BOUGHT WITH THE INCOME 
 PROM THE 
 
 SAGE ENDOWMENT FUND 
 
 THE GIFT OF 
 
 189X 
 
 hJM7t^ g///^.^ 
 
V.I -2 
 
 Geometrical drawing. 
 
 „ 3 1924 031 307 832 
 olin.anx 
 
The original of tliis book is in 
 tine Cornell University Library. 
 
 There are no known copyright restrictions in 
 the United States on the use of the text. 
 
 http://www.archive.org/details/cu31924031307832 
 
Geometrical Drawing 
 
 FOR THE USE^ q^ CA^IDATES FOR ARMY 
 
 EXAMINATIONS, AND AS AN INTRODUCTION 
 
 TO MECHANICAL DRA WING 
 
 BY 
 
 W. N. WILSON, M.A. 
 
 ASSISTANT MASTER AT RUGBY SCHOOL 
 
 THIRD EDITION 
 
 LONDON 
 LONGMANS, GREEN, AND CO. 
 
 AND NEW YORK 
 1895 
 
PREFACE 
 
 My aim in writing this book has been twofold : to 
 provide a School Text-book which shall contain aU that is 
 necessary for the Army Examinations, and be an Intro- 
 duction to the Mechanical Drawing of Engineers; and 
 to place Geometrical Drawing on a more sure basis 
 as a study of great educational value. The teaching 
 of Geometrical Drawing is often relegated to the Drawing 
 Master, who might as reasonably be expected to teach 
 Euclid; for it is distinctly a branch of Mathematics, 
 many of the problems requiring considerable geometrical 
 skill in their solution. It is a subject which I have 
 found most useful and suggestive to boys in the study 
 of Eiders in Geometry, and which lends a practical 
 interest to Euclid, especially to the more obscure pro- 
 positions of the Sixth Book. I therefore venture to 
 hope that Geometrical Drawing may become a subject 
 more generally taught by Mathematical masters, and 
 with this object in view, I have deviated from the custom 
 
iv PREFACE. 
 
 of existing works, by adding to each problem solved the 
 proof, either directly, or by reference to the correspond- 
 ing proposition in Euclid. 
 
 Special attention has been devoted to accuracy in the 
 Answers to the Exercises, and they have been carefully 
 
 verified by calculation. 
 
 W. N. WILSOK 
 Rugby, Januiwy 1888. 
 
 PREFACE TO THIRD EDITION. 
 
 Chaptee IV. has been re-written, and separate construc- 
 tions are given for the Eegular Figures up to eight sides. 
 Two of these are, I believe, original. In place of the 
 Geometrical Patterns in Chapter xi. I have given instruc- 
 tions for the use of Ink and Marquois. Scales, and have 
 added a number of selected Examples from Army Papers. 
 Otherwise no material changes have been made. 
 
 I propose to include the Geometrical Patterns in a 
 Second Course, which is to be published shortly, and which 
 is intended to carry out the scheme of the earlier part of 
 this work, — to give general principles rather than isolated 
 constructions. It is the result of ten years' continuous 
 teaching, and is intended to cover the chief difficulties, 
 which boys find, in a practically inexhaustible subject. 
 
 W. N. WILSON. 
 
 BnGBT, September 1895. 
 
CONTENTS 
 
 Introduction, . . , , 
 
 I. PEBLtMINAEy CONSTRUCTIONS, . 
 
 11. Triangles and Pakallelogbams, 
 
 III. Rbgulab Figures, .... 
 
 IV. Figures inscribed in Symmetrical Figures, 
 V. Proportion, ..... 
 
 VI. Scales, ...... 
 
 VII. Irregular Figures and Areas, 
 VIII. Sums and Differences of Areas : Square Root, 
 IX. Circles and Tangents, 
 X. Fractions of Areas, . , . . 
 
 XI. Inking in — Marquois Scales. 
 Examples, . 
 
 I 
 Miscellaneous 
 
 EXAMINATION PAPERS, . 
 ADDITIONAL SCALES, 
 ANSWERS, .... 
 ANSWERS TO ADDITIONAL SCALES, 
 
 PAGE 
 
 vii 
 1 
 
 11 
 28 
 34 
 42 
 53 
 63 
 73 
 80 
 100 
 
 112 
 119 
 132 
 135 
 143 
 
INTRODUCTION 
 
 The greatest care and accuracy are required in con- 
 structing the figures in Geometrical Drawing, and there- 
 fore the best instruments must be used. A complete case 
 of mathemaitical instruments, unless required for future 
 use, is unnecessary ; but a box containing all that is 
 required, viz. a bow-pencil, a pair of dividers, two vulcanite 
 set squares, and a scale and protractor, is specially con- 
 structed for this work by Messrs. Negretti and Zanibra, 
 London. 
 
 Pencils. — Two pencils, an HH and an H, are required 
 for drawing lines. The HH, which is used for lines of 
 construction, should be cut so as to have a flat chisel- 
 shaped edge. The H, which is used for marking in final 
 results, and for writing in letters, may have an ordinary 
 point. But as fine lines are always required, both must 
 be kept very sharp, and it is well to have a piece of glass- 
 paper in the box for this purpose. 
 
 (vii) 
 
viii INTRODUCTION. 
 
 The Dividers. — These are used for taking off lengths 
 from the scale, or for measuring the distance between 
 two points on the paper. They must be held nearly flat, 
 so that the points may not make marks on, and so 
 destroy, the scale. In taking off a length with the 
 dividers, open them to a wider distance than is required, 
 and press the legs towards one another. As there is a 
 certain amount of spring in the legs, be sure that you 
 have removed all pressure with your fingers, before you 
 take the dividers off the scale, and then hold them only 
 by the head. 
 
 The dividers are also useful in joining two points. 
 Hold one of the points of the dividers on one of the points 
 to be joined. Press a set square, with which all lines are 
 ruled, up against it, and move the set square round till 
 it is exactly up to the other point. Then remove the 
 dividers and rule the line. 
 
 The Diagonal Scale. — Distances are generally given in 
 Geometrical Drawing in inches ("), and decimals of an 
 inch, and can be measured to two places of decimals, 
 that is hundredths of an inch, by using a Diagonal Scale. 
 
INTRODUCTION. ix 
 
 Distances measured to the left of the line are inches, 
 those to the right, on the top parallel, are tenths. Thus, 
 
 ScaXt of Inches and decimals of an Inch. 
 
 (>- 
 
 ()- 
 
 a distance of 2' 6" can be measured by putting one end of 
 the dividers on the left hand 2, and the other on the right 
 hand 6. 
 
 If, however, a distance is to be measured to two places 
 of decimals, it is to be taken on one of the lower parallels. 
 For instance, if 2-64" is required, it is measured on the 
 fourth parallel, from the point where the diagonal through 
 6 cuts it. 
 
 The student must acquire facility in taking off and 
 reading distances in this way. The distances marked on 
 the parallels by the small circles are 
 
 2-64", 1-36", and 2-09". 
 
 A scale showing twelfths of an inch is also marked on 
 the protractor, and is used when fractions of an inch such 
 as i, ^, T, I are required. 
 
X INTRODUCTION. 
 
 Thx, Protractor. — The protractor is used for drawing 
 angles. If it is required to draw an angle of 70° at the 
 
 
 point -4 of a straight line AB, place the bottom edge of the 
 protractor along the line AB so that the centre mark is 
 at the point A. Make a mark on the paper at the 70 of 
 the protractor, remove it, and join the mark to A. 
 
 The figures 110 indicate that the same point gives an 
 angle of 110° at the point A of the straight line AB pro- 
 duced to the left. 
 
 Set Squares. — A set square, and never the protractor, is 
 to be used in ruling a straight line. Take care that the 
 light falls towards the edge which is being used, and not 
 across it so as to make a shadow, and that the flat edge of 
 the pencil is close up against the edge of the square. 
 
 The two set squares can be used together for drawing 
 parallel straight lines, lines perpendicular to one another, 
 and angles of 60°, 45°, and 30°. 
 
 The accompanying figures will best indicate the method 
 to be adopted in each case, but the student should acquire 
 readiness in the different constructions before proceeding 
 to the problems which follow. 
 
INTRODUCTION. xi 
 
 lAnes 'parallel to AB. One of the set squares is placed 
 with one of its sides along AB. The other is held against 
 
 either of the other edges of the first. The first can now 
 be moved along parallel to itself, as shown by the dotted 
 lines, and parallels can be drawn where required. 
 
 Lmes perpendicular to AB. One of the edges of the 
 
 9or 
 
 first, which contain the right angle, is placed along AB. 
 The other is held against the long edge of the first, which 
 can now be moved, and the perpendicular drawn, where 
 required. 
 
xii INTRODUCTION. 
 
 Angles of 60°, 45°, 30°. The set squares are cut 
 
 so that one contains angles of 60° and 30°, and the other 
 
 two angles of 45°. The method of using the set squares 
 
 to draw these angles will be obvious from the figures. 
 
EXERCISES 
 
 0% fhs use of the Set Squares and the Compass. 
 
 Note. — The term straight edge is used to denote the edge of a ruler, 
 held immovable on the paper so that the side of a set square 
 may slide along it. Thus, on pages xi. and xii., the shaded set 
 squares are used for straight edges. 
 
 For short, the terms " the 60 " and " the 45 " are here used for 
 the set squares which contain angles of 60° and 45° respectively. 
 
 All of the following exercises should be repeated several times, 
 so that facility may be acquired in the use of set squares. At 
 first continuous lines only should be used, but afterwards the 
 constructions may be repeated, using dotted lines. Specimens 
 of dotted lines are given on p. xv. 
 
 1. To draw an equilateral triangle, using set squares only. 
 Also to draw the perpendiculars to the three sides through the 
 angular points. 
 
 Place the 60, so that the angle of 
 30° is towards the top of the paper, 
 and along the shortest side draw the 
 line AB (see figure) about 2|" long. 
 
 Before moving the set square place 
 
 the 45 against the longer side of the 
 
 right angle; and then move the 60 
 
 B down a little way, parallel to itself. 
 
 Now, keeping the 60 fixed, transfer the 45 to the shortest 
 
 side of the 60, and, for the rest of the construction, use it 
 
 unmoved for a straight edge. 
 
 Thus, by moving the 60 along the straight edge, one of the 
 sides AC or B€ can be drawn through A or B, while the 
 other is given by turning the 60 over and sliding along, still 
 with the short side against the straight edge. 
 
 Lastly, the perpendicular CF is found by again sliding the 
 60 along, till the longer side of the right angle covers the 
 point G ; while a second perpendicular, AD or BE, can be 
 
 (xiii) 
 
EXERCISES. 
 
 drawn by turning the 60 rownd till the longer side is in con- 
 tact with the straight edge, and sliding as before ; and then 
 the third, by turning over. 
 
 The test of accuracy is that the three perpendiculars should 
 meet at the point 0. 
 
 Note. — Notice in this, and the next and following Exercises, how, 
 starting with an initial line, one set square is moved only when 
 the other is a straight edge in contact with it. Sometimes the 
 straight edge has to be transferred from one side to another of 
 the set square which is moved ; sometimes one set square is to 
 be used for straight edge, sometimes the other, according as the 
 angles to be drawn are 60° and 30°, or 45°. 
 
 2. Eepeat the previous figure : then with the 45, working 
 on the 60 as a straight edge parallel to BG, draw BG, bisecting 
 
 C the right angle ADB, and draw GS 
 
 perpendicular to AB. 
 
 Now use a pair of compasses to 
 mark off OJ, OK, each equal to OG. 
 Then, with a radius equal to GH, 
 draw the three circles of which the 
 centres are G, J, K, 
 A F H B The test of accui'acy is that the three 
 
 circles should touch each other and the sides of the triangle. 
 
 3. To draw a square, using the set squares only. 
 ^ C Start as in No. 1, but use the 45 
 
 instead of the 60, to draw a straight 
 line AB, about 2J" long, and use the 
 60 for a straight edge parallel to AB. 
 By sliding the 45 on the straight, 
 edge the lines BO, AG, AB (see figure) 
 can be drawn. Then slide the 45 
 back till it covers BO, and move the 
 B straight edge so as to be at right 
 angles to its former position. Then the line OB can be drawn. 
 
EXERCISES. 
 
 K^ 
 
 ^ 
 
 % 
 
 
 ^sJ 
 
 4. Repeat the square ABGD, and, still using the straight 
 J) c edge, draw, first, the second diagonal 
 
 BD, cutting AC at 0; second, paral- 
 lels through to the sides of the 
 square ; third, the diagonal EF of one 
 
 F \^y> <CX!!> <C I of the small squares so formed; 
 fourth, the perpendicular GH. 
 
 Mark off OJ, OK, OL, all equal to 
 OG, and, with a radius equal to GE, 
 
 A E E B draw the four circles as in the figure. 
 
 These circles should touch one another and the sides of 
 
 the square. 
 
 5. On a line, ahout 2J" long, and on opposite sides of it, 
 draw two isosceles triangles with each of the hase angles of 
 the one 60°, and of the other 30°. 
 
 Bisect each of the right angles thus formed, by means of 
 the 45, by lines meeting at 0. From draw a perpendicular 
 to one of the sides of one of the triangles. 
 
 With this perpendicular for radius, and for centre, draw 
 a circle. This circle ought to touch each of the sides of the 
 two triangles. 
 
 6. Measure with the dividers a distance of 1", and step this 
 out twice along a straight line AC, so that AB=\", BC=\". 
 
 With A for centre, and AB, AC for radii, draw two circles. 
 
 With set squares draw through B a perpendicular to AB, cut- 
 ting the outer circle at D, E. At the points B, E draw straight 
 lines within the circle, making angles of 60° with the line DE. 
 
 These two straight lines ought to touch the inner circle and 
 intersect each other at a point on the outer circle. 
 
 7. The following are the most useful forms of dotted lines. 
 The first is that most frequently used, but it is important to be 
 able to draw any of them quite as well as they are printed here. 
 
THINGS TO BE REMEMBERED. 
 
 Draw each line the same thickness throughout. 
 
 Avoid going over a line twice. 
 
 Eule from left to right, when possible. 
 
 Do not make holes through the paper. 
 
 To prevent a set square from slipping, hold it with three 
 fingers. 
 
 Do not make small constructions ; for any error is larger 
 compared with a small construction than with a large one. 
 
 Hold your compasses by the head only. 
 
 The position of a point is indicated by the intersection of 
 two lines ; if other lines are to be drawn passing through it, 
 do not draw them quite up to the intersection of the two lines. 
 
 . The Student who has not read the Third, Fourth, and Sixth Books 
 of Euclid will find a knowledge of the following propositions useful : — 
 
 Book III. 1, 11, 12, 14, 18, 21, 27, 29, 31, 32, 33, 37. 
 
 Book IV. 4, 5, 15. 
 
 Book VI. 2, 4, 19. Cor., 31. 
 
 All the figures in each Chapter should be copied exactly, 
 before the Exercises are attempted; but they should be drawn 
 on a larger scale, — about four to sheet of foolscap paper. All 
 lines drawn must be as fine as it is possible to make them. 
 
 Problems marked with a * may be omitted on a first 
 course. 
 
 (xvi) 
 
CHAPTER I. 
 PRELIMINARY CONSTRUCTIONS. 
 1. To bisect an angle BAG. 
 
 With centre A, and any radius, draw the arc of a circl^ 
 cutting AB at D, and AC at K With centres D, JH, and 
 the same radius, draw two arcs intersecting at F. Join 
 AF. 
 
 Then AF bisects the angle. 
 
 Peoof. In the triangles DAF, EAF, 
 
 DA=FA 
 AF=AF 
 base I>F=heise FF , 
 .: lDAF= lEAF. (Eucl. i. 8.) 
 
GEOMETRICAL DRA WING. 
 
 [Chap. I. 
 
 2. To construct an aingle of 60°. 
 
 With centre A, and any convenient radius, draw an arc 
 cutting AB at C. With centre G, and the same radius, 
 cut off CD. Join AD. 
 
 Then the angle CAD=%Qt\ 
 
 Peoof. AGD is an equilateral triangle (Eucl. i. 1). 
 
 The angles of a triangle together = two right angles, or 
 180°. (Eucl. I. 32.) 
 
Chap.l.] PRELIMINARY CONSTRUCTIONS. 3 
 
 3. To bisect a given straight line, or arc of a circle, AB. 
 
 With A, B for centres, and with the same radius, draw 
 arcs intersecting at G, D. Join GB, cutting AB at E. 
 
 Then AB is bisected at E. 
 
 Peoof. The angle AGB is bisected by GD (Prob. 1), 
 and in the triangles AGE, BGE, 
 
 AG=BG 
 
 GE=GE 
 lAGE=^lBGE 
 .: AE^BE. (Eucl. i. 4.) 
 
 Note 1. — This construction is also used for bisecting a line AB by 
 another OE at right angles to it. 
 
 Note 2. — Only small portions of the arcs at G, D need be drawn, 
 and should, if possible, cut at an angle of about 90°. 
 
4 GEOMETRICAL DRA WING. [Chap. L 
 
 4. To divide a straight line AB into any number of 
 equal parts (say six). 
 
 Througli A draw any line AO, making an angle of 
 about 30° with AB. From A, along AG, step out with 
 the dividers any six equal lengths Aa, ab, be, cd, de, ef. 
 Join Bf, and draw with the set squares e5, di, c3, 62, al, 
 parallel to Bf. 
 
 Then AB is divided into six equal parts at 1, 2, 3, i, 5, 
 
 Proof. Chap, v., Prob. 4. 
 
 5. To draw a line parallel to a given straight line AB, 
 through a given point 0. 
 
 ■ 
 
 IE? 
 
 V ""-. ; 
 
 /' 
 
 A F D B 
 
 With centre G, and any radius, draw the arc BE. 
 With centre B, and the same radius, draw the arc GF. 
 With centre B, and radius GF, cut off BE. Join GE. 
 Then GE is parallel to AB. 
 
Chap. I.} 
 
 PRELIMINARY COMSTRtrCTIONS. 
 
 Peoop. In the triangles CDE, BGF, 
 
 CE=BF • ; 
 
 CD=DG 
 base i)-ff= base CJ' 
 .-. lDGE= lCDF 
 :. GE is parallel to AB. 
 
 (Eucl. I. 8.) 
 (Eucl. I. 27.) 
 
 NoU. — Parallel straight lines should be drawn by means of set 
 squares, unless the construction is required. 
 
 6. To draw a line perpendicular to a given straight line 
 AB/rom a point G in it. 
 
 ^k 
 
 'D 
 
 ^ 
 
 e' ^ 
 
 First Gonstruction. — With centre G, cut off equal lengths 
 GD, GE. With centres D, E, and with equal radii, draw 
 arcs intersecting at F. Join GF. 
 
 Then GF is perpendicular to AB. 
 
 PfiOOF. Eucl. I. 11. 
 
GEOMETRICAL DBA WING. 
 
 [Cbap. t 
 
 Second Gonsti-uction. — With centre G, draw the arc BEF, 
 
 D B 
 
 With the same radius, cut off BE, EF. With centres 
 E, F, and the same radius, draw arcs intersecting at G. 
 Join GG. 
 
 Then GG is perpendicular to AB. 
 
 Proof. LZ)a£'=60°, and lEGF=60° (Prob. 2); also 
 LEGFis bisected by GG (Prob. 1). 
 
 .-. lBGG=60°+30° = 90''. 
 Third Gonstruction. — With any centre B, and radius 
 
 BG, draw an arc greater than a semicircle, cutting AB 
 at E. Join EB, cutting the arc again at F. Join GF. 
 Then GF is perpendicular to AB. 
 
Chap. I.] 
 
 PRELIMINARY CONSTRUCTIONS. 
 
 Proof. "The angle in a semicircle is a right angle" 
 (Eucl. III. 31). 
 
 'NoU 1. — Only so much of the semicircle and diameter need be 
 drawn as will give the points B, P. 
 
 Note 2. — The first of these constructions cannot be used if C is 
 near one end of the line AB. The second should always be used, 
 when other angles have to be constructed in the same figure. 
 
