtuse, the triangle is called an OBTUSE TRIANGLE (Fig. 63, III.). In a right triangle the side opposite to the right angle is called the HYPOTENUSE, and the other sides are called the LEGS. 3. If two angles of a triangle are known, the third angle will be found by subtracting the sum of the two known angles from 1 80. 66.] CHAPTER IV. TRIANGLES. 79 4. If two angles of one triangle are equal respectively to two angles of another triangle, then the third angle of the one is equal to the third angle of the other. 5. In a right triangle the sum of the acute angles is equal to a right angle or 90. Hence, if one of the acute angles in a right triangle is known, the other can be found by subtracting the first from 90. Definition. If the sum of two angles is 90, each is called the COMPLEMENT of the other. The acute angles of a right triangle are always complements of each other. Exercises. 1. Can a triangle have one right aw^/one obtuse angle? 2. If one leg of a right triangle is taken as base, what is the altitude? 3. Two angles of a triangle are, O) 37 and 71; (V) 45 32' 18" and 62 50' 57"; () 82 and 48; O) 64 47' 33" and 77 18' 41"; (<) 40 28' and 18 57'; (/) 179 o' 54" and o 59' 5"; find the third angle in each case. 4. If one acute angle of a right triangle is (a) 30; () 45; (c] 63; ( b ; also let the angle opposite to a be 70. Make an angle CAB = 70, and lay off A C b \ this determines two corners, A and C, of the triangle. The third corner must lie in the line A B, and its distance from C must be equal to a ; therefore, it must also lie in a circumference described from C as centre with a radius equal to a. Hence it must be the intersection of this circumference with the line AB. Now this circumference cuts A B in two points, B and B' 9 so that we obtain two triangles, A B C and A B' C. Of these, however, only A B C has the three given parts; A B' C has, it is true, the given sides, but not the given angle ; therefore, it is not the triangle required. 8 74'] CHAPTER IV. TRIANGLES. 87 If we construct another triangle, M ' N O, having the same three jarts, it will be equal to the triangle ABC. Proof. Place A MN O on A A B C, making the equal sides M O and A C coincide. MN will coincide in direction with A B because OMN= CAB = 70 ; therefore N will fall in AB. But N must also lie in a circumference having (now) C for centre, and radius = a ; therefore N will coincide with B. And ON will coincide with CB, and the triangles will coincide in all their parts. Hence we see that by two sides and the angle opposite to the greater side a triangle is completely determined ; and that, Theorem (III. Law of Equality). If in two triangles two sides and the angle opposite to the greater side are equal each to each, the triangles are equal. Case (//.). Let a and b (Fig. 70) be the given sides, and let a < b ; also let the angle op- posite to a = 42. By proceeding as in Case (/.) we obtain two triangles, AB C and A B'C, both hav- ing the three given parts, yet differing in size and in shape. A '\ .-~''B Therefore, by two sides and Fi - 7> the angle opposite to the less side, a triangle is not determined. There are, however, only two solutions, the triangles ABC and AB'C. . Exercises. 1. In the triangles A BC and AJB'C {Fig. 69) point out the equal parts ; also the unequal parts. 2. What relation exists between the angles B AC and &A C (Fig. 69) ? 3. What kind of triangles are the triangles BC '' and N ON 1 (Fig. 69) ? 4. What equal parts have the triangles AC B and ACB 1 (Fig. 70) ? 5. Draw a triangle with the sides 6o cm and 9O cm , and the angle opposite the greater side 76. 6. Construct a right triangle the hypotenuse of which is 8 dm , and one leg is 6^. What are the three given parts ? GEOMETRY FOR BEGINNERS. [75- 7. Construct a triangle with the sides 8 dm and 4 dm , and the angle opposite to the greater side 80. 8. When are the data in Case ('.) of this section such that no triangle is possible? Ans. When the less side is shorter than the perpendicular from C to A B (Fig. 70) ; also when the given angle is either right or obtuse. 9. When the less side is just equal to the perpendicular from C to A B, what kind of a triangle is obtained? 75. Problem. To construct a triangle having given its three sides. Let a b c (fig. 71) be the given sides. Make A B a ; then A and B are two corners of the triangle, and A B is one side. Since the third corner must be at the distance b from A, it must lie in a circumference described from A as centre with the ra- dius b. For a like reason it must also lie in a circumfer- ence described from B as cen- tre with the radius c. 1 Therefore it must be at the intersection of these circum- ferences. Now these circum- ferences cut each other in two points, C and C', one above the other below A B ; so that we obtain two triangles, ABC and ABC', both having the three given sides. These triangles, however, are equal. For, if we fold over ABC' on A B, C' will be brought above A B, and, as it remains the inter- section of circumferences having A and B for centres, and b and c for radii, it will fall on C, and the triangles ABC and ABC 1 will then coincide. 2 1 It is obvious that in solving this problem entire circumferences are not re- quired ; short arcs near the points of intersection are sufficient. 2 See Exercise 5, page 92. Fig. 71. 76.] CHAPTER IV. TRIANGLES. 89 If we construct another triangle with the same three sides, it will also be equal to ABC. For we can place it on A B C so that one of its sides will coincide with AB, and then its third corner must fall on C for the same reason that C' falls on C when ABC* is folded over on A B. From what precedes we conclude that, by the three sides a tri- angle is completely determined ; also that, Theorem (IV. Law of Equality). If in two triangles the three sides are equal each to each, the triangles are equal. Exercises. 1. The sides of a triangular garden measure 8o m , 65, and 45 m . Draw to scale a plan of the gai'den. 2. The sides of a triangular field are 2Oo m , 240, and 300. Draw to scale a plan of the field. 3. Construct a triangle whose sides are 2O cm , 4O cm , and 6o cm . 4. Construct a triangle whose sides are 2O cm , 3O cm , and 6o cm . 5. When is the solution of the problem of this section impossible? (See 62.) 6. Construct an isosceles triangle whose base is 7O cm , and the sum of whose other sides is 2 m . 7. Construct an equilateral triangle one side of which is i m . 8. Construct a triangle with the sides 3O cm , 3O cm , and 3O CU1 . What kind of a triangle is it? 9. Construct a triangle with the sides 25 cm , 25 cm , and 40'". What kind of a triangle is it? 10. Construct a triangle with the sides 2O cm , 3O cm , and 4.o cm . What kind of a triangle is it? 11. Are two triangles equal if they have equal perimeters ? (Compare the last three exercises.) 76. Problem. To make a triangle equal to a given triangle. Take any three parts of the given triangle that completely de- termine a triangle, and construct with them a new triangle ; it will be equal to the given triangle. The best parts to take for this pur- pose are the three sides. Show hew to construct the triangle with them ( 75). 90 GEOMETRY FOR BEGINNERS. [ 77. Exercises. 1. In what four ways can three parts be chosen with which to construct a triangle? 2. Draw a triangle; then construct another equal to it. Fig. 72. 77. Problem. At a given point D (Fig. 72) of a given line D G, to make an angle equal to a given angle BAG. Take on the sides of the given angle A M = AN, and join MN. Then construct (by the last problem) a triangle D E F equal to the triangle AMN. The angle D, corres- ponding to the angle A, will be equal to it. Exercises. 1. What kind of a triangle is^ M N~! Is it necessary for the solution of the problem to make A MN a. triangle of this kind? 2. Make an angle, and then make another equal to it. 3. Make an angle equal to a given angle with j the protractor. 4 Make an angle equal to a given angle with the ruler and the square. 5. Make (with ruler and compasses) an angle of 60; an angle of 120. 78. Problem. Through a given point C (Fig. 73) to draw a line parallel to a given line A B. Through C draw any straight line CD, cutting AB in D. Then draw (by the last problem) C P so that the angle DCP = CDB. CP is parallel to AB ( 57). Exercises. 1. Draw a line, and then \ through a point draw another line par- \ allel to the first. 2. The same exercise, using ruler and square instead of ruler and compasses. 3. Through the vertices of the angles of a triangle draw lines parallel to the opposite sides, and prolong them until they meet. What kind of a figure is thus formed? / D 73- 79-] CHAPTER IV. TRIANGLES. $1 79. A PROBLEM is a construction to be made according to geometrical laws. There is a general agreement that in the Science of Geometry no instrument shall be employed in making a construction except ruler and compasses ; if the problem cannot be solved by these means it is not regarded as belonging to the science. In the Art of Geometry that is, in the applications of Geometry to prac- tical purposes, such as Drawing and Land-measuring various other instruments which are found convenient are used, as divided rules, protractors, theodolites, etc. Thus, in practice, the problem of 77 is usually solved with the protractor, or with the ruler and the square, and the problem of 78 with the ruler and the square. In making a construction, all lines which are merely interme- diate steps between the data and what is required should be drawn dotted instead of full (see Figs. 72 and fj) . Such lines are called auxiliary lines. V. Some Consequences of the Equality of Triangles. 80. Let A B C (Fig. 74) be an isosceles triangle having A C = B C, and let CD be a line bisecting the angle C. This line divides the triangle ABC into two triangles, A C D and B C D. Com- pare these two triangles ; they have a common side CD, the side A C = BC (why ?), and the angle c=d (why?). Therefore the triangles are equal (II. Law of Equality), and among the equal parts we have a = b ; hence, Theorem. /;/ an isosceles triangle the angles opposite to the equal sides are equal. 92 GEOMETRY FOR BEGINNERS. [ 80. Corollary. In an equilateral triangle all the angles are equal ; therefore each angle = 1 80 : 3 = 60. Why does this corollary follow from the theorem? Remark. A theorem consists of three parts : (/.) The HYPOTHESIS (or hypotheses), or that which is given, or assumed as true, at the start. (ii.) The CONCLUSION (or conclusions), or that which is to be proved. (Hi.) The PROOF. In the above theorem the hypothesis is : A C B C (Fig. 74) , and the conclusion is : a b. The proof consists in showing that the triangles A C D and BCD are equal, and hence inferring that a = b. Exercises. 1. In an isosceles triangle the bisector of the angle at the vertex is perpendicular to the base, and bisects the base. Hints. Use Fig. 74. The two hypotheses are: A C = B C, and CD bisects A CB. The two conclusions are : C> A B, and CD bisects A B. Proof: show that &ACD^ /\BCD\ hence ;w = = 90 (why ?), and AD BD. 2. The perpendicular let fall from the vertex of an isosceles triangle to the base bisects the base, and also the angle at the vertex. Hints. Here the hypotheses are : A C = B C, and CD _L A B. The conclu- sions are : CD bisects A B, and CD bisects A C B. Prove by showing that A A CD ^ A B CD. 3. The line joining the vertex of an isosceles triangle to the middle of the base is perpendicular to the base, and bisects the angle at the vertex. Hints. In this case what are the hypotheses ? the conclusions ? Prove by showing (by IV. Law of Equality) that A A CD ^l\B CD. 4. What three conditions does the line CD {Fig. 74) fulfil? 5* Prove (with the aid of the theorem of this section) that the triangles A B C and A B C' {Fig. 77, p. 88) are equal. Hint. By drawing C C 1 we obtain two isosceles triangles, A C C' and B C C'. 6. Find the base angles of an isosceles triangle if the angle at the vertex is (*.) 27 35', (it.} 75 18', (Hi.} 124 40'. 7. Find the angle at the vertex if one of the base angles is (i.) 15 I4 ; > (ii.} 48 7 ', (Hi.} 83 4'. 8. In an isosceles right triangle find the base angles. 3 1.] CHAPTER IV. - TRIANGLES. 93 9, Can you devise a way of finding the height of an object (tree, tower, etc.) by means of an isosceles right triangle? 10 Show that in an isosceles triangle the exterior angles made by prolong- ing the base are equal, and that the exterior angles at the vertex are also equal. 11. Find the three angles of an isosceles triangle if the exterior angle at the vertex is 146 19'. 12. Find the three angles of an isosceles triangle if the exterior angle at the base is 120 50'. 13. Show that the exterior angles of an equilateral triangle are all equal, and find the value of one of them. 14. Construct an isosceles triangle having given, (a) The base and an adjacent angle ; (<) The base and the opposite angle ; (AB. Now compare the opposite angles. Since ABD = ABC + C BD, therefore ABD > ABC. And since A DB = ACB-CBD ( 66), therefore ADB < A CB, andhence<^f^C. Therefore.^ Z> > A D B. Here it is quite clear that the two conditions, A D > A B, and A B D > A D B, must always be ful- filled together ; hence, Theorem. I. Of two angles of a triangle the greater is oppo- site to the greater side. Theorem II. (Converse of Theorem I.). Of ftvo sides of a triangle the greater is opposite to the greater angle. Corollary. In a right triangle the hypothenuse is the greatest side. Fig. 76. CHAPTER IV. TRIANGLES. 95 Exercises, 1. From which theorem does the corollary follow? Why does it follow? 2. What similar corollary follows for an obtuse triangle? D F Fig. 77, 83. Let CD JL AB (Fig. 77), and let C E, C F, C G, be any other lines drawn from C to the line A B. These lines are the hy- pothenuses of the right triangles CDF, CDF, CD G, respectively, and therefore are, each of them, greater than CD ( 82, Corollary) ; therefore, Theorem I. Of all lines which - A can be drawn from a point to a straight line the perpendicular is the shortest. Corollary. Hence, the distance from a point to a straight line is the length of the perpendicular let fall from the point to the line. If DE = DF, then A CDE ^ A CDF (why?), therefore C E = CF; that is, Theorem II. Two oblique lines equally distant from the foot of the perpendicular are equal. Again, D G > D F. Likewise the angles CFD and CGD are each acute (why?). And CFG, being the supplement of CFD, must be obtuse. Therefore, in the triangle CFG we have CFG > CGF; hence C G > CF ( 82, Theorem II.). That is, Theorem III. Of two oblique lines unequally distant from the foot of the perpendicular, the more remote is the greater. Exercises. 1. In each of these three theorems what is the hypothesis? What the conclusion? 2. State and prove the converse of Theorem II. 96 GEOMETRY FOR BEGINNERS. [8 4 . 84. Let AB C and ABD (Fig. 78) be two isosceles trian- gles having a common base AB. Join C D. The triangles A CD and B CD are equal (IV. Law of Equali- ty). Let the triangle A CD be folded over CD until it coincides with the triangle C B D ; then A will fall on B, and all the lines and angles which fall upon each other will be equal. Hence we have, (*'.) # = and ; = BD (why?), therefore AD> BD. Hence, Theorem. Every point equidistant from the ends of a straight line is in the perpendicular which bisects the line. 88. Problem. To erect a perpendicular at a given point C (Fig. 83) of a given line A B. Analysis. If we find in A B two points, M and N, equally dis- tant from C, and then find any other point D equidistant from M and N, it is evident from 87 that the line passing through C and D will be the perpendicular required. Construction. With the centre C describe an arc cutting A Bin M and N. With M and N as centres M u ~~N and equal radii, describe arcs inter- secting at a point D. Join CD. / 83. Note. In practice this construction is usually effected by means of the ruler and the square (see 49). Exercises. 1. In the above construction what is the least value which the equal lines M D and A 7 D can have? 2. Erect a perpendicular at the end of a line without prolonging the line. Construction. Let A C (Fig. 84} be the line. Construct on A C as base an equilateral triangle ABC (how is this done?), and prolong A B to D, making B D = A B. CD is the perpendicular required. p roo f. Since A ABC is equilateral, ABC= 60; therefore, in A BDC=fo (66). Now A B CD is isosceles (why?); 89.] CHAPTER IV. TRIANGLES. 99 hence B CD = B D C- 30. Therefore A CD ~ A C B + B CD = 60 + 30 = 90. 3. Construct an equilateral triangle having the f altitude 4O cm . / Hints. Draw a line 40 long, erect at one end / a perpendicular, at the other end make two angles of 30 / each. 4-. Construct an isosceles triangle having given X (<7) the base and the altitude ; (b) a side and the altitude. 5. Bisect the three sides of a triangle, and then erect perpendiculars at the points of bisection. In how many points do they intersect? A Fig. 84. 89. Problem. To let fall a perpendicular from a point C (Fig. 85} to a given line A B. The analysis and the construction are the same as in the last problem, the only difference being that in this case the point C is not on the line A B. Give the analysis and the construction in full. Note. This construction is also usually ****.. effected in practice by means of the ruler and the square ( 49). Fig. 85. Exercises. 1. Choose a point on each side of a line, and let fall per- pendiculars from each point to the line. 2. Draw a triangle, and from the vertex of each angle let fall a perpen- dicular to the opposite side. In how many points do the perpendiculars intersect? 3. Through a given point draw a parallel to a given line by constructing two perpendiculars. 90. Theorem. Two parallel lines are everywhere equally distant from each other. Proof. From any two points A and B (Fig. 86) of one of the 100 GEOMETRY FOR BEGINNERS. [ 91. lines let fall perpendiculars meeting the other line in C and D ; then A _ B A C II B D ( 57, Corollary) . Join / B C. The triangles ABC and CD are equal ( 56 and I. Law of Equality ) . Therefore A C = B D ; that is, any two points of one line c Fig. 86. are equally distant from the other line : hence the lines are every- where equally distant from each other. 91. Let the problem be proposed : to find a point at a given distance from a given point. It is evident that any point in the circumference of a circle, de- scribed with the given point as centre and the given distance as radius, will satisfy the required condition, and will, therefore, be a solution of the problem. It appears, then, that there are an inde- finite number of solutions to the problem; hence the problem is termed indeterminate. The required point is limited in position to a certain line ; this line is called the locus of the point. Definition. The line (or lines) in which a point must be, in order to satisfy a given condition, is called the Locus * of the point. Again, let it be proposed : to find a point equidistant from two given points. It follows from 87 that the required point may be anywhere in the perpendicular which bisects the straight line joining the given points. The problem, therefore, is indeterminate, and the locus of the point required is the above-mentioned perpendicular. Generally speaking, when a point has to satisfy only one geomet- rical condition, the problem is indeterminate, and the solution con- sists in finding the locus of the required point. 1 Plural loci. 92.] CHAPTER IV. TRIANGLES. 10 1 Exercises. 1, In the cases above considered, the points were supposed to be confined to one plane : what would the loci be if this restriction were removed ? Ans. In the first case, the surface of a sphere with the given point as centre and the given distance as radius ; in the second case, the plane perpendicular to the line which joins the given points, and bisecting it. 2. Find the locus of a point at a given distance from a given straight line. 3. Find the locus of a point at a given distance from a given circumference. 4. Find the locus of a point at a given distance from a given plane surface. 5. Find the locus of a point equidistant from two given parallel lines. 6. Find the locus of a point equidistant from two given intersecting lines. Hints. The locus consists of the bisectors of the angles made by the lines. To prove this, let Pbe any point on either of the bisectors, P M and PN perpen- diculars from P to the two lines, O the point where the lines meet; then show that &POM^/\PON, and therefore that PMPN. Why is it that, although there are four angles to be bisected, the locus will consist of only two straight lines ? 92. When a point has to be found which satisfies two condi- tions, the problem (if possible at all) generally has one or two solutions ; in other words, the problem is determinate. Such problems are solved by constructing the loci of points which satisfy each condition separately ; then the point, or points, where the loci intersect, being common to both loci, will satisfy both conditions, and will be the solution of the problem. For example : find a point which shall be at given distances from two given points. The required point must be in a circumference having the first given point for centre and the first given distance for radius ; and it must also be in a circumference having the second given point for centre and the second given distance for radius ; hence it must be at the intersection of these two circumferences. 1 Since, in general, these circumferences intersect in two points, there will be in general two solutions of the problem. When will there be only one solution ? When will there be no solution at all ? In the following exercises, after having given the solution, state under what conditions (if any) the problem is impossible. 1 This reasoning has been already employed in \ 75. 102 GEOMETRY FOR BEGINNERS. [ 93. Exercises. 1. Find a point in a given straight line at a given distance from a given straight line. 2. Find a point in a given straight line at equal distances from two other straight lines. 3* Construct an isosceles triangle having a given base, and each of its sides equal to three times the base. 4. Find a point at a given distance from a given point, and at the same distance from a given straight line. 5. Construct a triangle having given the base, the sum of the other sides, and one of the angles at the base. 6. Construct a triangle having given the base, the difference of the other sides, and one of the angles at the base. 7. Find a point which shall be equidistant from three given points. 8. Find a point which shall be equidistant -from three given intersecting straight lines (see 91, Exercise 6). Show that there are four solutions, and construct the four points. How is the solution modified if two of the given lines are parallel? 93. The complete solution of a problem consists of three parts : (*.) The ANALYSIS, in which we explain how a correct construc- tion can be based upon known geometrical truths. (it.) The CONSTRUCTION, in which, guided by the analysis, we make the required lines, angles, etc., with the help of the ruler and the compasses. (in.) The DISCUSSION, in which we consider (a) whether there may be more than one solution ; (b) whether the data may have such values that no solution is possible ; (c) whether the solution has any other peculiarities for particular values of the data. The analysis of a problem that can be solved by the method of loci has already been given ( 92) ; in all other cases a theorem (or theorems) must first be found on which a correct construction can be based. In this search nothing but experience and ingenu- ity will ensure success ; but there is one general rule which will be found very useful ; namely : 93-] CHAPTER IV. TRIANGLES. 103 Rule. Suppose the solution effected, and draw a suitable figure ; then trace the relations among the parts of the figure until some re- lation is discovered which will give a clue to the right construction. We will give an example to illustrate the application of this rule. Problem. Through a given point to draw a line which shall cut off equal lengths from the sides of a given angle. Analysis. Let P (Fig. 87) be the given point, B A C the given angle. Suppose that O P Q were the line required ; then we know that A O Q must be an isos- celes triangle (why ?) , and hence, that if we bisected the angle A, O P Q would be perpendicular to the bi- sector (80, Exercise 1). This suggests to us the correct construc- tion. Construction. Bisect the angle Fig. A, let fall from P a perpendicular to the bisector, and produce it until it meets the sides of the given angle in O and Q : O P Q is the line required. Discussion. Construct the figure for the case where P is not between the sides of the given angle. For what position of P is the solution impossible? Exercises. 1. Draw a line which shall pass through a given point and make equal angles with two given intersecting lines. 2. In a given straight line find a point which shall be equidistant from two given points. 3. In one side of a triangle find a point which shall be equidistant from the other two sides (see 91, Exercise 6). 4. In one side of an angle a point is given ; find in the same side another point which shall be equidistant from the first point, and from the other side of the angle. 5. From a given point without a given straight line draw a line making a given angle with the given line. 6. Trisect a right angle (that is, divide it into three equal parts). 104 GEOMETRY FOR BEGINNERS. [ 94, VI. Applications. 94. If I wish. to find the distance from A to B (Fig. 88) it is plain that the intervening pond will prevent me from using a chain or other direct means of measurement. In numerous instances direct measurement would be very troublesome or quite impossible. On the ground obstacles, such as houses, water, swamps, are often met with ; if we want to meas- ure the distance from the earth to the moon, only one end of the line to be measured is accessible ; while in the case of the distance between the sun and a planet, the line is wholly inaccessible. Here Geometry comes to our aid by giving us methods of indi- rect measurement ; in which, for example, by measuring one line we are able to learn the length of another. How this can be done by the help of the laws of equal triangles, we are now prepared to understand. All cases of the indirect measurement of a line may be reduced to four, (/.) Both ends of the line are accessible ; ('.) Only one end is accessible ; (tit.) Both ends are inaccessible ; (tv.) The line is wholly inaccessible. 95. Problem. To measure a line the ends of which only are accessible. Method I. Let A B (Fig. 88) be the line to be measured. Choose a point C from which A and B are both visible. Measure A C, B C, and the angle A C B. Prolong A C to D, making C D = C A, and prolong B C to E, making C E = C B. &DEC^&ABC (why?). There- fore DE = A B, and its length (which can be measured directly) is the distance required. 96-] CHAPTER IV. TRIANGLES. 105 Method II. Measure A C, B C, and A C B, as before. Then construct on paper an angle acb ACB; choose a suita- ble scale of reduction, and lay off on the paper the reduced lengths a c and b c of the lines A C and B C. Join a b : its length gives, to the reduced scale, the distance A B required. For example : if A C = 8oo m , Fig. 88. D B C iooo m , and the scale of reduction is i : 5000, then a c = i6 cm , and b c = 2O cm . If now we find that a b = 24 cm , then A B = 24 cm x 5000 = 1200 metres. Remark. The first method is preferable, because any error made in measuring a b is multiplied 5000 times in the result. How can the length of the pond (Fig. 88) be found ? 96. Problem. To measure a line one end only of which is accessible. Method I. Let A B (Fig. 89) be the line to be measured, B the end which is accessible. At a point C, in A B pro- longed, place a signal (a pole or flag). Then place a signal at a convenient point D, and measure the distances BD and CD. Prolong CD to E, making DE = DC, and prolong B D to F, making DF = DB. Place signals at E and F. Then proceed in the direction EF until a 106 GEOMETRY FOR BEGINNERS. [97- point G is reached which falls in the line A D. Measure F G : its length is equal to the distance A B required. Proof: A D E F^D B C (II. Law of Equality). Therefore angle DEF = D CB ; hence E G II A C ( 57, Remark). In the triangles DAB and D F G, D B = D F, ADB = FD G (why?), and AB D = DFG (why?) ; therefore A DAB^ A DF G (I. Law of Equality), and F G = A B. Method II. (By an isosceles triangle.) Choose a convenient direction B H for running a straight line from B, measure the angle A B H, and find in B H a point H at which the angle A HB shall be equal to (180 ASH). Then B H = AB (why?). What will be the value of A HB ifABH= 9 o? Remark. In both methods three things have to be measured directly : in Method I., three lines ; in Method II., one line and two angles. But, if the means of measuring an angle are at hand, Method II. is preferable, because it is easier to measure an angle than a line. 97. Problem. To measure a line when both ends of it are inaccessible. Let A B (Fig. go) be the line, and let us suppose that a part of the line between A and B is accessible, as shown in the Figure. Method I. At a point C in A B erect a perpendicular C D ( 88), and take DE = CD. At E also erect a perpendic- ular FG\ then F G will be parallel to A B (why ?) Find in FG the point F which falls in the line B D, and the point G which falls in the line A D. Now &ACD g* A ED G (why?), and A Fi g . 90. B CD ^ A EDF (why?). CHAPTER IV. TRIANGLES. 107 Therefore FE = B C, and E G = A C ; and hence FE + E G = BC + AC = AB,m the line to be measured. Method II. (By isosceles triangles.) In C E find the points H and K from which the directions of A and B respectively make with C E the angle 45. Then C H -f C K = AS. Explain fully why. 98. Problem. To measure a line which is wholly inacces- sible. Method I. Choose a convenient point C from which A and B are both visible ; measure A C and B C as in 96 ; then measure A'B'as in 95. Method n. Choose C as before ; also a point F from which A is visible, and a point G from which B is visible. Find E, the intersection of A F and B C, and D, the intersection of B G and A C. Measure all the angles at C, and also the lines C F, C E, C G, CD. Then represent on paper the measured angles ^ and lines (the latter to a reduced scale) . By prolonging E F and D C till they meet, we find the point on the paper which corresponds to A on the ground ; and by prolonging in like manner D G and E C, we find the point which corresponds to B. The length of A B on the paper gives to the scale employed the required distance. 108 GEOMETRY FOR BEGINNER^, REVIEW OF CHAPTER IV. QUESTIONS. 1, Define a plane figure; a polygon; its perimeter. 2. Define a triangle, and explain how it is named. 8. What six parts has every triangle? 4:. Distinguish between exterior and interior angles'* 5. The sum of any two sides of a triangle is greater than the third side. 6. Define equilateral, isosceles, and scalene triangles. 7. Define the terms base, vertex, and altitude. When will the altitude lie outside the triangle? 8. The sum of the angles of a triangle is equal to 180. How can this truth be illustrated? 9. Why can a triangle have only one right or one obtuse angle ? 10. Define acute, right, and obtuse triangles. 11. If two angles of a triangle are known how can the third be found? 12. In a right triangle what is the sum of the acute angles? 13. What is the complement of an angle? 14r. An exterior angle of a triangle = the sum of the opposite interior angles. 15. Define similar, equivalent, and equal magnitudes, and give examples of each kind. 16. In two triangles what are corresponding side% (or angles) ? 17. What is the least number of parts which determine the size and shape of a triangle? Of these how many at least must be sides? 18. In what cases do three parts fail to determine the triangle? 19. When are two triangles mutually equiangular? 20. Construct a triangle, having given, (*'.) a side and two angles ; (.) two sides and the included angle ; (m.) two sides and the angle opposite to the greater side ; (zz/.) the three sides. 21. State the four laws of equal triangles. 22. Make a triangle equal to a given triangle. 23. Make an angle equal to a given angle. 24. Draw a line parallel to a given line. 25. What is a problem ? What are auxiliary lines, and how distinguished from others in the construction? CHAPTER IV. REVIEW, 109 26. In an isosceles triangle the angles opposite to the equal sides are equal. Corollary. 27. If two angles of a triangle are equal the triangle is isosceles. Corollary. 28. What is the converse of a theorem? Give an example. 29. In a triangle the greater of two angles is opposite to the greater side, and conversely. Corollary. 30. The perpendicular is the shortest line from a point to a straight line. Corollary. 81. Oblique lines equally distant from the foot of the perpendicular are equal. 32. Of two oblique lines unequally distant from the foot of the perpendicu- lar, the more remote is the greater. 33. If two isosceles triangles have a common base the line joining their vertices (z.) bisects the angle at the vertices, (zY.) bisects the base, (in.) is perpendicular to the base. 34:. Bisect a given angle. 35. Bisect a given straight line. 36. Every point equidistant from the ends of a straight line is in the perpen- dicular bisecting the line. 37. Erect a perpendicular at a given point of a given line. 38. Let fall a perpendicular from a given point to a given line. 39. Two parallel lines are everywhere equally distant from each other. 40. Give an example of an indeterminate problem. 41. Define the locus of a point, and illustrate the definition. 42. Give an example of a determinate problem. 43. How are problems solved by the method of loci? 44. What are the three parts of the complete solution of a problem? 45. What general rule is often useful in solving problems? 46. What four cases may occur in the indirect measurement of a line? 47. Measure a line the ends only of which are accessible. 48. Measure a line one end only of which is accessible. 49. Measui'e a line neither end of which is accessible. 50. Measure a line which is wholly inaccessible. EXERCISES. 1. In an isosceles triangle the exterior angles at the base are equal. 2. In an isosceles triangle the bisector of an exterior angle at the vertex is parallel to the base. 110 GEOMETRY FOR BEGINNERS. 3. In an isosceles triangle the bisectors of the base angles (prolonged until they meet the sides) are equal. 4. In what cases are we able, knowing one angle of a triangle, to find the other two angles? 5. If in a right triangle the acute angles are 30 and 60, the hypotenuse is equal to twice the smaller leg. 6. State and prove the converse of the preceding theorem. 7. Divide an isosceles triangle into two equal right triangles. 8. Make an angle equal to three times a given angle. 9. Make an angle of 15; also an angle of 150. 10. In what two ways can an angle of 22^ be constructed? 11. Name various angles between o and 180 which can be constructed by means of the theorems and problems already given. 12. A straight railway passes within two miles of a town. A place is de- scribed as four miles from the town and one mile from the railway. How many places satisfy the conditions? 13. Place a line of given length between the sides of an angle so as to be parallel to a given line. 14. Through a given point between two lines not parallel draw a line which shall be bisected at that point. 15. Three lines meet in a point ; draw a line such that the parts of it inter- cepted by the three lines shall be equal. 16. What axiom is implied in the last line of the proof of the theorem in 87? 17. The perpendiculars erected at the middle points of the sides of a triangle meet in one point. , Hints, Erect two of the perpendiculars ; show (by II. Law of Equality) that their intersection is equidistant from the three corners of the triangle, then make use of 80, Exercise 2. 18. The bisectors of the three angles of a triangle meet in one point. Hints. Draw two bisectors; show by $ 91, Exercise 6, that their intersection is equidistant from the sides of the triangle, and (by III. Law of Equality), that the line joining the intersection with the third corner bisects the angle at that corner. 19. Trisect a given straight line. 20. Illustrate by an example the application of the rule in 93. 21. Choose two stations on the ground, and find their distance by means of (i.), 95; (.), 96; (iii^ 97; <>.), 98. 22. Can you find the height of an object (a tree, a church-spire, etc.) by means of truths established in this chapter? How? 23 How many and what parts are sufficient to determine a right triangle? an isosceles triangle ? an equilateral triangle ? an isosceles right triangle ? 99-] CHAPTER V. QUADRILATERALS. Ill CHAPTER V. QUADRILATERALS. CONTENTS. I. Sides and Angles of a Quadrilateral ( 99, 100). II. Different Kinds of Quadrilaterals ( 101-105). HI- Construction of Quadrilaterals ( 106-111). IV. Subdivision of a Lane ( 112-114). J. Sides and Angles of a Quadrilateral. 99. A plane figure bounded by four straight lines is called a QUADRILATERAL (A B CD, Fig. 92). A straight line A C, which joins two opposite corners of a quadrilateral, is called a DIAGONAL. Exercises. 1. How many sides and how many angles has a quadrilateral? 2. How is a quadrilateral named or de- noted? 3. What is the perimeter of a quadrilateral ( 59) ? 4. Into how many triangles is a quadrilateral divided by a diagonal? 5. How many diagonals can be drawn in a quadrilateral? 100. If we divide a quadrilateral AB CD (Fig. 92) into two triangles by drawing a diagonal AC, it is evident that the sum of the four angles of the quadrilateral is the same as the sum of the six angles of the two triangles. Now the sum of the angles of the two triangles is 360 (why?) ; hence, Theorem. The sum of the angles of a quadrilateral is equal to four right angles, or 360. Exercises. 1. If the angles of a quadrilateral are all equal, what is the value of each? 2. Three angles of a quadrilateral are 70, 40, and 120; find the fourth. 112 GEOMETRY FOR BEGINNERS. [ IOI. II. Different Kinds of Quadrilaterals. 101. With respect to the position of the sides there are three kinds of quadrilaterals. A quadrilateral which has no sides parallel is called a TRAPEZIUM (Fig. 93,1.). A quadrilateral which has only two sides parallel is called a TRAPEZOID (Fig 93, II.). A quadrilateral which has its opposite sides parallel is called a PARALLELOGRAM (Fig. 93, III.). Exercises. 1. Draw (free-hand) a trapezium, a trapezoid, and a paral- lelogram. 2. Draw a trapezoid having two right angles. 102. Let AB CD (Fig. 94) be a parallelogram. Draw the diagonal BD. The alternate angles m and n are equal, and likewise the alter- nate angles / and q ( 56). Therefore A ABD ^ A CBD (I. Law of Equal- D c ity), and AB = CD, and AD = BC. Also the angle A equals the angle C ; and, since m n and p q, therefore m -\- n = p -\- q; that is, the angles at B and D are equal. Hence. Theorem I. Every parallelogram is divided by a diagonal into two equal triangles. I03-] CHAPTER V. QUADRILATERALS. 113 Theorem II. The opposite sides of .a parallelogram are equal. Theorem III. The opposite angles of a parallelogram are qual. Corollaries. From Theorem II. it follows that, 1. Parallels between parallels are equal. 2. Perpendiculars between parallels are equal ( 57, Corol- lary). 3. Parallels are everywhere equally distant (see 90) . Exercises. 1. If two adjacent sides of a parallelogram are equal, what follows as to the other sides? 2. What relation exists between two adjacent angles of a parallelogram? Why? 3. If one angle of a parallelogram is a right angle, what must the others be? Illustrate with a figure. 4 If one angle of a parallelogram is acute, what can be inferred of the other angles? Illustrate with a figure. 5. One angle of a parallelogram is (*'.) 45, ('.) 54 1 8', (m.) 1 21 1 6' 44" ; find the other angles. 6. Prove the converse of Theorem II. 103. The sides of a parallelogram may be either equal or -unequal, and the angles may be either right or oblique ; so there are in all four kinds of parallelograms : the oblique unequal-sided IV. parallelogram or RHOMBOID (Fig. 95, I.), the oblique equal- sided parallelogram or RHOMBUS (Fig. 95, II.), the right unequal-sided 114 GEOMETRY FOR BEGINNERS. [ 104. parallelogram or RECTANGLE (Fig. 95, III.), and the right equal- sided parallelogram or Square (Fig. 95, IV.). Exercises. 1. Give examples of parallelograms (surface of a table, roof of a house, etc.), and state the kind in each case. 2. Draw (free-hand) a parallelogram of each kind. 3. One angle of a rhombus is 60; find the other angles. 4. One side of a rhombus is 24; find its perimeter. 5* The sides of a rectangular park are 400'" and 240. How many trees must be set out in its perimeter if they are to stand io m apart? 104. In the parallelogram ABCD (Fig. 96), draw the diagonals AD and BC ; how many triangles are thus formed? A AOB ^ C O D (why ?) ; therefore the sides opposite to the equal angles are equal; that is, AO = DO and BO = CO : hence, z?* Ji L ^ Theorem. The two diagonals of a parallelogram mutually bisect each other. Exercises. 1. Find the angles which the diagonals of a square make with the sides of the square. 2. In a square the diagonals are equal and perpendicular to each other. 8. In a rectangle the diagonals are equal and inclined to each other. Hint, Draw the diagonals, and make use of $ 73. 4. In a rhombus the diagonals are perpendicular to each other. Are they equal or unequal? 5. In a rhomboid the diagonals are inclined to each other. Are they equal or unequal? 6. What kind of parallelograms are divided by the diagonals into isosce- les triangles? Into right triangles? 105. Either side of a parallelogram may be regarded as the base ; then a perpendiculai let fall to the base from any point in the opposite side is the altitude of the parallelogram. io6.] CHAPTER V. QUADRILATERALS. 115 In the parallelogram A B C D (Fig. 97) A B is taken as the base, and either DE or FB or CG is equal to the altitude. Why are .DE, FB, and CD all equal? In a rectangle whatever side be taken as base the adjacent side is the altitude. In a square the base and altitude are equal. In a trapezoid one of the parallel sides is taken as the base. In a trapezium the terms base and altitude are not used. III. Construction of Quadrilaterals. 106. Problem. To construct a square, having given one side, a (Fig. 98). Construction. At A make a right angle ; take AB = AD = a' t then from B and D as centres, with a radius a equal to a, describe arcs intersecting at a point C. AB CD is the square required. Proof 1 The four sides are each equal to a by construction. Join B D, and make use of 102, Theorem I. Construct another square with the same side a then prove (by division into triangles, and IV. Law of Equality) that the two squares are equal. Hence, a square is completely determined if one side is known ; and, Theorem. Two squares are equal if they have a side equal. Exercises. 1. Construct a square whose side shall be i dm . 2. Construct a square whose perimeter shall be i m . 3. Draw a rectangle ; then construct a square with the same perimeter as that of the rectangle. Fig. 98. 1 If the construction of a problem is given without a previous analysis, a proof of the construction (unless it is quite obvious without it) ought to follow. 116 GEOMETRY FOR BEGINNERS. [ 107. 4r Construct a square, having given its diagonal. 5. Construct a square, and upon each side an equilateral triangle. 6. Construct an equilateral triangle, and upon each of its sides a square. 107. Problem. To construct a rectangle, having given two adjacent sides, a and b (Fig. 99) . Construction and proof similar to those of the last section. Construct another rectangle with ~~~ b the same sides, and prove that the two rectangles are equal. Hence, a rectangle is determined if we know two adjacent sides ; also, Theorem. Two rectangles are equal if they have tivo adjacent sides equal each to each. Exercises. 1. Construct a rectangle with the sides 4O cm and 6o cm . (If on paper, use a reduced scale.) 2. Make a plan of a rectangular field whose sides measure 640 and 360. 3. How many rectangles can be constructed upon a given line as the diagonal? 108. Problem. To construct a parallelogram, having given two adjacent sides and the included angle. Construction. Let a and b (Fig. 100) be the given sides, 70 _a the included angle. Make an angle A = 70, then proceed as in the last section. A B C D is the required parallelogram. Proof similar to that in 106. Construct a second parallelogram with the same given parts, and prove that it is equal to the first. Therefore, tivo sides and the in- Fig. ioo. eluded angle completely determine a parallelogram ; and, further, lOp.] CHAPTER V. QUADRILATERALS. 117 Theorem. Two parallelograms are equal if two sides and the included angle of one are equal respectively to two sides and tlce included angle of the other. Exercises. 1. Construct a parallelogram, the two sides being 44 cm and 66 cm , and the included angle 1 10. 2. Construct a rectangle, having given (a) A side and a diagonal ; (b} A side and the opposite angle of the diagonals ; (c) A diagonal and the angle between the diagonals. 3. Construct a rhombus, having given (a) A side and an angle ; (b) A side and a diagonal ; (r) The diagonals ; (^/) An angle and the diagonal through its vertex. 4. Construct a rhombus whose angles shall have the ratio 1 : 3. 109. Problem. To construct a trapezoid, having given three sides (two being the parallel sides} and one angle. Construction. Let a, b, c {Fig. a JQI) be the given sides, 68 the given angle. Make an angle A = 68 ; take AB = a, A D = b through D draw a line parallel to A B, and on it take D C = c. Join B C. A B C D is the trapezoid required. Exercises. 1. How many and what parts determine a trapezoid? A r-,. B rig. 101. 2. Construct a trapezoid with the sides 6o cm , 7O cm , 8o cm , and the angle between the first two sides equal to 75. 3* Construct a trapezoid, having given (a) The two parallel sides and the altitude ; (3) The two sides not parallel and the altitude ; (c) The two parallel sides and one of the other sides ; ( r6. The triangle and the quadrilateral are polygons ; but, as they have been already studied, we shall now apply the term chiefly to those figures which have more than four sides. A polygon of five sides is called a PENTAGON, of six sides a HEXAGON, of seven sides a HEPTAGON, of eight sides an OCTAGON, of nine sides a NONAGON, of ten sides a DECAGON, of twelve sides a DODECAGON, of twenty- sides an ICOSAGON. Which of these polygons are represented in Fig. 106 ? Every polygon has as many angles as sides ; each side has two adjacent angles ; each angle is formed by two intersecting sides. A line joining two corners not in the same side is called a DIAG- ONAL. By drawing as many diagonals as possible from one corner of a polygon we divide the polygon into triangles ; and Exercise i, below, leads to the following general law : The number of diagonals which can be drawn from one corner of a polygon is THREE less than the number of sides ; and the num- ber of triangles into which the polygon is divided is TWO less than the number of sides. Exercises. 1. Draw polygons of 3, 4, 5, 6, 7, 8 sides, and from a corner of each all the diagonals possible ; then prepare three vertical columns, in the first writing the number of sides of the several polygons, in the second the number of diagonals, in the third the number of triangles. What general conclusions follow from the results? 2. How many diagonals can be drawn from one corner of a polygon of 60 sides, and how many triangles will be formed? 3* Divide an octagon into three quadrilaterals and a triangle. 4. If you cut off a triangle from a hexagon with a diagonal, what kind of polygon is. left? 5. How many diagonals in #//can be drawn in a quadrilateral? a penta- gon? a hexagon? a decagon? 116. The angles of a polygon may be acute, right, obtuse, or even convex. In the last case, they are usually termed reentrant 1 1 6.] CHAPTER VI. POLYGONS. 