 7. To draw a line perpendicular to a given straight line 
 AB, from a point G outside it. 
 
 -^ 
 
 ,.^1. 
 
 D'-:, 
 
 ,.....'■'£ 
 
 >K? 
 
 First Construction. — With centre G draw an arc cutting 
 AB at B, E, -With centres D, B, and the same radius, 
 draw arcs below AB intersecting at F. Join GF. 
 
 Then GG is .perpendicular to AB. 
 
 PiiooF. See Prob. 3. . 
 
8 GEOMETRICAL DRA WING. [Chap. I. 
 
 Second Constriction. — Take any point D in AB. Join 
 
 OR Bisect CD at E (Prob. S). With centre E, and radius 
 C!^, draw an arc cutting AB at F. Join CJ". 
 Then CF is perpendicular to ^.B. 
 
 Proof. "The angle in the semicircle CFB is a right 
 angle " (Eucl. III. 31). 
 
 Note 1. — The second construction must be used if the point C is 
 nearly over one extremity of AB. 
 
 Note 2. — A pe^endicular can also be drawn by means of set squares, 
 unless the problem given includes the construction of the perpendicular. 
 
 8. To construct an angle of 45°. 
 
 A B 
 
 Draw the angle BAG= 90° (Prob. 6, Second Construc- 
 tion). 
 
 Bisect the angle BAG by AD (Prob. 1). 
 Then the angle B-42? = 45°. 
 
Chap. L] 
 
 PRELIMINARY CONSTRUCTIONS. 
 
 9. To construct angles of 30° and 15°. 
 Draw the angle BAC=60°. 
 
 (Prob. 2). 
 Bisect the angle BA G hy AD. 
 
 (Prob. 1). 
 Then the angle BAD =30'. 
 Bisect the angle BAD by AM 
 Then the angle BAH =15°. _ 
 
 Note 1. — Similarly, by bisecting an angle of 15° we can construct 
 7^°, but there is no simple geometrical construction by which an angle 
 containing a whole number of degrees less than 15° can be constructed ; 
 consequently, when whole numbers of degrees are required, only the 
 angles 
 
 15°, 30°, 45°, 60°, 75°, 90°, 105°, etc., 
 can be constructed ; and any angle which cannot be divided into the 
 sum or difference of any of these angles must be drawn with the 
 protractor. 
 
 Note 2. — In constructing angles, it is best to mark them on a 
 
 X 
 
 >r 
 
 primary circle, as in the accompanying figure, in which aU the angles 
 are measured from the same base line ; when once an angle has been 
 
10 GEOMETRICAL DRA WING. [Chap. 1. 
 
 constructed, it can then be used for constructing others, by stepping 
 out the distance which it measures on the circle, at any point required. 
 Thus, first the angles 60° and 120° are marked off, then the angle 
 90°. Then the angle 75° is formed by bisecting. This gives, between 
 60° and 75°, the measure of the angle 15°, which can be stepped out 
 all round the circumference. 
 
 EXEECISES— I. 
 
 1. Draw a straight line 3 '68" long, and by continually bi- 
 
 secting, find an eighth part of it. 
 
 2. Draw a straight line 3-68" long, and divide it into eight 
 
 equal parts by the method of Problem 4, 
 
 3. Construct the angles 60°, 30°, 75°. 
 
 4. Construct the angles 45°, 1^°, 135°. 
 
 5. Construct the angles 52 J°, 97^°, 165°. 
 
 6. Draw ^5=3"; at A make the angle £^C=45°, and 
 
 mark ofiF^C'=l"; draw, by construction, OD parallel 
 to AB, making 00=2"; draw BE perpendicular to 
 AB. Measure BE, BE. 
 
 7. Draw a triangle, about 2^" across, and bisect the angles. 
 
 8. Draw a triangle, and draw, by construction, perpendiculars 
 
 from the three vertices to the opposite sides. 
 
 9. Draw a triangle ; bisect the sides, and join the points so 
 
 found to the opposite vertices. 
 
 10. Draw a line parallel to any line AB, about 3" long, at a 
 
 distance of J" from it. 
 
 11. Take any two points and any straight line, and find the 
 
 point in the line equidistant from the two points. 
 
CHAPTER II. 
 
 TRIANGLES AND PARALLELOGRAMS. 
 
 A TRIANGLE OT parallelogram can be drawn when three 
 conditions are given, except when angles are the only- 
 conditions given. 
 
 For shortness, the angles of the triangle ABG will be 
 denoted by A, B, G, and the sides opposite to them, a, I, c 
 respectively. 
 
 Problems. 
 
 \. To draw a triangle when the three 'sides are given 
 {a, b, c). 
 
 A 
 
 Draw AB=c. With centre A, and radius b, and with 
 centre B and radius a, draw two arcs intersecting at 0. 
 Join AG, BG. 
 
 Then ABG is the triangle. 
 
 Pkoof. Eucl. I. 22. 
 
 11 
 
t2 
 
 GEOMETRICAL DRA WING [chap. M. 
 
 2. To draw a triangle, given two sides and the included 
 angle (b, A, c). 
 
 Construct the angle BAC=A. Mark off the lengths 
 AB= c, and A0= b. Join £G. 
 
 Then ABC is the triangle. 
 
 Proof. Eucl. i. 4. 
 
 3. To draw a triangle, given two sides and the angle 
 opposite to one of them (a. A, c). Ambiguous Case. 
 
 B 
 
 Construct the angle BAO=A. Mark off AB=e, and 
 with centre B and radius a, draw a circle, cutting AG 
 in a 
 
Chap. II.] TRIANGLES AND PARALLELOGRAMS. 13 
 
 Then ABC is the triangle. 
 Peoof. Obvious. 
 
 NoU.—\xi. general two triangles ABO, ABC can be drawn satisfy- 
 ing the conditions. Hence the name Amhiguous Oase. 
 The triangle is not ambiguous — 
 
 (i.) If a is greater than, or equal to, c. 
 
 (ii.) If BC is perpendicular to AG, for then the circle touches 
 AC. 
 
 ,,-"'C 
 
 4. To dravj a triangle, given two angles and the side 
 adjacent to both (A, B, c). 
 
 
 
 Draw AB=e, At A, make the angle BAO=A; and 
 at B, the angle ABG=B. 
 
 Then ABG is the triangle. 
 
 Proof. Obvious. 
 
14 
 
 GEOMETRICAL DRA WING. 
 
 [Cbap. II. 
 
 5. To draw a triangle, given two aisles and, the side 
 opposite to one of them (A, B, 6), 
 
 A. —-^ B 
 
 First Construction. — Draw AC= h. At A, draw the angle 
 CAB=A; and at G, the angle AGB=l?,(i°-(A+B). 
 
 Then ABC is the triangle. 
 
 Proof. The three angles of a triangle = 180°. Therefore 
 by construction ABC=B. 
 
 Second Construction. — Draw AC=b. At A, draw the 
 C 
 
 angle CAB— A. At any point B, in AB, make the 
 angle ABE= B. Draw CB parallel to BE, 
 Then ABC is the triangle. 
 
 Peoof. " The exterior angle ABE = the interior oppo- 
 site angle ABC" (Eucl. I. 29). 
 
Chap. II.] TRIANGLES AND PARALLELOGRAMS. 15 
 
 6. To draw a triayigle, given the base, a hose an^le, and 
 the sum, of the sides (c, A, a+h). 
 
 y 
 
 Draw AB—c, and the angle BAO=A. On AG mark 
 off AD = a + i. Join DB. Bisect BB at right angles by 
 EO (Chap. I. Prob. 3). Join BO. 
 
 Then ABC is the triangle. 
 
 Proof. In the triangles BEG, BEG, 
 
 BE=DE 
 
 EG=EG 
 
 right angle jB^C= right angle BEG 1 
 
 .-. base ^C=base BG (Eucl. i. 4.) 
 
 .\BG+GA = BA 
 = a + h 
 
i6 
 
 GEOMETRICAL DRA WING. icaap. 11. 
 
 7. To draw a triangle, given the 
 angle (c, G). 
 
 base and the vertical 
 
 Draw AB=c, BAD=^Q°-G. Bisect AB at right 
 angles by EF. With centre F, radius FA, draw a circle. 
 
 Any point on this circle joined to A, B will give the 
 triangle. 
 
 Proof. The construction is a modification of that in 
 Euc. III. 33. Here the angle BAD is drawn equal to 
 90° — 0, instead of drawing the angle dX A below AB, 
 and AD perpendicular to the line thus drawn. 
 
 Note 1. — If the vertical angle is greater than 90°, the line AD is to 
 be drawn hdow AB. 
 
 Note 2. — Only two conditions have been given so that a third can be 
 satisfied. Thus — 
 
 (i.) If the triangle is to be isosceles. 
 
 Take the point H, where UF cuts the circle. 
 (iL) If the altitude is given. 
 
 Mark off the altitude EG on MF, through G draw GO 
 parallel to AB, and take one of the points G where it 
 cuts the circle. 
 
Chap. II.] TRIANGLES_AND PARALLELOGRAMS. 17 
 
 8. To dravJ a right-angled triangle when the hypotenuse 
 and one side are given, (c, a, C= 90°.) 
 
 X 
 
 Draw AB=e, the hypotenuse. On AB draw a semi- 
 circle. With B for centre, and radius a, cut off BG=a. 
 
 Join AG, BG. 
 
 Then ABG is the triangle. 
 
 Peoof. By Euc. iii. 31, " the angle in a semicircle is a 
 right angle." 
 
 Cqk. " To draw a rectangle when a diagonal and one 
 side are giwen," draw a circle of which, the diagonal is 
 diameter, and apply' the above construction. 
 
 B 
 
1 8 
 
 GEOMETRICAL DRA WING. 
 
 [Chap. II. 
 
 * 9. To draw a triangle, the perimeter being given, and 
 the sides p-oportional to 2, 3, 4 (or any other numbers). 
 
 Draw AB equal to the perimeter. Divide it at L, M, so 
 that AL : LM : JO : : 2 : 3 : 4. {See Chapter v. Prob. 7.) 
 
 Draw the triangle LMN, with the sides equal to 
 AL, LM, MB (by Prob. 1). 
 
 * 10. To draw a triangle, the perimeter being given, and 
 the angles proportional to 2, 3, 4 (or any other numbers). 
 
 B A 
 
 Draw a semicircle, and, by trial, divide the circum- 
 ference into 9 (i.e. 2 + 3 + 4) equal parts. Join ^0, ^2, 
 Af>. Draw any line BG parallel to A5. 
 
 * These problems, and all others marked with a *, can be omitted 
 in a first course. 
 
CHap. II.] TRIANGLES AND PARALLELOGRAMS. 19 
 
 Then, by Prob. 9, draw a triangle whose perimeter is 
 that given, having its sides proportional to those of the 
 triangle ABG. 
 
 Proof. 
 
 y 9 
 
 and L CBA = l 5AQ = -| x 180°. 
 
 Also, since the sides of the second triangle are propor- 
 tional to those of ABG, therefore the triangles are similar 
 (Euc. VI. 5) ; that is, the angles in both are equal. 
 
 * 11. To draw a triangle, given the base, the vertical 
 angle, and the sides in a given proportion, (c, G, a:h.) 
 
 E 
 Draw the angle EBF= G. Step out with the dividers 
 DG, DA in the proportion required. Join AG. Mark 
 off AB=e along AG. Draw BG parallel to DG. 
 Then ABG.ia the triangle. 
 Pboof. ABG is similar to AGD ; (Ch. v. § 2.) 
 
 ■. lBGA= lGDA; 
 and BG:AC::DG:AD 
 : : a : 6. 
 
20 
 
 GEOMETRICAL DRA WING 
 
 [Chap. n. 
 
 12. To draw a triangle, given the altitude and the base 
 angles. {A, B, altitude.) 
 
 At a point C of a straight line BE draw the angles 
 EGA=A,DGB=B, in opposite directions. Draw CH^ei- 
 pendicular to DE, making GH equal to the altitude given. 
 Draw BHA perpendicular to GH. 
 
 Then ABG is the triangle. 
 
 Proof, AB is parallel to DE; 
 
 .: lBAG=lAGE=A; 
 and lABG=lBGD=B. 
 
OUap. II.] TRIANGLES AND PARALLELOGRAMS. 21 
 
 13. To draw a parallelogrcum ABG, given two sides AB, 
 AD, and the included angle BAD. 
 
 Draw the angle BAD. Mark off the lengths AB, AD. 
 With centre B and radius equal to AD, and with centre 
 D and radius equal to AB, draw two arcs intersecting at 
 a Join BC, CD. 
 
 < Then ABGD is the parallelogram. 
 
 Peoof. Join AG. AB= GD 
 
 AG=GA 
 base 5C=base AD 
 .: lBAC= lDGA; 
 .: AB is parallel to GD ; (Eucl. i. 27.) 
 So also BG is parallel to AD. 
 
! GEOMETRICAL DRA WING. 
 
 14. To inscribe a circle in a triangle ABO. 
 A 
 
 [Cbap. IL 
 
 Bisect two of the angles B, 0, by BO, CO. Draw OD 
 perpendicular to EG. 
 
 Then is the centre of the inscribed circle, OD the 
 radius. 
 
 Proof. Eucl. iv. 4. 
 
 15. To draw the escribed circles of a triangle ABO. 
 
 A-4 
 
 Bisect the exterior angles at B, 0. Draw OD perpen- 
 dicular to BO. 
 
Chap. II.] TRIANGLES AND PARALLELOGRAMS. 23 
 
 ■ Then is the centre of one of the escribed circles, and 
 OB is the radius. 
 
 Proof. Similar to Eucl. iv. 4. 
 JVofe. — There are three eacribed circles to every triangle. 
 
 16. To describe a circle about a triangle ABC. 
 
 Bisect two of the sides JBC, GA by the perpendiculars 
 DO, EO. 
 
 is the centre of the circumscribing circle, and OA or 
 OG the radius. 
 
 Proof. Eucl. iv. 5. 
 
24 
 
 GEOMETRICAL DRA WING. 
 
 rchap. II. 
 
 * 17. To draw a triangle, given the radivs of the inscribed 
 circle, and the three sides or angles in a given proportion. 
 
 Draw any triangle ABG, whose sides or angles are in 
 the given proportion (Probs. 9, 10). Find 0, the centre of 
 the inscribed circle of this triangle (Prob. 14). Draw the 
 perpendiculars OD, OE, OF. With centre 0, and the 
 given radius, draw a circle. Produce OD, OE, OF to 
 meet this circle in d, e, f respectively. Draw perpen- 
 diculars to these lines through d, e,f. 
 
 Then LMN is the triangle. 
 
 Pboof. The triangles are similar (Ch. v. § 2), and 
 therefore the sides of LMN are in the given proportions ; 
 also defis the inscribed circle. 
 
Chap. II.] TRIANGLES AND PARALLELOGRAMS. 25 
 
 * 18. To draw a triangle, given the radius of the circum- 
 scribing circle, and the sides or angles in a given proportion. 
 
 Draw any triangle ABC, whose sides or angles are in 
 the given proportion (Probs. 9, 10). Find 0, the centre 
 of the circle circumscribing this triangle. With centre 0, 
 and the given radius, draw a circle, and join OA, OB, 00, 
 meeting it at Z, M, iV. 
 
 Then ZJfiV is the triangle. 
 
 Peoof. OA = OB, OL = OM; 
 
 :. OA:OL::OB: OM; (Chap. v. § 1.) 
 .-. AB is parallel to LM. 
 Similarly, the other sides' are parallel, and the triangles 
 are similar. 
 
26 GEOMETRICAL DRA WING. [Chap. II. 
 
 EXERCISES— II. 
 
 1. Draw a triangle, given that a= 2 -28", 6 = 1-86", c = 2'03", 
 
 and in it inscribe a circle. Measure the radius. 
 
 2. Draw a triangle, given that a = 1-53", c = 2', .(4 = 44°. 
 
 Measure h. 
 
 3. Draw a triangle, given that a=l-l", ^ = 30°, .8=60°, and 
 
 about it describe a circle. Measure the radius. 
 
 4. The perimeter of a triangle =4-22"; c=M7", ^ = 55°. 
 
 Draw the triangle. Measure a. 
 
 5. Draw a triangle whose base is r86", altitude 1'2", and 
 
 vertical angle 30°. Measure the sides. 
 
 6. Draw an isosceles triangle whose base is 1"04", and 
 
 vertical angle 30°. Measure a side. 
 
 7. The perimeter of a triangle is 5", and the sides are in the 
 
 proportion of 2:3:4. Draw the triangle. Measure 
 the longest side. 
 
 8. Draw a triangle similar to that in the last question, the 
 
 radius of its inscribed circle being f". Measure the 
 shortest side. 
 
 , 9. Draw a right-angled triangle, given the hypotenuse 2", 
 and one side •1f>". Measure the third side.. 
 
 10. Draw a right-angled triangle, given the hypotenuse 2", 
 and the sides in the proportion of 2 : 3. Measure the 
 shortest side. 
 
CHAPTER III. 
 
 REGULAR FIGURES. 
 
 A Regular Figure is one in which all the sides and 
 angles are equal. Every Regular Figure can be inscribed 
 in a circle, which is thus divided into as many equal arcs 
 as there are sides of the figure. 
 
 Problems. 
 
 1. To inscribe an Equilateral Triangle in a circle. 
 The radius AO can be stepped 
 
 out exactly six times round the 
 circumference, giving the six 
 equal arcs AB, BG, CD, etc. 
 
 Join alternate points A, 0; 
 G, E; E, A. 
 
 Then AGE is an equilateral 
 triangle. 
 
 Proof. The arcs AG, GE, EF 
 are all equal. 
 
 "Equal arcs are subtended by equal straight lines" 
 (Eucl. III. 29). .-. The lines AG, GE, EF are all equal. 
 
 2. To draw anEquilateral Triangle onagiven straight line. 
 ■S With A, B, the extremities of the 
 
 given straight line for centres, and 
 with AB for radius, draw arcs inter 
 secting at G. 
 
 Join AG, BG. 
 
 Then ABG is an equilateral tri- 
 
 A B angle. 
 
 Note. — The equilateral triangle can also be drawn with set squares. 
 See p. xiii. 
 
 Proof. Eucl. i. 1. 
 
 27 
 
28 
 
 GEOMETRICAL DRA WING. 
 
 [CUap. III. 
 
 3. To inscribe a Squa/re in a circle. 
 
 With set squares draw two 
 diameters AG, BB perpendi- 
 cular to one another. 
 
 Join AB, BG, GB, BA. 
 Then ABGB is a square. 
 Pkoof. Eucl. IV. 6. 
 
 Note. — If, in this and the other regular figures, further constructions 
 are to follow, it will often be unnecessary to put in all the lines of 
 construction. Thus, here, the dotted diameters would then be omitted. 
 Short lines through A, B, O, D and O should, however, always be 
 drawn, so as to show how the construction has been effected. 
 
 4. To draw a Square on a given line, (See p. xiv.) 
 
 Using the 60 set square for straight 
 edge, draw, with the 45, perpendiculars 
 AB, BG to the given straight line AB ; 
 and also AG, making an angle of 45° 
 with AB. Through G draw GB paral- 
 lel to AB. 
 
 Then ABGB is the square. 
 Pboof. Obvious. 
 
Cbap. III.] 
 
 REGULAR FIGURES. 
 
 29 
 
 K 
 
 5. To inscribe a regular Pentagon in a circle. 
 
 Through the centre draw 
 the radius OA perpendicular to 
 a diameter £0. Bisect OB at 
 B. With centre D, and radius 
 DA, draw an arc cutting OC 
 \b at R 
 
 Then AH is the side of the 
 inscribed pentagon, which may 
 be stepped out round the cir- 
 cumference. 
 
 Pkoof. The proof depends upon Trigonometry. 
 
 5 (o). To draw a regula/r Pentagon on a given line. 
 
 Bisect the given line AJB by 
 a perpendicular through C. 
 
 Find D, on the perpendicular, 
 by marking off the length AG 
 twice. 
 
 Join BD and produce it, mak- 
 ing BU equal to AG. 
 