125 angles. In Fig. 107, A is a polygon having one reentrant angle, and B a polygon having two reentrant angles. Fig. 207. But whatever be the values of the angles, their sum in any poly- gon always bears a simple relation to the number of sides. If a polygon could be divided into as many triangles as it has sides, then the sum of all its angles would evidently be equal to twice as many right angles as there are sides ( 64). But since by the last section it appears that the number of these triangles is two less than the number of sides, it follows that the sum of all the angles will be twice as many right angles as there are sides less two. Hence, Theorem. The sum of the angles of a polygon is equal to twice as many right angles as there are sides less two. Thus the hexagon (Fig. 108) consists of 6 2 = 4 triangles ; and the sum of its angles = 4X2 = 8 right angles = 720. In general, if n be the number of sides (n being any number), the following equation or FORMULA enables us to find the sum of the angles : Sum of the angles (in degrees) = 180(n-2). [i.] ' p . Exercises. 1. What polygon cannot have reentrant angles? 2. Find the sum of the angles of polygons of 3, 4, 5, 6, 8, 10, 12, 20 sides. 3* Find by division into triangles the sum of the angles of the polygon A {Fig. 707). Also find the sum by means of Formula [i]. 126 GEOMETRY FOR BEGINNERS. [ 117. 4. Find in the same two ways the sum of the angles of the polygon B (Fig. 707). 5. Of how few sides can you make a polygon with one reentrant angle? with two reentrant angles? 6. Make a hexagon with as many reentrant angles as possible. 7. Make a hexagon having only right angles or their multiples. 8. Draw a polygon such that one of its diagonals shall lie wholly outside the polygon. 117. Problem. To make a polygon equal to a given poly- gon ABCDEF (Fig. /op). Analysis. Divide the given polygon into triangles by drawing diagonals from one corner ; then construct a series of new triangles equal respec- tively to those of the given polygon, and similarly placed. It is evident that Fig. log. these taken together will form a new polygon equal to the given polygon (Axiom II.). Draw the figures and give the construction in full. Exercises. 1. Construct a pentagon equal to a given pentagon. 2. Construct an octagon equal to a given octagon. 3. When are two polygons equal to each other? II. Regular Polygons. 118. Definitions. A polygon which has equal sides is termed EQUILATERAL (Fig. no, A) ; a polygon which has equal angles is termed EQUIANGULAR (Fig. no, B} ; and a polygon which has equal sides and equal angles (Fig. no, C) is termed REGULAR. Since the angles of a regular polygon are equal, each angle is equal to the sum of all the angles divided by their number ; or, in other words, divided by the number of sides of the polygon. If we let n denote the number of sides, the sum of the angles 1 1 9.] CHAPTER VI. POLYGONS. 127 (by 116) = 1 80 (n 2) = 180 w 360. Dividing this value by n we obtain (Axiom V.) the formula, RAA Each angle of a regular polygon = 180 - . [ 2 .] Fig. no. Exercises. 1, What is the regular triangle called? the regular quadri- lateral? 2. Can a triangle be equilateral without being equiangular? 3* What quadrilateral is equilateral but not equiangular? equiangular but not equilateral? 4. Make three hexagons : the first equilateral but not equiangular ; the second equiangular but not equilatei'al ; the third regular. 5. Find each angle of a regular polygon of 3, 4, 5, 6, 8, 10, 12, 20 sides, and arrange the results in a vertical column. As the number of sides increases how does the angle change? 6. Find the angle of a regular polygon of 360 sides. 7. Express Formula [2] in words. 8. One side of a regular dodecagon is i8 m ; find its perimeter. Give a rule for finding the perimeter of a regular polygon when a side is known. 119. Regular polygons possess a very important property stated in the following theorem : Theorem. The bisectors of the angles of a regular polygon meet in a point equally distant (i.) from all the corners, and also (ii.) from all the sides. Proof. Let the bisectors AO, BO (Fig. in or 112) of the angles A, B, meet in <9, and join OC, OD, etc. The triangles AOB, BOC are isosceles ( 81); therefore OA=OB. The 128 GEOMETRY FOR BEGINNERS. [ "9- same two triangles are also equal ( 73) ; therefore OC= OA. Also, OCB= OAB \ one of the equal angles of the polygon; therefore OC bisects the angle BCD. In like manner we can prove that O>= OA, and bisects the angle CDE ; etc. D A G B Fig. in. A G Fig. JI2. (ii.) The distances from O to the sides of the polygon are the perpendiculars let fall from O to the sides ( 83, Corollary) . Take any two of these perpendiculars OG and OH; they are equal, because A OGB ^ OHB (why?). In like manner we can show that OH OJ = OK, etc. ; in other words, these perpen- diculars are all equal. Definitions. This remarkable point O in a regular polygon is called its CENTRE ; the distance from the centre to a corner is called the GREATER RADIUS ; the distance from the centre to a side is called the LESS RADIUS ; the angle between two radii is called the ANGLE AT THE CENTRE. Corollaries. I. Every regular polygon can be divided into as many equal isosceles triangles as it has sides (how ?) . 2. Every regular polygon can be divided into twice as many equal right triangles as it has sides (how ?) . 3. The greater radii of a regular polygon are all equal, the less radii are equal, and the angles at the centre are equal. 4. The angle at the centre is equal to 360 divided by the num- ber of sides ; in other words, to the last term 4,ooo,oooi mm . 7. Subtract i a from i ha , and give the answer in square meters. 8. How many house-lots, each containing 2 a , are there in a field which contains 8 ha ? 9. A man bought 3 ha of land at $200 per hectar, and sold it for $2.50 per ar. Did he gain or lose, and how much? 10. Which is the greater farm, 200 hectars or 500 acres? 11. Divide 6 ha into 64 equal lots ; how many square meters in each lot? 12. What is the difference between 3 square meters and 3 meters square? II. Areas of Polygons. 126. Definition. The number of times a unit of surface is contained in any surface, followed by the name of the unif, is called the AREA of the surface. NOTE. Hence units of surface are often called units of area, as at the begin- ning of this chapter. In order to find the area of a surface, it might at first thought seem necessary to apply a unit of area to the surface over and over again as many times as possible, just as we measure a line directly by using a yard-stick or a meter-rule. But this method would be very tedious, and in many cases (for instance, a pond, swamp, forest, etc.) utterly impracticable. Fortunately, however, Geometry supplies us with indirect meth- ods of measurement which are applicable to all surfaces. It teaches that the value of areas depends upon the lengths of certain lines ; whence it follows that areas can be found by performing certain 127.] CHAPTER VII. AREAS. 139 simple operations upon the numbers which express the lengths of these lines. In explaining these indirect methods we begin with the square and the rectangle, because they can be readily decomposed into square meters, or into squares larger or smaller than the square meter. 127. The SQUARE. If the side of a square contains a whole number of meters, as six, it is easy to see that (by proceeding exactly as in 125) we can first divide the square into six equal rectangles, then each rectangle into six square meters; so that the square will contain in all 6 X 6 = 36 square meters. The same mode of reasoning may be used when the side con- tains meters and subdivisions of a meter. Let, for instance, the side = 6.35 ra ; then, in decomposing the square into smaller squares, we may take as the unit of length one centimeter. In place of six rectangles, there will be 635 ; and in place of 6 X 6 = 36 square meters, there will be 635 X 635 = 403,225 square centimeters = 40.3225 square meters, a result obtained simply by multiplying 6.35 by itself. In short, the number of units of area in a square is always found by multiplying by itself the number of the corresponding units of length in one of its sides. Hence, when we multiply a number by itself, we are said to square it ; and it is usual to express the preceding rule more briefly thus : The area of a square is equal to the square of one of its sides. Conversely, if the area is known, in order to find the value of one side, we must find a number which, multiplied by itself, will give the numerical value of the area ; in other words, we must extract the square root of the given area. For example : if the area of a square = 8.1225 square meters, me side = V8.I225 = 2.85 meters. Remark. For the sake of convenience we use the expression 140 GEOMETRY FOR BEGINNERS. [ 128. "the square of AB" and we write AB*- These expressions do not mean that the line AB is multiplied by itself, but that the number of linear units in the line is multiplied by itself, the result being the number of units of area in the square constructed upon AB as a side. 128. The RECTANGLE. A rectangle can be decomposed into units of area in the same way as a square, the only difference being that the number of units of length in its adjacent sides is not the same. Thus, the rectangle ABCD (Fig. 122), the sides of which are A= y m , AC=4 m , contains 7x4=28 square meters. The side AB is the base of the rect- D Fig. 122. angle, and the side AC is the altitude ( 104) ; so that the number of units of area in a rectangle is found by multiplying the number of linear units in the base by the number of linear units in the altitude. Or, more briefly : The area of a rectangle is equal to the pro- duct of its base by its altitude. If the area and base are known, how can the altitude be found? If the area and altitude are known, how can the base be found? Remark. In practice we must be careful to express both base and altitude in terms of the same linear unit ; the area will then be expressed in terms of the corresponding unit of area. For example : the base of a rectangle = i2 m , the altitude = 8o cm ; find the area. Here we must not multiply 12 by 80 ; but, if we choose the meter as the linear unit, we must multiply 12 by 0.8, because 8o cm = 0.8. Answer, 9.6 square meters. If we take the centimeter as the linear unit, what will the answer be? CHAPTER VII. AREAS. 141 Fig. 123. 129. The PARALLELOGRAM. It is easy to transform a paral- lelogram into an equivalent rectangle. In the parallelogram ABCD (Fig. 123) erect perpendiculars at the points A and B of the base ; they cut, one of them the opposite side, the other the opposite side pro- longed, in the points F and E ; and the figure ABEF is a rectangle having the same base and altitude as the parallelogram. The right triangles AC E and B DP are equal (why?) ; hence, whichever of these triangles we add to the quadrilateral ABCF, we shall obtain the same area. If we add AAC, we obtain the rectangle ABEF] and if we add ABFD, we obtain the paral- lelogram ABCD: therefore, the parallelogram ABCD is equal to the rectangle ABEF. Theorem. A parallelogram is equivalent to a rectangle having the same base and altitude. NOTE. This transformation maybe easily shown to the eye by cutting from stiff paper the trapezoid A CBF {Fig. 123), and the right triangle A E C, and then placing the triangle in the two positions shown in the figure. Corollaries. i. By combining this theorem with 128, we see that the area of a parallelogram is equal to the product of its base by its altitude. 2. All parallelograms having equal bases and equal altitudes are equivalent. Is the converse of Corollary 2 true ? Draw a diagram to illustrate. Construct several equivalent parallelograms differing in shape. Remark. In a rhombus the diagonals bisect each other at right angles ( 104). Construct about the rhombus ABCD (Fig. 124) a rectangle PQMN, by drawing through the corners of the Fig. 124. 142 GEOMETRY FOR BEGINNERS. [ 130. rhombus lines parallel to its diagonals. It is easy to show that this rectangle is composed of eight equal -right triangles, of which the rhombus contains four (how?). Now, since the base and altitude of the rectangle are equal to the two diagonals of the rhombus, its area is equal to the product of these diagonals. Hence, the area of the rhombus is equal to half the product of its diagonals. 130. The TRIANGLE. A triangle ABC (Fig. 125) may be regarded as half a parallelogram having an equal base and equal altitude ( 101). We have only to draw through two corners, B and C, of the triangle lines parallel to the op- posite sides ; then, since &ABC=% ABDC, and ABDC = AB X CE, therefore A ABC = %AxC. That is, - The area of a triangle is equal to half the product of its base by its altitude. Fi Corollary. All triangles having equal bases and equal altitudes are equivalent. NOTE. In finding the area of a triangle, we may either multiply the base by the altitude, and then divide by 2, or we may multiply half the base by the altitude, or we may multiply half the altitude by the base. For example : if the base = 14"', and the altitude = 8 m , the area = 14X8 14 X - = 56 square meters. 131. The TRAPEZOID. A trapezoid ABCD (Fig. 126) is divided by the diagonal BC into two triangles, ABC and BCD, whose bases, AB and CD, are the parallel sides of the trapezoid, and whose common altitude CE is the altitude of the trapezoid. Since A ABC = AB X CE, and A BCD = } CD X CE; therefore, the trapezoid ABCD = %(AB+ CD} X CE. That is, the area of a trapezoid is equal to half the product of the sum of the parallel sides by the altitude. CHAPTER VII. AREAS. 143 132. Any QUADRILATERAL. To find the area of any quadri- lateral, divide it by a diagonal into two triangles, find the areas cf these triangles, and add the results together. 133. A REGULAR POLYGON. We know ( 119, Corollary i) that a regular polygon ABCDEF (Fig. 127) is composed of as many equal isosceles triangles as there are sides. Therefore its area may be found by adding together the areas of these triangles. The altitude of each triangle is equal to the less radius of the polygon, so that the area of each triangle is half the product of its base by the less radius. Therefore the area of the polygon is half the product of the sum of all the bases (that is to say, the perimeter of the polygon) by the common altitude (that is to say, the less radius of the polygon) . The area of a regular polygon is half the product of its perimeter by its less radius. 134. Any POLYGON. The area of any polygon may be found by dividing it into triangles. Fig. 128 shows two ways of doing this. In either case, if we compute the area of each triangle separately, and then add the results, we shall find the area of the \ \ / / \ / polygon. In each of these cases Fig. I2J. Fig. 128. how many lines require to be measured directly in order to com- pute the area? But it is generally easier to find the area of an irregularly-shaped 144 GEOMETRY FOR BEGINNERS. [ 134- polygon (as a field, farm, etc.) by division into right triangles and trapezoids, as shown in Fig. 129. Join the two most dis- tant corners of the polygon by a straight line, and let fall from the- other corners per- pendiculars to this line taken as a common base. The polygon is thus divided into right triangles and trapezoids, the areas of which are then computed separately, and their sum is obviously equal to the area of the polygon. For example : let (Fig. 129) Bb = 6.8 m ; Cc = io.6 m ; Dd = io.i m ; Ff = 8.3 m ; Gg = 6.2 m ; Nh = 9.2 : also let Ab = 5.6; th = 2 .6 m ; /^ = 4 .2 m ; ^= 4 .6 m ; gf = 3 m ; fd = 2.8 m ; dE The work of computation may be arranged as follows : FACTORS. PARTS OF THE POLYGON. BASES, OR SUMS PRODUCTS. OF PARALLEL SIDES. ALTITUDES. &ABb Bb= 6.8 Ab= 5.6 38.08 Trap. Bbc C Bb + Cc= 17.4 bc= 6.8 118.32 Trap. CcDd Cc -}- Dd= 20.7 c d= 10.4 215.28 l\DdE Dd 10.1 <*= 5-8 58.58 &FfE Ff= 8.3 /= 8.6 7I-38 Trap. Ffg G *f+ Gg = 14.5 /^r= s 43-50 Trap. GghH Gg+Hh~- 15.4 gh= 8.8 I35-5 2 kAhH Ah- 8.2 ^>^= 9.2 75-44 2)756.10 Polygon ABCDEFGH = 378.o5qm 135-3 CHAPTER VII. AREAS. 145 NOTE i. In the preceding table, in place of dividing each product by 2, we first add the products, and then divide their sum by 2. NOTE 2. In case the interior of the polygon is inaccessible, Fig. ijo illustrates one way of meeting the difficulty. Construct a rectangle which shall completely enclose the polygon, and, from all the corners of the polygon which are not in the Fig. 130. sides of the rectangle, let fall perpendiculars to the sides of the rectangle. In this way a number of right triangles and trapezoids are formed, the areas of which can be found in the usual way ; and it is evident that if the sum of their areas is sub- tracted from the area of the rectangle, the remainder will be equal to the area of the polygon. 135. It will now be clear in what way Geometry supplies us with indirect methods of finding areas. In the first place, it teaches that the area of a rectangle (the square is simply a rectangle with equal sides) depends entirely on the lengths of its two sides, and can therefore be found as soon as the lengths of these sides are known ; it then shows that any parallelogram can be transformed into an equivalent rectangle, that a triangle is half a parallelogram of the same base and altitude, and that a trapezoid consists of two triangles of equal altitude ; lastly, it shows that any polygon what- soever may be decomposed into triangles and trapezoids. Remark. In some cases the rule for finding an area may be very briefly expressed by a formula in which letters are used to represent the values of the various quantities. Let 6" denote in all cases the area, a the side of a square, a and b the adjacent sides of a rectangle, a and h the base and altitude 146 GEOMETRY FOR BEGINNERS. [ 136. of a parallelogram or triangle, a and c the parallel sides, h the altitude of a trapezoid, p and r the perimeter and less radius of a regular polygon. Then, For the square, For the rectangle, For the parallelogram, For the triangle, For the trapezoid, For the regular polygon, S = a 2 , S = a X b, S = a x h, S = \ a X h, S = Ka + c) X h, S=fpxr, [3-1 [4-] T5-] [6.] L7-] [8.] III. Practical Exercises and Applications. NOTE. In solving most of the exercises, it is well to draw (free-hand) a figure representing as nearly as possible the true shape of the polygon. 136. THE SQUARE (S = a*). In a square, given, 1. The side = 23; find the area. 2. The side 35; find the area. 3. The side = 4| ft ; find the area. 4. The side = 5fy ds ; find the area. 6. The side = 6.25; find the area. 6. The side = 48.72; find the area. 7. The area =6259; find the side. 8. The area = 515. 29}. In a rectangle, given, 20. Length = 6o m , breadth = 24; find the area. 21. Length = 2 m , breadth = 8o cm ; find the area. 22. Length = 8.5, breadth = 40; find the area. 23. Length - 92 dkm , breadth = i6.37 dkm ; nnc1 - tne area in hectars. 24. Length = 28f ft , breadth = i"jl ft ; find the area in sq. yds. 25. Length = 627?^, breadth = 435^'^; find the area in acres. 26. Area = 3601, length = 30; find the breadth. 27. Area = 5.6% length = i6 m ; find the breadth. 28. Area .= 50^, length = f km ; find the breadth. 29. Area = 6 acres, breadth = 44^ ; find the length. 30. Area = 767.24 find the sides (see page 50, Exercise 1 8). 16. If the legs of a right triangle are as 3:4, and the hypotenuse = 40, find the legs 17. A rope 41 long is tied by one end to the top of a mast, and by the other end to a ring on the deck 9 m from the foot of the mast. How high is the mast ? 18. The longest side of a meadow in the shape of a right triangle cannot be measured directly, by reason of a swamp. Compute its length, if the other sides are 8o.6 m and i6o m . 19. How many meters of wall must be built to fenc'e in a garden in the shape of a right triangle, the legs being 1 2O m and 35 ? 20. A garden has the shape of a right triangle. The hypotenuse i3O m , one leg = 87.5. Find (z.) the area of the garden; (w.) the distance from the right angle to the hypotenuse. 21. A cannon-ball, fired into the air at an angle of 45 with the horizon, passes over 345 in one second. How high is it at the end of the second ? 22. Two travellers start together from the same place and travel, each 4 miles an hour, one due north, the other due east. When will they be 60 miles apart ? How far apart will they be after each has travelled 80 miles ? 23. Find the area of an isosceles triangle, if one of the equal sides = io m and the base = 8o m (see 80). 24. In an isosceles triangle, the base = i8.4 m , the perimeter 64. Find the area. 25. Prove that in an isosceles right triangle the altitude is equal to half of the base. 26. The area of an isosceles right triangle = 84 the &BCD. Therefore construct the &CED ^ &BCD ( 130, Corollary) , and take it from A A CD, f_ leaving A AE D, which is the trian- gle required. Give in full the construction re- quired. Exercises. 1. Solve this problem for the case where the given base b is less than the base of the given triangle. 2. Construct a triangle with the sides 4 dm , 6*, and 8 dm , and then trans- form it, (z.) Into an isosceles triangle with the base 5 dm . (z'z.) Into a right triangle with a leg equal to 4 dm . (z'z'z.) Into an equilateral triangle, (zz/.) Into a triangle with the angle 50 and the base 3 dm . 3. Transform a given parallelogram into another having a given side. 147. Problem. To transform a triangle ABC (Fig. ij6) into another having a given altitude h. Analysis and construction to be given by the learner (with the aid of Fig. ij6). Exercises. 1. Solve this problem for a case in which the new altitude h is less than that of the given triangle. 2. Construct a triangle with the sides Fig. 136. 4 dm , 6 dm , and 8 dm , and transform it, (z.) Into a triangle, the altitude (z'z.) Into a triangle with the angle 70 and the altitude 4 dm . E 148. Problem. To transform a triangle ABC (Fig. 137) into a A rectangle. Fig. 137. Construction. At A and B erect perpendiculars to AB, and 160 GEOMETRY FOR BEGINNERS. [ 149' through Z>, the middle of A C, draw EF II AB. Rectangle ABEF = &ABC. Proof. Show that each is half of the rectangle ABHG. . Exercises. 1. Transform a triangle into a parallelogram with the an- gle 60. 2. Construct a rectangle with the sides 6 dm and 8 dm , and then transform it into (?'.) a triangle ; ('.) a right triangle ; (m.) an isosceles triangle ; (z'z/.) an equilateral triangle. 149. Problem. To transform a rectangle ABCD (Fig. 138) into a square. Construction. Make BE = BC, and about O, the middle point of B E, describe an arc cutting DA prolonged in F. Join FB, and the square con- structed upon FB will be equal in area to the rectangle ABCD. Proof. EBF is a right triangle. For the triangles OEF and OBF are isosceles (why?) ; hence EFO = FEO, and OFB = OBF ( 80). By addi- tion, EFO + OFB = EFB = FEO -f OBF. But EFB + FEO + OBF = 180 ( 64) ; therefore EFB = 90. l&L. Fig. 138. Since EBFis a right triangle, it follows from 142, Corollary 3, that ~FJ? = the area of the rectangle ABCD. Exercises. 1. Construct a rectangle with the sides 9 dm and 4 dm , and transform it into a square. Measure the side of the square. What is its length ? What ought its length to be ? 2. Construct a triangle with the sides 4 dm , 6 dm , and 8 dm , and transform it into a square. 3. Transform a given parallelogram into a square. 150. Problem. To transform any polygon ABCDE (Fig. 139} int a triangle. I5 1 -] CHAPTER VII. AREAS. 161 Construction. Draw a diagonal AD. Through E draw a line parallel to AD and meeting A B prolonged in F. Join DF. Then is the quadri- lateral BCDF equivalent to the pentagon ABODE. Repeat this construction by drawing the diagonal JSD, a parallel through C, and then joining DG ; this reduces the quadrilateral F &*39* BCDE to the equivalent triangle FGD. Proof. Left for the learner, with the aid of Fig. ijp and the analysis of 146. Corollary. Hence, any polygon may be transformed into a square. By what three steps ? This supplies another way of finding the area of a polygonal field. Construct (to scale) the polygon on paper, transform it into a square, and multiply one side of this square by itself. Exercises. 1. Transform a hexagon into a triangle. 2. Transform a pentagon into a square. 3* Construct three equal octagons. Find the area of the first and second by two of the methods given in 134, and the area of the third by transform- ation into a square. VI. Partition of Figures. 151. In buying and selling land it often becomes necessary to run lines of division through estates, forests, fields, etc., dividing them into smaller parts in some assigned way. This is done either (/.) by computation (arithmetical partition) or (".) by construction (geometrical partition) . The first method, when practicable, is generally adopted. Let us examine some cases. 