 With centre B and radius BB 
 draw an arc, cutting GD at F. 
 With A, F, £ for centres, and 
 radius equal to AB, draw arcs intersecting at G, H. 
 Join AG, GF, FH, SB. 
 Then AGFEB is the regular pentagon on AB. 
 Pboof. This depends on Trigonometry. 
 
 X 
 
3° 
 
 GEOMETRICAL DRA WING. 
 
 [Chap. III. 
 
 6. To inscribe a regular Hexagon in a circle. 
 
 Peoof. Eucl. IV. 15. 
 
 From any point A the radius 
 can be stepped out round the 
 circumference of the circle ex- 
 C actly six times. 
 
 Join consecutive points A, B; 
 B, G, etc., forming the regular 
 hexagon ABGDEF. 
 
 Note. — It is well to use pencil compasses for marking off the arc, 
 AB, BO, etc., so as to show by small arcs at B, O, etc., how the con- 
 struction has been performed. 
 
 6 (a). To dravj a regular Hexagon on a given line. 
 
 I' l^ ^.n With A, B, the extremities 
 
 of the given line, for centres, 
 and with AB for radius, draw 
 two arcs to intersect at 0. 
 
 With for centre, and with 
 the same radius, draw the circle 
 ^j 7; passing through A, B. 
 
 Step out the radius round the circumference and 
 join up. 
 
 Pkoof. Eucl. IV. 15. 
 
Cbap. III.] 
 
 REGULAR FIGURES. 
 
 31 
 
 7. To inscribe a regular Heptagon in a circle. 
 
 Bisect a radius OCf of the 
 circle by the perpendicular AB, 
 cutting the circle at A. 
 
 Then AB is the side of the 
 regular heptagon, which can be 
 stepped out on the circumference. 
 Peoof. The construction is 
 approximately true. When AJ) 
 has been stepped out on the 
 circle, the seventh point is short of A by less than ^^th 
 of the circumference. 
 
 7 (a). To draw a regular Heptagon on a given line. 
 
 Draw, with set squares, ani/ equilateral triangle LMN, 
 and its perpendicular ZB. 
 
 On ZB mark off ZS equal to AB, the given line. 
 
 Draw ST parallel to MJST to meet ZM at T. 
 
 Then, ZT is the radius of the circle circumscribing the 
 heptagon. 
 
 A B M Ji N 
 
 With A, B for centres, and radius ZT, draw arcs inter- 
 secting at 0. 
 
 With for centre, and the same radius, draw a circle. 
 
 Step out AB round the circumference. 
 
 Peoof. The construction is approximately true. 
 
32 
 
 GEOMETRICAL DRA WING. 
 
 [Chap. lU. 
 
 8. To inscribe a regular Octagon in a drele. 
 
 Draw the diameters AE, CG 
 perpendicular to one another. 
 Bisect the arc JEIG at F. 
 
 Then draw the diameters 
 1^ FB, DE perpendicular to one 
 another. 
 
 Join AB, BG, GD, etc., form- 
 ing a regular octagon. 
 Peoof. The arcs AB, BG, etc., 
 are all equal, and " equal arcs subtend equal straight lines." 
 
 8 (a). To draw a regular Octagon on a given line. 
 
 Bisect the given line AB by 
 a perpendicular through G. On 
 the perpendicular mark off GD 
 equal to^C.andDO equal to DA. 
 
 Then is the centre of the 
 circle which will circumscribe 
 the octagon. Draw this circle, 
 with radius OA ; step out AB 
 round the circumference, and 
 join up. 
 
 Peoof. If AD, ^0 be joined then 
 
 AG=GD, .: lADG= lDAG=4:5''. 
 AD=DO,.: lAOD= lDAO: 
 
 but L ADG= L AOD+ l DAO. 
 :. lA0D='L1\°. 
 :. Z.^05=45° = |X360° 
 i.e. L AOB is the angle subtended by the side of a regular 
 octagon inscribed in the circle of which is the centre. 
 
 X 
 
Chap. III.] REGULAR FIGURES. 33 
 
 5. To describe a regular figure about a circle. 
 
 Find the points on the circle belonging to the inscribed 
 figure of the same number of sides, and draw perpendicu- 
 lars through these points to the radii of the circle. 
 
 EXEEOISES— III. 
 
 {Note. — In each case measure the side of the figure or the radius 
 of the circle; which has to be determined.) 
 
 In circles of 1"25" radius inscribe the following regular 
 figures : — 
 
 1. A square. 2. A pentagon. 
 
 3. A hexagon. 4. A heptagon. 
 
 5. An octagon. 
 
 6. On a line of 1"25" long, draw a regular pentagon, 
 
 7. On a line 1" long, draw a regular hexagon. 
 
 8. Describe a regular pentagon about a circle of 1" radius. 
 
 9. Draw a regular hexagon, given that the diagonal joining 
 
 two opposite points is 1 -84:". 
 
 10. Inscribe an equilateral triangle in a circle of 1'2" radius. 
 
 11. Draw a regular heptagon on a line 1-5" long. 
 
 12. Construct a regular octagon with a side of 1". 
 
CHAPTER IV. 
 
 FIGURES INSCRIBED IN SYMMETRICAL 
 FIGURES. 
 
 A Symmetrical Figure is one which is of the same 
 form on either side of a certain line, which is called the 
 Axis of Symmetry. All Regular Figures are particular 
 cases of Symmetrical Figures. 
 
 Problems. 
 
 1. To inscribe a circle in a Symmetrical Figwre. 
 
 Bisect two of the angles A, B, by lines meeting at 0. 
 Draw OD perpendicular to AB. 
 
 S4 
 
Chap; IV.] FIGURES IN SYMMETRICAL FIGURES. 35 
 
 Then is the centre of the inscribed circle, OD the 
 radius. 
 
 Pkoop. l 0AD= l OAE s 
 
 right L ODA= right l OEA t 
 side 0^ = side OA ) 
 
 .: OB=OE. (Euc. I. 26.) 
 
 Similarly, OD=OF. 
 
 Note. — It is not possible to inscribe a circle in a rectangle, with 
 adjacent sides unequal, or in some other symmetrical figures. 
 
 2. To inscribe a square in a Symmeirical Figure (e.g. in 
 a Begular Pentagon). 
 
 B 
 
 Draw a diagonal AG, cutting off one vertex, B. On 
 the side oi AG remote from B, draw the perpendicular 
 GE, equal to AG. Join EB, cutting GB in F. Draw FG 
 parallel to GE, and EG parallel to AG, and complete the 
 figure FGHK, which is a square. 
 
36 GEOMETRICAL DRA WING. [Chap. IV. 
 
 * Pkoof. FG, GH are parallel to EC, GA respectively. 
 .-. lFGE= Ll!GA = ^(i°. 
 Also from the similar triangles BGH, BGA, 
 
 EG:AG::BG:BO. (Chap. v. § 2.) 
 And from the similar triangles BGF, BOB. 
 
 BG -.BG-.-.GF-.GE. 
 .:HG:AG::GF:GB. (Chap. V. § 3.) 
 
 But AG= CE, by construction ; 
 :.EG=GF. 
 
 Hence, by symmetry, the figure is a square. 
 
 COK. I. " To inscribe a square, in a semicircle" is only 
 
 a particular case of inscribing a square in a sector of 
 
Chap. IV.] FIGURES IN SYMMETRICAL FIGURES. 37 
 
 a circle the diagonal being AG, and the vertex, B, the 
 centre of the semicircle. 
 
 Tte 
 
 A n B G c 
 The proof in both cases is identical with that given above. 
 
 3. To inscribe a square in a triangle ABC. 
 
 E. 
 
 The construction and proof are identical with Pro- 
 blem 2 ; the side AG being taken for diagonal. 
 
38 
 
 GEOMETRICAL DRA WING. [OLap. IT. 
 
 4. To mscribe an Hquilateral Triangle in a Si/mmetrical 
 Figure (e.g. in a Square). 
 
 The construction and proof are identical with Pro- 
 blem 1, except that the angle ACH is made equal to 60°, 
 the angle of an equilateral triangle. 
 
 * 5. To inscribe a Triangle, similar to a given one, LMN, 
 in a Symmetrical Figure (e.g. in a Regular Pentagon). 
 
 At Cin AC make the angle ACE equal to the angle at 
 M. Make OF so that — 
 
 NM:ML::AG: OF. (Chap. V. Prob. 1.) 
 
 Join BF. Draw FG parallel to OF, and &E parallel 
 to AC. Join HF. 
 
Chap. IV.] FIGURES IN SYMMETRICAL FIGURES. 39 
 
 Then FGE is similar to LMN. 
 5 
 
 Proof. EG, GF are parallel to AC, GE respectively. 
 
 .-. lHGF=lAGE 
 =M. 
 Also EG:AG:.BG:BC 
 -.-.GF -.GE 
 .: HG:GF::AG:GE (Chap. V. § 2, Note 2.) 
 but AG -.GE:: NM: ML 
 .: EG: GF::NM:ML 
 
 by Eucl. VI. 6, the triangles FGE, LMN, which have 
 the sides about equal angles proportional, are similar. 
 
40 GEOMETRICAL JDRA WING. [CHap. IV. 
 
 NoU 1. — If the inscribed figure is equilateral, it is obvious that 
 ^hen one side has been found by the preceding method of construction, 
 the rest can be stepped out by means of the dividers. 
 
 NoU 2. — In this way any regular figure can be inscribed in another 
 of a greater number of sides, by making the angle AGE equal to the 
 angle of th@ required figure. 
 
 Note 3. — A regular figure can be inscribed in another of the same 
 number of sides by this method, or, more conveniently, by joining the 
 middle points of the sides of the given figure. 
 
 EXEKCISES— IV. 
 
 (Note. — In every case measure the length of the sides of 
 the inscribed figure.) 
 
 1. On opposite sides of a line 1 '6" long draw an equilateral 
 
 triangle, and an isosceles triangle whose sides are 2" 
 long. In the figure so formed inscribe a square, and 
 a circle. 
 
 2. On a line 2" long draw an equilateral triangle, and in it 
 
 inscribe a square. 
 
 3. On a line 1" long draw a regular hexagon, and in it 
 
 inscribe a square. 
 
 4. Inscribe a regular pentagon in a circle of 1" radius, and 
 
 in the pentagon inscribe a square. 
 
 5. Inscribe a square and an equilateral triangle in the quad- 
 
 rant of a circle of 1'5" radius. 
 
 6. On a line TS" long draw a regular pentagon, and in it 
 
 inscribe an equilateral triangle. 
 
Chap. IV.] FIGURES IN SYMMETRICAL FIGURES. 41 
 
 7. On a line 1 '2" long draw a square, and in it inscribe an 
 
 equilateral triangle. 
 
 8. Draw a triangle whose sides are 2-4'', 1'9', and 1'3". On 
 
 the shortest side draw a square, and in the square 
 inscribe a triangle similar to the first. 
 
 9. In the same triangle inscribe a rectangle, with one side 
 
 double the first, and having one of its longer sides in 
 the longest side of the triangle. 
 
 10. Inscribe a regular pentagon in a regular hexagon of 1-6" 
 side. 
 
CHAPTER V. 
 
 PROPORTION. 
 
 § 1. When four quantities, a, 6, c, d, are proportional — 
 
 that is, so that 
 
 a is to 2) as c is to d 
 
 or, a : h : -.c : d 
 
 then, by Arithmetic and Algebra, we know that 
 
 a _c 
 T~~d 
 
 that is, JlfL=^ 
 second fourth 
 
 or, ad=hc 
 
 i.e., "product of extremes =jproduct of means." 
 
 § 2. Fractions and products are not geometrical concep- 
 tions, and, though these conditions are useful in checking 
 results, another definition is necessary. 
 
 Geometkioal Definition of Peopoetion. — "In two 
 similar tria/ngles the sides opposite to one pair of equal 
 angles are in proportion with the sides opposite to another 
 pair." 
 
Cl»ap- v.] PROPORTION. 
 
 43 
 
 Similar triangles are those which have all their angles 
 respectively equal. 
 
 A B n 
 
 Thus, in the figure, since BG is parallel to BE, 
 
 .: lACB= lAEB, and lABG= lABE; 
 and AB, AB, and AG, AE, 
 
 are the sides respectively opposite to them, 
 
 .-. AB : AB :: AG : AE. 
 So also it follows, from the definition, that 
 AG:AE::BG: BE, 
 and that AB : AB : : BG : BE. 
 
 Note 1. — In forming a proportion from two similar triangles, the 
 first. and third proportionals must be taken from one triangle, and 
 the second and fourth from the other. 
 
 Ifote 2. — From any proportion another may be formed, by inter- 
 changing the second and third proportionals, or the firit wndfowrtK. 
 
 Thus, if a :h :: c id 
 then a : c : -.h : d 
 and d :}>•.■. e : a. 
 
 The truth of this can be seen by applying the conditions of § 1. 
 
44 GEOMETRICAL DRA WING. [Chap. V- 
 
 Nate, 3. — Results derived from proportion are of frequent occur- 
 rence in Geometrical Drawing, and it is necessary that the student 
 should acquire facility in stating the proportions derived from 
 similar triangles. 
 
 The correctness of a proportion should be tested in the following 
 manner : — 
 
 Write the names of the two triangles, so that the letters denoting 
 the equal angles are in the same order in both. Thus, in the above 
 figure, 
 
 ABG and ABB, or GBA and BBA. 
 
 Then the letters denoting the first and second proportionals will 
 occur in the same order in these arrangements, and so also those 
 denoting the third and fourth. Thus : — 
 
 1 2 
 
 So Ide 
 
 AB:AD::AG: AE. 
 
 § 3. It can also be proved in the figure of § 2, that 
 AB:BD::AG:GE 
 and that AD : BD : : AE : GE. (Eucl. vi. 2.) 
 
 And it follows at once, from the Algebraical definition of 
 proportion, that if 
 
 a -.h :: c -.d 
 
 and also a -.1 :: e -.f 
 
 then c : d ■.: e :f. 
 
Chap, v.] 
 
 PROPORTION. 
 
 45 
 
 § 4. The proportions obtained from the accompanying 
 figure are of the greatest importance, and should be 
 verified by the student. 
 
 If ABG is a right-angled triangle, and BD be perpen- 
 dicular to AG, then 
 
 since L ABD+ L DBO = 90° 
 and L ABD+ l DAB= 90° ; - 
 .-. lBBG -= lBAB. 
 ^osy^oLABB= lBCB. 
 
 Hence the three triangles ABG, ABB,BBG 
 are similar to one another. 
 
 Therefore, in the triangles ABB, BBG, 
 AB:BB::BB:BG\ 
 
 in the triangles ABG, ABB, 
 CA:AB::AB:AB; 
 
 and in the triangles ABG, BBG, 
 AG:GB::GB:GB. 
 
46 GEOMETRICAL DRA WING. [Cbap. V, 
 
 § 5. If, as in the three preceding proportions, the second 
 and third proportionals are the same, that is, if 
 
 a:6: :6 :c 
 or ac = 6^ 
 
 6 is called the Mean Proportional between a and c, and c 
 is called the Third Proportional to a and &. 
 
 Problems. 
 
 1. To find, the Fowrth Proportional to three given straight 
 lines A\, A2, A3. 
 
 A .. 12 
 
 Draw two straight lines, making any convenient angle 
 about 30°. Mark A\, A2 along one, and AS along the 
 other. Join 1, 3, and draw 2 4 parallel to 1 3. 
 
 Then A 4:ia the fourth proportional. 
 
 Pkoof. /See § 1. 
 
 Note. — In the same way, any one of four proportionals can be 
 found, when the other three are given, remembering always that the 
 first and second are to be measured along one line, and the third and 
 fov/rth along the other, and that the line joining the ends of the first 
 a/nd third is pwrallel to the line joining the ends of the second and 
 fov/rth. 
 
Chap, v.] 
 
 2. To find i 
 lines Al, A2. 
 
 PROPORTION. 47 
 
 TMrd, Proportioned to two given straight 
 
 A 1 
 
 By the definition of § 5, this is to find a fourth pro- 
 portional to Al, A2, A2. 
 
 Thus the construction and proof are the same as in 
 Problem 1. 
 
 3. To find the Mean Proportional between two given 
 straight lines Al, A3. 
 
 x 
 
 Mark off Al, A3 in opposite directions along the same 
 straight line. Bisect 1 3 at 0, and on 1 3 describe a semi- 
 circle. Draw A 2 perpendicular to 1 3. 
 
 Then A2 is the Mean Proportional. 
 
 Proof. Join 1 2, 2 3. Then, by § 4, 
 
 ^1 :^2::^2 ■.A3. 
 
48 GEOMETRICAL DRA WING. [Chap. V. 
 
 4. To divide, a given straight line AB into amy numher 
 of equal parts (say six). 
 
 A 1 
 
 Through A draw any other straight line AC. Step out 
 along AC six equal parts of any convenient length Aa, 
 ah ef. Join Bf. Draw e5, di al parallel to £f. 
 
 Then 1, 2, 3, 4, 5 are the points required. 
 
 Pkoov By § 1, 
 
 4l:AB::Aa:Af, 
 
 Ai^_Aa_ Aa 1 
 * J^'^Z/" 672^ = 6' 
 
 :.A\=^.AB. 
 Similarly A^-"^ AB, and so on. 
 
 COE. a\= \. fB, 62 = |. fB, etc 
 By§l, al:fB::Aa:Af, 
 
 al_Aa_ Aa _1 
 fB~Af~J:ZI^~6' 
 .: al =^.fB, and so on. 
 
 This is the principle employed in Diagonal Scales 
 (Chap. VI. § 6). 
 
Chap, v.] 
 
 PROPORTION. 
 
 49 
 
 5. To cut off a given fraction (say ^ths) of a given 
 straight lime AB. 
 
 D 
 
 B 
 
 Mark off five equal lengths along AG. 
 Draw 2i) parallel to 55. 
 
 Va.'SQ.AJ)=\.AB. 
 Proof. AB : AB -. -. A1 -. A^ 
 
 '' ab~a^~i~a\~'^' 
 
 .•.AJ)=l.AB. 
 
 Join f)B. 
 
 6. To divide a line AB into two 
 proportion (say 3 : 6). 
 
 segments m a given 
 
 AD B 
 
 Mark off 8 (that is 3 + 5) equal parts along AC. 
 Join 9>B. Draw 3i? parallel to 85. 
 
 Then AB : BB : : Z : h. 
 Proof. § 1. 
 
so 
 
 GEOMETRICAL DRA WING. 
 
 [Cbap. V. 
 
 7. To divide a straight line into three segments in a given 
 ^proportion (say 2:3:4). 
 
 A P E 
 
 Mark off 9 (that is, 2 + 3 + 4) equal parts along AG. 
 Join 9^. Draw W, bE, parallel to 9^. 
 
 Then AJ) : DS : HB : : 2 : 3 : 4:. 
 Peoof. § 1. 
 
 * 8. To divide a line AB into "Extreme and Mean 
 Ratio" thai is, so that the whole is to the greater part as the 
 greater part is to the less. 
 
 X 
 
 Draw AI) perpendicular to AB, making AD equal to 
 
Chap, v.] PROPORTION. ji 
 
 half AB. Join BB. With centre B, draw the arc AE. 
 With centre B, draw the arc EG. 
 
 Then C is the point required. 
 Proof. The construction is a modification of Eucl. ii. 
 11, from which it follows that 
 
 AB.AG^BU'; 
 or, by § 5, AB:BG::BG: AG. 
 
 9. Note. — When a proportional has been found by any of the above 
 
 methods, it is well to test the accuracy of the result by arithmetic. 
 
 For example, in finding a fourth proportional (Prob. 1), since 
 
 first X fourth = second x third ; 
 
 » , v second x third 
 
 tourth = 2-— ; 
 
 first 
 
 and, taking the actual measurements of the figure, we have — 
 
 fourth=i:z^iii:?? 
 
 1-25 
 = l'e8 inches, 
 which is the length found by measurement. 
 
 EXERCISES— V. 
 