162 GEOMETRY FOR BEGINNERS. [ 152. 152. TRIANGLES. Case 1. A line CD (Fig. 140) joining the vertex of a triangle ABC to the middle point of the base AB divides the triangle into two equal parts ( 130, Corollary). Again, it AE=. \AB, and EB = \ AB, then &ACE = \kABC, and &ECB = f &ABC (why?). And, A ACE:& ECB = 1:2. In gen- eral, A line joining the vertex of a triangle to the base divides both the triangle and its base into parts which have the same ratio. What ratio have the triangles A EC and ABCt the triangles ECBvrAABCt Case 2. Line of division DE (Fig. 141) passes through two sides of the triangle. = I AC] then i),and &ADE = $ A ADC (Case i) ; therefore A ADE = J X \&ABC= %t\ABC. Again, if CE = CA, and CF=% CB, then (by the same reasoning) A CEF \ X %&ABC^ %&ABC; that is, A CEF \&ABC=2\*i. In general, The ratio of the triangle cut off to the entire triangle is equal to the product of the ratios between the parts of the sides that meet at the common vertex. Case 3. Hence, if a definite part say, one-fourth of a tri- angle is to be cut off by a line from side to side, this can be done in various ways, because the fraction may be obtained as the product of various pairs of factors; for example, and J, \ and |, f and |, I- and ft, etc. If, however, the line of division has to pass through a giver) point in one side, then its position is determined. B F Fig. 141. 1 5 3-] CHAPTER VII. AREAS. 163 For example, if AD (Fig. 142} = \AB, and the AADE is to be made equal to \AABC, then the line of division will pass through a point in the side AC, since A ADC would be too large (how large is it?) ; and, since | -j- = |, the line will pass through a point E such that How must a line DF be drawn if A B DF is to be made equal to \AABC1 153. PARALLELOGRAMS. Case 1. Line of division joins two opposite corners. See 102. Case 2. Line of division AE (Fig. 143) joins a corner to a point in one side. Let (Fig. 143} CE = J. CD, then A ACE = \AACD ( 152, Case i). But AACD = the parallelogram ABCD ( 102);" therefore A ^ C Figm I43 ^ ~~& = % the parallelogram ABCD. CE must be what part of C.Z2, if A ACE = f the parallelogram? Case 3. Line of division joins two opposite sides and is parallel to the other sides. This case is left for the learner to investigate. How can a rectangle be divided into six equal parts ? Case 4. Line of division EF (Fig. 144) joins two opposite sides, and is not parallel to the other sides. Let (Fig. 144) CE = CD, AF %AB, then AACE = \AACD, and A AEF = AADF (why?) = AABD ; therefore A A CE -f- A A E F = the trap- *^ P ^44'* \ of half the parallelogram -f f of half the parallelogram = J the parallelogram. 164 GEOMETRY FOR BEGINNERS. [ 154. Case 5. Line of division FE (Fig. 145) joins two adjacent sides. Let AE = AB, AF = f AC, then ( 152, Case 2) &AEF = X f X &ABC= AAC= T V the whole parallelogram. A B Fig. i 45. 154. We will now give two or three cases in which the partition is effected by direct construction. The proofs of the constructions are left to the ingenuity of the learner. Problem I. To divide a triangle ABC (Fig. 146) from a point D in one side into two equal parts. Construction. }^ CD, bisect AB in E, draw EF II CD, join DF, DFis the line of division required. D E Fig. 146. Problem II. To divide a triangle ABC (Fig. 147) from a point D in one side into three equal parts. Construction. Join CD, trisect AB in E and F, draw EG and FH both II CD, join DG and D H "; these last are the lines of division required. Problem III. To divide a trapezoid A B C D (Fig. 148) from B, C, and D are in proportion or are proportionals. Of the four terms a, b, c, and d which express the proportion, a and d are called the extremes, b and c the means. If b = c, b is called the mean proportional between a and d. 157. A proportion, expressed, as just explained, by the num- bers which are the numerical values of the magnitudes forming the proportion, is subject to certain laws, among which the following are useful for our purposes : 1 . If we write the proportion a : b = c : d in the form -=-,, and then divide the equation i = i by the equation - = -, we o a obtain (Axiom V.) - = -, or b : a = d\ c. a c Law I. The truth of a proportion is not affected if the terms of each ratio are transposed. 2. If we multiply both sides of the equation - = - by -, we b d a obtain (Axiom IV.), after cancelling common factors, - = -, or d : b = c : a. If we multiply - = - by -, we obtain - = -, or b d c c a . a : c = b : d. Law II. The truth of a proportion is not affected, if the ex- tremes are transposed, or if the means are transposed. 3. If we multiply both sides of the equation - = - by the 158.] CHAPTER VIII. - SIMILAR FIGURES. 173 product b x //, we obtain, after cancelling common factors, a X d = b X c. Law III. In every proportion the product of the extremes is equal to the product of the means. If in a proportion the two means and one extreme are given, how can the other extreme be found ? If in a proportion the two extremes and one mean are given, how can the other mean be found ? Exercises. 1. Verify the above laws with the lengths of the lines in Fig. 150. 2. Verify the above laws in the case of the proportion 3:9 = 5:15. 8. In a proportion, the first, third, and fourth terms are 12, 25, 200; find the second term. 4. Find a fourth proportional to the lengths 24, 50, and 8o m . 5. Find a length which shall be to 6o m as 5 : 6. 6. What relation exists between the side of a square and the sides of the equivalent rectangle ? 7. Find the mean proportional between 9 and i6 m . 8. Show that the mean proportional between a and b is equal to "^Tab. 158. Let P and Q denote the areas of two squares, two par- allelograms, or two triangles ; a and b the sides of the squares, or the bases of the parallelograms or the triangles ; h and k the alti- tudes of the parallelograms or of the triangles ; then, for Two squares, P= a\ Q = P' } hence ^= -[ Two parallelograms, P= aKh,Q = bKk] hence = a X h Q b X k Two triangles, P= 2**, Q = *^- } hence 1= **JL 2 2 Q b x k Or, in words, Two squares are to each other as the squares of their sides. State in words the results for parallelograms and for triangles. 174 GEOMETRY FOR BEGINNERS. [ 158. In the cases of parallelograms and triangles, if (i) h = k, and (2) a b, the preceding equations become, (0 --- (2) - h () Q *' Q-J That is, ( T ) Two parallelograms or two triangles, with equal altitudes, are to each other as their bases. (2) Two parallelograms or two triangles, with equal bases, are to each other as their altitudes. Exercises. 1. The legs of a right triangle are 24 and i8 m . In another triangle, the base = 24 m , the altitude = i8 m . Compare the areas of the tri- angles (that is, find the ratio of the areas). 2. The bases of two triangles of equal altitude are 25 and 3o m . Com- pare the areas. 3. If the areas of two triangles of equal altitude are i6oi m and 2OO&DEFis read, the triangle ABC is similar to the triangle D E F. In order to discover the chief properties possessed by similar tri- angles, take on one side AK (Fig. 151) of an angle KAL any number, say five, equal parts, AB, BD, D F, FH, HK, and through the points B, D, F, H, K, draw parallel lines cutting A L in the points C, E, G, /, L ; then ( 112) AC=CE=EG= GI=IL. The triangles ABC, ADE, etc., differ in size, but have the same shape ; hence are similar. Now compare the angles and sides of any two of these triangles, as A DE and AKL. The angles are equal; K Fi L for the angle A is common ; and of the other angles, ADE = AKL, and AED^ALK ( 56). As for the sides, it is evident from the construction that AD \AK = 2:5, and likewise that A E : AL = 2 15. The same ratio also exists between the sides D E and KL ; for, if we draw through B, D, F, and PI lines parallel to AL, these lines will divide (by 112) DE into two equal parts, and KL into five equal parts, 176 GEOMETRY FOR BEGINNERS. [ 159. and the parts of DE will be equal to those of KL ( 102, Corol- lary i) ; therefore, DE:XL = 2 15. Hence, A D : A K = A E :AL = DE\ KL ; that is to say, the sides opposite equal angles in the triangles are proportional. The sides opposite equal angles are termed corresponding or homologous sides. We should obtain like results if the angle A had any other value, or if the parallels BC, DE, etc., had any other direction, and also whatever be the length or number of the equal parts AB, BD, etc. Therefore, - Theorem I. Two similar triangles are mutually equiangular, and have their homologous sides proportional. Corollary. By transposing ( 157, Law II.) the means of the proportion AD:AK=AE:AL, we obtain AD:AE = A K : A L. This shows that, in similar triangles the sides which contain equal angles have the same ratio. Two other theorems also follow from the preceding investiga- tion : Theorem II. If a line is drawn through two sides of a tri- angle, parallel to the third side, a smaller triangle is formed which is similar to the given triangle. Theorem III. A line drawn through two sides of a triangle, parallel to the third side, divides those two sides proportionally. Exercises. 1 . The converse of Theorem III. is also true. State it, and illustrate it by means of numbers. 2, The three sides of a triangle are 84, i2O m , and ioo m . Through a point in the first side, 28 from the intersection of the first and second sides, a line is drawn parallel to the third side. Find where it will cut the second side. Find, also, the ratio between the triangle cut off and the entire triangle ( 152). 3* If the sides of a triangle are l6 m , 24, and 32, and a line is drawn from the first side to the second side, connecting points distant 5 m and 7, respectively, from the intersection of the first and second sides, will this line be parallel to the third side? 4. Draw two similar triangles. Also, two equal triangles. Can two triangles be equal without being similar? Can they be similar without being equal? l6o,] CHAPTER VIII. - SIMILAR FIGURES. 177 160. It appears from the last section that the similarity of two triangles involves six conditions : the equality of three pairs of angles and the equality of the ratios between three pairs of sides. What six conditions does the equality of two triangles involve ? We have seen ( 72-75) that, in general, from the equality of three parts in two triangles, we can conclude that the triangles are equal in all respects. We shall now proceed to show that if two triangles fulfil certain conditions out of the six above mentioned as belonging to similar triangles, then these triangles must fulfil the remaining conditions, and hence must be similar to each other, 161. In the triangles ABC and DEF (Fig. 152) let the angle A == D, B = E, and therefore C^F. Make CG ~ F>, and draw GH\\AB. Then&GJ?C^ADF (I. Law of Equality). But A ABC ^ A GHC ( 159, Theorem II.). Therefore A ABC ~ A DEF. Hence >- Fig. 15*. Theorem (I. Law of Similar- ity) . Two triangles are similar if they are mutually equiangular. Exercises. 1. Show that two triangles are similar if two angles of one' are equal respectively to two angles of the other. 2. What condition suffices to make two isosceles triangles similar? two right triangles? 8. Why are two equilateral triangles always similar? 4. Construct three similar triangles, each having the angles 60 and 80. 5. Construct upon a given line A B a triangle similar to a given triianglej 6. Two triangles are similar if their sides, taken two by two, are parallel. 7. Two triangles are s-imilar if their sides, taken two by two, are perpen- dicular to each other. Point out the homologous sides. Hints. Revolve one of the triangles about a corner through 90 ; then its sides Become parallel respectively to the sides of the other triangle. See Exercise 6, 178 GEOMETRY FOR BEGINNERS. [ 162. 162. In the triangles ABC and DBF (Fig. is 2), let AC: DF=BC\EF, and let also the angle C = F. Make CG = DF, and draw .#11 AB\ then AC:GC= BC\HC ( 159, Theorem III.) Since the first three terms of this proportion are the same as the first three of the proportion assumed at the start, the fourth terms must also be equal; therefore CH = EF. Then A GHC ^ &DEF (II. Law of Equality). But A ABC - A GHC (why ?): therefore A ABC ~ A DEF\ and hence, - Theorem (II. Law of Similarity). Two triangles arc similar if they have an angle equal, and the sides which include the angle proportional. 163. In the triangles ABC and DEF (Fig. 752), let AC -.DF=BC:EF, AOBC, DF>EF, and B = E. Make CG = DF, and draw GH\\AB; then AC-. GC = BC-.HC. This proportion and the proportion assumed at first have the first three terms equal, therefore their fourth terms must be equal, or CH = EF. Whence, A GHC ^ DEF (III. Law of Equality) . ButA^C^ A GHC, therefore kABC^&DEF. Hence, - Theorem (III. Law of Similarity). Two triangles are similar if two sides of one are proportional to two sides of the other, and the angles opposite the greater sides are equal. 164. In the triangles ABC and DEF (Fig. 152], let AC-.DF = BC-.EF, and AC\DF= AB-.DE. Make CG = DF, and draw GH II AB, then- In .the first and third of these proportions, the first three terms are equal, therefore the fourth terms must also be equal ; that is, CH EF. For the same reason it follows from the second and fourth proportions that GH =>E. Hence, A GHC ^ A DEF 165.] CHAPTER VIII. SIMILAR FIGURES. 170 (IV. Law of Equality) ; but A ABC ~~ A GNC, therefore &ABC ^DEF. That is, - Theorem (IV. Law of Similarity). Two triangles are similar if tJicir sides, taken in order, are proportionals. Corollary. If each side of a triangle is two, three, four, etc., times the homologous side of a similar triangle, then the sum of the sides, that is, the perimeter, of one triangle is two, three, four, etc., times the perimeter of the other triangle ; that is, In similar triangles the perimeters are to each other as any two homologous sides. Exercise. Verify the truth of this Corollary by .two triangles, the sides of one being 2 m , 4, 5, and those of the other 6 m , 12, 15. 165. Let the parallels AC, DF (Fig. 153), be cut by lines OA, OB, OC, drawn through a common point O. &OAB ^ &ODE, and ( 56, and I. Law of Similarity). There- fore,- / / _ V ^ = :d^Land = <* B> OE DE^ OE EF Fig.i 53 . Whence The same reasoning might be applied to any number of parallels and lines intersecting them. Therefore, Theorem, Lines drawn through a common point divide par- allels into proportional parts. Exercises. 1. If one of the parallels is divided into equal parts, how will the others be divided ? 2. If a parallel to AC (Fig. 153) is drawn above O, cutting the lines through O prolonged, will the theorem still hold true ? Can you prove ? 180 GEOMETRY FOR BEGINNERS. [ i6 7 . 166. In the right triangle ABC, right angled at C (Fig. draw CD _L A B ; it divides A B into two parts or segments. The smaller right triangles A CD and BCD are each equiangular with respect to the triangle ABC (why?); therefore they are equi- angular with respect to each other. Hence (I. Law of Similarity), AACD- therefore AB \ AC = AC : AD. therefore AB : BC ' = BC : BD. 154- Theorems. If in a right triangle a perpendicular is drawn from the vertex of the right angle to the hypotenuse : 1. Either leg is a mean proportional beeween the hypotenuse and the adjacent segment. II. The perpendicular is a mean proportional between the two segments of the hypotenuse. Exercises. In a right triangle let a and b denote the legs, c the hypote- nuse, h the altitude with hypotenuse for base, / and q the segments of the ihypotenuse. Find the other quantities, if, .1. a = 4-5 m , h = 3.6. 3. a = 9.6, / o.784 m . 2. /& = o.8 m ,/ = i.5 m . 4. c= 167. Let (Fig. 155) kABC^&DEF, and draw the alti- tudes CG and FH corresponding to two homologous sides AB and DE taken as bases. By hypothesis, AB\DE = AC; DF. The right triangles ACG and DFH are similar (why?), there- fore A C : DF = CG : FH. Hence AB-.DE = CG'.FH. Thatis,- Theorem. In two similar triangles, any two homologous sidet taken as bases are to each other ax the corresponding altitudes* 1 68.] CHAPTER VIII. SIMILAR FIGURES. 181 168. Let P and Q denote the areas, a and b the bases, h and k the corresponding altitudes of two similar triangles. Then (, 158), But (| 167) If for the fac the equation becomes, If for the factor - in the first equation we substitute its equal -, k b = y f = <2 J S 2 ' Since and / may be any two homologous sides taken as bases, therefore, Theorem. Similar triangles are to each other as the squares of their homologous sides. Exercises. 1. If the homologous sides of one triangle are each one-fifth those of a similar triangle, compare the areas of the triangles. 2. The areas of two similar triangles are 4OO < i m and 1 296i m ; find the ratio of two homologous sides. 8* Two triangular gardens are similar in shape. What is the easiest way (z.) to compare their areas ? (zY.) to compute both their areas ? 4. A field has the shape of an equilateral triangle. From one corner I wish to cut off a similar triangle equal in area to one-sixteenth of the whole ; how must the line of division be run ? Draw a triangle ; then construct a similar triangle equal in area to 5. 4 times that of the first triangle. 6. 2 times that of the first triangle. 7 \ that of the first triangle. 8. T 9 g- that of the first triangle. 9. Can you prove the theorem of this section by a method suggested by 152, Case 2 ? Hints. Let a line cut i, , |, or any other fraction, from each of two sides of a triangle; and use 159, Exercise i, and Theorem II. 182 GEOMETRY FOR BEGINNERS. [ I6 9 . IIL Problems and Applications. 169. Problem. To find the fourth proportional to three given lines a, b, c {Fig. 156). Construction. Make any angle O ; on its sides take ^ 1 I O DC Fig. 156. 4. b > a, and c > b. 5. b > a, and c > a. 6. 7. and draw BD II AC. OD is the fourth proportional re- quired. Proof. Apply 159, Theorem III. Exercises. Give the con- struction for the following cases : 1. b a. 2. b a, and c a. 7, b>a, and c = b. 170. Problem. 71? find a mean proportional to two given lines a and b (Fig. 157} . 169 furnishes one solution (see Exercises 3 and 7). The _2 more common solution is as fol- _ lows : Construction. Draw a straight Xf> x line, and take A B = a, B C = b. At \\ B erect a perpendicular. Bisect A C \\ in O, and with a radius equal to O A \\ describe an arc cutting the perpen- ~~0 B ' c dicular in D. BD is the mean pro- portional required. Proof. Prove, as in 149, that ADC = 90; then apply 166. ,*'' 1 7 i.] CHAPTER VIII. SIMILAR FIGURES. 183 171. Problem. To reduce or enlarge given lines a, b, c, etc. (Fig. 158) , in a given ratio. Construction. Let the ratio be 3 : 4 ; that is, let it be required to reduce the lines each to three fourths its present ~~~ length. Draw a straight line and ' take on it four equal parts, of which OA contains four, OB three. At A and B erect perpendiculars ; make etc. etc. Join OC, OD, OE, BF, BG, BH, etc., O B A Fig. 158. are the reduced lengths required. Proof. &OBF<^ AOAC, AO3G ^ AOAD, etc. (why?). Then apply 159, Theorem I. Exercises. 1. Draw four lines and reduce them in the ratio 3: 5. 2. Draw four lines and enlarge them in the ratio 5 : 2. 172. Problem. To divide a given line into equal parts. One solution of this problem has been given ( 113) ; but, since that solution requires the construction of several parallels, it is tedious and liable to lead to errors. The following construction is more simple : Construction. Let it be required to divide MN (Fig. 159) into five equal parts. Draw any line MO through M-, then through a point P of this line draw a line PQ II MN, and take on this parallel any five equal parts ending at a point Q. pig. M N 184 GEOMETRY FOR BEGINNERS. [Si7> Through TV 7 arid Q draw a line cutting MO at O. Lastly, draw lines through O and the points of division of PQ. These lines will divide MN into five equal parts. Proof. Apply 165. Exercises. 1. Draw a line and divide it into eight equal parts. 2. Divide the line AB {Fig. 166} into ten equal parts. Solution. Here the parts are so short that if obtained in the way above de- scribed they would be indistinct. In this case pro- C, , -D ceed thus : Erect at A and B perpendiculars, lay off upon each ten equal parts, and through the points of division draw lines parallel to A B. If now we draw the diagonal AD, then the distances from the points marked i, 2, 3, 4, etc., to the line BD will be equal to fa, fa, T 3 o To> et c. of. the line A B. The proof is left to the learner. 3. When (in making maps, plans, etc.) many lines have to be very much reduced in length, it becomes often convenient to employ scales of equal parts, known among surveyors as PLAT- TING SCALES. Explain how such a scale (shown in Fig. 161}, containing one thousand parts, di- vided decimally, is constructed and used. Solution. Draw a straight line, and take on it ten equal parts, A [B = B C= CD, etc., 1 each of these lengths representing one hundred units of length (meters). Divide AB into ten equal parts, each Pip 160. part representing therefore ten units of length. Last- ly, to represent single units of length, draw above AB ten equidistant lines parallel to A B t erect perpendiculars at the points A, B, C, 9 8 7 6 c Fig. 161. 1 Only a part of the scale can be shown in the figure. 1 73-] CHAPTER VIII. SIMILAR FIGURES. 185 etc., and, after having subdivided FO into ten equal parts, join the points of division of FO with those of A D by lines drawn obliquely, as shown in the figure. The perpendicular OB, divided into ten equal parts by the parallels is num- bered from L to 10, beginning at O\ then it follows (see Exercise 2) that the dis- tance from i to the first oblique line measured on the parallel = i unit of length, the distance from i to the second oblique line = II units, etc. ; and the distance from 2 to the first oblique line = 2 units, the distance to the second oblique line = 12 units, etc.; and similarly for the other points of division of OB. The point marked O is the zero of the scale ; hundreds are read off to the right, tens to the left, and units on the vertical line. From this explanation it is easy to see that a length of 348 meters measured on the ground is represented upon the scale by the line mn ; also, that the length p q corresponds to a distance of 216 meters, the length rs to 183 meters, etc. 4. Construct, with the aid of the scale in Fig. 161, a triangle whose sides are 137, i6o m , and 225"*. 5. Draw a triangle, and determine the lengths of its sides by means of the scale in Fig. 161. 6. Construct a scale of one hundred parts, divided decimally. 173. The measurement of the distance of an inaccessible object is one of the most interesting applications of the laws of Geometry. It gives us a good idea of the nature of those methods which have enabled astronomers to measure the distance from the earth to the moon, to the sun, and even to some of the fixed stars. We have already seen ( 95-98) how distances may be meas- ured indirectly by means of equal triangles. Now it is clear that, if equal triangles are employed, the greater the distance to be measured the larger the triangle which must be constructed ; so that, when the distance is considerable, the construction of the triangle becomes very inconvenient, and in many cases (for in- stance, the distance from the earth to the moon) utterly impos- sible. In such cases other methods must be found ; and these methods are supplied by the laws of similar triangles. In fact, the second method given for solving the problems of 95 and 98 is based upon the properties of similar triangles. In 95, Method II., we construct on paper to a reduced scale a fac-simile of the actual lines on the ground. What is \kasfac-simik but a small triangle similar to the much larger one which contains the distance sought? In other words, the triangles ABC and abc 186 GEOMETRY FOR BEGINNERS. [ 174. (Fig. 88) are similar; hence ac : ab =AC : A B, a proportion which, by making (as on page 105) ac = i6 cnv , ab = 24 cm , AC = 8oo m , becomes 16:24 = 800 : AB ; whence AB = 24 X 8oQ = i 2 oo ra . 16 Representing lines to scale on paper, all angles remaining un- changed, always gives a figure similar to that formed by the unre- duced lines, and is termed PLATTING TO SCALE. Some examples will now be given of the indirect measurement of distances by means of similar triangles. 