 1. Find a fourth proportional to 1", 2", 1'5", and measure it 
 
 2. Find x and y, so that 
 
 1" : If : : l|" : x 
 and r : y : : If : If". 
 Measure x and y, and test the results by Arithmetic. 
 
 3. Draw and measure the third proportional to, and the 
 
 mean proportional between, the lines 1-25", 2'25". 
 
 4. Draw a line 3f long, and cut off and measure yths of it. 
 
 5. Divide a line 2" long into ten equal parts. 
 
52 GEOMETRICAL DRA WING. [Chap. V. 
 
 6. A line AB is 247" long. Divide it at C, so that 
 
 ^C : ^a : : 3 : 4. 
 
 7. Find the mean proportional between 2" and J". 
 
 8. Divide a line 3" long into three parts, so that thfe first is 
 
 half the length of the second, and the second half that 
 of the third. 
 
 9. Divide a line 1 '5" long into extreme and mean ratio. 
 
 10. The perimeter of a triangle is 5", and the sides are pro- 
 
 portional to 3, 4, 5. Find the sides, and measure 
 them. 
 
 11. ^5= 1-26". Find AC, so that AB : AC ::% -.h. 
 
 12. Find, by a geometrical construction, the value of 
 
 — — — - — , and test the result by Arithmetic. 
 
CHAPTER VI. 
 
 SCALES. 
 PLAIN SCALES. 
 
 1. Ordnance maps are usually drawn on the scale of 
 an inch to a mile, and therefore a line 10" long divided 
 into 10 equal parts will enable us to measure any number 
 of miles from 1 up to 10. 
 
 Similarly any scale can be constructed, when the actual 
 distance between two points is known, as well as the 
 distance between the two points representing them on 
 the map. For instance — 
 
 " If the distance 5*72 miles is represented on a map by 
 1-33 inches, how long will a scale be which represents 
 10 miles?" 
 
 This is a question in sipaple proportion. If ic represents 
 the number of inches for 10 miles, then 
 
 miles miles 
 
 5-72: 10:: 1-33: a; 
 
 _ 10Xl-33 
 
 •■■ "" 5W 
 
 = 2-325 inches, nearly. 
 
 If then a line 2-325 inches long be drawn and divided 
 
 into ten equal parts (Chap. i. Prob. 4), each part will 
 
 represent one mile. 
 
 as 
 
54 GEOMETRICAL DRAWING. [Chap. VI, 
 
 iVoie. — The length of the Scale must be calculated in inches to two 
 places of decimals, for you can always measure such distances by 
 means of a diagonal scale, which is divided into tenths of inches. 
 The length should be calculated to three places of decimals, and the 
 closest approximation for two places then taken ; 
 
 <i.g. 2-32 for lengths between 2-320 and 2-324 
 and 2-33 „ „ 2-325 „ 2-329. 
 
 2. To figure, a Scale: Primary Divisions— Secondary 
 Divisions. 
 
 Pkimaey Divisions. If the total length of the Scale 
 is between 10 and 100 feet, or yards, or miles, etc., it must 
 be divided so that each division is 10. If the total length 
 is between 100 and 1000, each division must be 100; and 
 if between 1000 and 10,000, each division must be 1000. 
 
 no W 1\0 glO SlO no Si O 
 
 1 100 10 l\oo Hjoo swo u on 
 10\00 \o io\oo moo nof nn 
 
 These are called Primary Divisions. The cipher is 
 put at the end of the first division, as in the examples above, 
 and the tens, hundreds, or thousands figured right and left. 
 
 SECONDARy Divisions. After the Scale has been figured 
 in this manner, smaller distances can be shown by divid- 
 ing up only the first Primary Division into the number of 
 parts required, and figuring, if necessary, from right to left. 
 
 By dividing and figuring the Scale in this manner, it 
 will be clear that any length can be measured with the 
 
Chap. VI.] SCALES. SS 
 
 dividers, by putting one end on a Primary Division and the 
 other on a Secondary Division. 
 
 Scale of Yards and Feet. 
 
 ^-T-^ l " I' f \ ' 
 
 Scale of SO Miles, and single Miles. 
 ^ 1 1 ^ 1^ l-^ i 1 ^ I'' V 
 
 _2iO 
 
 X X 
 
 Scale of SOO Miles showing tens. 
 X X 
 
 For example, in the three scales given the crosses repre- 
 sent the distances taken by the dividers to show 2 yds. 2 ft., 
 27 miles, and 150 miles. 
 
 3. Representative Fraction. (K. F.) 
 
 To find what fraction a plan is of full size, it is necessary 
 to form a fraction in which the numerator is the distance 
 between known points on the plan, and the denominator is 
 the actual distance between the same two points, expressed 
 in the same denomination. 
 
 The.Eepresentative Fraction is this fraction, when the 
 numerator is made 1, by dividing the numerator and the 
 denominator by the numerator and omitting decimal 
 points in the denominator. 
 
 Thus, in the example of § 1 of this Chapter — 
 Since 1-33 inches represents 5-72 miles, 
 
 1-33 1-33 1-33 
 
 5-72x1760x3 xl2~5-72x63366~362419-2 
 
 ~ 272495-6" 272496 '^^^^•^' 
 The meaning is that 1 inch, foot, yard ... on the plan 
 represents 272,496 inches, feet, yards . . respectively, in 
 reality. 
 
56 GEOMETRICAL DRA WING. [Chap. VI. 
 
 NaU. — In order to check error in calculating a Bepresentative 
 Fiaction, it will be well to notice the number of figures in the deno- 
 minators of the following fractions : — 
 
 -^ represents the scale 1 inch to 1 foot. 
 
 ^ 1 inch to 1 yard. 
 
 ■j^^ 1 inch to 1 furlong. 
 
 4. To draw a Scale when the Bepresentative Fraction is 
 given. 
 
 Suppose it is xequired to draw the Scale whose K. F. is 
 ^T to show single yards up to 20. 
 
 Here 316 yards are represented by 1 yard or 36 inches, 
 and we have to find the length represented by 20 yards. 
 
 yards yards 
 
 316: 20:: 36" 103 
 20X36 
 
 
 
 
 • 
 
 
 316 
 
 = 2-278 
 
 = 2'28 inches nearly. 
 
 And the complete scale is as follows — 
 
 
 Scale of $0 Yards. R.F. -^ 
 
 I 
 
 
 
 8 
 
 e 
 
 U 
 
 » 
 
 1 
 
 
 -1^ 
 
 uu 
 
 LU 
 
 UJ 
 
 LU 
 
 
 Note. — This figure wiU serve as a type for drawing plain scales. In 
 each case proceed as follows : Draw three parallel straight lines at 
 convenient distances apart. On the lowest mark the primary and 
 secondary divisions (Chap. i. Prob. 4), using the point of a small pair 
 of compasses to mark each on the paper. Then, with set squares, 
 draw the perpendiculars through these points. Bub out aJl un- 
 necessary lines, and finally rule a thicker line below the lowest line, 
 figure the scale, write above it what the scale represents, and the 
 representative fraction. 
 
Chap. VI.] SCALES. 57 
 
 EXERCISES— VI. Plain Scales. 
 
 Draw Scales from the following data, show all your worlang,- and 
 find the Bepreseniative Fraction in thefwstfmr Examples. 
 
 1. 100 yards = 8-46". Show yards up to 50. 
 
 2. 1 foot = 1". Show feet and inches up to 6 feet. 
 
 3. 16 miles = 1". Show miles up to 100. 
 
 4. 1000 miles = 9". Show, miles by twenties up to 600. 
 
 5. E. r. = ^. Draw a scale of feet showing 20 feet. 
 
 6. R. F. = iiii^aaao . Draw a scale showing miles by tens 
 
 up to 200. 
 
 7. R. F. = a aae '^oooTF- Draw a scale of hundreds of miles up 
 
 to 5000. 
 
 COMPARATIVE' SCALES. 
 
 5. Sometimes the known distance is given in one 
 measure, and the Scale is required to be drawn in 
 another. 
 
 The method is, with slight modifications, similar to 
 that indicated above. For example — 
 
 "A distance of ^1 English miles is represented on a 
 map hy 4 '3 inches. Draw a scale of Irish miles for 
 the map showing 20 Irish miles^ given that 1 Irish mile 
 = 2200 yards." [Woolwich, July 1884.] 
 
 Express the known distance in Irish miles ; thus — 
 
 1 English mile = -—-=- Irish miles. 
 
 .: 37 English miles = - x 37 = -^ Irish miles. 
 5 5 
 
58 GEOMETRICAL DRA WING. [Chap. VI. 
 
 Then the question becomes, " If 4-3 inches represents 
 
 — Irish miles, what is the. number of inches for 20 Irish 
 5 
 
 148 
 miles 1 " 
 
 il«:20::4-3:«,. 
 
 
 
 _ 5 X 20 X 4-3 
 
 ••^ 148~" 
 
 107-5 
 
 37 
 
 = 2-91" 
 and the scale is as follows : — 
 
 Comparative Scale of 20 Irish Miles. 
 
 ) a 
 
 6 i 
 
 S 9 m 
 
 ^^ 
 
 zn'-iz 
 
 -i- 
 
 
 EXERCISES— VI. Comparative Scales. 
 
 {Show ail yowr working, wnd calculate the Bepresentative 
 Fraction in Examples 8, 9, 10.) 
 
 8. lOJ Eussian versts = 1-34". Draw a scale of English 
 
 miles showing 30. [1 verst = -661 English miles.] 
 
 9. 18 French leagues = 3 English inches. Draw a scale of 
 
 English miles up to 100. [1 French league = 4262-83 
 yards.] 
 
 10. 40 miles = 2-41'. Draw a scale of 30 Geographical 
 
 leagues. [1 Geographical league = 6072 yards.] 
 
 11. E. F. = iti^ii i. Draw a scale of 20 Spanish leagues. 
 
 [1 Spanish league = 2-63 English miles.] 
 
Chap. VI.] 
 
 SCALES. 
 
 59 
 
 DIAGONAL SCALES. 
 
 6. When still smaller measures than those given by the 
 Secondary Diyisions of a Scale are required, these smaller 
 divisions are denoted in the following manner by a 
 Diagonal Scale. 
 
 The Scale below shows yards and feet. Suppose that 
 it is required to show inches as well. 
 
 Draw perpendiculars through the primary divisions 
 3, 0, 1, 2, 3. Draw twelve lines parallel to the length 
 of the scale at equal distances apart. 
 
 Scale to show J^ Yards, Feet, and Inches. 
 
 t— V — V — 
 
 
 
 
 \ \ \ 
 
 
 
 
 ^^=^v^ 
 
 
 
 
 \ ^y \ 
 
 
 ^ 
 
 ") 
 
 
 
 
 
 \ \ \ 
 
 
 
 
 — \ A — 1 
 
 
 
 
 i 1 
 
 
 
 1 
 
 3 S 
 
 Join the point 12 to the last of the secondary divisions, 
 here 2, and draw parallels to this through the other 
 secondary divisions. 
 
 Then a consideration of Chap, v., Prob. 4, Cor., will 
 show that each of the diagonal lines cuts off so many 
 twelfths on each of the twelve parallel lines, when mea- 
 sured from the perpendicular, in addition to the amount 
 of the secondary division to which the diagonal Hue belongs. 
 
 Thus the distance measured by the two small circles is 
 2 yards 1 foot 6 inches. 
 
6o 
 
 GEOMETRICAL DRAWING. 
 
 [Cliap.yi. 
 
 EXEECISES— VI. Diagonal Scales. 
 
 ((SiAozc all the working, and, where necessary, jmd the 
 Rejpresentative Fraction.) 
 
 12. 10 miles = 1". Show 50 miles, miles, and furlongs. 
 
 1 3. E. F. = ■^. Show 5 yards, feet, and inches. 
 
 14. 153 yards = l-iQ". Show single yards up to 400. 
 
 15. R F. = -^. Show single feet up to 150^ 
 
 16. 29 chains = 32". Show chiEiins and links up to 5 chains. 
 
 [1 chain = 100 links = 22 yards.] 
 
 17. E. F. = ■^. Show metres, tenths of metres (or deci- 
 
 metres), and hundredths of metres (or centimetres), 
 up to 3 metres. [1 metre = 39".] 
 
 18. E. F; = ag\^ . Show distances of two yards up to 20 
 
 chains. [1 chain = 22 yards.] 
 
 SCALE OF CHORDS. 
 
 * 7. This scale, which is usually marked (7 or CffO in a 
 box of instruments, is used for the construction of angles. 
 
Chap. VI.]. SCALES. 6i : 
 
 A Scale of Chords, for angles up to 90°, can be con- 
 structed in the following way : — 
 
 Draw two radii, AO, BO, of a circle at right angles to 
 one another, and divide the arc AB into nine equal parts, 
 each measuring 1 0°. Join AB, and with centre A, and with ' 
 radii up to each of the points of division in turn, draw arcs 
 cutting AB, at the points 10, 20, etc. This is a Scale of 
 Chords for angles at intervals of ten degrees up to 90°. 
 
 Its use depends upon the fact that the radius of the 
 arc, AQ, is the same as that of the original circle. 
 
 If it is required to construct angles 20° and 50' at the 
 point of a straight line AO, with centre and radius 
 equal to the chord 60, draw a circle cutting AO at A, and 
 with centre A, and radius equal to the chords 20 and 50, 
 cut off the arcs ^20, AbO. Join 020, 050, which give 
 the angles required. 
 
 l\0 I S\0 I S\0 I 7\0 
 
 Similarly, a Scale of Chords can be drawn to measure 
 angles as great as 180°, and at smaller intervals than 10°. 
 
62 GEOMETRICAL DRA WING. [Chap. VI. 
 
 EXEKCISES— VI. Scales of Chords. 
 
 19. Draw a Scale of Chords, with radius 2", to show 90°, by 
 
 intervals of 10°. Construct the angle 40° from the 
 scale. 
 
 20. Draw a Scale of Chords, with radius 2", to show 120°, by 
 
 intervals of 10°. Construct the angle 110°. 
 
 21. Draw a Scale of Chords, with radius IJ", to show 180° by 
 
 intervals of 5°. Construct the angle 135°. 
 
CHAPTER VII. 
 
 IRREGULAR FIGURES AND AREAS. 
 
 1. To draw an irregular figure the position of each point 
 must be determined separately. 
 
 A point can be found when we know — 
 (i.) Its distance from two known points. 
 
 For, then, it is the intersection of the two 
 circles drawn with the two known points for 
 centres, and the given distances for radii, 
 (ii.) Its distance from a known point, and an angle. 
 For, then, it is the intersection of the side of 
 the angle with the circle, drawn with the known 
 point for centre, 
 (iii.) Two angles ; 
 
 that is, the angles which lines, joining it to 
 two fixed points, make with known lines. 
 
 For, then, the angles can be constructed, 
 and the point of intersection will be the point 
 required. 
 
 The method to be employed will be readily understood 
 from the following example : — 
 
H GEOMETRICAL DRA WING. [Chap. VIL 
 
 To draw thefigwre ABODE, given that — 
 
 ^-5=2-84" ^2?= 2-38" lBGD=&0°. 
 
 AC= 3-72" LGDE=2iO'. 
 
 BC=\&' i. BAB =60'. 
 
 Draw AR 
 
 (i.) To find C. With centres A, B, and radii AC, 
 BG respectively, draw arcs intersecting at G. 
 Join BG. 
 (ii.) To find D. Make lBGI)=60'. With centred, 
 
 and radius AD, cut off the point i?. 
 (iii.) To find B. At A make the angle BAB, and at D 
 the angle GDB. 
 
 Note.— When an angle is given, it is supposed to be measured start- 
 ing from the same side of the line as the previous points of the 
 figure, e.g. L BCD is measured to the left, and not the right, and 
 ^ GDE is measured up, not down. 
 
Chap, vn.] IRREGULAR FIGURES AND AREAS. 65 
 
 2. Heights and distances between inaccessible points 
 can be determined by this method. 
 
 For example: — Th& distance heiweeri' two points of db'ser- 
 vCbtion A,B is 2 -9 miles. C, B are two' indcceiiihh pbints 
 ;in the same plane with A, B. Having given — ''■ ' 
 
 ; LBAG=Qky L BAB- 4:2° . - 
 
 lABO=38'' LABB=i2\ ' ' ' 
 
 draw a plan -on the scale tif 1 ifuih to a mile, and adtehnine 
 the distance between G and B. ■ 
 
 ^ 111 1 1 1 11 1 11° 1^ 
 
 Draw a scale to show 4, or some other coiavenient 
 number of miles ; and draw the plan. 
 
 Measure CJ5,on the sjcale, by means of the dividers. 
 
 Ci? = l ■94 miles. ;■ 
 
 Note.— To the student of Trigonometry, the practical advantage of 
 this method of finding heights and distances will be obvious. 
 
 E 
 
66 
 
 GEOMETRICAL DRA WING. 
 
 [Chap. vn. 
 
 AREAS. 
 
 3. Tha Area of a Rectangle = The length XtJie breadth. 
 
 Generally, to find the area of a rectangle, it will be suf- 
 ficient to measure the length and breadth, and multiply ; 
 but since multiplication is not a geometrical conception, 
 the student must be prepared to find the area as in the 
 following example : — 
 
 The sides of a reotangle are 1'84" avd "88"; find Us area 
 by a strictly geometrical method. 
 
 A li E 
 
 
 t 
 
 C 
 
 ^^^ 
 
 
 i 
 
 ,^^ 
 
 ( 
 
 Draw the rectangle ABGB. Produce one of the sides 
 AB, so that BE== 1". Join EG, and produce it to meet 
 AD produced in F. 
 
 Then BF ( = '1 "6) represents the number of square inches 
 in ABCB. 
 
 Pkoqi;, Complete the parallelogram AFGE. 
 Then the complement ^C=the complement CG; 
 that is, area ABGD= GHy. EG 
 =BE^BF 
 = \y.BF 
 = BF 
 
Chap. Vli;] IRREGULAR FIGURES AND AREAS. 67 
 
 CoE. The area of any parallelogram = length x breadth, 
 the breadth being the perpendicular from any point of bhe 
 length on to the opposite side, (Eucl. i. 35). 
 
 4. The Area of a Triangle = - x base x altitiide. 
 
 Pkoof. a ABC = I X rectangle AE (Eucl. i 4 1 . ) 
 = \y.AGY.GE 
 = \y.AGY.BB. 
 
 Hence, to find the area of a triangle, draw a perpen- 
 dicular from one of the angular points on to the opposite 
 side, bisect either the perpendicular or the base, and 
 multiply ; or, which is the same thing, multiply the 
 base by the perpendicular and divide by 2. 
 
68 
 
 GEOMETRICAL DRA WING. 
 
 [Chap. VIL 
 
 5. Tq find a Bedangle and a Square each equal in area to 
 a Triangle, ABG. 
 
 B 
 
 
 Draw the perpendicular BD. Bisect it at E. Complete 
 the rectangle ACFG. 
 
 This rectangle is obviously equal to the triangle. 
 To find the square, produce AG to IT, making Cff= OF. 
 Find €K, the mean proportional between AU, CM. 
 (Ch. V. Prob. 3.) 
 
 Then GK is a side of the square. 
 Pboof. .AG:GK::GK: GE; 
 
 .•.GK^=AG.GE 
 =AG.FG 
 = l.AG.BD. 
 
Chap. VII.] IRREGULAR FIGURES AND AREAS. 69c 
 
 6. To redme an irregular Foitr-sided i'igwe toatria/ngle 
 of equal area, having its vertex' at any point of the figwre, 
 and its base in any side produced.. 
 
 Let ABCD be the figure. Let A be taken for the 
 Vertex of the equivalent triangle whose base is in CD 
 produced. Join^C?. Draw ^^ parallel to -4 C Join^jK 
 
 Then the triangle ADE =th.e figure ABGB. ^■•'■ 
 
 Proof. aAGI!= ^ABG. (Eucl. i. 37.) 
 
 Add to each aACB. 
 .: A ABB = ABGB. 
 
70 
 
 GEOMETRICAL DRAWING. 
 
 [Chap. VII. 
 