174. Problem. To measure the distance of two points A, X, sepa- rated by a river. At the point A (Fig. 162), sup- posed accessible, erect (with a sur- veyor's cross or other means) a line _L AX, and measure its length to any point B. Suppose, for exam- ple, that AB i20 m ; take BC = io m , and at C erect a second per- pendicular CD, meeting the line B X at D. Measure CD ; let its length be i8 m . Then it follows that, in all right triangles similar to BCD that is to say, in all right triangles having an acute angle equal to DBC the ratio of the legs must be 18 : IO. 1 Among these similar right triangles is the triangle ABX; there- fore, AX-.AB = 18: 10; whence, AX=^2i6 m . Exercises. 1. In Fig. 162, BC is made equal to io m solely for con- venience of computation. If B C= 15, what would D C be found to be ? 2. If A is on the edge of the river, can you find A X by this method ? What change in the position of the triangle B CD is necessary ? Give the construction for this case. 3. How can AX be found by means of two equal right triangles ? 4. How can A X be found by means of an isosceles right triangle ? 1 In Trigonometry, 18 : 10, or {, is called the TANGENT of the angle DBC. I75-] CHAPTER VIII. SIMILAR FIGURES. 187 175. Problem. To find the distance between two inac- cessible points, X, Y. Case 1 (Fig. i6j). Choose an accessible point, A, and meas- ure the distances AX, AY, by the preceding problem. Upon AX measure a length Ax equal to any fractional part (as T V, 5^, etc.) of AX, and upon A Y meas- ure Ay equal to the same frac- tional part of A Y. Lastly, meas- ure xy, which is parallel to X Y (why ?) . Then the distance X Y can be found from the Fig. 163. If, for example, Ax = -fa AX, then XY = 10 xy. Case 2. Where the line X Y prolonged is accessible (fig. 164) . Erect at M. an accessible point, a line J_ MX, and mark the points N where this line intersects the perpendicular let fall from Y, and P, where it cuts XY prolonged. Measure MP and NP; let, for example, MP = 360, NP = 200. Take PO = io m ; at O erect OQ _L PM, and measure the hypotenuse PQ of the right triangle POQ. Sup- pose, for example, that PQ = 24 ; then, in all right triangles having an acute angle equal to OPQ the ratio of the leg corresponding to OP to the hypotenuse is 10 : 24.* PMX and PNY are two such tri- angle? therefore, N Fig. 164. M 10 : 24 = PM, or 360 10 : 24 = PN, or 200"= and XY= PX - PY = 384* PX, whence PX = 86 4 m ; PY, whence PY = 48o m ; 1 In Trigonometry, the ratio 10: 2.4, or j 5 ^, is called the COSINE of the angle OPQ. iss GEOMETRY FOR BEGINNERS. [ Exercises. 1. What is the advantage in Case 2 of making P = io m rather than any other number ? 2. Explain how, with the aid of an instrument for measuring angles, the distance X Fin Fig. 163 may be found by platting a similar triangle on paper. By this method how many lines must be measured on the ground ? How many by the other method ? 3. Given (Fig. 163} A X = 6oo m , A Y= 720, Ax = 5 AX, Ay = -fa AY, xy-2i.iS m , XAY= 36. Find XY (z.) by a direct proportion; (.) by platting a similar triangle to scale on paper. How nearly do the results agree ? Why do they not exactly agree ? 4. Fig. zdj illustrates another method of measuring an inaccessible line XY, a method by which only one line AB has to be measured on the ground. Can you explain the method ? Fig. 165. 176. Problem. To measure the vertical height of an ob- ject (a tower, tree, church- spire, etc.) 1st Method. By means of a shadow. Suppose that we wish to find the height A X of a tree (Fig. 166) which casts the shadow 'A B on the ground. Fix a staff ver- tically in the ground near the tree, and measure its height mn. Measure, also, the lengths of the shadows n o cast by the staff, and AB cast by the tree. Now the ray of light XB is parallel to the ray mo : whence it follows that ABX and mno are similar trian- 1 76.] CHAPTER VIll. SIMILAR FIGURES. 189 gles (why ?) . In other words, the shadows cast at the same by different objects are proportional to the heights of the objects. Therefore no : AB = m n : A X. If no = 0.9, AB = 24. 9 m , mn = i.i7 m , then, 0.9 : 24.9 =1.17 : AX, whence AX =32. 37. NOTE. We may also reason thus : if 9O cm of shadow correspond to H7 cin of height, then l cm of shadow corresponds to -U_7 = 1.3^1 o f height ; therefore 249o cm of shadow will cor- respond to i.3 cm X 2490 = 3237 = 32.37 m . 2d Method. With the aid of an instrument for measuring angles. Let, for example, the height of the tower (Fig. 167) be required. Place the instrument at a point B, some distance from the foot of the tower, and measure the angle abX, or angle of elevation, as it is termed. Measure AB, the distance from the instrument to the foot of the tower. Then construct on paper a right triangle similar to abX, and the side of this triangle homologous to a X will give to the scale employed the height aX, to which we must add A a, in order to obtain the entire height. pnnnnc Fig. 166. H 4- " Fig. 167. Fig. 168. 3d Method. When the base of the object is inaccessible, choose 190 GEOMETRY FOR BEGINNERS. [ 177. a convenient place near the object, and measure a base line AB (Fig. 1 68). Then measure the angles at the base, B and B A X, and also the angle of elevation of the object XAO, and construct on paper a triangle similar to XB A. This will give to scale the length A X. Then, knowing A X and the acute angle X A O, con- struct on paper the right triangle OA X, which will give to scale the height OX required. Here the unknown height OX depends on the construction of two similar triangles, the first similar to the triangle ABX, the second to the right triangle OAX. If the scale of reduction in both cases is i mm to i m , the number of millimeters contained in the line on paper which represents OX will be the same as the number of meters in OX. Exercises. 1. A monument casts a shadow of 44 at the same time that: a rod 2 m high casts a shadow of i.33 m ; find the height of the monument. 2 In the 2d Method, why is the construction of a triangle on paper avoided by placing the instrument so that the angle of elevation shall be equal to 45 ? 3. What instruments are required in the 3d Method ? 4. Find the height of the building (Fig. 168} if A B = 250, angle B = 25 30', angle XA B = 104, and angle XA = 41 30'. IV. Similar Polygons. 177. Two polygons are said to be SIMILAR if they have the same shape. In order to see what are the properties of similar polygons, divide the polygon A B CD E (Fig. 169) into triangles by drawing the diagonals AC, AD ; then begin at any point M of AB, and make a new polygon by drawing MN II J3C, NO II CD, OP II DE. The new polygon A MNO P is smaller than the original polygon, but has the same shape, and is therefore similar to it. By the same process, we can make a polygon AQRST larger than the polygon ABCDE, and similar to it. I 77-] CHAPTER VIII. SIMILAR FIGURES. 191 Now the triangles AMN, ABC, AQR, are similar (why?); so are the triangles AON, ADC,ARS\ and lastly, the triangles A OP, ADE, A S T. Whence it follows that, Theorem I. Similar polygons are composed of the same number of triangles similar to each other and similarly placed. From this it follows, that similar polygons are mutually equiangular. Why, for example, is the angle A M B o AMN equal to the angle ABC? Fig. 169. the angle MNO equal to the angle BCD? Compare now the sides in any two of the polygons, say the two smaller ones. From the similar triangles AMN and ABC, AM:AN = MN-.BC, and also, AN: AC = MN: BC-, and from the similar triangles A NO and AC D, AN-. AC = NO: CD. From the second and third proportions, it follows (Axiom I.) that In this way, whatever be the number of sides, it may be shown that the sides similarly placed are always proportional. Theorem II. Similar polygons are mutually equiangular, and have their homologous sides proportional. Corollaries. I . Two regular polygons of the same number of sides are similar. Why? 2. If each side of a polygon is 2, 3, 4, etc., times as great as the homologous side of a similar polygon, then the sum of the sides, that is, the perimeter of the first polygon, must be 2, 3, 4, etc., times as great as the sum of the sides, or the perimeter, of the second polygon. That is to say, the perimeters of similar polygons are to each other as any two homologous sides. 192 GEOMETRY FOR BEGINNERS. [ 178. 178. If two similar polygons, the sides of one of which are double those of the other, are divided into triangles, by drawing diagonals from the vertices of two equal angles, then each triangle in the first polygon is (by 164) four times as great as the corre- sponding triangle on the second polygon ; hence the sum of all the triangles in the first polygon in other words, the area of the first polygon is four times as great as the sum of all the triangles, or the area, of the second polygon. Hence the areas of the two polygons are to each other as 4 : i. But this is the ratio of the squares of any two homologous sides. A like result would follow if the sides had any other ratio than 2:1. Hence, - Theorem. Similar polygons are to each other as the squares of their homologous sides. Therefore, if we have a polygonal figure on the ground, and construct a similar figure on paper with its sides reduced to , ^, |-, T V, etc., of their lengths, the area of the figure on the paper will b e i> FJ TB> T etc '> the area of the larger and similar figure on the ground. Exercises. 1. Compare the areas of two similar polygons with their perimeters (see 177, Corollary 2). 2* Prove that the areas of regular polygons of the same number of sides are to each other, (z.) as the squares of their sides ; (zY.) as the squares of their perimeters. 3. If in two hexagonal parks a side of one is four times a side of the other, how much larger is one park than the other ? Find, also, the ratio of their perimeters. 179. Problem. To construct a polygon similar to a given polygon. There are various methods of solving this problem. 1st Method. When one side F G of the required polygon is given. Solution. By means of 177, Theorem I. Divide the given polygon ABODE (Fig. 170) into triangles by diagonals; then 1 79-] CHAPTER VIII. SIMILAR FIGURES. 193 construct upon the given length FG, homologous to AB, a series of triangles similar to those of the given polygon respectively, and similarly placed. The poly- gon FGHKL is the polygon required. Give the construction in full. 2(1 Method. One side F G being given, as before. Solution. On one side AB of the given polygon ABCDE (Fig. 777) take AAf=FG, construct the polygon AMNOP^ ABCDE (how is this done?), and then construct (by 117) upon the side FG the polygon FGHKL & the polygon AM NO P. Give the construction in full. H Fig. 170. F 3d Method. When the homologous sides of the two polygons are to be to each other in a given ratio. Solution. Choose any point O, either within or without the given polygon (see Figs. 172 and //j), and join it to all the cor- ners by straight lines. Divide one of these lines at a point Af, so that O M\ OA the given ratio, and then draw MN II AB, NP II BC, etc. MNPQR is the required polygon.. 194 GEOMETRY FOR BEGINNERS. [ 179. In Fig. 172, OM\ OA = i : 3 ; in Fig. 173. OM\ OA = 3 : 5. Give the construction and proof for both cases. Fig. 172. Fig. 173. Exercises. 1. Given a hexagon; construct a similar but larger hexagon. 2. Given an octagon; construct a similar but smaller octagon. 3. Upon the lengths 2O cm and 30 as homologous sides construct two similar pentagons. 4. Draw a polygon, and then construct another whose sides shall be, (z.) one-half those of the first polygon ; (it.} two-thirds those of the first polygon ; (tit.') twice those of the first polygon ; (zz/.) three and one-half times those of the first polygon. 5* Draw a polygon, and then construct a similar polygon, (z.) one-ninth as large as the first ( 174) ; (z'z.) nine twenty-fifths as large as the first ; (z'z'z.) four times as large as the first ; (zV.) twice as large as the first. 6. Construct two similar pentagons which shall be to each other as 4: 9. 7. Construct two similar regular pentagons, one having three times the perimeter of the other. What is the ratio of their areas? 8. Construct two regular hexagons about the same point as centre, making the less radius of one = l\ times that of the other. 9. Construct two regular octagons, making the less radius of one = f that of the other. What is the ratio of their perimeters ? of their areas ? CHAPTER VIII. REVIEW. 195 REVIEW OF CHAPTER VIII. SYNOPSIS. 1. A proportion is an equation between two equal ratios. 2. The truth of a proportion is not affected if the terms of both ratios are transposed. 3. The truth of a proportion is not affected if the extremes are transposed, or if the means are transposed. 4. In every proportion the product of the extremes is equal to the product of the means. 5. Two squares are to each other as the squares of their sides; two paral- lelograms, or two triangles, as the products of their bases by their alti- tudes. 6. If two parallelograms, or two triangles, have equal bases, they are to each other as their altitudes ; if they have equal altitudes, they are to each other as their bases. 7. Similar triangles are triangles that have the same shape. 8. Similar triangles are mutually equiangular, and have their homologous sides proportional. 9. In two similar triangles, the sides that include equal angles have the same ratio. 10. A line drawn through two sides of a triangle parallel to the third side cuts off a smaller triangle similar to the entire triangle. 11. A line drawn through two sides of a triangle parallel to the third side divides those sides proportionally. 12. I. Law of Similarity. Two triangles are similar if they are mutually equiangular. 13. II. Law of Similarity. Two triangles are similar if they have an angle equal, and the sides including this angle proportional. 14. III. Law of Similarity. Two triangles are similar if two sides of one are proportional to two sides of the other, and the angles opposite the greater sides are equal. 15. IV. Law of Similarity. Two triangles are similar if their sides, taken in order, are proportional. 16. Lines through a common point divide parallels proportionally. 17. If in a right triangle a perpendicular is drawn from the vertex, of the right angle to the hypotenuse, 1.96 GEOMETRY FOR BEGINNERS. (1) each leg is a mean proportional between the hypotenuse and the adjacent segment ; (2) the perpendicular is a mean proportional between the segments of the hypotenuse. 18. The homologous sides of similar triangles are to each other (i) as the corresponding altitudes, (2) as the perimeters. 19. Similar triangles are to each other as the squares of their homologous sides. 20. Problem. To find the fourth proportional to three given lines. 21. Problem. To find the mean proportional between two given lines. 22. Problem. To reduce or enlarge given lines in a given ratio. 23. Problem. To divide a given line into equal parts. (Two ways.) 24. Problem. To measure the distance from an accessible to an inaccessi- ble point. 25. Problem. To measure the distance between two inaccessible points. (Two cases.) 26. Problem. To measure the height of an object. (Three methods.) 27. Similar polygons are polygons that have the same shape. 28. Similar polygons are composed of the same number of similar triangles similarly placed. 29. Similar polygons are mutually equiangular, and have their homologous sides proportional. 30. Regular polygons of the same number of sides are similar. 31. The perimeters of similar polygons are to each other as any two homo- logous sides. 32. Similar polygons are to each other as the squares of their homologous sides. 33. Problem. To construct a polygon similar to a given polygon. (Three methods.) EXERCISES. 1. A square and a rhombus have equal areas : find the ratio of their perimeters, if the altitude of the rhombus is one-fourth that of the square. 2. What single condition will make two right triangles similar ? also two isosceles triangles ? 3. Given a triangle, two of whose angles are 55 and 80, construct a simi- lar triangle one-fourth as large. CHAPTER VIII. - REVIEW. 197 4. Given a right triangle, an acute angle of which = 35; construct a similar triangle four times as large. 5. Construct an isosceles triangle with the angle at the vertex = 30, and one side = 6o cm . Then construct a similar triangle one-ninth as large. 6. Construct a triangle similar to a triangle whose sides are to each other as the numbers 7, 8, II. 7. If a triangle is to be made three times as large as a given triangle, but similar to it, what values must its angles have ? its sides? 8. Prove that parallels divide all lines that intersect them into proportional parts. Hints. This is an extension of 112. Compare also this theorem with the theorem of 165. 9. The sides of a right triangle are 21, 28, and 35; find (z.) the seg- ments of the hypotenuse made by a perpendicular let fall from the vertex of the right angle, (z'z.) the length of this perpendicular. 10. Prove that the bisector of an angle of a triangle divides the opposite side into parts that have the same ratio as the adjacent sides. Hints. If ABC is the triangle, B D the bisector, prolong AB till it is met by a parallel to the bisector throuh C. A A B D A CBE\s, isosceles. at E by a parallel to the bisector through C. A A B D ~ A ACE, and 11. The three medians of a triangle meet in one point. (A median is a line drawn from a vertex and bisecting the opposite side.) &ints. LetACbe the triangle. Draw two of the medians AD and BE meeting at a point O, and then draw DF\\ AC meeting BE in F. Show that A D F O ^ A A O E ; whence show that A O : O D =2:1. Draw the third median CG, cutting AD in P, and show in like manner that AP : PD II 2 : i ; therefore P must coincide with O. The point O is called the centre of gravity of the triangle. 12. Construct a right triangle having given, (.) The hypotenuse 70, and the ratio 3 : 2 of the legs. (z'z.) The altitude 40, and the ratio 3 : 5 of the legs. 13. Construct on the diagonal of a rectangle an equal rectangle. 14. What is the ratio of the area ojf a field to that of its plan on paper, (z.) if the scale of reduction is i cm to the metei ? (zz.) if the scale of reduction is i mm to the meter ? 15. Wishing to find the distance from a point A on the mainland to an island B, I erect at A a line A C JL A B, measure upon it a length A C = 900, and also measure the angle ACB, which I find to be 80. From these data find the distance A B. 16. In Fig. s6j, let A X 542, A Y= 735, and the angle A = 57 30'; find the distance XY. 198 GEOMETRY FOR BEGINNERS. 17. A vertical rod 5 long throws upon a horizontal plane a shadow 5.7 long. At the same time a tower throws a shadow 130 long. How high is the tower ? 18. Explain how to find the height A B (Fig. 174) by means of the right triangles 4CandAD. 19. Explain how, without measuring any angles, the distance X Y (Fig. *75) can be found by means of observations and measurements performed on this side of the water 20. The sides of a pentagon are 12, 20, n m , 15, and 22 m ; the perimeter of a similar pentagon is i6 m ; find its sides. Fig. ///. 21. Construct a polygon similar to a given polygon, and having to it the ratio 4 : 9. 22. Construct a rectangle similar to one given rectangle and equal to another given rectangle. l8o.] CHAPTER IX. THE CIRCLE. 199 CHAPTER IX. THE CIRCLE. CONTENTS. I. Sectors, Angles at the Centre, Chords, Segments ( 180-184). II. Inscribed Angles (185-188). III. Secants and Tangents ($ 189-192). IV. Two Circles ( 193, 194). V. Inscribed and Circumscribed Figures ( 195-207). VI. Length of a Circumference ( 208-211). VII. Area of a Circle ( 212-218). I. Sectors, Angles at the Centre, Chords, Segments. 180. Review 26, 27, 28. Some new definitions will now be given : Definitions. I. The parts into which a circle (Fig. 176) is divided by two radii are called SECTORS. II. The angle between two radii is called an ANGLE AT THE CENTRE. III. A straight line joining two points of a circumference is called a CHORD. IV. A chord passing through the centre is called a DIAMETER. V. A chord divides the circumference into two arcs ; if equal, these arcs are called SEMI-CIRCUMFERENCES ; if unequal, the GREATER , to O. Then O D> OA (why?). That is, OA is shorter than any other line which can be drawn from O to the line BC; therefore ( 83), OA i_ BC and BC _L OA. Theorem. A tangent to a circle is perpendicular to the radius drawn to the point of contact. Fig. 187. Exercises. 1 . Find the locus of the centres of circles which touch a given line in a given point. 2. Find the locus of the centres of circles which have a given radius and touch a given line. (Two lines. See 90.) 3. The radius perpendicular to a tangent bisects every chord parallel to the tangent. 191. Problem. Through a given point to draw a tangent to a given circle. There are three cases ; for the given point may lie, (*.) within, (ii.) in, (Hi.) without, the circumference. Case (.) cannot be solved. Why ? Case (ii.}. 190 supplies the means of solution. How? Give the construction. 210 GEOMETRY FOR BEGINNERS. [ 191. Case (iii.). Construction. Let A (Fig. / the diameter of the circle). 7. Describe a circle that passes through a given point and touches a given line in a given point. (Where cannot the first point be ?) 8 To a given circle draw a tangent which, (a) is parallel to a given line. () is perpendicular to a given line. (c) makes a given angle with a given line. 9. Describe a circle that touches two given intersecting lines, and, (a) whose centre is in a given line. () which touches one of the lines in a given point. (f) which has a given radius. CHAPTER IX. THE CIRCLE. 211 10. Describe a circle that touches two given parallel lines, one of them in a given point. 11. Describe a circle that touches two given parallel lines, and passes through a point between them. (Two solutions.) 12. Describe a circle that touches three given lines. (Four, two, or no solution.) A Fig. 189. 192. Let (Fig. 189) AD be a tangent to a circle at A, and AB any chord drawn from A. Draw the diameter AC, and join B to the centre O of the circle. CAD = 90 (why?) = arc ABC (measured in degrees) . CAB=COB (why ?) = arc C B (measured in degrees). The angle between the tangent and the chord BAD = CAD CAB. Therefore, BAD = arc ABC - % arc C B = arc BD. Theorem. The angle between a tangent and a chord drawn through the point of contact, is measured in degrees by half the arc included between its sides. Exercises. 1. Prove that the obtuse angle BAR (Fig. 189) made by the tangent and the chord is measured by half the arc included between its sides. 2. The angle made by two secants that meet without the circle is measured by half the difference of the arcs included between its sides. Hint. Join two alternate points in which the secants meet the circle. Use 66 and 130. 3. The angle made by two tangents drawn through a point without a circle is measured by half the difference of the arcs included between its sides ( 66 and the Theorem above). 4:. The less arc between the points of contact of two tangents drawn through a point without a circle = 120. Find the angle between the tan- gents. 5. The angle made by a tangent and a secant that meet without the circle is measured by half the difference of the arcs included between its sides. 212 GEOMETRY FOR BEGINNERS. [ 193. IV. ~ Two Circles. 193. The relative position of two circles depends upon the positions of their centres and the lengths of their radii. Definitions. I. Two circles that have the same centre (Fig. 190) are said to be CONCENTRIC. II. The plane figure lying between their circumferences is called a RING. III. Two circles that have different cen- tres are said to be ECCENTRIC. IV. The line joining their centres is called the LINE OF CENTRES. 194. Let (Fig. 191} O and P be the centres of two eccentric circles, with the radii r and r 1 respectively, and let us proceed to examine the different positions which these circles will have, as the distance of their centres OP = c is supposed continually to in- crease. There are five different cases. Case (i.) (Fig. 192} . c < r r'. If the distance of the centres is less than the difference of the radii, the smaller circle lies wholly within the larger. Fig. 191. Fig. 192. 193- Case (*.) (Fig. 193) . c=r r 1 . If the distance of the cen- tres is equal to the difference of the radii, the circles touch, or are tangent, internally. 1 94.] CHAPTER IX. THE CIRCLE. 213 Case (Hi.'