 7. To find the Area of any Rectilinear Figure hy, reduping 
 it to a triangle of equal area. 
 
 Take A for the vertex of the triangle, BE for the base 
 line. 
 
 As in the preceding Problem, reduce the triangle ABG 
 to one of ec^ual area A\G, having its base in CD. Then 
 reduce the triangle A\I) to one of equal area AiJD, 
 having its base in BE. 
 
 Then A'iE is a triangle equal to the given figure, and 
 the area can be found by Prob. 4. 
 
 Note 1. — Each of the reductions diminishes the number of sides of 
 the figure by one. The rule to remember is, to take the first two 
 sides, starting from the vertex A, and reduce on to the third side, 
 repeating this until the base line is reached. 
 
 Note 2. — The vertex of the triangle can be any point in a side of 
 the figure, and the reduction can be performed on either side of the 
 vertex. 
 
Chap. VII.] IRREGULAR FIGURES AND AREAS. ji 
 
 EXEECISES— VII. 
 
 {Cmstruct figwres frrnn the following data. All the angles are to 
 he dratm by construction.) 
 
 1. A triangle. AB=2-32", 5(7=2-05", OJ=i-96". 
 Find its area, and draw a square equal to it in area. 
 
 2. A quadrilateral. JB=2", BC=1-2S'', lBOD = 90% 
 
 AC=2-U", OD=-75". 
 Measure AD. 
 
 3. A pentagon. AB=1 -6", B0= 1", 
 
 AC=l", CD=\", lADE=135% 
 CE=l-b",DA = l". 
 Measure AE. 
 
 4. A quadrilateral. AB=2". 
 
 lABC=90\ lABD=75\ 
 lBAD=W, LBAC=i5°. 
 Measure GD, and find the area of the figure, 
 
 5. A pentagon. AB=r, BC=-7", 
 
 AO=2-25", CD=V5"; lODE=105°, 
 BD=2", AE=l-37". 
 Measure DE. 
 
 6. A hexagon. BO=2-6", CD=l-2", DE=l-5", 
 
 cABG=90°, lBGD=60°, EF=l-7", 
 
 L DEF= 75°, L EFA ^226°. l ODE =2iO°. 
 Measure AB. 
 
 7. Draw a regular hexagon of 1" side, and find its area, 
 
 8. Draw a regular hexagon of 1" side. Cut out a triangle, 
 
 by joining the extremities of one of the sides to the 
 centre of the figure, and find the area of the remainder, 
 
72 
 
 GEOMETRICAL DRA WING. ' [CJliap. VII.' 
 
 9. Draw a plain scale of yards having a Representative 
 Fraction of ^^^, and draw the following figure to 
 that scale. 
 ^5=375 yards. a^5C=105° 
 
 £C=525 „ ^£Ci> = 120° 
 
 00=575 „ lCDE=U5' 
 
 DE = 525 „ lDEF=90° 
 
 EF =650 „ LEFd=1b° 
 
 FG =475 „ 
 Join ..^ff, and write down its length. 
 10. Draw a scale of 7 metres to the inch, showing decimetres 
 by the diagonal method, and by means of this scale 
 draw the figure ABODE ; given that 
 AB = 24-5 metres, lBAC= 75°, l ABC= 30° 
 ■\bJD=60°, lABI)=60' 
 lBAE=SO°, -ABE=90° 
 Scale and write down the lengths oi AC, AD, AE. 
 (1 metre=10 decimetres.) 
 
CHAPTER VJII. 
 
 SUMS and! differences of areas. 
 
 SQUARE ROOT. 
 
 Problems. 
 
 1. To find a Sgua/re equal to the sum of two given squares. 
 
 Draw two lines AB, AC at right angles to one another. 
 Mark off along them Al, A2, equal to sides of the two 
 given squares. Join 1, 2. 
 
 Then 12 is a side of, thei square required. 
 
 Pkoof. Eucl.' I. 47. 
 
 73 
 
74 
 
 GEOMETRICAL DRA WING. [Chap. VIII. 
 
 2. To find a Square equal to the sum of three or more 
 given squares. 
 
 Draw two lines' ^5, AO at right angles to one another. 
 Mark off along them Al, A2, equal to sides of the first 
 and second squares. Join 1, 2. Step out AD, equal to 1 2, 
 along A£. Mark off A3, equal to the side of the third 
 square, along AO. Join 2)3. 
 
 Then D3 is a side of the square required, which can be 
 drawn by stepping out AS, equal to D3, along A£, and 
 completing the square. 
 
 Proof. By Eucl. i. 47. 
 
 m^=I)A^+A3^ 
 = 12^+A3^ 
 = Al^+A2'^+A3\ 
 
Chap. VIII.] SUMS AND DIFFERENCES OF AREAS. 75 
 3. To find a Sgua/ra equal to the difference between two 
 
 B 1 A 
 
 Draw twp fines, AB, AG, at right angles to one another. 
 Mark off, along AB, A\ equal to the side of the smaller 
 square. With centre 1, and radius equal to the side of 
 the larger square, cut off A^. 
 
 Then ^2 is a side of the square required. 
 Peoof. ^12+^2^=122 (Eucl, I. 47.) 
 
 .-. ^22=122-A12. 
 
 4. To find a Figure equal to the mm or difference of two 
 similar fibres and similar to ijhem. 
 / 
 
 For example, " To draw an equilateral triangle equal to 
 the sum of two other equilateral triangles:." 
 
76 .'A>,' ■ GEOMETRICAL DRAWli<IG/-\aia.v.^TSl. 
 
 Or, " 2*0 draw- a. circle equal in area to the .difference 
 between two given circles." 
 
 This is only an extension of the preceding problems, 
 using corresponding sides or radii in the place of sides of 
 the squares. | 
 
 Proof. Eucl. ¥i. 31. 
 
 • EXERCISES— V-III. 
 
 1. Draw a square equal in area to the difference of the 
 squares on the lines 5" and 4". 
 
 3. Draw a square equal in area to the sum of the squares on 
 the lines 1", IJ", If". 
 
 3. Draw an equilateral ; triangle equal to the sum of two 
 
 equilateral triangles, each of 1" side. 
 
 4. Draw a circle equal in area to two circles, whose radii are 
 
 2" and 1 "5" respectively. 
 
 5. The two diameters of a circular ring are 2'5" and 2". 
 
 Draw the ring and a circle equal to it in area. 
 
Chap. VIII.] 
 
 SQUARE ROOT. 
 
 77 
 
 SQUARE ROOT. 
 
 The Square Eoot of a number can be drawn, correct to 
 two places of decimals, in two ways. 
 
 5. To find the value of Va. First Method. 
 
 iff 
 
 B 1 A 
 
 Find two numbers, such that the sum or difference of 
 their squares = a. 
 
 Proceed as in Problems 1 'or 3. >, 
 
 Theathe line found in inches is Va. 
 
 Peoof. The square on the line found = a. 
 
 the line found = Va. 
 
 For example. To find VB. 
 
 5=lH2l ' ' 
 
 Measure 1" and 2".along A£, AG) the line joining the 
 extremities is VB. 
 
 Or, again, 5 = 9 — 4 . 
 = 3^-22. 
 
 Proceed as in the second figure. 
 
78 
 
 GEOMETRICAL DRA WING. [Chap. Vlli: 
 
 Note 1. — If the number cannot be expressed, as the sum or differ- 
 ence of two perfect squares, we can proceed as follows : — 
 Example— "To find Vs." 
 
 Here 3=1 + 2 
 
 = l2 + («/2)2 
 Now 2 = 1' + 1". 
 Hence, we can first find >J\ and use this length to find Vs. 
 Note 2. — Somfitimes the number can be found as the sum of three 
 or more squares. In this case we can proceed as in Prob. 2. 
 For example, to find ^14. 
 
 Here 14=1 + 4 + 9 
 
 = l''' + 22 + 3l 
 
 6. To find -Ja. Second Method. 
 
 X 
 
 Draw a line AB=a" Produce it to C so that BC= 1 
 ¥ind the mean proportional BD between AB, BO. 
 Then BD= Va. 
 
 Proof. AB -.BD-.-.BD: BO 
 
 .:BI)^=ABxBO 
 
 = ax 1 
 .-. BB = Va. 
 
Chap. VIII.] SQUARE ROOT. 79 
 
 Note 1. — If the quantity a can be factorised, we can find the root 
 as the mean proportional between the two factors. 
 
 For example, Vl2 is the mean proportional between 4 and 3, or 
 6 and 2. 
 
 Note 2. — For small numbers the second method has the advantage 
 in simplicity ; but for large numbers, shorter lines are, as a rule, 
 required in the first method. 
 
 EXERCISES— VIII. 
 
 6. Find by both methods the value of ^\ isj\ 4/6 correet 
 
 to two places of decimals. 
 
 7. Find the value of Vfl, ^12, Vl5. 
 
 8. Find the value of Vl+ V2, 2 '>/2,-^ 
 
CHAPTER IX. 
 
 CIRCLES AND TANGENTS. 
 
 A CIRCLE can be drawn to satisfy three conditions, and, 
 in general, no more. Por instance, a circle can.be drawn 
 to pass through three given points (Chap. II. Proh. 16), 
 or to have a given radius' and touch two given straight 
 lines. ■' ■" ' 
 
 In the majority of problems on the drawing of circles, 
 to satisfy certain' condition's, it is necessary to find the 
 centre of the circle as the intersection of two lines. The 
 construction of these lines depends on one or more of the 
 following Principles : — 
 
 I. The centre of the circle is on the straight line 
 which bisects at right angles the line joining 
 any two points on the circle. (Eucl. ill. 1.) 
 
 II. A tangent is the line through a point on the 
 circumference at right angles to the radius 
 through that point; or, which is the same 
 thing, the centre is on the line at right angles 
 to a tangent, through the point of contact. 
 III. The centre of a circle is on the bisector of the 
 angle between two tangents. 
 
Chap. IX.] CIRCLES AND TANGENTS. 81 
 
 IV. The two tangents from a point are equal. 
 
 V. If two circles touch one another, the two centres 
 and the point of contact are in one and the 
 same straight line. (Eucl. Iil. 11, 12.) 
 
 VI. The distance between the centres of two circles 
 which touch one another is equ^l to the sum or 
 difference of the radii, according as the circles 
 touch externally or internally. (See, Prob. 6.) 
 
 NoU. — In the following problems, when no ambiguity is likely to 
 arise, the circles whose centres are A, B, G will be called " the circles 
 A, B, G," and their radii will be denoted by a, b, respectively. 
 
 Problems. 
 
 1. To find the Centre of a Circle, or Arc of a Circle. 
 
 Draw any two chords AB, CD. Bisect them at right 
 angles, by lines meeting at 0. 
 
 Then, is the centre of the circle. 
 Peoof. Principle i. 
 
82 GEOMETRICAL DRA WING. [Chap. IX. 
 
 2. To draw a Tangent from a point to a circle A. 
 
 Join OA. On OA describe a semicircle, cutting the 
 given circle at £. Join 0£. 
 
 OB is a tangent through 0. 
 
 Pboof. Principle ii. — AB is perpendicular to OB, 
 because " the angle in a semicircle is a right angle." 
 (Eucl. III. 31.) 
 
 Note. — In practice a tangent must always be drawn, by construct- 
 ing the point of contact, and not simply by using a ruler. 
 
 If a tangent has to be drawn at a given point on a circle, it must be 
 drawn at right angles to the line joining the point to the centre. 
 
Chap. IX.] CIRCLES AND TANGENTS. 83 
 
 3. To draw a Lmefrom a point to a circle A cuttiv^ 
 off a given chord in the circle. 
 
 Place a line BG in the circle equal to the chord re- 
 quired. Draw AB perpendicular to BG. With centre 
 A, radius AB, draw a circle touching BG at D. Draw 
 OFG to touch this circle. 
 
 OFG is the line required. 
 
 Peoof. AB=AE; 
 
 :. GF, BG, at equal distances from the centre, are equal. 
 (Eucl. III. 14.) 
 
84 GEOMETRICAL DRA WING. 
 
 4. To draw a common Tangent to toiich two 
 A, B, (1) externally ; (2) internally. 
 
 [Chap. IX. 
 
 Circles 
 
 (1.) If it touches the circles externally. 
 
 Join the centres A, B, and on AB describe a semicircle. 
 On the radius of A, the larger circle, mark off XT=i, 
 so that ^ !F is the difference of the radii. With centre A, 
 and radius A T, draw an arc cutting the semicircle at G. 
 Join AG, cutting the circle A at D, and draw BE parallel 
 to AB. Join B, E. 
 
 Then, BE touches both circles. 
 
 (2.) If it touches the circles internally. 
 
 Make the same construction, but mark off XY, so that 
 ^ F is ^^ sum of the radii. 
 
 .0 
 
Chap. IX.] 
 
 Proof. 
 
 CIRCLES And tangents. 
 
 85 
 
 on = AD -AC, or AG -AD 
 = a — (a — b),or {a+b) — a 
 = h. 
 Hence CD, BE are equal and parallel. 
 
 .". BCDEis a parallelogram ; (Eucl. i. 33.) 
 
 also L ACB is a right angle. (Eucl. Iir. 31.) 
 
 .-. L ADE, and l BED are right angles. (Eucl. I. 29 ) 
 
 .*. DE is a tangent to both circles. (Principle 11.) 
 
 Note. — There are, in general, four common tangents to two 
 
 circles. 
 
 5. To draw ttoo Tangents to a Circle, inclined to one 
 another at a given angle. 
 
 Draw any tangent AB. Draw any line BDE, making 
 the required angle with AB, and meeting the circle in 
 D, E. Draw CF, perpendicular to DE, meeting the circle 
 at F. Draw FG parallel to BE. 
 
 Then, FG is the second tangent. 
 Pjroof. lAGF== lABE, (Eucl. i. 29.) 
 
 and GF is a tangent. (Principle 11.) 
 
86 GEOMETRICAL DRA WING. [CHap. IX. 
 
 Problems in which the Radius of the Circle to 
 be constructed is known. 
 
 6. To draw a Circle C, of given radius c, to tovxsh two 
 given circles A, B, (1) hath externally ; (2) A, eternally, and 
 B, internally ; (3) hoth internally. (Principles v., vi.) 
 
 (1.) With centre A, radius c + a, and with centre B, 
 radius c + &, draw two arcs intersecting at G. 
 
 G is the centre of the circle, and if GA, GB be joined, 
 CD or CE is ladius. 
 
 Pboop. CJD=CA-AD 
 
 = (c+a) — a = c. 
 CE=CB-BE 
 
 = (c+6)-& = c. 
 
Chap. IX] CIRCLES AND TANGENT^. 87 
 
 (2.) With centre A, radius c+a, and with centre B, 
 radius cr-6, draw two arcs intersecting at 0, 
 
 C is the centre of the circle. 
 
 Peoof. CD=0A-AD 
 
 = (G+a) — a = e. 
 CE=CB+BE 
 = (c-b)+h=c. 
 
 (3.) With centre A, radius c—a, and with centre B, 
 radius c—b, draw two arcs intersecting at 0. 
 
 Proof. 
 
 C is the centre of the circle. 
 
 CB=GA+AD 
 = (G — a) + a = e. 
 
 €B=CB+BB 
 = (e-6) + 6 = c. 
 
88 
 
 GEOMETRICAL DRA WING. 
 
 [Chap. IX. 
 
 GoR. By considering a point as a circle whose radius 
 = 0, we have, a simUar construction for the two following 
 problems : — 
 
 (L) " To draw a circle, of given radius, to touch a given 
 circle A, and pass through a given point B." 
 
 The construction is made identical, by putting 6 = 0; 
 while (1) and (2) give the same construction. 
 
 (ii) " To draw a circle of given radius, to pass through 
 two given points A and B." 
 
 For (1), (2), and (3) reduce to the same construction on 
 putting a = 0,1 = 0. 
 
 7. To draw a Circle G of given radium c, to touch a given 
 straight line, and a given circle A, (1) externally, (2) inter- 
 nally. (Principles II., vi.) 
 
 (1.) Draw^^ perpendicular to the line. Mark off BD 
 equal to c. Draw BG parallel to the given line. With 
 centre A, radius c+ a, draw an arc,- cutting this line in G. 
 
 A 
 
 A 
 
 ^-^ 
 
 "'''^■- \ 
 
 
 -- I -/g 1 
 
 
 \ 'J 
 
 i ~ B 
 
 v^ y 
 
 Then G is the centre required. 
 Join AG, and draw the perpendicular GF. Then the 
 circle can be drawn. 
 Peoof. GF=BB=c. 
 
 GE=GA-AE 
 = {c+a) — a = c. 
 
Chap. IX.] CIRCLES AND TANGENTS. 
 
 89 
 
 (2.) The same construction, except that a radius e— a 
 is taken. 
 
 Cor. " To draw a circle of given radius to touch a given 
 straight line, and pass through a given poini," Put a = o. 
 
 8. To draw a Circle G, of given radius c, to touch two 
 given 1 
 
 Draw parallels to the lines, at a distance c from each, 
 intersecting at G. 
 
 C is the centre required. 
 
90 GEOMETRICAL DRA WING. [Chap. IX. 
 
 Draw the perpendiculars CA, CB. Then the circle can 
 be drawn. 
 
 Proof. By construction, CA=c, GB=c. 
 
 Note 1. — Obviously, four such circles can be drawn, one in each of 
 the four angles fonned by the intersection of the, given lines. 
 
 Note 2. — If the lines are parallel, the problem will be impossible, 
 unless the distance between the lines is equal to the diameter of the 
 circle C, in which case its centre is found by bisecting the common 
 perpendicular to the two lines. 
 
 9. To draw three Circles, whose radii a, b, c are given, 
 to touch one another. 
 
 Draw a triangle ABC, with the sides opposite A, B, C 
 equal to h+c, c+a, a+b, respectively. 
 
 Then A, B, C are the centres. 
 
 Pkoof Principles v., vi. 
 
Chap. IX.] 
 
 CIRCLES AND TANGENTS. 
 
 91 
 
 Problems in which the Radius of the Circle to 
 be drawn is not known. 
 
 10. To draw a Circle to pass through three given points. 
 This is to describe a circle about the triangle formed by 
 
 joining the points. {See Chap. 11. Prob. 16.) 
 
 11. To draw a Circle to pass through two given points, 
 A, B, and toiich a given line. 
 
 ^--/^ 
 
 Join AB, and produce it to meet the given line in D. 
 Find DB, a mean proportional, between AB, BD. Mark 
 oSDF=DE. 
 
 Since AD. BB=BE^ (See Chap. v. Prob. 3.) 
 
 .-. BF is a tangent to the circle required at the point 
 
 F, by Eucl. III. 37. 
 
 The centre C is found by applying Principles i., 11. 
 
 NoU.—li AB is parallel to the line, bisect AB at right angles to 
 meet the line in E, and draw a circle through A, B, E. 
 
92 
 
 GEOMETRICAL DRA WING. 
 
 [Chap. \%. 
 
 12. To draw a Circle to pass through one given point 
 A,- and to touch two given straight lines. 
 
 Bisect the angle between the lines by £D. 
 The centre is on this line (Principle III.). 
 Draw any circle D, to touch the given lines (Prob. 8). 
 Join A£, cutting the circle D at S, F. Join SH, DF. 
 Through A, draw AG, AC, parallel to BE, DF. 
 Then G is the centre of one circle, and C" is the centre 
 of another satisfying the conditions. 
 
 Peoof. Draw DT, GX perpendicular to one of the 
 lines. Then, by similar triangles — 
 
 DE:GA::BB:GB 
 
 •.-.BY-.CX 
 
 but BE=BY :. GA = GX. 
 
 13, To droflo a Circle to touch three given straight lines. 
 
 This is to inscribe a circle in the triangle formed by 
 the lines. {See Chap. Ii. Prob. 14.) 
 
Chap. IX.] CIRCLES AND TANGENTS. 
 
 93 
 
 14. To draw a Circle to pass through a given point A, 
 and touch a given line in a given point B. 
 
 Join AB. Then apply the Principles i., ii. to find the 
 centre C. 
 