} (Fig. 193) . c > r r 1 and < r + /. If the distance of the centres is greater than the difference, but less than the sum of the radii, the circles intersect in two points. Case (if.) (Fig. 194) . c = r + /. If the distance of the cen- tres is equal to the sum of the radii, the circles touch, ^ or are tangent, externally. Fig. 194. Fig. 195. Case (v.) (Fig. 195}. c>r+r'. If the distance of the cen- tres is greater than the sum of the radii, the circles lie wholly outside, each other. Exercises. 1. Can you give a reason why two circles cannot meet in more than two points ? 2. What different positions can two circles have relatively to each other ? 3. Let r and r 1 be the radii of two circles, c the distance of their centres. What relative positions have the circles for the following values of r, r' , and c ? ()r=s, ^=3, c = 8. 0) rT, r'-^, c = 2. (c} r = 6, r r =2, ; therefore, ^/> + BP=BD + BC = CD-, that is, 77/ = BE ; that is, The ends of the minor axis are equidistant from the foci, the distance being equal to half the major axis. V. Since A A OE ^/^AOF, therefore OE = OF ; that is, The centre of an ellipse bisects the minor axis. Exercises. 1. To what are the three sides of the right triangle AOE {Fig. 2ii\ respectively equal ? 2. What kind of plane figure \&AEBF (Fig. 211} ? Why ? 3. Given the axes of an ellipse, find the positions of the. foci. 4. Given the major axis and the positions of the foci, find the minor axis. 5. Semi-major axis = 18; semi-minor axis= I2 m . Find the eccentricity. 6. Semi-major axis = 1. 3 m ; eccentricity = 0.34. Find the semi-minor axis. 1. Semi-minor axis= 2.9; eccentricity = 0.9. Find the semi-major axis. 8. Major axis = 58 dm ; minor axis = 45 dm . How far apart are the foci ? 9 The path of the earth about the sun (at one focus) is an ellipse whose axes are 20,657,70x3 and 20,655,100 geographical miles. How can the least and the greatest distances of the earth from the sun be computed ? 221. AREA OF AN ELLIPSE. If we describe circles about the axes of an ellipse as diameters (see Fig. 213), it is obvious that the area of the ellipse is less than that of the circle with the semi- major axis as radius, and greater than that of the circle with the 221.] CHAPTER X. THE ELLIPSE. 243 semi-minor axis as radius. Now it can be proved that the area of the ellipse is exactly equal to that of the circle whose radius is a mean proportional between the semi-axes of the ellipse. If a and b are the semi-axes of an ellipse, r the radius of the circle equal in area to the ellipse, then a : r r : b, or r 2 a b. The area of this circle = -r 2 ( 213) ; therefore the area of the ellipse = TC r 2 ; or, since r 2 = a b, Area of the ellipse = irab. [12.] That is to say, the area of an ellipse is equal to the continued product of its semi-axes and the number TT. Exercises. Find the area of an ellipse, if, 1. The semi-axes are I2 dm and 8 dm . Find the area. 2. The axes are 8.5 and 4.5. Find the area. 3. The major axis = 9, the eccentricity = 1.5. Find the area. 4. The minor axis = 2 m , the eccentricity = i m . Find the area. 5. The major axis = 5, the focal distance = 3. Find the area. 6. The area= 681, the major axis = I2 m . Find the minor axis and the eccentricity. 7. How much cloth will cover an elliptical table 2.4 long, i.3 m wide ? 8. The famous amphitheatre at Verona, built by the Emperor Domitian, and which seated 24,000 spectators, has for its base an ellipse whose axes are 133 and 105. Find the area of this ellipse. 9. The section of an arch has the shape of a semi-ellipse i6 m long, 4.8 m high. Find the area of the section. 10. The diameter of a circle = 7. Upon this diameter as major axis an ellipse is constructed half as large as the circle. Find its minor axis. 11. A pond in the shape of an ellipse, with the axes 6o m and 44, is changed to a circle with the radius 6o m . How much larger is it than before ? 12. Find the radius of a circle equal in area to an ellipse whose axes are I3 m and 9. 13. Find the side of a square equal in area to an ellipse whose axes are I7 m and io m . 14. Find the area of the largest ellipse that can be made out of a board 2 m long and o.6 m wide. 15. The axes of an ellipse are 4 m and 3. Upon these axes as diameters circles are described. Find the areas of all three figures. 244 GEOMETRY FOR BEGINNERS. [ 222. II. Construction of Ellipses. 222. Problem. To construct an ellipse, having given the major axis and the eccentricity. Let {Fig. 212} CD be the given major axis, O its middle point. 3^3 Make OA = OB the given ec- centricity ; then A and B are the foci. Find the points E and F, } D such . that the distance from each of them to A and to B is equal to the semi-major axis ; E and F are the ends of the minor axis. In order to find more points of the ellipse, assume any points m, 11, p, etc., lying in the major axis between A and O. With a radius equal to Cm, describe arcs from each focus as centre, both above and below CD\ and likewise do the same with a radius equal to Dm ; the four points marked i, where these arcs intersect, are four points of the ellipse. For, by this construction, the sum of the lines drawn from either of the points marked i to the two foci is equal to the major axis. (See 220, III.) In like manner, by the use of n, the points marked 2, and by the use of/, those marked 3, are found. By drawing (free-hand) through all these points a curved line, making the curvature as nearly uniform as possible, we obtain a curve which approaches the nearer to the exact ellipse required, the greater the number of points that are previously determined. Exercises. 1. How can an ellipse be described by continuous motion ( 219). (Gardeners sometimes use this method to trace the outline of an elliptical flower-bed.) 2 Construct the ellipse whose major axis = 8o cm and eccentricity = 3O cm , by finding sixteen points in the ellipse. i 1 * Construct the ellipse whose major axis = 6o cm and eccentricity = i6 cm , fourteen points. 22 3 .] CHAPTER X. THE ELLIPSE. 245 223. Problem. To construct an ellipse, having given the axes, by the aid of circles described about the axes as diameters. Let AB and CD {Fig. 213) be the given axes intersecting at O. Describe circles about A B and CD as diameters. Through O draw any radii O E, OF, etc. Through the points where these radii cut the inner circle draw lines II AB, and through the points where the radii cut the outer circle draw lines II CD. The intersection of each pair of these lines is a point in the required ellipse ; so that nothing remains but to draw (free-hand) through all the points of intersection a smooth curve. Fig. 213. Exercises. 1. Construct an ellipse having for its axes 72 and 4O cm , and find the positions of the foci. 2. Construct an ellipse whose major axis is to the minor axis as 5 : 4, and find the positions of the foci. 224. Problem. To construct with arcs of circles a curve that resembles an ellipse. Draw a line A D (Fig. 214) , and divide it into three equal parts, AB, BC, CD. With a radius equal to one of these parts describe about B and C as centres two circles, cutting each other in E and F. Through these A\ points and the two centres draw the lines EG, EH, FI, FK. Then de- scribe about E and F as centres, with a radius equal to EG, the arcs Fig. 214. IK. The curve AGPID IK is the required curve. 246 GEOMETRY FOR BEGINNERS. [ 22 4 . The elliptical curves employed in the arts are usually constructed in this or a similar way from arcs of circles. By compounding arcs of circles, in different ways, a great variety of curves may be constructed. Some examples are added below, in Figs. 215222. Explain how each figure is constructed. Fig. 215. Fig. 216. Fig. 217. Fig. 218. Fig. 219. Fig. 220. 22 4 /! CHAPTER X. THE ELLIPSE. 247 Fig. 22T. Fig. 222. Exercises. 1. Construct an elliptical qirve with arcs of circles. 2. Construct the curve shown in Fig. 215. 3. Construct the curve shown in Fig. 216. 4. Construct the curve shown in Fig. 217. 5. Construct the curve shown in Fig. 218. 6. Construct the curve shown in Fig. 219. 7. Construct the curve shown in Fig. 220. 8. Construct an aval like that in Fig. 221. 9. Construct a spiral like that in Fig. 222. 10. Find the area of an oval like that of Fig. 221, if the diameter AB = 4. Explain each step of the solution, stating the truth on which it depends. SYNOPSIS OF CHAPTER X. 219. Definition of the ellipse, its foci, its centre, its eccentricity. Relation of the ellipse to the circle. 220. Definition of the axes of an ellipse. Relations between the ends of the axes, the foci, and the centre. Five general relations. 221. Area of an ellipse. Formula for the area. 222. Construction of an ellipse, given the minor axis and the eccentricity. 223. Construction of an ellipse with the aid of circles, given the axes of the ellipse. 224. Construction of a curve resembling an ellipse by means of arcs of circles. Other instances of the use of circular arcs in constructing curves. 248 GEOMETRY FOR BEGINNERS. [ 225. CHAPTER XI. PLANES. CONTENTS. I. Straight Lines in Space ( 225-227). II. A Plane ( 228-230). III. A Plane and a Straight Line ($$ 231-235). IV. Two Planes ($$ 236-239). V. Three Planes ($ 240). VI. Solid Angles ( 241). I. Straight Lines in Space. 225. Thus far the figures studied (triangles, polygons, circles, etc.) are such as can be represented in their true shape on paper or the blackboard ; they are figures which lie wholly in one plane. We shall now proceed to consider some of the properties of figures which are not confined to one plane. Every actual body which we can see or handle is an example of such a figure. We shall begin by examining the relations of lines and planes in space to one another. NOTE i. In this part of Geometry i\\e first and cAief difficulty encountered by beginners lies in conceiving and holding firmly in the mind the true relative posi- tions of the lines and planes under consideration. Without the power to do this the proofs of theorems amount to mere words ; with it the proofs are easily mastered ; in fact many of the theorems will seem self-evident. Accordingly, in the brief space here devoted to the subject, rigorous proofs will not always be given, but care will always be taken to set forth the truth so that it shall be completely realized in thought. The geometric imagination must be exercised and strengthened before the mind is prepared to study the subject with all the rigor of logic. NOTE 2. An additional difficulty arises from the fact that in representing magnitudes in different planes upon a single plane surface (that of paper or the blackboard) these magnitudes must, in general, be changed in size or shape or both. Lines and angles must be altered in size, plane figures in both size and shape. The rules for doing this correctly are based on the principles of Geometry, and constitute the Art of Perspective, of Isometric Drawing, etc. There is one general rule of Isometric Drawing which it is useful to bear in mind : a square, not in the plane of the drawing, is represented in that plane by a rhombus, a rectangle by a parallelogram, and a circle by an ellipse. Beginners should make a habit of comparing the figures in the book, and those which they draw themselves, with actual lines and planes placed in the proper relative positions, until they are able to recog- nize at once the true relative positions of the parts of a drawing, and to make similar drawings themselves. For lines, wires and stretched threads may be employed; for planes, paper, cardboard, a thin piece of board, glass, the floor and sides of the 226.] CHAPTER XI. PLANES. 249 room ; for bodies, suitable models made of wood or cardboard (see figs. 1-6) or constructed simply of stiff wire for the edges. NOTE 3. Some of the relations of lines and planes have been noticed in Chap- ter I. and in 24, 25. The learner should refresh his memory, if necessary, by reviewing them. 226. Two straight lines in space may have either of three relative positions, as follows : (/.) They may be parallel to each other. (//.) They may intersect each other. (///.) They may be neither parallel nor intersecting. In the first two cases the lines must lie in the same plane ( 29). Give examples of both cases. If the lines intersect, they may be either perpendicular to each other or inclined to each other ( 48). In the third case the two lines do not lie in the same plane. A line from north to south on the table, and a line from east to west on the floor, is an example of two such lines. Let AB and CD (Fig. 22j) represent two lines which are not parallel and do not intersect, and let E be the point of A B which is nearest to CD, and F the point of CD which is nearest to AB. Then the line E F is the shortest distance between the lines ABA E . B and CD, and is a common perpendicular Fig, 223. to both lines. NOTE. In Fig. 223, A B is to be regarded as in the plane of the paper, CD as nearly perpendicular to this plane. 227. Let ABC and DEF (Fig. 224) be two angles in space, and let AB II DF, and BC II F F. Conceive the angle ABC to move so that its vertex B keeps in the line BE, and AB and CB to remain parallel to their first positions. When B reaches F, it is evident that A B will coincide with DE, and CB with 250 GEOMETRY FOR BEGINNERS. [ 228. FE, and therefore the angle ABC will coincide with the angle DEF. Therefore, the theorem of Plane Geometry ( 58) that angles, whose sides are respectively parallel and directed the same way from the vertex, are equal, also holds true of any two angles .in space. II. A Plane. 228. The surface of still water, the surface of a table, the floor, or the blackboard, are examples of plane surfaces, or planes. Give another example. How can a plane be generated by motion? (See 28.) How was a plane denned in 28 ? And what is the test of a plane surface? The following is the more abstract but more precise definition commonly given by mathematicians : Definition. A PLANE is a stirface such that the straight line which joins any two points in the surface lies wholly in the surface. A plane (like a straight line) may be conceived to extend indefi- nitely. The surface of the table may be extended in thought as far as we please, and in such a way that the surface which we imagine will be everywhere a part of the same plane as the actual surface which we see. Exercise. Give an instance of a vertical plane, a horizontal plane, an inclined plane. 229. When we conceive a plane as containing a given point or straight line, we are said to pass the plane through the point or the line. Through one point an endless number of planes in all conceiva- ble positions may be passed. Through two points, also, we may pass as many different planes 230.] CHAPTER XI. PLANES. 251 as we please ; for if we pass a plane through the line joining the points A and B (Fig. 22 5) , and then rotate the plane about this line as an axis, it will assume different positions, in all of which it contains the line, and therefore the points A and B. But if a third point C be given, through which the plane must also pass, it is evident that as we rotate the plane there is only one position in which it will contain all three points. Through three points not in the same straight line only one plane can be passed ; and the same is obviously true of a straight line and a point not in the line. It is likewise true of two intersecting lines and of two parallel lines ; for if we pass a plane through one of the lines, and then turn the plane about this line, there is only one position in which it can contain the other line. Therefore, a plane is determined, 1 . By three points not in the same straight line. 2. By a straight line and a point not in the line. 3. By two intersecting lines. 4. By two parallel lines. 230. Any two planes which are passed through a straight line AB (Fig. 2 25} have this line for their intersection ; and it is quite clear that through a curved line only one plane at most can be passed. The intersection of two planes is always a straight line. i ^^^ Exercises. 1. How many planes can Fig. 225. intersect in the same straight line ? 2. How many planes can be passed through a curve drawn on the black- board ? 3. Make (with a wire) a curve such that no plane can be passed through it 252 GEOMETRY FOR BEGINNERS. [ 231. III. A Plane and a Straight Line. 231. A straight line may be either, (/.) Parallel to a plane. ('.) Perpendicular to a plane. (Hi.) Inclined to a plane. An edge of the ceiling is parallel to the floor ; a plumb-line is perpendicular to the surface of still water ; a rafter of a common roof is inclined to the level ground. Give other examples. In Fig. 226, the line AB is parallel to the plane M. 1 The line PO is perpendicular to M t and the lines PC and PD are inclined to M. Exercises. 1. Draw a figure like Fig. 226, to illustrate the different positions which a straight line may have relative to a plane. 2. Hold a pencil, (z.) parallel, (.) perpendicular, (iii) inclined, to the table. 232. Definition. A straight line and a plane are parallel, if they cannot meet however far extended. This condition will be fulfilled if the straight line A B (Fig. 226) is parallel to a straight line C E in the plane ; for, since A B can never meet CE, and can never leave the plane determined by AB and CJl ( 229,4), it can never meet the plane M in which C E lies. Hence, I. A straight line is parallel to a p- 22 fr plane if it is parallel to a straight line drawn in the plane. II. If two lines are parallel, every plane passed through one of the lines will be parallel to the other line. What exception is there to this last statement ? 1 In naming a plane, a single letter, as M or N, is usually sufficient. 233-] CHAPTER XL PLANES. 253 Exercises. 1. How many lines parallel to a plane can be drawn through a point outside the plane ? 2. If two lines are parallel to a plane, are they parallel to each other ? Give illustrations. 233. A straight line, perpendicular or inclined to a plane, will (prolonged, if necessary) meet the plane in a point. This point is called the foot of the line. Definition. A straight line is perpendicular to a plane, if it is perpendicular to every straight line that can be drawn through its foot in the plane. It is also said to be normal to the plane. A line having this property can always be drawn from a point P (Fig. 227) to a plane M. For, among all the lines that can be drawn from P to meet the plane, there will be one, PO, shorter than any other. If we draw through its foot O any straight lines AB, CD, etc., PO will be the shortest line from P to each of these lines ; there- fore ( 83), perpendicular to each and all of them. Hence, a perpendicular measures the distance from a point to a plane. Through the intersection O (Fig. 228) of two lines, AB and CD, draw any line O E _L CD, and revolve OE about CD, keeping OE _L CD ; there is one, and only one, position in which the line will also be J_ AB. Let OP be this position ; then O P, being the only line through O perpendicular to both A B and CD, must be perpendicular to the plane passed through A B and CD. Therefore, 227. Fig. 228. 254 GEOMETRY FOR BEGINNERS. [ 234. If a straight line is perpendicular to two straight lines drawn through a point of the line, it is perpendicular to the plane contain- ing those two lines. This proposition enables us to solve the problem : To make a plane perpendicular to a straight line. We have only to draw through any point of the line two perpendiculars, and then to pas:; a plane through these perpendiculars. Explain how a carpenter would proceed in order to saw squarely in two the beam of wood shown in Fig. 229. Exercises. 1. Give examples of a line perpendicular to a P lane ' 2. When a right triangle revolves about one of its legs, what is generated by the other leg ? 3. At a point of a straight line in space, how many perpendiculars can be erected ? From a point not in the line, how many perpendiculars can be let fall to the line ? 4. Find the distance from a point P to a plane, if its distance from a point A in the plane = a, and the distance of this point from the foot of the perpendicular let fall from the first point to the plane = b. Solve for the case where a = 17, b = 8. 5. Prove that lines drawn from a point meeting a plane at equal distances from the foot of the perpendicular let fall from the point are equal ; and that of two lines drawn to unequal distances from the foot of the perpendicular, the more remote is the greater. 6. Prove the converse of the first part of the last exercise. 7. Find the locus of points in a plane equidistant from points not in the plane. 8. Find the locus of points equidistant from two given points. 9. Find the locus of points equidistant from three given points. 234. Let A B and CD (Fig. 236) be normal to the plane J/, and suppose AB to move along AC, keeping parallel 1 ) its first position. Then its position relative to the plane will not change, that is, it will remain normal to the plane ; and' when A coincides '350 CHAPTER XI. PLANES. 255 with C, AB will coincide with CD. Therefore, AB II CD. That is, Two lines that are normal to a plane are parallel to each other. The converse is also true. State it. Of Fig. 2jo. Fig. 231. 235. Lines neither parallel nor perpendicular to a plane are inclined to the plane. Let A O (Fig. 2ji) be inclined to the plane M. Draw AB _L M, and join OB ; then OB is called the PROJECTION of the line AO upon the plane. When is the projection of a line equal in length to the line ? less than the line ? When is it a point ? The acute angle A OB, which A O makes with its projection BO, is the smallest angle which A O makes with any line drawn through its foot on the plane M. To see this more clearly, revolve BO about O in the plane M. The angle which the revolving line makes with AO will continually increase, becoming equal to a right angle AOC after one-fourth of a revolution, and attaining its greatest obtuse value AOD after half a revolution. During the remaining half of the revolution what changes does the angle undergo ? The angle which a line inclined to a plane makes with its pro- jection in the plane, is called the inclination of the line to the plane. NOTE. What is said above may be illustrated by drawing lines through a point on a sheet of paper, and placing the end of the pencil held in an inclined position against this point. If in this position we let the pencil drop, the angle which it describes in falling is the inclination of the pencil to the table. 256 GEOMETRY FOR BEGINNERS. [ 236- IV. Two Planes. 236. Two planes may be either, (/.) Parallel to each other/ ('.) Perpendicular to each other. (V.) Inclined to each other. The floor and ceiling of the room are parallel planes ; the floor and a side of the room are perpendicular to each other ; the floor and the roof of the house are inclined to each other. Give other examples. Exercises. 1. In Fig. 232, which planes are parallel? perpendicular? inclined ? 2 Draw two planes of each kind. 3. Hold two planes (books, thin sticks of wood, etc.) parallel ; perpen- dular ; inclined. Fig. 232. Fig- 233. 237. Definition. Two planes are parallel if they cannot meet however far extended. This condition is fulfilled if the planes M and N (Fig. 2jj) are both perpendicular to the same line AB. For if we pass a third plane through AB, it will intersect M in a line CD \-AB (why?), and N in a line EF ^LAB. Therefore, CD and EF t being in the same plane, are parallel ( 57), and can never meet. The same reasoning may be applied to all planes that can be passed through AB ; therefore, the planes J/and TV can never meet. 238.] CHAPTER XI. PLANES. 257 Therefore, I. Planes that have a common normal are parallel. The converse is also true. State it. It is also evident from what precedes, that, II. The intersections of two parallel planes by a third plane are parallel lines. If, also, CE II DF (Fig. 233), then CDFE is a parallelogram ( 101), and AC = BD ( 102) ; that is, - III. Parallel lines between parallel planes are equal. Among these equal parallel lines are the common normals ( 234), which measure the distance apart of the planes; there- fore, IV. Parallel planes are everywhere equally distant. 238. Definition. Two planes are perpendicular to each other if a normal to either plane, erected at any point of their intersection, lies in the other plane. The angle made by the normals to the two planes, erected at the same point of the intersection, is a right angle (why ?) . Let the planes M and N (Fig. 234} intersect in the line AB, and let CD be normal to M and lie in N. Draw D E normal to N\ then C D E is a right angle (why?) ; therefore DE must lie in M, the plane containing all lines perpen- dicular to CD at D. Further, let G be any other point of the intersection, and draw GF nor- mal to My and G H normal to N. Then GF II DC ( 234) ; and since DC lies in the plane N, and G also lies in N, therefore GF must lie in N. For like reasons, GH lies in M. Comparing these results with the definition, we see that, 258 GEOMETRY FOR BEGINNERS. [ 239. 1. Two planes must be perpendicular, if a single normal to one of the planes erected at a point of the intersection lies in tJie other plane. II. Every plane passed through a normal to a plane is perpen- dicular to that plane. Exercises. 1, How can a plane perpendicular to a given plane be passed through a given point ? 2. How can a plane perpendicular to a given plane be passed through a straight line, (z.) when the line lies in the given plane ? (zY.) when it is inclined to the given plane ? (zYY.) when it is parallel to the given plane ? 239. Two inclined planes M and N (Fig. 235} intersect each other and form what is called a DIHEDRAL ANGLED This angle may be conceived as generated by the revolution of one of the planes M about the line A B until it comes into the position of the other plane N. The planes M and N are called the faces of the angle ; the intersection rAB is called the edge of the angle. A dihedral angle is always measured by the angle of two lines. If OC is J_ A~ ~~~ AB, during the revolution of M about AB, OC describes the angle COD, which has the same value wherever the point O is situated in the line AB ( 227). This angle may therefore be used to measure the dihedral angle made by the planes. That is, A dihedral angle is measured by the angle made by two lines drawn through any point of the edge, perpendicular to the edge, one line in one plane, the other in the other plane. Exercises. 1. Illustrate, by the leaves and edges of a book, dihedral angles, and how they are measured. 2. If the angle of the lines which measure a dihedral angle is 90, what position do the planes have ? 1 From Greek words meaning two-faced. 240.] CHAPTER XI. PLANES. 259 V. Three Planes. 240. Three planes may have five different relative positions, shown in Figs. 236-238. (/.) In Fig. 236, the planes are parallel to one another. 7 V Fig. 236. Fig. 237. (.) In Fig. 237, the two planes J/and N are parallel, and the third plane P (or Q) intersects M and N in parallel lines. (See 237.) (///.) In Fig. 237, the planes N, P, and Q have a common line of intersection. (iv.) In /3g-. 2J7, J/, P, and a, CO and DO will coincide in a line OF, either in front of the plane of the paper (Fig. 242), or behind it (Fig. 243), according as the motion is forward or backward '; and a trihedral angle is formed. And, since a is supposed the greatest of the three angles, it follows that a + b > c, and that a -f- c > b. Hence, in every trihedral ang -he sum of any two of the plane angles is greater than the third. CHAPTER XL PLANES. 261 If the motion is forwards, the trihedral angle shown in Fig. 242 is formed ; if backwards, the trihedral angle shown in Fig. 243. These two angles are composed of equal plane angles, taken in Fig. 241. Fig. 242. Fig. 243. order ; yet they cannot be made to coincide, because their equal parts follow in reverse order, in one, from left to right; in the other, from right to left. They stand to each other in the same relation as an object to its image in a mirror, or as the right hand to the left. Two such solid angles are said to be symmetrical. Two solid angles are equal, if they are composed of equal plane angles taken in the same order. We may conceive two equal solid angles placed one upon the other so as to coincide in all their parts. Whatever be the number of plane angles which compose a solid angle, their sum, estimated in degrees, must be less than j6o; if it were equal to 360, the plane angles would necessarily lie in one plane, and would not form a solid angle. Exercises. 1. Give examples of solid angles. 2. In a solid angle, how is the angle which two faces make with each other measured? 3. Draw a solid angle having three faces ; four faces ; six faces. 262 GEOMETRY FOR BEGINNERS. REVIEW OF CHAPTER XL I. Straight Lines in Space. 225. Nature of Solid Geometry. 226. Two straight lines in space. 227. Two angles in space. II. A Plane. 228. Definition of a plane. 229. Determination of a plane. 230. Intersection of two planes. III. A Straight Line and a Plane. 231. Positions of a straight line with respect to a plane. 232. The straight line parallel to a plane. 233. The straight line perpendicular to a plane. 234. Two normals to a plane. 235. The straight line inclined to a plane. IV. Two Planes. 236. Relative positions of two planes. 237. Two parallel planes. 238. Two perpendicular planes. 239. Two inclined planes. V. Three Planes. 240. Relative positions of three planes. VI. Solid Angles. 241. Solid angles. Symmetrical solid angles. Equal solid angles. 242.'] CHAPTER XII. GEOMETRICAL BODIES. 263 CHAPTER XII. GEOMETRICAL BODIES. CONTENTS. I. Polyhedrons ( 242,243). II. The Prism, Cylinder, Pyramid, and Cone ( 244-250). III. The Sphere ($ 251, 252). J. Polyhedrons. 242. A limited portion of space regarded as possessing -solely form and magnitude is called a GEOMETRICAL BODY, or a SOLID (see 7). In this sense a room is a body as much as a stone or piece of wood. Speaking exactly, it is the space occupied by the air, etc., in the room, and limited by the floor, sides, and ceiling of the room. A geometrical body bounded on all sides by planes is called a POLYHEDRON. Give examples of polyhedrons. The bounding planes are called the faces of the polyhedron, their intersections are called the edges, and the intersections of the edges are called the corners. The faces that meet at each corner form a solid angle, composed of as many plane angles as there are faces. A polyhedron must have at least four faces ; for it takes at least three planes to form a solid angle, and a fourth plane is required to close the opening between the three planes. A polyhedron with four faces is called a TETRAHEDRON ; with five faces, a PENTAHEDRON ; with six faces, an HEXAHEDRON ; with eight faces, an OCTAHEDRON ; with ten faces, a DECAHEDRON ; with twelve faces, a DODECAHEDRON ; with twenty faces, an ICOSAHEDRON. A very simple relation exists between the faces, corners, and edges of a polyhedron. 264 GEOMETRY FOR BEGINNERS. [ 243. If we examine the polyhedrons in Fig. 2, page 4, we find that, No. I. has 5 faces, 6 corners, and 9 edges. II. has 6 " 8 " " 12 " III. has 8 " 12 " " 1 8 " IV. has 5 " 5 "8 " In each of these cases, the number of faces 4- number of cor- ners = number of edges + 2 - This relation is perfectly general, and is called Euler's Theorem. 1 Let F denote the number of faces, C the number of corners, E the number of edges ; then we may state Euler's Theorem as follows : In every polyhedron, F + C = E + 2. [13.] 243. A polyhedron whose faces are equal regular polygons, and whose solid angles are all equal, is called REGULAR. It is easy to show that only five regular polyhedrons are possible. We know already that at least three faces are necessary to form a solid angle ; also, that the sum of the plane angles at each corner must be less than four right angles ( 241) . The angle of the equilateral triangle = 60, and 3 X 60 180 ; therefore, three such triangles may form a solid angle. Likewise, four such triangles may be used, since 4 X 60 = 240 ; also five such triangles, since 5 X 60 = 300. But 6 X 60 = 360, so that six, or more than six, equilateral triangles cannot form a solid angle. Therefore, only three regular polyhedrons are possible whose faces are equilateral triangles. The angle of a square = 90. Since 3 X 90 = 270, while 4 x 90 = 360, three is the only number of squares that can form a solid angle. The angle of a regular pentagon = 108. Since 3 x 108 324, while 4 x 1 08 = 432, only three regular pentagons can be used to form a solid angle. 1 Euler was a famous German mathematician, born 1707, died 1783. 2430 CHAPTER XII. GEOMETRICAL BODIES. 265 The angle of a regular hexagon = 1 20. Since 3 X 120 = 360, three hexagons placed together with a corner common would lie in one plane and could not form a solid angle. Therefore, no solid angle can be formed with regular hexagons ; still less with regular polygons having more than six sides. Fig- 247- Therefore, only five regular polyhedrons are possible. They are, (1) The regular Tetrahedron (Fig. 244), bounded by four equilateral triangles. (2) The regular Octahedron (Fig. 245}, bounded by eight equilateral triangles. (3) The regular Icosahedron (Fig. 246), bounded by twenty equilateral triangles. (4) The regular Hexahedron (Fig. 247}, bounded by six squares. (5) The regular Dodecahedron (Fig. 248), bounded by twelve regular pentagons. 266 GEOMETRY FOR BEGINNERS. [ 243- To DEVELOP 1 the surface of a solid is to draw in one plane the various parts of the surface, so that when folded up they shall form the solid. The surface of the solid, when drawn in one plane, is called the DEVELOPMENT of the surface. The surfaces of many solids cannot be developed ; but when this can be done, it is easy to construct by this means models of the solids. The surfaces of the regular polyhedrons are all developable, and their plane models are shown in Figs. 249-253. Fig. 249. Fig. 250. Fig. 251. Fig. 252. 253- To make models of these solids, draw on cardboard the diagrams shown in these figures, cut them out of the cardboard, and then cut half through the edges of the adjacent polygons. The figures will then readily fold up into the shape of the solids, and can be retained in shape by means of glue or paste. Literally, -unroll. 2 44-] CHAPTER XII. GEOMETRICAL BODIES. 267 Exercises. 1. Give examples of polyhedrons. 2. How many edges have each of the regular polyhedrons? 3. How many corners have each of the regular polyhedrons? 4. Verify Euler's theorem for each of the regular polyhedrons? 5. Construct models of the regular polyhedrons. NOTE. In making the model of the regular dodecahedron, the diagonals, shown dotted in Fig. 253, may be used to advantage. II. The Prism, Cylinder, Pyramid, and Cone. 244. If the polygon ABODE (Fig. 254) move along the line AF t keeping parallel to its first position, its sides will describe parallelograms, and the polygon will gener- ate a solid called a Prism. Definition. A PRISM is a polyhedron bounded by parallelograms and by two Fig. 254. equal and parallel polygons. The parallelograms are called the lateral faces, their lines of intersection the lateral edges, and the two polygons the bases of the prism. The height of the prism is the distance (HL) between the bases. A prism is called triangular, quadrangular, etc., according as the base is a triangle, quadrilateral, etc. A prism is oblique if the lateral edges are inclined to the bases (Fig. 255); right, if they are normal to the bases (Fig. 256); and regular, if it is a right prism with a regular polygon for its base. A parallelepiped is a prism whose bases are parallelograms. It is, therefore, a solid bounded by six parallelograms. Fig. 255. Fig. 256. Fig. 257. Fig. 258. Fig. 259. 268 GEOMETRY FOR BEGINNERS. [ MS- What is an oblique parallelepiped (Fig. 257) ? A right parallele- piped (Fig. 258} ? A rectangular parallelepiped is a right parallelepiped, all of whose faces are rectangles. A cube is a right parallelepiped, all of whose faces are squares (Fig. 259). Exercises. 1. Can you give an example of a prism? What kind of a prism is it ? 2. In Fig. 234, point out lateral faces, lateral edges, the bases, the height. 3. Name a prism with a base of five sides, six sides, eight sides. 4. What are the lateral faces of a right prism? Why is a lateral edge equal to the height of the prism? 5. How many faces, edges, corners, in a prism with a base of three sides, four sides, six sides, eight sides, twenty sides? Apply Euler's formula to each one of these cases. 6. Draw the following : (*'.) Oblique triangular prism. (z>.) (z'z.) Right pentagonal prism. (&*) (z'z'z.) Right hexagonal prism. (z/z'z.) (zV.) Regular hexagonal prism. (viii.} Oblique parallelepiped. Right parallelepiped. Rectangular parallelepiped. Cube. 245. If a circle (Fig. 260) move in the direction normal to its own plane, keeping always parallel to its first position, it generates a solid called the right circular Cylinder. The same solid is also generated by a rectangle ABCD (Fig. 257), which revolves about one side CD. Definition. A right circular CYL- INDER is the solid produced by the revolu- tion of a rectangle about one of its sides. NOTE. In future, by the word "cylinder" the right circular cylinder will be Fig. 260. meant. A cylinder is bounded by a curved surface called the convex surface, and by two equal and parallel circles called the bases. The line joining the centres of the bases is called the axis ; its length is equal to the height of the cylinder. 246.] CHAPTER XII. GEOMETRICAL BODIES. 269 246. If the polygon ABODE (Fig. 261} move along the line AO, keeping parallel to its first position, but diminishing in size at a uniform rate, without al- tering in shape, until it is reduced to a point O, its sides will describe trian- gles, and the polygon will generate a solid called a Pyramid. Fig. 261. Definition. A PYRAMID is a polyhedron bounded by triangles that have a common vertex, and by a polygon whose sides are the bases of the triangles. The triangles are called the lateral faces ; their lines of intersec- tion, the lateral edges ; their common vertex, the vertex of the pyramid ; and the polygon, the base of the pyramid. The heightQi the pyramid is the distance (OF) from the vertex to the base. A pyramid is triangular, quadrangular, etc., ac- cording as its base is a triangle, quadrilateral, etc. A regular pyramid is a pyramid which has a regular polygon for its base, and in which the perpendicular from the vertex passes through the centre of the base (Fig. 262). The distance OM from the vertex to a side of the base is called the slant height. Fig. 262. Exercises. 1. In Fig. 261. point out lateral faces, lateral edges, the base, the vertex, the height. 2. Name a pyramid with a base of five sides ; 6 sides ; 8 sides. 3. What kind of a pyramid is the tetrahedron (Fig. 244) ? 4. What kind of triangles are the faces of a regular pyramid ? 5. Draw (*'.) a triangular pyramid ; (it.} a regular hexagonal pyramid. 270 GEOMETRY FOR BEGINNERS. [ 247. 247. If a circle (Fig. 263) move in a direction normal to its own plane, keeping parallel to its first position, but diminishing in size at a uniform rate till it is reduced to a point O, it will gener- ate a solid called a right circular Cone. The same solid is also generated Fig 263 by a right triangle AOB, which revolves about the leg OB. Definition. A right circular CONE is the solid produced by the revolution of a right triangle about one of its legs. NOTE. In future, by the word " cone " a right circular cone will be meant. A cone is bounded by a curved surface called its convex surface, and by a circle called its base. The point O is the vertex of the cone, and the line OB joining the vertex to the centre of the base is the axis ; its length is equal to the height of the cone. The length of the line AO (Fig. 263} is called the slant height. 248. From the mode of generation of the four bodies now described, Prism, Cylinder, Pyramid, Cone, it follows that a section made by a plane parallel to the base is, \ ': Fig. 264. Fig. 265. ~ Fig. 266. In a prism, a polygon equal to the base. In a cylinder, a circle equal to the base {Fig. 264). In a pyramid, a polygon similar to the base (Fig. 271} . In a cone, a circle smaller than the base (Fig. 272). The section of a cylinder made by a plane inclined to the base 249-] CHAPTER XII. GEOMETRICAL BODIES. 271 is an ellipse (Fig- 265} . Such sections are made when two cylin- ders intersect each other, as in the case of two pipes meeting at right angles (Fig. 266) . In a cone, a section inclined to the base cuts the surface in an ellipse (Fig. 268} , if it does not meet the base ; in a parabola (Fig. 269), if it meets the base and is parallel to the side of the cone ; in a hyperbola (Fig. 270) , if it meets the base and is not parallel to the side of the cone. Fig. 267. Fig. 268, The four sections shown in Figs. 267-270 namely, circle, ellipse, parabola, hyperbola are sometimes called the conic sections. NOTE. These sections may be shown to the eye by holding a cone partly under water, at first with the axis vertical, then with the axis more and more inclined. If the water is colored red, the effect is better seen. Exercises. What is the section made by a plane passing through, ^ 1. Two lateral edges of a prism ? 3. The axis of a cylinder ? 2. Two lateral edges of a pyramid ? 4. The axis of a cone ? 249. A plane parallel to the base of a pyramid, or of a cone, divides the body into a smaller pyramid or cone, and a body called the frustum of the pyramid or cone (Figs. 271 and 272) . The base and the section parallel to the base are called the lower and upper bases of the frustum ; their distance from each other is the height of the frustum. If we know the height PQ (Fig. 271} of a frustum of a pyramid, and two homologous sides AB and EF of the bases, we can find the height OP of the entire pyramid, as follows : Through OP and the edge OA pass a plane intersecting the 272 GEOMETRY FOR BEGINNERS. [ 250. bases of the frustum in the parallel lines APand EQ (why paral- lel?), and draw E M II OB. By 159, = _ d OE = ' ' PQ EA' EA AM Therefore, OQ EF = And PQx EF PQxEF = AUT OP=OQ+QP. Fig. 271. Fig. 272. Exercises. 1. What are the faces of the frustum of a pyramid ? 2. If (Fig. 271} PQ = S m , A=4 m , EF= 2 m , find OQ and OP. 3. Given the height PQ (Fig. 272} of the frustum of a cone, and the radii of the bases ; find the height of the cone cut off, and also the height of the entire cone. Hint. Draw BC \\ OP, and use the similar triangles OBQ andAC. 4. If (Fig. 272} PQ=4 m , AP = 3 m , BQ_ = i m , find OQ and OP. 250. The surfaces of the prism, the pyramid, the cylinder, and the cone are all developable. Fig. 273 shows the development of a regular hexagonal prism ; Fig. 2J4 that of a regular pyramid whose base is a square. The auxiliary lines in the two figures serve to illustrate how they are to be constructed. Fig. 273 is the development of a regular hexagonal pyramid. How is it constructed ? 250.] CHAPTER XII. GEOMETRICAL BODIES. 273 The development of the convex surface of a cylinder is a rect- angle ; for we can roll up a rectangular sheet of paper, for exam- ple, into the shape of a cylindrical surface ; and conversely, a cylin- drical surface can be unrolled into a rectangle (Fig. 276). Fig. 274. 275- Fig. 276. Fig. 277. Fig. 278. The development of the convex surface of a cone is a circular sector whose arc is the circumference of the base of the cone, and whose radius is the slant height of the cone (Fig. 278). Fig. 276 shows the development of the entire surface of a cylin- der ; Fig. 277, that of the entire surface of a cone ; Fig. 277, that of the entire surface of the frustum of a cone. Exercises. Make models of the following solids : 1. A regular triangular prism. 2. A regular hexagonal prism. 3. A right pi-ism, not regular. 4:, A rectangular parallelepiped. 5* A regular hexagonal pyramid. 6. A cube. 7 A cylinder. 8. A cone. 9. A frustum of a pyramid. 10. A frustum of a cone. 274 GEOMETRY FOR BEGINNERS. [ 25L Fig. 279. III. The Sphere. 251. Definition. A SPHERE is the solid produced by the revolution of a semicircle (or circle) about the diameter (Fig. 279). NOTE. Spin a piece of money rapidly on the table, and it will present the appearance of a sphere. The semi-circumference PMQ describes the surface of the sphere ; and it is clear that this surface must be everywhere equally distant from O, the centre of the semicircle This point O is called the centre of th< sphere. Therefore, a sphere may also be de- fined as a solid bounded by a surface, which is everywhere equally distanl from a point called the centre. What is a radius of a sphere ? a diameter ? An orange may be taken as a good example of a sphere. If we cut an orange into a series of parallel slices, the planes of division between the slices are circles differing in size, the largest being that which passes through the centre of the orange. Every section of a sphere made by a plane is a circle. If the section passes through the centre of the sphere, it is called a great circle ; in all other cases, it is called a small circle. The equal parts into which a great circle divides the sphere are called Hemispheres. In Fig. 2 So, the sections PAQB and AGB H are great circles ; the sections CEDF2cs\& MS NT me small circles. All great circles of a sphere are equal; they have the same centre, the same radius, the same diameter, as the sphere itself. 251.] CHAPTER XII. GEOMETRICAL BODIES. 275 All great circles divide each other into two equal parts ; every great circle divides the sphere and its surface into two equal parts. Small circles are the smaller the further they are from the centre of the sphere. Every small circle divides the sphere and its surface into two unequal parts. The centre of any circle of the sphere must lie in the diameter of the sphere which is normal to the plane of the circle ; the two points where this diameter meets the surface of the sphere are called the Poles of the circle. In Fig. 280, the points P and Q are the poles of what circles ? On the earth (or terrestrial sphere) , parallels of 'latitude are small circles, the equator is a great circle, and the north and south poles are the poles of all these circles. All points in the circumference of a circle (great or small) are equidistant from either one of its poles, whether this distance be measured in a straight line or along the arc of a great circle pass- ing through the point and the pole. Since the latter way of meas- uring distances is much more convenient on a sphere, it is always employed. Thus, in Fig. 280, the distance from the point C to the pole Pis the length of the arc PC. Of all lines that can be drawn on the surface of a sphere from one point to another, the arc of the great circle passing through the two points is the shortest. In this respect, the arc of a great circle is to the surface of a sphere what the straight line is to the plane. By means of arcs of great circles (never small circles) , spherical triangles and spherical polygons may be drawn on the surface of a sphere. The portion of the surface of a sphere comprised between the circumference of two parallel circles is called a Zone. 1 The dis- 1 From a Greek word, meaning belt or girdle. 276 GEOMETRY FOR BEGINNERS. [ 252. tance between the planes of the circles is the height of the zone, and the circumferences of the circles are its bases. If one of the planes touches the sphere without cutting it (or is a tangent plane), the zone is called a zone of one base. Point out zones in Fig. 280. Exercises. 1. Give examples of spheres (globes, balls). 2. Draw a figure of a sphere with lines in it to illustrate the definitions of this section. 3. How will a sphere be divided by three great circles mutually perpen- dicular to each other ? 4. In what zones is the surface of the earth divided ? Which are zones of one base ? 252. An orange, after the peel is removed, is easily divided into wedge-shaped slices, separated by natural planes of division. These planes are planes of great circles intersecting in a common diameter. In the fruit, the number of the planes is limited ; but, in any sphere, we may imagine as many such planes as we please, all passed through a common diameter, and each cutting the sur- face of the sphere in the circumference of a great circle. Fig. 281 shows several such planes passing through the diameter PQ. They intersect the surface in the circumferences, the front parts of which are the semi-circumferences PA Q, PBQ, etc. These semi-circumferences are called Meridians, and the portion of the sur- face of the sphere comprised between two meridians is called a Lune. Point out lunes in Fig. 281. Terrestrial Meridians are the circum- ferences of great circles on the earth's surface made by planes passing through the earth's axis. The surface of a sphere is not developable, but a close approxi- CHAPTER XII. GEOMETRICAL BODIES. 