 15. To draw a Circle to pass through a given point A: 
 and touch a given circle B, i/n a given point D. 
 
 Join BI), and produce it to meet the line hisecting 
 AD at right angles, at C, the centre required. 
 Proof. Principles i., v. 
 
94 
 
 GEOMETRICAL DRA WING. 
 
 [Cbap. IX. 
 
 16. To draw a Circle to tamh a given straight line in a 
 given point B, and a given circle A, (1) externally, (2) inter- 
 
 Draw BO at right angles to the given line, and mark 
 off BuD equal to a, 
 
 (1) on the opposite side to the centre A ; 
 
 (2) on the same side as the centre A. 
 
 Join AD. Bisect it at E at right angles, to meet BD 
 ate. 
 
 C is the centre of the circle. 
 
Chap. IX.1 CIRCLES AND TANGENTS.. 95 
 
 Pkoof. The triangles AEG, BEG are equal in all 
 respects. 
 
 .-. AG=BG; 
 but AF=BD; 
 .•.CF=GB. 
 
 Also Principles 11., v. are satisfied. 
 
 17. To inscribe a Circle in the sector of a circle ABC. 
 
 B F 
 
 Bisect the ailgle AGB by GB. Draw the tangent EBF, 
 meeting CA, GB, in E, F. Bisect the angle GEB, by 
 EG meeting GB at G. 
 
 Then Q is the centre of the circle, GB the radius. 
 
 Pkoof. G is the centre of the circle inscribed in the 
 triangle GEF. (Chap. 11. Prob. 14.) 
 
 Therefore the circle touches AG, BG, and EF, or, which 
 is the same thing, AB. 
 
 CoE. Similarly a circle can he dravm to touch the two 
 radii produced, and the circle externally. (Chap. Ii. 
 Prob. 15.) 
 
96 
 
 GEOMETRICAL DRA WING. 
 
 [Cbap. DC. 
 
 18. To draw a Cirde to touch a given circle A, and two 
 of its tangents. 
 
 Join the centre A to B, the intersection of the tangents. 
 At D, where it cuts the circle, draw the perpendicular 
 EDF. Bisect the angle BED hy EO, meeting AB at C. 
 
 Then G is the centre of the circle required, and CD its 
 radius. 
 
 Proof. G is the centre of the circle inscribed in the 
 triangle BEF. The circle therefore touches EF, that is, 
 the circle A. (Chap. ii. Prob. 14.) 
 
Chap. IX.] CIRCLES AND TANGENTS. 
 
 97 
 
 19. To dram three . Circles, whose centres A, B, C, are 
 given, to touch one another. 
 
 In the triangle ABC, inscribe a circle (Prob. 1 3) toucli- 
 ing the sides BG, GA, AB, at D, E, F, respectively. 
 
 Then D, E',F, are the points where the circles toueh, 
 and therefore the circles can b& drawn. 
 
 Peoof. ^^, 4-?', touch the inscribed circle at 1'', jR' 
 .-. AE=AF ' (Principle IV.) 
 
 So also BF= BD, CD = GE. 
 
 EXEECISES— IX. 
 
 1. Draw two circles, of 1" and |" radius, to touch a given 
 
 straight line, and one another. 
 
 2. At. the ends of a straight line 2|." long, draw Wo circles 
 
 of Y radius to touch the line. Draw a third circle to 
 touch these two and the straight line. 
 G 
 
98 GEOMETRICAL DRAWING. [Chap. n. 
 
 3. Draw two circles of -95" and "45" radius, with their 
 
 centres 2 '6" apart, and draw a third circle of radius 
 1"15" to touch the first externally, and the second 
 internally. 
 
 4. Draw the same two circles as in Ex. 3, and draw a 
 
 common tangent touching both externally. 
 
 5. From a point 2" from the centre of a circle of 1" radius 
 
 draw a tangent, and a line cutting ofif a length of 1 "h" 
 within the circle. 
 
 6. In a quadrant of a circle of 1 J" radius inscribe a circle to 
 
 touch the tWo radii and the circumference. 
 
 7. Draw a circle of '5" radius, and draw all the circles of radius 
 
 I'l" to touch it, and pass through a point l"4"from its 
 centre. 
 
 8. Dfaw three circles of radius •^\", '86", and 1'04" to touch 
 
 one another. 
 
 9. Draw two circles of "7" radius, with their centres 2-8' 
 
 apart, and draw lines to touch both internally. 
 10.. Draw two straight lines inclined to one another at an 
 angle of 60°. Draw a circle of 1 '5" radius to touch 
 them, and draw' a second circle, between the first circle 
 and the point of intersection of the lines, to touch it 
 and the two given lines. 
 
 11. Draw two straight lines inclined at an angle of 60°. On 
 
 one of them mark off points distant 1" and 1'75", from 
 the intersection of the two lines. Draw a circle to 
 pass through these two points and to touch the other 
 line. 
 
 12. Draw a circle of •75" radius with its centre 1*55" from a 
 
 given straight line, Mark a point on this straight 
 
Chap. IX.] CIRCLES AND TANGENTS. 99 
 
 line 2'45" from the centre of the circle, and draw a 
 circle to touch the straight line at this point, and 
 to touch the circle externally. 
 
 13. Draw a triangle with sides 2-6", 2-4", and 1-4". With the 
 
 vertices of the triangle for centres, draw three circles 
 to touch one another. 
 
 14. Draw a triangle with sides 3'2", 2-72", and 1". Draw the 
 
 circle which passes through the vertices ; and draw a 
 second circle of "75" radius to touch it and the first 
 side. 
 
1. To 
 
 vertex A. 
 
 CHAPTER X. 
 FRACTIONS OF AREAS. 
 
 Problems. 
 
 a triangle ABO by a line drawn th/rcnigh a 
 
 Bisect the side BO at D. Join AD. 
 
 Then AD bisects the triangle. 
 
 Proof. By Eucl. i. 38, " triangles on equal bases and 
 between the same parallels are equal." 
 
 100 
 
Chap. X] FRACTIONS OF AREAS. loi 
 
 • ' 2. To hisect a triangle ABG ly a line dravm through a 
 given point 0, in the side BO. 
 
 B O 
 
 Bisect BG in D. Join AO. Draw DE parallel to AO. 
 Join OE. 
 
 Then the triangle ABG is bisected by OE. 
 
 Proof. Join AD. Then ^ABD= A AGD (Eucl. i. 38.) 
 
 =^-AABG.: 
 Also A AOE= AAOB. (Eucl. i. 37.) 
 Add to each the AAOB. 
 .: figure ABOE= A ABB 
 = ^-AABG. 
 
GEOMETRICAL DRAWING. 
 
 [Chap. X 
 
 3. To lisect a triaTigU ABG hy a line perpendicular to 
 the hose BG. 
 
 MS DO 
 
 \ 
 
 Draw AD perpendicular to BO, and bisect BG at E. 
 Produce BB, the greater segment of the base, so that 
 BF=BE. Find the mean proportional, BG, between BF, 
 BB. On BG mark off BH= BG. Draw HJ perpendicular 
 to^C. 
 
 Then HJ bisects the triangle. 
 
 Peoof. BD-.BGwBG: BF, by construction. 
 
 .-. BB : BR: : BR: BF, by construction. 
 .-. ABBA : ABRJ ::BD:BE (Eucl. VI. 19. Cor.) 
 
 .'. ABRJ= ABBA X fj 
 BJj 
 
 = l-5i?x2)^xg(Ch.yn.,§4.) 
 
 = l-BB.BA.^ 
 
 = hy.{\-BG y^DA) 
 = J ■ ABAG. 
 
Chap. X.J FRACTIONS OF AREAS. 103 
 
 4. To bisect a quadrilateral ABGJD fry a line dravm 
 through an angular point A, 
 
 Draw the diagonals AG, BD. Bisect BD, the one 
 which does not pass through A, at E. Draw EF parallel 
 to^C. 3am AF. 
 
 Then ^i^ bisects the quadrilateral. 
 
 Proof. LACF= LACE. (Eucl. i. 37.) 
 
 Add to each LABO. 
 
 .-. AFCB=AECB. 
 Also AECB^i'ABCB', 
 toT LAEB =AAED <Euc1. I. 38.) 
 
 andAGEB = AGED; ._ ■ 
 
 .-. AEGB=AEGD; 
 .; AEGB=^-ABGD, 
 
I04 
 
 GEOMETRICAL DRA WING. 
 
 [Cliap. Z. 
 
 5. To divide a triangle ABG into any nuwier of 
 equal parts (e.g. five) hy lines drawn tJvrvu^h an angular 
 point A. 
 
 Divide the base ^Cinto five equal parts at 1, 2, 3, 4. 
 Join Al, A2, A3, Ai. 
 These lines divide the triangle into five fequal parts. 
 Pboof. Eucl. I, 38. 
 
 6. Ih divide any figwre into any number of equal, parts 
 {e.g. four) hy lines drawn through an angular point, or a 
 point in a side. 
 
 \-' 
 
 With for vertex', reduce the figure to a' triangle OFQ' 
 
Chap. X.] FRACTIONS OF AREAS. loj 
 
 of equal area. (Chap. "Wi., Proh 7.) Divide FG into 4, 
 the required number of equal parts, at 1, 2; 3. ' Join the 
 point 1, which is between B, G, to 0. Of the rest, draw 26, 3c 
 parallel to OG. Join h,, which is between C, B, to 0. From 
 c, which is outside GB, iva,w ed parallel to OB. Join Od. 
 
 Then, the figure is divided into four equal parts by 01, 
 Oh, Od. 
 
 Peoof. , , OAB\ = \-ABGBE. 
 
 For ^OAB=^OFB (Eucl. i. 37.) 
 
 add to each A 0:51 
 ::OABl=AOFl 
 
 = ^ ■ A OFG, by construction 
 ! ( = i .4^(7Z>^, byiconstruction. 
 
 Similarly A0bG=A02G.: 
 
 .•.OhGl = A02l 
 ■■■' '• ' ' =\-AOFG 
 
 = l-ABGBE. 
 
 Similarly Old may be shown to be equal to \ ■ ABGBE, 
 and, therefore, so also is the remainder OdE. 
 
io6 GEOMETRICAL DRA WING. [Chap. X. 
 
 7. To divide a triangle ABC into any nwmler of eguai 
 parts (e.g. three) by lines dravm parallel to one of the 
 sides BG. 
 
 Divide AB into three equal parts at 1, 2. On AB 
 describe a semicircle, and draw \a, 26 perpendicular to 
 AB. With centre A, and radii Aa, Ah, draw. arcs cutting 
 AB at B, E. Draw Dd, Ee parallel to BG. 
 Then Dd, Be trisect the triangle. 
 
 Proof. Aa is a mean proportional between Al, and 
 AB. (Chap. V. § 4.) 
 
 .■. Al :Aa: :Aa:AB. 
 .■.A\:AD::AD:AB. 
 .•.A1:AB::A ADd : A ABG. (Eucl. VI. 1 9. Cor.) 
 
 .■./i,AI)d=4L-AABG 
 AB 
 
 = iAABG. 
 
 Similarly A ABe = § • A ABO. 
 .: A APd = BdeB= BeGB. 
 
Chaj). X,] 
 
 fra{:tion.s of areas. 
 
 ro; 
 
 8. To divide a circle, A into any nwmber (e.g. three) of 
 concentric rings of equal area,. 
 
 Draw the radius .45, and divide' it into three equal 
 parts at 1, 2. Oil AB describe a semicircle, and draw la, 
 2& perpendicular to AB. 
 
 Then Aa^^ Ab are the radii of the circles required. 
 
 Proof. Identical with Prob. 7. 
 
 9. To divide a square (or any parallelogram) info equal 
 parts (e.g. three) by lines drawn through a vertex A. 
 
 , Join .4 to the opposite vertex G; divide each of the 
 sides BC, CD, into three equal parts, at 1, 2, a, b. Join 
 ^ to the alternate points 2, 6. 
 
 Then the figure is trisected. 
 
 Peoof. The triangles ^IB ACb are aU equal 
 
 (Eucl. I. 38) ; .•, the figures AB2, A2ib, AbD are equal. 
 
io8 
 
 GEOMETRICAL DRAWING. 
 
 [Chap. X 
 
 10, To divide a square (or awy parallehgram) into 
 equal parts (e.g. three) by lines drawn parallel to a 
 diagonal AG. \ 
 
 /\\ 
 
 \\ 
 
 i \l \ 
 
 ^ i 
 
 Divide each of the triangles AGB, ACD into three 
 equal parts by lines parallel to the diagonal AC 
 (Prob. 7). 
 
 Then, the. figures formed by taking the alternate lines 
 will divide the iBgure as required. 
 
 Proof. See Prob. 9. 
 
Chap. X.] 
 
 jF-Ji ACTIONS OF AREAS. 
 
 t09 
 
 11. To draw a figure similar to a given figv/re and equal 
 to a given fraction of it (e.g. ^ths) in area. 
 
 Produce a side AB to P, so that BP=l- AB. Find at 
 mean proportional BQ between AB, BP. Mark off Ba 
 equal to BQ. Join BE, BD, and draw de, ed, de parallel 
 to AE, ED, BQ respectively.. ; 
 
 Then aBcde is the figure required. 
 
 Peoof. The figures are similar, because the sides are 
 all parallel. • • : : 
 
 MsoBP-.B^-.-.Bd-.BA; 
 .: BP:BA:: aBede -.ABOBE. . (Eucl. VI. 20.) 
 
 .: aBcde=~- ABCBE 
 BA 
 
 = 1- ABODE. 
 
no GEOMETRICAL DRAWING. [Chap. i. 
 
 12. To draw an equUdteral triangle eqUaJ, in area to a 
 given triangle ABO. 
 
 X 
 
 Describe an ecfujlateral triangle ASD on any side AB. 
 Produce D£ or BA; and draw OB parallel to AB, to 
 meet the line produced in E. Find BF a mean propor- 
 tional between BB, BE. 
 
 Then, BF is a side of the equilateral triangle required. 
 Peoof, DB:BF::BF:BE; 
 
 .-. BB -.BE:: ABBA : ABFG (Eucl. vi. 19. Cor.) 
 
 But BB :BE:: AABB : AABE (Eucl. vi. 1.) 
 
 And AABE=AABO; (Eucl. i. 37.) 
 
 .-. BB:BE:: AABD -lAABO; 
 but DB:BE:: AABB {ABFG ; 
 .-. ABFG = AABO. 
 
Chap.X.] FRACTIONS OF AREAS. ni 
 
 EXERCISES— X. 
 
 Draw a triangle with sides 3", 2-3", and 1-4", and bisect 
 
 it by a line drawn through a point in the longest side 
 
 1" from the end of that side. 
 Draw an equilateral triangle, whose side is 2", and divide 
 
 it into four equal parts by lines drawn parallel to one 
 
 of the sides. 
 Draw an isosceles triangle with base 1"25", and sides each 
 
 2", and bisect it by a line perpendicular to one of the 
 
 4. Draw a regular pentagon, whose side is 1-5", and divide 
 
 it into five equaf parts by lines drawn through one of 
 the vertices. 
 
 5. Draw a rectangle, with sides 1" and r73", and divide it 
 
 into four equal parts by lines drawn through one of the 
 vertices. 
 
 6. Draw a regular hexagon, whose side is 1'2", and draw 
 
 another regular hexagon equal to half of it in area» 
 
 7. Draw a triangle, whose sides are 2-28", 2-36", and 1-62", 
 
 and find an equilateral triangle equal to it in area. 
 
 8. Draw a regular hexagon, whose side is 1", and divide it 
 
 into 3 equal parts by lines drawn through the middle 
 point of one of the sides. 
 
 9. Draw a square whose side is 2", and trisect it by means 
 
 of lines drawn parallel to one of the diagonals. 
 
 10. Draw an isosceles triangle with sides 1'82" and base 2-16", 
 
 and divide it into four equal parts by lines drawn per- 
 pendicular to the base. 
 
 11. Draw a circle of 1 '6" radius, and divide it into four equal 
 
 concentric rings. 
 
 12. Draw a circle of 3" diameter, and cut out half of its area 
 
 by means of another circle touching the first. 
 
CHAPTER XL 
 
 INKING. IN, MARQUOIS SCALES. 
 
 Before proceeding further with the subject) it will be 
 well to revise the preceding chapters, and at the same 
 time to practise Inhiiig in and Prmting, and learn the 
 use of Marquois -Scales. 
 
 For Inking in, it will be sufficient to ink over the 
 Problems which are done in revision ; while, for practice 
 in Printing, the enunciations of the Problems should be 
 exactly copied — two or three at a time will soon give the 
 necessary facility. The instructions below' must be care- 
 fully followed.. 
 
 Inking. — The bow-pen, for straight lines, and ink- 
 compass for circles, should be filled by means of a smaU 
 camel's hair brush, drawn between the two nibs. Care 
 must be taken that there is no ink on the outside of the 
 nibs, or blots will inevitably occur when lines are ruled. 
 Indian ink must be used, for ordinary ink will corrode 
 and spoil the, pen. It is more convenient, in Geometrical 
 Drawing, to use the specially prepared liquid Indian ink 
 than to prepare it for oneself from the . solid cake. The 
 figure must always be drawn in pencil first, so that any 
 mistakes may be corrected before the inking in is done ; 
 
 112 
 
Chap. XI.] INKING IN. MARQUOIS SCALES. 113 
 
 and the pencil drawing should be done as carefully as 
 possible, for, in examination, there may not be time 
 enough to ink in. If, however, it is certain that the 
 figure will be inked over, then dotted pencil lines need not 
 be drawn : for continuous straight lines drawn in pencil 
 can be converted into dotted lines when the ink is 
 applied. All circles and arcs of circles should be inked 
 in first, and straight lines added afterwards. 
 
 The screw of the pen should not be touched after the 
 inking is begun, or the lines drawn will not be of the 
 same thickness throughout. If some lines are required 
 thicker than others, such as the final lines or circle con- 
 structed, then, when everything has been inked in, the 
 pen should be opened a little, and the darker parts should 
 be drawn over again. Ordinary ink-lines should be so 
 fine that the ink dries immediately the lines are drawn. 
 
 The pen must never be put away in the box until all 
 superflous ink has been removed. It should be unscrewed 
 till the nibs are well apart, and then every trace of ink 
 should be rubbed off with a piece of paper. If the screw 
 has been removed in cleaning, care must be taken that 
 the threads are not crossed in putting it back again, or 
 the pen will be destroyed. The nibs should not quite 
 touch one another when the pen is put away in the box." 
 
 When a pen does not mark properly, the cause is 
 generally due to want of care in cleaning. Pens, to draw 
 well, must be sharp. If your pens are blunt, send them 
 to the makers to be put right : do not tamper with them 
 yourself. ^ - j 
 
114 GEOMETRICAL DRAWING. [Chap. XI. 
 
 Pkinting. — ^A m-ow-quill or mapping pen should be 
 used for printing in ink. Printing must always be done 
 between parallel lines ruled in pencil, which may, if 
 necessary, afterwards be rubbed out. 
 
 Four lines are ruled by means of Marquois Scales, as 
 explained below, the spaces between the lines being 
 respectively 
 
 3 4 3 
 
 ^Trths, ^hs, and ^7;ths of an inch. 
 60 '60 60 
 
 The following is an example showing the use of the 
 
 four lines, which, of course, should never actually be 
 
 inked in : — 
 
 Maequois Scales. — The primary use of Marquois 
 Scales is for drawing parallel straight lines at known 
 distances apart. The scales are numbered 20, 25, 30, 
 35, 40, 45, 50, and 60 ; and the meaning of these numbers 
 is, that the scales are so divided up that with them you 
 can draw 20, 25, 30, etc. lines to the inch, respectively. 
 To do this you take the triangular ruler and place it so 
 that the index in the middle of the longest side is against 
 the point on the scale marked 0. Draw a line along the 
 bevelled edge of the triangle, and then move the index 
 through one division of the scale and draw a second line ; 
 then, if you are using the 60 scale, these two lines will be 
 ^Vths of an inch apart. So, again, two lines drawn with 
 the index first at the and then at the 60, will be one 
 
Chap. XI.] INKING IN. MARQUOIS SCALES. 115 
 
 inch apart; while, to draw the four lines required for 
 printing, you use the same 60 scale and rule lines with 
 the index of the triangular ruler at the divisions 0, 3, 7, 
 and 10, so as to get lines /^ths, ^ths, and ■i^'Csis, apart. 
 
 When you have to decide for yourself which of the 
 Marquois Scales to use, select that, if possible, on which 
 the index will have to move through 5 or 10 divisions 
 at a time, for these are the easiest to read, and counting 
 is avoided. Thus — 
 
 For lines ^" apart use the 40 scale, for then 10 
 divisions will give a distance of ^ths, or a quarter of 
 an inch ; 
 
 For decimals of an inch use the 50 scale, for then 5 
 divisions give ^^th of an inch ; 
 
 For the scale of 1" to a foot use the 60 scale, for then 
 5 division^ represent ^ths or j^th of a foot, or one 
 inch. 
 
 EXERCISES— XL 
 Examples taken from Aemy Papers. 
 
 Additional Scales and Irregular Figures vnll be found on page 132. 
 
 1. On a base of 3" describe a triangle having a perimeter of 
 
 7J inches, and one of the base angles 35°. 
 
 2. The angles of a triangle are in the ratio of 2, 3, i, and 
 
 its perimeter is 7 inches. Construct the triangle. 
 
 3. The perimeter of a triangle is 10", and its sides are in the 
 
 ratio of 4, ,5, 7. Construct the triangle, and bisect it 
 by a straight line parallel to its shortest side, 
 
ii6 GEOMETRICAL DRAWING. [Chap. XI. 
 
 4. Construct a triangle of base AB=i-T\ one side AG=%-\", 
 
 and the angle AGB=^\°. 
 
 5. On a base of 2J" describe an isosceles triangle having a 
 
 vertical angle of 40°, and on the opposite side of the 
 base describe an equilateral triangle. In the quadri- 
 lateral figure thus obtained inscribe a circle. 
 
 6. Two sides of a triangle are 3 '5" and 2-4" respectively, and 
 
 the angle opposite the shorter side contains 37J°. 
 Construct two triangles to fulfil these conditions. Cir- 
 cumscribe each with a circle. Construct the angle 37|-°. 
 
 7. Construct the triangle ABC, given the angle ABC=GO°, 
 
 and ACB =55°, and radius of inscribed circle =1 •4". 
 
 8. On a base '75" long draw a regular octagon. 
 
 9. Describe a regular hexagon equal in area to an equilateral 
 
 triangle of 2" side. 
 
 10. In a circle of 1-75" radius inscribe a regular heptagon, 
 
 and about the same circle describe a regular pentagon. 
 Make the methods of construction perfectly evident. 
 
 11. With a radius of 2" draw a quadrant, and inscribe a square 
 
 with two corners in the arc of the quadrant. 
 
 12. In a triangle whose sides are 2-5", 2", and 3" respectively, 
 
 inscribe the largest square possible. 
 
 13. Draw a hexagon of 1-75" side, and inscribe a square. 
 
 14. Describe a square equal in area to a triangle whose sides 
 
 are 2^, 3, and 3| inches respectively, and in the square 
 inscribe an equilateral triangle. 
 
 15. Obtain by construction the greatest and least fourth 
 
 proportionals to three lines If, 2J, and 1| inches long. 
 
 16. Draw a straight line AB, 2-7" long, and produce it to a 
 
 point G, so that AG may be to AB at 7 to 4. 
 
 17. Three straight lines A, B, and G are 1 J, 2^ and 2f inches 
 
 long respectively. Determine by geometrical con- 
 struction X, y, and z, so that 
 
Chap. XI.] INKING IN. MARQUOIS SCALES. 117 
 
 {y)x:A::B:G. {2) y:B::G -.A. 
 (3)A:z::z:C. 
 Figure each line x, y, and z separately, and write 
 down its length. Test your results by Arithmetic. 
 
 18. Draw three concentric circles whose radii are to one 
 
 another as 1:2:3, the radius of the largest being 
 equal to the diagonal of a square of 1^" side. 
 
 19. Find (by working on a diminished scale) a mean pro- 
 
 portional between two lines 168 and 2-9" long. 
 
 20. Describe a triangle having its sides 2f, 2 J, and If inches 
 
 respectively ; and construct a square and an isosceles 
 triangle each equal to it in area. 
 
 21. The sides of a triangle are 2f, 3, and 4 J inches respec- 
 
 tively. Draw the triangle and obtain, by geometrical 
 construction, (1) a parallelogram of equal area, having 
 one of its angles 60°, (2) a square of equal area. 
 
 22. On a straight line 2J" long, and on the same side of it, 
 
 construct an equilateral triangle and a regular pentagon. 
 Beduce the space outside the triangle and inside the 
 pentagon to a triangle of equal area. 
 
 23. Construct a line „ VS inches long. 
 
 24. Taking a line of \\" long as unity, construct a rectangle 
 
 whose sides are Vs and iJQ. 
 
 25. The line M represents unity. Find lines representing 
 
 n/5, n/7, Vf. M 
 
 26. Describe a circle of 3" diameter, and inscribe three circles, 
 
 touching one another and the circumference. 
 
 27. Two lines meet at an angle of 25°. , Place three circles 
 
 between these, each touching the two lines and the 
 next circle, the smallest being J" radius. 
 
ii8 GEOMETRICAL DRAWING. [Chap. XL 
 
 28. Eound a circle of Y i'9'<lius draw six equal circles ; each 
 
 touching two and the given circle. 
 
 29. With radii f", f", and J", describe three circles touching 
 
 each other, and about them describe a triangle, each 
 side to touch two of the circles. The method of 
 finding the points of contact must be clearly shown. 
 
 30. At A, in a line of indefinite length, erect a perpendicular 
 
 AD, 1 J" long: with D as centre, and a radius of f, 
 describe a circle. Describe a second circle which shall 
 be touched internally by the first, and by the line in. a 
 given point 2J" from the centre of the circle. 
 
 31. Draw a straight line ABoi indefinite length and describe 
 
 a circle of ^" radius touching it at A. Describe a 
 second circle of 1" radius, touching AB, and the first 
 externally, and describe a third circle of IJ" radius 
 touching AB and the second circle. All three circles 
 are to be on the same side of AB. 
 
 32. Draw a circle of 3" diameter, and draw two tangents to 
 
 it, including an angle of 50°. Protractor allowed. 
 Explain briefly your method of construction. 
 
 33. Draw a circle of 2" radius, and, by two different methods 
 
 divide its area into 3 equal parts. 
 
 31. Draw a triangle ABO whose sides are 2^", 3J", and 4". 
 From a point P, in AC, f" from A, draw two straight 
 lines dividing the triangle into three equal parts. 
 
 35. Draw a circle of 2" radius and a second of two-thirds its 
 
 area. Show your construction clearly. 
 
 36. The sides AB, BG, and CA, of a triangle are 4-2", 3-7", 
 
 and 2-9" respeptively. Draw the triangle, and divide 
 it into four equal parts by straight lines drawn from 
 a point in AB, r5" from A. 
 
EXAMINATION PAPERS. 
 
 SANDHURST. 
 
 (No Proof is required in these Papers.) 
 
 1. Construct angles of 15°, 22J°, 30°, 75°, and 105°, without 
 
 the aid of a protractor. 
 
 2. The following measurements of a five-sided field ABODE 
 
 are given : — 
 
 Side A£=iO yards. Diagonal AC=i5 yards. 
 „ BO^U „ „ BD=iO „ 
 
 „ CD=30 „ Angle CBE=1Q5°. 
 
 „ BE =28 „ 
 Draw a scale of 15 yards to an inch showing exactly 50 
 yards, and draw a plan of the field to that scale. Measure 
 AE. 
 
 3. Describe two circles of 1 inch and J inch radius respectively, 
 
 and having their centres 2^ inches apart. Describe a 
 third circle of 1^ inch radius that shall touch the first 
 externally and be touched internally by the second or 
 smaller one. 
 
 4. Construct a diagonal scale of chains and links for a plan 
 
 on which 27 chains are represented by 31 inches. The 
 scale to be properly figured and all calculations shown. 
 Show 3 chains. 
 
 (One chain =100 links.) 
 
120 EXAMINATION PAPERS. 
 
 5. The base of a triangle is 2-5 inches, its altitude 1-8 inches, 
 
 and its vertical angle 60° ; construct the triangle. 
 
 6. Taking one inch as the unit, determine by two different 
 
 geometrical constructions a line equal to a/^- 
 
 7. Describe a circle of 2J inches diameter, and from a point 
 
 3 inches from its centre draw two tangents to the 
 circle, one on either side of it, showing how the 
 tangent points are determined. .Within the angle 
 formed by these two tangents describe a second circle 
 touching the first and each of the tangents. 
 
 II. 
 
 1. Construct a diagonal scale of -^g- to measure feet and 
 
 inches. Show 40 feet. Figure the scale properly and 
 show your calculations. Show, by two small marks 
 on the scale, the points you would take in order to take 
 off a distance of 17 feet 8 inches. 
 
 2. Draw a straight line 4:*4; inches long and trisect it. 
 
 3. A man walks from a point A, 3| miles in a straight line 
 
 due north, to B, when he turns to the right through 
 an angle of 45°, and walks in the new direction 1| 
 miles to 0, where he again turns to the right through 
 an angle of 75° and walks 2f miles to B. From B he 
 walks due south for 2^ miles to B. Draw a scale of 
 I mile to an inch, and draw a plan of his walk to that 
 scale. Write down the distance in miles from E to 
 A. A protractor is not to be used. 
 
 4. Draw a straight line AB 3-5 inches long, and from A 
 
 and B as centres, with radii 1 inch and 1'5 inches 
 respectively, describe circles. 
 
SANDHURST. 121 
 
 (1) Draw a common tangent to these two circles, 
 touching them both on the same side of AB. 
 
 (2) From the point A draw a tangent to the circle 
 whose centre is B. In each case show clearly 
 how the tangent point is determined. 
 
 5. A distance of 2 '4 miles is represented on a map by 7 '8 
 
 inches ; construct a comparative scale of yards for the 
 map, showing 3000 yards, and divided into distances 
 of 100 yards. Show all your calculations, figure your 
 scale properly, and write above it the representative 
 fraction. 
 
 ( 1 mile= 1760 yards.) 
 
 6. Draw a triangle whose sides are 4^ inches, 2\ inches, and 
 
 3 inches respectively, and bisect it by a line drawn 
 perpendicular to its longest side. What is the area of 
 the whole triangle ? 
 
 7. Describe a circle of 2 inches radius, divide it into quad- 
 
 rants, and in each quadrant inscribe a circle. 
 
 III. 
 
 Construct a diagonal scale of y^ to show single yards. 
 Let your scale represent 150 yards, figure it properly, 
 and show by two small marks on the scale the points 
 you would choose in order to take ofif a length of 67 
 yards. 
 
 Find, by construction, a third proportional to two straight 
 lines 1^ and If inches long; and a fourth proportional 
 to three straight lines IJ, 2\, and If inches long 
 Write down the length of each line so found. 
 
1 22 EXAMINA TION PAPERS. 
 
 3. The sides of a quadrilateral figure ABOD are as follows : 
 
 AB = 45 yards, BG = 38 yards, CD = 33 yards, and 
 AD=3i& yards, and the diagonal AG = 6& yards. 
 Make a scale of 15 yards to an inch, and draw the 
 figure to that scale and bisect it by a straight line 
 drawn from the point B. Determine the area of the 
 whole figure in square yards. 
 
 4. Construct an isosceles triangle equal to the sum of four 
 
 squares whose sides are respectively ^, -f, ^, and 1| 
 inch. 
 
 5. On an Austrian map a distance of 2^ Austrian miles is 
 
 represented by 1'15 inches. Draw a scale of English 
 miles for the map. Show 30 English miles. Figure 
 your scale properly, write above it its representative 
 fraction, and show all the necessary calculations. 
 
 1 Austrian mile = 3'3312 English miles. 
 
 1 English mile = 1760 yards. 
 
 6. With a radius of 2J inches describe a semicircle, and in 
 
 it inscribe a square having two of its corners in the 
 arc of the semicircle, and the side non-adjacent to 
 these upon the bounding diameter. 
 
 7. On one side of a straight line 2^ inches long as base 
 
 describe an isosceles triangle having its angle at the 
 vertex 55°, and on the other side of the line as base 
 describe an isosceles triangle having its angle at the 
 vertex 35°. In the quadrilateral figure so obtained 
 inscribe a circle. 
 
 IV. 
 
 1. Construct a diagonal scale of 4^ inches to a mile to show 
 1000 yards, to measure distances of 5 yards. Show 
 by two small marks on the scale the points you 
 
SANDHURST. 123 
 
 would choose in order to take oflf a length of 735 yards. 
 Figure your scale properly, and write above it its 
 representative fraction. (1 mile=1760 yards.) 
 
 2. Take two points, A and B, 4 inches apart, and a third 
 
 point G, 2f inches from A and 2| inches from B. 
 With A, B, and G as centres, describe three circles 
 touching each other. 
 
 3. Draw a straight line AB, and from a point in it con- 
 
 struct on the same side of AB the following angles : — 
 A0G=^i5°, AOD=75'', A0E=n'2,y, A0F=157^\ 
 and make OA = 33 feet, 0C=42 feet, 0D=25 feet, 
 OE =57 feet, OF=84: feet, and OB=60 feet. Join 
 AC, GD, BE, EF, and FB. Scale and write down the 
 lengths in feet of these five lines. 
 
 Scale, which should be drawn, 15 feet=l inch. 
 
 . Determine the side of a square of 7 inches area ; draw 
 the square, and trisect it by lines drawn from one of 
 its angles. 
 
 5. Two places on a French map are known to be 17 French 
 
 leagues apart, and this distance is represented on the 
 map by 2 -45 English inches. Draw a scale of English 
 miles for the map, showing 100 miles. Show all your 
 calculations, figure the scale properly, and write above, 
 it its representative fraction. 
 
 (1 French league =4262 -84 English yards.) 
 
 6. The sides of a triangle are 3^ inches, 2f inches, and 4 
 
 inches respectively. Draw the triangle, and determine 
 an equilateral triangle of equal area. 
 
 7. Describe a regular pentagon of 2| inches side, and in it 
 
 inscribe a square. 
 
124 EXAMINATION PAPERS. 
 
 1. A distance of 134 yards is represented on a plan by 2 J 
 
 inches. Construct a diagonal scale for the plan by 
 which single yards may be measured. Show 300 yards. 
 Show all your calculations, figure your scale properly, 
 and write above it the representative fraction. Show 
 by two small marks on the scale the points you would 
 take in order to measure off a distance of 183 yards. 
 
 2. Draw a square of If inches side, and about it describe a 
 
 circle. About this circle describe a second square 
 having its sides parallel to the diagonals of the first. 
 About the second square describe a circle, and about 
 this circle describe a third square having its sides 
 parallel to the sides of the first. 
 
 3. The four sides of a field ABGD measure as follows : — 
 
 ^£=280 yards, 5(7=320 yards, GD=ZiO yards, 
 1)^=360 yards, diagonal 1)5=300 yards. Draw a 
 scale of 120 yards to an inch, draw a plan of the field 
 to that scale, and calculate its area in yards. 
 
 4. The sides of a triangle are in the ratio of the number 
 
 7, 10, and 12, and the radius of the inscribed circle it 
 1 inch. Draw the triangle. 
 
 .5. A distance of 154 geographical miles is represented on a 
 map by 3*37 inches. Construct a scale of English 
 statute miles for the map, showing 250 statute miles, 
 and by which distances of 10 statute miles may be 
 measured. 
 
 (60 geographical miles =69^ statute miles.) 
 
 6. Draw an arc of a circle of 3 inches radius, and describe a 
 circle of | inch radius touching it externally. Describe 
 a second circle of f inch radius, touching the first and 
 
SANDHURST. 125 
 
 the arc externally, and a third circle of 1 inch radius, 
 touching the arc externally and the second circle. 
 Explain briefly your construction. 
 7. Construct an isosceles triangle equal in area to the sum of 
 the areas of four squares whose sides are J inch, f inch, 
 1 inch, and \\ inches respectively. 
 
 VI. 
 
 1. Draw a scale of ywrds for a map on which 3-28 miles are 
 
 representpd by 4J inches. Show 10,000 yards and 
 distances of 200 yards. Show all your calculations, 
 figure the scale properly, and write' above it its repre- 
 sentative fraction. (1 mile=1760 yards.) 
 
 2. Describe a square of \\ inches side, and on each side 
 
 describe an isosceles triangle having each of its equal 
 sides If inches. About the figure thus formed describe 
 a regular octagon, so that the apex of each of the 
 isosceles triangles may be in one of the angles of the 
 octagon. 
 
 3. Construct a scale of chords to read to 5°. (Radius 4".) 
 
 By means of the scale plot an angle of 65°. 
 
 4. Draw a diagonal scale of y^ to measure single inches. 
 
 Show 50 feet. Show by two small marks on the scale 
 the points you would take in order to measure oflF a 
 distance of 27 feet 8 inches. 
 
 5. Describe three equal circles of 1 inch radius, each 
 
 touching the other two, and about them describe a 
 circle touching all three. 
 
 6. Describe a regular hexagon of 1^ inches side and a second 
 
 regular hexagon three-fourths of the area of the first. 
 
 7. On a base of i\ inches describe a triangle having its 
 
 vertical angle 50 and its altitude 2^ inches. 
 
1 26 EXAM IN A TION PAPERS. 
 
 VII. 
 
 1. A distance of 31 miles 5 furlongs is represented on a map 
 
 by 11 inches. Draw a, diagonal scale of miles and fur- 
 longs for the map, showing 15 miles. Show all your cal- 
 culations, figure your scale properly, and write above 
 it the representative fraction. Show, by two small 
 marks on the scale, the points you would take in order 
 to measure off a distance of 7 miles 3 furlongs. (1 
 mile=8 furlongs=1760 yards.) 
 
 2. Draw a straight line AB, 4 inches long, and at A, without 
 
 producing the line, erect a perpendicular to it AC. At 
 B make the angle ABE on the same side of ^J8=60°. 
 Make BE =5 inches long, and bisect it by construction 
 in 2). Determine by geometrical construction a point 
 F in AC equidistant from jD and E. 
 
 3. A map is drawn to a scale of feet whose representative 
 
 fraction is xttst- Draw a scale of paces for the map 
 ■by which distances of 5 paces to 200 paces may be 
 measured. (A pace=32 inches.) 
 
 4. Three straight lines A, B, and C are 1 J inches, 2 J inches, 
 
 and 2f inches long respectively. Determine, by 
 geometrical construction : — 
 
 (1) The length of a straight line x that has the 
 
 same ratio to A that B has to C. 
 
 (2) The length of a straight line y that has the 
 
 same ratio to B that C has to A. 
 
 (3) The length of a straight line z such that A has 
 
 the same ratio to z that z has to 0. 
 Figure each line x, y, and z separately, and write 
 down its length. 
 
SANDHURST. 127 
 
 5. Two points A and B are 660 yards apart. It is required 
 
 to ascertain the distance of two inaccessible points 
 and D, both on the same side of the line joining^ and 
 B. The angle BAGis measured and found to be 105°, 
 the angle BAD 75°, the angle ABB 60°, and the angle 
 ABG 45°. Draw a scale of 200 yards to an inch, 
 dividing it to show distances of 20 yards, and make a 
 plan showing the position of the four points to that 
 scale. Scale and write down the distance of G and D 
 from each of the points A and B. The angles are 
 to be obtained by construction. 
 
 6. The angles of a triangle are in the ratio of the numbers 
 
 3, 5, and 10, and its perimeter is 4^ inches. Construct 
 the triangle. 
 
 7. Describe an equilateral triangle of 2 J inches side, and on 
 
 one of its sides describe an isosceles triangle, having 
 its two equal sides each 3 inches long. In the quadri- 
 lateral figure thus obtained inscribe a square. 
 
 VIII. 
 