277 mation to it can be constructed by drawing a series of equal lunes in one plane, as follows : Divide a line AB {Fig. 282), which must be made equal to 3^ times the diameter of the proposed sphere, into twelve equal parts, and add nine of these parts to each end of the line. Then, with a U A V A A \l i ii in t-M- Fig. 282. radius equal to ten of these parts, describe arcs from the points marked i, 2, 3, etc. ; and, likewise, arcs from the points marked I., II., III., etc., as shown in the figure. These arcs will inclose lunes which, cut out and properly folded up, will very nearly form a complete spherical surface. It is by a process of this sort that the surfaces of balloons, base- balls, globes, etc., are made. Exercises. 1. Describe how a hemisphere would be generated by motion. 2. Make a model of a sphere. 278 GEOMETRY FOR BEGINNERS. REVIEW OF CHAPTER XII. I. Polyhedrons. 242. Definition of the Polyhedron. Specific names of polyhedrons. Euler's theorem. 243. The five regular polyhedrons. Development of their surfaces. How models of them are made. II. The Prism, Cylinder, Pyramid, and Cone. 244. Definition of the Prism. Different kinds of prisms. The parallele- piped. 245. Definition of the Cylinder. 246. Definition of the Pyramid. Different kinds of pyramids. 247. Definition of the Cone. 248. Sections of the prism, cylinder, pyramid, and cone. 249. Frustum of a pyramid or a cone. 250. Development of the surfaces of these four solids. How models of them are made. III. The Sphere. 251. The Sphere defined. Section of a sphere made by a plane. Great and small circles. Spherical triangles and polygons. Zones. 252. Meridian lines. Lunes. How a model of a sphere may be made. 253-] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 279 CHAPTER XIII. SURFACES AND VOLUMES OP BODIES. CONTENTS. I. Surfaces of Bodies ( 253-258). II. Volumes of Bodies ($$259- 274). III. Exercises and Applications ( 275-279). J. Surfaces of Bodies. 253. THE PRISM. The lateral surface consists of parallelo- grams ; the bases are polygons. Therefore, I. Lateral surface = sum of the areas of the parallelograms of which it is composed. II. Total surface = lateral surface -\- areas of the bases. Exercises. Find the total surface of a cube if, 1. The edge = 8 m . 3. The edge = 4 ft . 2. The edge = 3.5 m . 4. The edge = 7f in . 5. If the edge of a cube contain a unit of length, how many units of area are there in the entire surface ? Ans. 6 a 2 . 6. The total surface of a cube 6ooi m . Find the length of the edge. 7 . If the total surface of a cube contains S units of area, how many units of length are there in one edge ? Ans. -\J S -+ 6. 8. Find the total surface of a rectangular parallelepiped if the length = 15, the breadth = 9 m , the height 6 m . 9. Find the total surface of a rectangular parallelepiped if the length = a, breadth = b, height c. Ans. 2 ab + 2 ac -f 2 be. 10. How many bricks, 8 in by 3!, are required to line the bottom and sides of a reservoir 60^ long, 24** wide, 12^- deep ? Find the total surface of the following regular prisms : 11. Triangular prism, height = 76 cm , side of base = 40. 12. Quadrangular prism, height = 6 ft , side of base = 27 in . 13. Hexagonal prism, height = 2.46 m , side of base 84 cm . Hints. Find less radius of base by use of $ 202, Case 2, and $ 142, and area of base by $ 133. 280 GEOMETRY FOR BEGINNERS. [ 254. 14, How would Exercise 13 be solved if the less (or the greater) radius of the base were given instead of a side of the base ? 15, The total surface of a regular hexagonal prism = 822 n Pi O"ni" ^* " tangular parallelepiped having the same height A ^ 2? and an equivalent base ( 129). There- Fig. 287. fore, by 261, - Volume of a right parallelepiped = base X height. 290 GEOMETRY FOR BEGINNERS. [ 263. Since area ABCD AB X B M, therefore, the volume of the right paraUelopipedAG = AB X B MX BF= AB X BF X BM = area AB EFx BM = a rectangular face taken as base x the corresponding height. Exercises. 1. Find the volume of a right parallelepiped if the height = Ii m , and the base is a parallelogram having the length I2 m and the altitude 5. 2. A wall 2O m long, 3 high, is o.8 1M thick at the bottom and 0.4 thick at the top. Find how many cubic meters it contains. Hint. Consider the wall a right parallelepiped having for base a trapezoid whose parallel sides are o.8 m and.o.4" 1 . 263. THE OBLIQUE PARALLELOPIPED. Let AG (Fig. 288) be an oblique parallelepiped. Through the corner E pass a plane perpendicular to the edge EF, cutting from the parallelepiped the solid AMDNEHO, and place this solid on the right in the position BPCQFGR. By this operation, the oblique par- allelepiped is transformed into an equivalent right parallele- piped, having the same height, and a rectangular base EFRO equivalent to the base EFGH of the oblique parallelepiped ( 129). Now the volume of the right parallelepiped = its base EFR O x the height ( 262) ; therefore, Volume of the oblique parallelepiped = base EFGH X height. 264. THE TRIANGULAR PRISM. If through the edges AB and GH of the oblique parallelepiped AG (Fig. 288) we pass a plane, it will divide the solid into two oblique triangular prisms. The same plane divides the equivalent right parallelepiped into two right triangular prisms. These right prisms are equivalent, each to its 265.] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 291 corresponding oblique prism, because each is formed by cutting off (by the plane through E _L the edge EP) a portion of the oblique prism, and transferring it to the right of the figure. Now these right prisms are also equivalent to each other ; for, if one of them is inverted, it is then clear that it has precisely the same shape as the other, and that its faces are equal and parallel to those of the other. Therefore, the two oblique prisms are equivalent, and hence each is half of the parallelepiped. Take ADEN as the base of the parallelepiped ; then the distance between the planes A D EH and B CFG is its height, the two prisms have the same height, and their bases are the triangles A EH and. ADH, each of which is half the parallelogram ADEN. So that the volume of either prism (half the volume of the parallelepiped) is equal to its base (half the base of the parallelepiped) X its height. In general, Volume of a triangular prism = its base X its height. Exercises. Find the volume of a regular triangular prism, having given, 1. Height X 75 cm , area of base = i4O ccm . 2. Height h, area of base = S. 3. Height = 8.5 m , one side of the base = 3. 4. Height = h, one side of the base = a. 5. The bases of a triangular prism 8o cm high are isosceles right triangles, a leg of which = I2 cm . Find the volume of the prism. 6. If the volume of a triangular prism with isosceles right triangles for bases = 6oo cbm , and the height = 14, find (z.) area of the base, (zY.) perime- ter of the base. 265. ANY PRISM. Any prism (Pig. 289) can be divided into triangular prisms by passing planes, as shown in the figure, through one of the lateral edges. The trian- gular prisms have the same height as the entire prisms, and the sum of their faces is equal to the base of the entire prism. Therefore, by 264 and by addition, it follows that the Volume of any prism = its base X its height. Hence, any two prisms having equivalent bases Fig. 289. and equal heights are equivalent. A 292 GEOMETRY FOR BEGINNERS. [ 266. Exercises. Find the volumes of the following regular prisms : 1. Hexagonal prism, height = I7 m , side of base = 2 m . 2. Hexagonal prism, height =: h, side of base = a. 3. Octagonal prism, height = 85 cm , side of base io cm 4. Octagonal prism, height //, side of base = a. 266. THE CYLINDER. If we inscribe in the bases of a cylin- der regular polygons of the same number of sides, so that their sides, taken pair by pair, are parallel, and then pass planes through each pair of sides, these planes with the polygons will form a prism inscribed in the cylinder (Fig. 290}. Now the greater the number of lateral faces in the inscribed prism, the nearer does its volume approach that of the cylinder ; that is to say, - Volume of cylinder = limit of volume of in- Fig. 290. scribed prism. The volume of the prism = its base X its height ; and its base has the base of the cylinder for its limit. Therefore, limit of the volume of the inscribed prism base of cylinder x its height. Hence (Axiom I.), Volume of the cylinder = its base X its height. Exercises. Find the volumes of the following cylinders : 1. -Height = 6 m , area of base = 421. 2. Height = 8 m , radius of base = 2 m . 3. Height = 64 cm , diameter of base = i6 cm . 4. Height = 7^, circumference of base = 44 in . 5. Height = h, radius of base = r. 6. Height = h, radius of base = d. 7. The volume of a cylinder = 2OO cbm , the height = 13"'. Find the radius of the base. 8. The volume of a cylinder = V, the height = h. Find the radius of the base. {). A cylindrical pail holding exactly one liter is i8 cm high. Find the diameter of the base. 267.] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 293 10. A hollow iron tube is 6 m long, the outer diameter = 92, the inner diameter = 68 cm . How many cubic decimeters of iron are there ? 11. How long is an iron tube, if the inner diameter = 3.5 cm , the outer cir- cumference = I5 cm , the volume = i2O ccm ? 12. A vessel in the shape of a cylinder is to be made from sheet-iron 2 cm thick. The vessel is to be i' n high, and to hold 100 liters. What must be its outer diameter ? 267. THE TRIANGULAR PYRAMID. Let S-ABC and S -MNO {Fig. 291) be two triangular pyramids standing on th: same horizontal plane, and having equivalent bases ABC and MNO, and equal heights. Divide the common height into any number of equal parts, and through the points of division pass planes parallel to the bases of the pyramids. Since the bases are equivalent, it follows, from the mode of generation of a pyramid ( 246), that any two sections, as DEF and PQR, situated at the same distance from the bases, will also be equivalent. In the two pyramids construct triangular prisms, such as DEFGH1 and PQRTUV, having for upper base one of the sections; for lateral edges, lines parallel to SA or SM-, and for a common height, the distance apart of two sections. It follows, from 265, that these prisms will be equivalent, taken pair by pair, in the two pyramids ; therefore, the sums of their volumes will be equal, no matter how many or how few of them there may be. But, the greater their number the nearer the sums of their volumes approach the volumes of the two pyra- mids ; that is to say, Volume of each pyramid =: limit of sum of volumes of the in- scribed prisms. N Fig. 291. 294 GEOMETRY FOR BEGINNERS. [ 267. Now the sums of these volumes remain always equal ; therefore their limits, or the volumes of the pyramids, must also be equal. Hence, two triangular pyramids having equivalent bases and eqital heights are equivalent. Now let S-ABC (Fig. 292) be any triangular pyramid. If we pass a plane through BC II SA, and a plane through S II ABC, we form a triangular prism A B CSP Q. This prism is divided by the plane SBC, and a plane passed through C and SP, into three trian- gular pyramids, S-ABC, P- SB C, and C - SPQ. Of Fig. 292. these, S-AB C and C-SPQ are equivalent because they have equivalent bases and equal heights. Also, P-SBC and C-SPQ are equivalent, because we may take *S for the common vertex, and then they have equal heights (the distance from S to the plane B C P Q) and equivalent bases (the equal triangles PBC and PC Q). Therefore, each pyramid = the prism. Now the volume of the prism = base x height ; therefore, the volume of the pyramid S-ABC (which has the same base and the same height as the prism) = % base X height. Therefore, Volume of a triangular pyramid '= % its base X its height. Exercises. 1. Height of a regular triangular pyramid I5 m , one side of the base = 2 m . Find the volume. 2. He.ight of a regular triangular pyramid = h, one side of the base - a. Find the volume. 3. A triangular pyramid 9O cm high has an isosceles right triangle for base, ote leg of which 4O cm . Find the volume of the prism. 4. A triangular pyramid with an isosceles right triangle for base is 3.4 high. Its volume = 8o cm . Find (?.) the area of the base ; (?V.) the perime- ter of the base. 5. The frustum of a regular triangular pyramid is 4 m high, and two homolo- gous sides of the bases are 2 rn and i.25 m . Find its volume. (See 249.) 268.] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 295 268. ANY PYRAMID. Any pyramid whatever can be divided into triangular pyramids by passing planes through one of the lateral edges, as shown in Fig. 293. These triangular pyramids all have the same height as the entire pyramid, and the sum of their bases = the base of the entire pyramid. Therefore, . Volume of any pyramid = -J base X height. Fig. 293. Exercises. 1. Find the volume of a regular hexagonal pyramid, if the height = 36 cm and a side of the base = 8 cm . 2. Find the volume of a regular octagonal pyramid, if the height = 44"* and a side of the base = 4 m . 3. Solve Exercise I, if the height = h and a side of the base = a. 4. Solve Exercise 2, if the height = h and a side of the base = a. 5. The frustum of a regular hexagonal pyramid is 2.7 high, and two homologous sides of the bases are i.2 m and 7O cm . Find the volume. 6. A vessel has the shape of the frustum of a regular four-sided pyramid. It is i8 cm high, and the outer edges of its bases are 24 cm and i6 cm . How many liters of water will it hold if the material of which it is made is 2 cm thick ? 269. THE CONE. Inscribe in the base of a cone (Fig. a regular polygon, and pass planes through the sides of this polygon and the vertex of the cone. These planes with the polygon form an in- scribed regular pyramid whose volume ap- proaches the nearer to that of the cone, the greater the number of sides in the regular poly- gon ; that is, Volume of cone = limit of volume of inscribed pyramid. But volume of any pyramid = its base X its height. And in this case the base has the base of the cone for its limit. Therefore, limit of volume of inscribed pyramid = base of cone X height. Therefore (Axiom I.), Volume of a cone = % its base X its height. Fig. 294. 296 GEOMETRY FOR BEGINNERS. [ 270. Exercises. Find the volumes of the following cones : 1. Height = 15, area of base = 6o c i m . 2. Height = I2 rn , radius of base ~ 3 m . 3. Height = h, radius of base = r. 4. Height 74 cm , diameter of base = 4.6. 5. Height h, diameter of base = d. 6. Height 6 ft , circumference of base = ijjft. 7. Slant height = 5, radius of base = 3. 8. Slant height = k, radius of base r. 9. Find the volume of a cone whose slant height = diameter of the base - 2. 10. Volume of a cone 8oo cbm , radius of base i.8 m . Find the height. 11. Find the volume of the frustum of a cone, if the radii of the bases are 7-5 cm and 5 cm , and the height 25 cm . (See 249.) 12. A vessel has the shape of the frustum of a cone i m high, and the inner circumferences of the bases are 5 and 4 m . How many cubic meters of water will it hold ? 13. What part of a cone remains if a cone two-fifths as high as the entire cone is cut off ? 14. How many gallons of water will a vessel hold which has the shape of the frustum of a cone, the height being 3 ft , the diameters of the bases i8 in and I2 in ? 15. A Dutch windmill in the shape of the frustum of a cone is I2 m high. The outer diameter at the bottom and the top are i6 m and I2 m ; the inner diameters I2 m and io m . How many cubic meters of stone were required to build it ? 16. The chimney of a factory is 32 high, and the outer circumferences of its bottom and top are 20 and I4 m , while the bore is, throughout, 2 m wide. How many cubic meters of brick does it contain ? 270. The results obtained in H 261-266 may be summed up in one general rule, as follows : Volume of a prism or a cylinder = base X height. And the results obtained in $ 267-269 may be summed up in one general rule, as follows : Volume of a pyramid or a cone = J base x height. 271.] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 297 271. THE SPHERE. In order to find the volume of a sphere, conceive a series of pyramids formed, having for a common vertex the centre of the sphere, and for bases the faces of a polyhedron circumscribed about the sphere (that is, a polyhedron whose faces touch the surface of the sphere, each in a single point). By in- creasing the number of these pyramids, we can make their sum approach the volume of the sphere as nearly as we please ; or, in other words, Volume of sphere = limit of sum of the pyramids. Now each pyramid = height X base, and their common height = radius of the sphere ; therefore, sum of the pyramids = radius of sphere x sum of the faces. The limit of the sum of the faces (when their number is increased more and more) = the surface of the sphere. Therefore, limit of the sum of the pyramids = radius X surface of the sphere. Therefore (Axiom I.), Volume of a sphere == radius X surface of sphere. Exercises. 1. Find the volume of a sphere if the radius = 6 m . 2. Find the volume of a sphere if the radius = r. 3. Find the volume of a sphere if the diameter 14. 4. Find the volume of a sphere if the diameter = d. 5. Find the volume of a sphere if the circumference of a great circle = io ft . 6. Required the volume of the largest sphere that can be turned from a cubical block of wood whose edge = i dm . How much of the wood is lost ? 7. The volume of a sphere = i8o cbm . Find the radius. 8. The volume of a sphere = V. Find the radius. 9. Find the volume of the largest sphere that can be turned from a cylin- der of wood whose diameter = its height I2 cm . 10. The diameter of a cylinder = its height = h. Compare the volume of this cylinder with that of the largest sphere that can be inclosed in it. Ans. Volume of the sphere = f volume of the cylinder. NOTE. This curious fact was discovered by Archimedes about 230 B.C. 11. Find the difference between the volumes of two concentric spheres whose radii are 4 m and 5. 12. In two concentric spheres, the volume of the larger = twice that of the smaller. If the radius of the smaller = i m , find the radius of the larger. 298 GEOMETRY FOR BEGINNERS. [ 272. 272. The volume of any body, whatever be its shape, may be found by weighing it, and then dividing the weight by the weight of one unit of volume. The quotient will be the number of units of volume in the body. In the metric system, the most common units of weight are, The GRAM (g.)= the weight of i ccm of pure water (at 4 C.). The KILOGRAM (kg.)= weight of i cdm of pure water (at 4 C.). Since i cdm = iooo ccm , i kilogram = 1000 grams. The ratio of the weight of any substance to the weight of the same volume of pure water is called the SPECIFIC GRAVITY of that substance. SPECIFIC GRAVITIES OF SOME COMMON SUBSTANCES. Platinum . . 22.00 Brass (sheet) . 8.40 Boxwood . 1.20 Gold . IQ 35 Steel .... 780 Ice ... O Q2 Mercury . 13.60 Iron (cast) . . / *** 7-25 Oak . . . . 0.90 Lead . . "-35 Zinc .... 7.19 Alcohol . . . 0.80 Silver . . . 10.51 Marble . . . 2.70 Pine . . . o-57 Copper . . . 8.80 Glass .... 2.60 Cork . . '. . 0.24 The gram and the kilogram being by definition the weights of unit volumes of water, it follows that the above numbers express the weights of i ccra of the substance in grams, or of i cdm (i liter) of the substance in kilograms. Whence it is clear that in the metric system, T7 / * t j Volume of a body = _ its specific gravity the volume being expressed (/.) in cubic centimeters, or (it.) in cubic decimeters (liters), according as the weight is expressed (*'.) in grams or ('.) in kilograms. The most important English units of weight are the POUND (Ib.) avoirdupois, the OUNCE (oz.) = jV 1 *, and the TON = 2Ooo lb . 272.] CHAPTER XIII. SURFACES AND VOLUMES OF BODIES. 299 They are connected (nearly enough for practical purposes) with the weight of unit volume of water by the relation, Weight of i cubic foot of 'water = 62^ lbs = iooo oz . The weight in pounds of a cubic foot of any other substance is found by multiplying its specific gravity by 62^, and the weight in ounces by multiplying its specific gravity by 1000. NOTE. 283g T.OZ (nearly); i k s = 2.2^3 (nearly). Exercises. 1. Find the volume of a piece of granite which weighs 4 k s. 2. A piece of lead weighs 24 k s. Find its volume. 3. What is the volume of 848 of silver ? 4. Required the edge of a cubical vessel which will hold 2OOO lbs (i ton) of mercury. 5. Find the edge of a marble cube weighing 184^. 6. An iron cylinder 2.$ m long weighs 68o k s. Find its diameter. 7. Find the edge of a cube of cork which weighs the same as an iron ball 22 cm in diameter. 8. How many liters will a vessel hold if the vessel weighs when empty I.5 k S, and when full of water i-5 k s ? 9 An oak cone 40 high weighs 2 k s. Find its diameter. 10. What is the diameter of a cannon ball which weighs 64 lbs ? 11. How can the weight of a body be found if its volume and its specific gravity are known ? 12. Find the weight of I M of water, i hl of alcohol, and i cub ft of mercury. 13. Find the weight of a rectangular parallelepiped of copper, o.34 m long, o.n m wide, o.O3 m thick. 14. Find the weight of an oak cone, if the radius of the base 32 cm and the height = 9O cm . 15. Find the weight of a hollow brass cube, if the inner diameter = ^ and the thickness of the brass = i cm . 16. If a cube, whose edge = 8 cm , weighs 508, find the weight of the largest cylinder that can be turned from it. Also the weight of the largest cone. Also the weight of the largest sphere. 17. How can the specific gravity of a body be found, if its weight and its volume are known ? 18. A cubical block of pine wood, an edge of which I2 cm , weighs i k . Find the specific gravity of the wood. 19. If a mass of ice containing 27O cbm is known to weigh 229,000^, find the specific gravity of ice. 20. Find the specific gravity of cast iron, if 2^ cutt weigh 953 lbs . 300 GEOMETRY FOR BEGINNERS, [ 273. 273. Equivalent solids have the same volume (size) ; similar solids, the same shape ; equal solids, the same volume and the same shape. Equal polyhedrons must fulfil three conditions : (/.) They must have the same number of faces, equal each to each. (i/.) Their homologous edges must be equal. (m.) Their homologous solid angles must be equal. Similar polyhedrons must also fulfil three conditions : (/.) They must have the same number of faces similar each to each. ('.) Their homologous edges must be proportional. (iii.) Their homologous solid angles must be equal. In both cases, there are the same number of edges and also of corners. Conversely, if all three of the above conditions hold true of two polyhedrons, then these polyhedrons are equal or similar, as the case may be. There are four important classes of similar solids : (/.) All cubes and regular polyhedrons of the same name are similar. (//.) In order that two regular prisms or two regular pyramids may be similar, they must have the same number of faces, and their homologous lateral edges must be proportional to the homologous sides of their bases. But the homologous lateral edges in the prisms are obviously proportional to the heights, and in the pyra- mids they are easily proved to be proportional to the heights, by reasoning as in 249 ; and the homologous sides of the bases are (by 167) proportional to the less radii of the bases. Therefore, two regular prisms or pyramids with the same number of sides are similar, if their heights are proportional to the less radii of their bases. 274-] CHAPTER xm. SURFACES AND VOLUMES OF BODIES. 301 (in.) By regarding the cylinder or the cone as the limit of the inscribed regular prism or regular pyramid, when the number of its sides is increased more and more ( 266, 269), it becomes evident that two cylinders or two cones are similar, if their heights ire proportional to the radii of their bases. (iv.) All spheres are similar. 274. If the edge of a cube = a, the (total) surface = 6 a 2 , and the volume a 3 . Now, if we have other cubes whose edges are 2 a, 3 a, 4 a, etc., their surfaces will be 4, 9, 16, etc., times that of the first cube, since (2tf) 2 = 4# 2 , (3?i-7,'33 YB 17293 800542 UNIVERSITY OF CALIFORNIA LIBRARY