 1. A distance of 729 yards is represented on a plan by 10'8 
 
 inches. Construct a diagonal scale of yards for the 
 plan, showing 500 yards, by which single yards may be 
 measured. Show all your calculations, figure your 
 scale properly, and write above it the representative 
 fraction. Show by two small marks on the scale the 
 points you would take in order to measure off a dis- 
 tance of 127 yards. 
 
 2. The measurements of an irregular six-sided field ABGDEF 
 
 are taken and found to be as follows : — 
 Sides. AB=^5Q yards, BG= 190 yards, 00= 140 yards, 
 
 DE=22Q yards, EF=lbO yards. 
 Diagonals. JfC=260 yards, BF=iOQ yards, C^=360 yards. 
 Angle. ^5C=110°. 
 
128 EXAMINA TION PAPERS. 
 
 Draw a scale of yards having a representative fraction of 
 ■j^g^, and draw a plan of the field to that scale. 
 
 Write down the length in yards of the side AF and the 
 diagonal AG, and calculate the area in yards of the 
 whole field. 
 
 3. Describe a circle of 1^ inches radius, and from a point 5 
 
 inches from its centre draw — 
 
 (1) A tangent to the circle, showing how the tangent 
 
 .point is determined. 
 
 (2) A straight line that shall cut the circle into two 
 
 segments, having a chord of 2 inches. 
 
 4. A map is drawn to a scale of 12 inches to the mile. 
 
 Draw a comparative scale of chains for the map, show- 
 ing 50 chains. Show all your calculations, and figure 
 your scale properly, and write above it the representa- 
 tive fraction. 1 mile=1760 yards. 1 chain=66 feet. 
 
 5. On a straight line 3 inches long describe two equal and 
 
 similar triangles, having an altitude of 2 inches, and a 
 vertical angle of 50°, and also an isosceles triangle 
 having the same vertical angle. 
 
 6. Draw a straight line AB, 5 inches long, and describe two 
 
 circles each of 1 inch radius, touching AB in A and B 
 respectively. 
 Describe a third circle touching the first two and the 
 straight line AB. 
 
 IX. 
 
 1. A map is drawn to a scale of six inches to a mile. Draw 
 a scale of yards for the map showing 2000 yards, and 
 divide it to show distances of 50 yards. Show all 
 your calculations, figure your scale properly, and write 
 above it the representative fraction. 
 
SANDHURST. 129 
 
 2. Construct a rectangle having its diameter 2>\ inches long, 
 
 and one of its sides 2f inches. 
 
 3. Draw a scale of yards having a representative fraction of 
 
 ^^jTj-. Draw a straight line AB representing 240 
 yards to this scale, and bisect it in G. On the same 
 side of AB construct the angles ACB=W, ACE=1h°, 
 and ACF=UO\ Make GD=250 yards, CE =1Q5 
 yards, GF= 285, and join ADEFB. Eeduce the figure 
 to a triangle of equal area, having its apex at E, and 
 its base in AB produced. 
 
 4. Draw a straight line AB and a circle of one inch radius 
 
 touching it at A. Describe a circle of 2 J inches radius, 
 touching the iirst circle and the straight line AB. 
 
 5. A length of three metres four decimetres is represented 
 
 on a French plan by two English inches. Construct a 
 diagonal scale for the plan showing 10 metres. Show 
 all your calculations, and figure your scale properly. 
 Show by two small marks on the scale the points you would 
 take in order to measure oif a length of six metres 
 eight decimetres. (1 metre =10 decimetres.) 
 
 6. Describe a triangle having its sides 4^ inches, 3f inches, 
 
 and 2J inches respectively, and in it inscribe a square, 
 
 X. 
 
 1. Two men start from the same point A, to walk by two 
 different routes to B, where the routes cross. One 
 walks due north from A, for a distance of 2300 yards, 
 and turns off to the right through an angle of 45°. 
 The other walks due east from A, for a distance of 
 3700 yards, and then due north to B. 
 Draw a scale of yards having a representative fraction of 
 ^^^^. Draw a plan of their walk to that scale, and 
 write down the total distance traversed by each. 
 I 
 
1 30 EX A MINA TION PAPERS. 
 
 2. Describe a square equal in area to a triangle whose sides 
 
 are 2 J inches, 3 inches, and 3 J inches respectively, and 
 in the square inscribe an equilateral triangle. 
 
 3. A length of 3 feet 6 inches is represented on a plan by 
 
 ■j^ of an inch , construct a scale for the plan to show 
 40 feet and inches by diagonal division. Show all 
 your calculations, iigure your scale properly, and find 
 the representative fraction. 
 Show, by two small marks on the scale, the points you 
 would take in order to measure ofif a length of 23 feet 
 5 inches. 
 
 4. Two points are each 3f inches from the centre of a circle 
 
 of 1:^ inch radius, and 2f inches from each other. 
 Describe a circle that shall pass through the two 
 points, and touch the given circle. 
 
 5. A Spanish plan is drawn to a scale of 16 palms to an 
 
 English inch. 
 Draw a scale of English feet for the plan showing 50 feet. 
 Show all your calculations, figure your scale properly, 
 and write above it its representative fraction. 
 1 Spanish palm = -eSS of an English foot. 
 
 6. About a circle of f inch radius, describe four equal circles 
 
 touching each other and the given circle. 
 
 WOOLWICH. 
 XL 
 
 1. Draw a scale of -j-J^ to measure feet. Show 200 feet and 
 
 divide the scale to show distances of 5 feet. 
 
 2. Draw a straight line AB 2-7" long, and produce it to a 
 
 point 0, so that AG may be to AB as 7 to 4. Your 
 solution must be strictly geometrical. 
 
WOOLWICH. 131 
 
 S. Describe a circle of 1"25" radius and take any point P in 
 its circumference. Determine a point Q, 1'75" from P 
 and 2-5" from the centre of the circle. Describe a 
 second circle touching the first in P and passing 
 through the point Q. 
 
 4. A distance of 1-35 miles is represented on a map by 2-15 
 
 inches. Construct a diagonal scale of chains for the 
 map by which single chains may be measured. Show 
 250 chains. Show all your calculations, figure the 
 scale properly, and write above it the representative 
 fraction. Show, by two small marks on the scale, the 
 points you would take in order to measure off a dis- 
 tance of 137 chains. 
 
 1 mile =80 chains. 
 
 5. Describe a circle of 2" radius and a second two-thirds of it 
 
 in area. Show clearly how the radius of the second 
 circle is obtained. 
 
 6. Draw a hexagon of 1-75" side and in it inscribe a square. 
 
 XII. 
 
 1. On a map 67-5 miles are represented by 9-3 inches.. Draw 
 
 a scale of miles for the map showing 30 miles. Show 
 furlongs by the diagonal method. Show all your 
 calculations, figure your scale properly, and write 
 above it the representative fraction. Show by two 
 small marks on the scale the points you would take in 
 order to measure off a distance of 17 miles 3 furlongs, 
 1 mile =8 furlongs =1760 yards. 
 
 2. Having given that — 
 
 (i) 3-5" : 2-75" : : 2-5" : AB 
 (ii) 2-8" -.BO-.-.BG: 1-3", 
 find AB and BG by geometrical constructions. 
 
132 EXAMINATION PAPERS. 
 
 3. The base of a triangle is 3-25", one of the angles at the 
 
 base 45°, and the sum of the other two sides 5-6". 
 Construct the triangle. 
 
 4. A distance of 37 English miles is represented on a map by 
 
 4'3 inches. Draw a scale of Irish miles for the map 
 showing 40 Irish miles (an Irish mile=2200 yards). 
 
 5. The centre A of a, circle of 1 -25" radius is 1 -75" from a 
 
 given straight line. Describe a circle that shall touch 
 the given straight line at a point P, 2-5" from A, and 
 shall be touched externally by the given circle. 
 
 6. Describe a quadrant with a radius of 2 -IS", and in it 
 
 inscribe a square having two of its corners in the arc 
 of the quadrant 
 
 ADDITIONAL SCALES AND IRREGULAR FIGURES, 
 TAKEN FROM ARMY PAPERS. 
 
 1. What is a Scale ? What are the different kinds of Scales, 
 
 and when is each to be used ? Explain the meaning 
 of the term Representative Fraction. 
 
 The two lines AB and CD are drawn to scale, and 
 represent distances of 1320 yards and 3 J miles respec- 
 tively. Give the representative fraction of the scale 
 to which each line is drawn. Show your calculations. 
 
 A B 
 
 C D 
 
 2. A distance of If miles is represented on a map by 2J". 
 
 Draw a scale of yards for the map showing 8000 
 yards, and divided to show distances of 200 yards. 
 
ADDITIONAL SCALES. 133 
 
 Show your calculations, figure the scale properly, and 
 write above it the representative fraction. 
 (1 mile= 1760 yards.) 
 
 3. Construct a plain scale of 2^ chains to 1", to show 10 
 
 chains ; and a comparative scale of feet showing dis- 
 tances from 10 feet up to 1000. Give all calculations, 
 figure the scale properly, and give the representative 
 fraction. (1 chain =66 feet.) 
 
 4. On a map 2 -5 chains are represented by 1"15". Draw a 
 
 scale of feet for the map showing 500 feet, and divide 
 it to show distances of 20 feet. Find the represen- 
 tative fraction. 
 
 5. Draw a scale of yards, to read distances of 5 yards, 
 
 comparative to a scale of 2 chains to an inch. Show 
 300 yards, and find the representative fraction, 
 (1 chain =22 yards.) 
 
 6. A distance of 20 Flemish miles is represented on a Flemish 
 
 map by 8J English inches. Draw a scale of English 
 miles for the map, showing single miles up to sixty. 
 Find the representative fraction. 
 
 (1 Flemish mile=3-6387 English miles.) 
 
 7. A body of troops takes 1 hour to march 2| miles. Con- 
 
 struct a time scale to show intervals of 5 minutes. 
 The scale of the map is 2" to the mile. 
 
 8. The measurements of a five-sided field, ABODE, are as 
 
 follows : — 
 
 Sides ^5=140 yards. 1^^= 75 yards. 
 
 BG= 70 „ Diagonals ^0=195 „ 
 
 C'i)=220 „• ^Z)=160 „ 
 
 BE== 90 „ 
 
 Construct a scale of 70 yards to an inch, and draw a 
 
 plan of the field to that scale. Calculate and write 
 
 down its area in square yards. 
 
134 EXAMINATION PAPERS,. 
 
 9. A point A is taken in line with two inaccessible points 
 
 P and Q {P being the nearer), which are known to be 
 200 yards apart. A line AB, 300 yards long, is set 
 off at right angles to APQ, and the angle ABQ found 
 to be 67J°. Determine by a geometrical construction 
 the distance of P from A, and measure with a pro- 
 tractor the angle ABP. The angle ABQ should be 
 constructed. 
 
 Scale, which is to be properly drawn and figured, 
 150 yards to an inch, and showing 1000 yards. 
 
 10. Draw, to a scale of 10 feet to an inch, the irregular 
 
 figure ABCDEF, of which the dimensions are as 
 follows : — 
 
 AB=2Q feet. l GBA = 105\ 
 
 BG=37 „ lBAF=135\ 
 
 AF=25 „ lAFE=UQ\ 
 
 FF=22 „ L FED=75°. 
 
 ED=22 „ 
 
 The scale need not be drawn. The angles are to be 
 
 constructed. Measure with the protractor the angles 
 
 EDO and DGB. Write down their values, as also the 
 
 length of CD, in feet. 
 
ANSWERS. 
 
 Note. — A measurement may be considered correct if it is right 
 to withi/n two figv/res jm the second decimal place. 
 
 Chapter I. (Page 10.) 
 
 1. -46". 6. BE=-2W'. DE=-n". 
 
 7, 8, 9. The test of accuracy is that the three lines meet at 
 the same point. 
 
 10. Take any two points in AB : erect perpendiculars each J" 
 
 long, and join the extremities. 
 
 11. Join the points : bisect the line so found by a perpendi- 
 
 cular, cutting the given line in the point required. 
 
 Chapter II. (Page 26.) ■ 
 
 1. -58". 2. 208" or -80". 3. 1-1". 
 
 4. 1-40". 5. 2-99", 1-49". 6. 2-01". 
 
 7. 2-22". 8. 2-32". 9. 1-85". 
 
 10. Ml". 
 
 Chapter III. (Page 33.) 
 
 1. 1-77". 2.. 1-47". 3. 1-25". 
 
 4. 1-08". o. -96". 7. 1". 
 
 8. 1-45". 9. -92". 10. 2-08". 
 
 11. 1-73" 12. 1-31". 
 
 ISB 
 
•36 
 
 ANSWERS. 
 
 Chapter IV. (Page 40.) 
 
 1. 1-07"; radius=-72". 2. -93". 
 
 3. 1-27". 4. 1-25". 
 
 5. -95", 1-10". 6. 1-63". 
 
 8. 1-69", 1-34", -92", or 1-33", 1-05", -72". 
 
 9. 1-11", -55". 10. l^l". 
 
 7. 1-24'. 
 
 Chaptee V. (Page 51.) 
 
 1. 3". 
 
 3. 4-05", 1-68". 
 5. Eachpart=-2" 
 7. 1". 
 
 9. -93", -57". 
 U. 2-1". 
 
 2. x=2-63",y-l-n". 
 
 4. 1-39''. 
 
 6. AC=l-0&', BG=l-4:l". 
 
 8. Prob. 7. -43", -86", V72". 
 10. 1-25", 1-67", 2-08". 
 12. Prob. 1. 1-77". 
 
 Chapter VI. (Pages 57, 58, 60.) 
 
 Note. — The B. F. must be calculated from the distance given in the 
 question, not from that represented on the scale. 
 
 1. Length=4-23". E.F.=j^g. 
 
 3. 
 
 6-25". 
 
 1 
 
 1013760- 
 
 5. 
 
 7-5". 
 
 
 7. 
 
 5". 
 
 
 8. 
 
 5-79". 
 
 1 
 328172- 
 
 in 
 
 6-24". 
 
 1 
 
 lyj. 
 
 1051618- 
 
 12. 
 
 • 633600' 
 
 14. 
 
 3-82". 
 
 1 
 3773- 
 
 16. 
 
 5-52". 
 
 1 
 
 2. 
 4. 
 6. 
 
 6". ^. 
 
 '' * • 704U000- 
 
 5-64". 
 
 9. 
 11. 
 
 6-88"- »7i- 
 5". 
 
 13. 
 
 4". 
 
 15. 
 
 4-46". 
 
 17. 
 
 4-18". 
 
 18. 5-43" 
 
ANSWERS. 137 
 
 Chapter VII. (Page 71.) 
 
 1. Area= -98 square inches. Side of square = -OQ". 
 
 2. 1-48". 3. 1-40". 4. -73", 3-00 square inches. 
 5. 1-03". 6. 4-71". 
 
 7. 2-60 square inches. 8. 2-17 square inches. 
 
 9. Scale to show 800 yards = 3 -33". AG = AO0 yards. 
 10. Scale to show 30 metres = 4-29". 
 
 ^C=12-7 metres, AD = U-5, AE=28-3. 
 
 Chapter VIII. . (Pages 76, 79.) 
 
 1. Side = 3". 2. Side = 2-31". 
 
 3. Side = 1-41". 4. Eadius = 2-50". 
 
 5. Diameter = 1-5". 6. 1-41, 1-73, 2-45. 
 7. 3-32, 3-46, 3-87. 8. 1-55, 2-83, -87. 
 
 Chapter IX. (Page 97.) 
 
 1. Draw a line perpendicular to the given line, and mark ofi 
 
 1" on it. Draw the first circle, and then Prob. 7. 
 
 2. The third circle must touch the given line at its middle 
 
 point; hence Prob. 16. Radius = -78". 
 
 3. Prob. 6. 4. Prob. 4. 
 
 6. Prob. 2; Prob. 3. 6. Prob. 17. Radius =-62". 
 
 7. Prob. 6. Cor. 1. 8. Prob. 9. 
 9. Prob. 4. 
 
 10. Prob. 8; Prob. 18. Radius = -5". 
 
 11. Prob. 11. Radius=-83". 12. Prob. 16. Radius = l-18'. 
 
 13. Prob. 19. Radii =1-8", -8", -6". 
 
 14. Prob. 10; Prob. 7. Radius of large circle = 1 -70". 
 
138 ANSWERS. 
 
 Chapter X. (Page 111.) 
 
 1. 1-50" or -90". 2. 1", 1-41", 1-73". 
 
 3. -94". 4. 2-11", 2-33", 233", 2-11'. 
 
 5. 1-80", 2", 1-32". 6. -85". 
 
 7. Side = 2-02". 
 
 8. 1'80", 1-80", each passing through a vertex of the 
 
 hexagon. 
 
 9. 2-31", 2-31". 
 
 10. 1-04", 1-47", 1-04". Prob. 3; and Prob. 3, or Prob. 7. 
 
 11. Eadii are -80", 1-13", 1-39". 
 
 12. Diameter = 1-41". Prob. 8, using diameters instead of 
 
 radii. 
 
ANSWERS TO PAPERS. 
 
 1. I. 9. 
 
 2. 50 yards=3'33". Make five primary divisions, each 
 
 showing 10 yards. Divide the first primary division 
 into 10 parts, each showing 1 yard. AE=23 yards. 
 
 3. IX. 6. 
 
 4. 3 chains=3'44". Divide into 3 equal parts, showing 
 
 single chains. Divide 1 chain into 10 parts, each 
 showing 10 links, and draw 10 parallels. 
 
 5. II. 7. 6. VIII. 5, 6; ^6 = 245. 
 7. IX. 2, 18. Radius =-51". 
 
 II. 
 
 1. 40 feet=5*45". 4 primary divisions, 10 secondary 
 
 divisions, 12 parallels. 
 Test of accuracy, 17 feet 8 inches = 241''. 
 
 2. V. 4. Each part= 1-47". 
 
 3. 4 miles =5 '33". Show quarters of miles. .<4jE = 3^ miles. 
 
 4. (1) IX. 4. (2) IX. 2. 
 
 5. 3000 yards = 5-54". R. F.=t7|^. 
 
 6. II. 1; X. 3. Bisecting line = 1-3 8". • Area=3-22". Test 
 
 by finding the area of both triangles. 
 
 7. IX. 17. 
 
 13» 
 
I40 ANSWERS TO PAPERS. 
 
 IIL 
 
 1. 150 yards = 6'82". 3 primary divisions, 5 secondary 
 
 divisions, 10 parallels. 
 Test of accuracy, 67 yards = 3 -OS". 
 
 2. V. 2r 2'30". V. 1. 3-03". 
 
 3. For the scale, 80 yards=5-33". x. 4. Length of bisecting 
 
 line = 37'6 yards, or 2-51". Area=1298 square yards. 
 It may be found by doubling the area of the triangle 
 formed by bisecting the quadrilateral, Vii. 4 ; or move 
 accurately by vii. 6. 
 
 4. VIII. 2. Side of square so formed =1-72". Produce one 
 
 side of the square, mark off its length, and join the 
 extremities of the whole line to the vertex of the 
 square. 
 
 5. 30 miles = 4-14". R F. = „^y. 
 
 6. IV. 2, CoK. 1. Side =2 -00". 
 
 7. 11. 7, IV. 1. Eadius = l-16". 
 
 IV. 
 
 1. 1000 yards = 2'41". 10 primary divisions, 2 secondary 
 
 divisions, lOparallels. 735 yards = 1-77". E.F. = j-j^^j^. 
 
 2. IX. 19. Eadii = 2'12", 1-87", -63". 
 
 3. For scale, 90 feet = 6". ^C=30 feet, CZ) = 24 feet, 
 
 Di? = 40feet, EF=&0 feet, FB = i1 feet. 
 
 4. VIII. 6. Side = 2-65". x. 9. Dividing lines = 3-18". 
 
 5. 100 miles = 5-95". E. F. = TijTj^xy. 
 
 6. X. 12. Side = 3-30". 
 
 7. III. 4 J IV. 2. Side of square = 2 -eS". 
 
ANSWERS TO PAPERS. 141 
 
 1. 300 yards = 5 04". R F.=3n:VT- 183 yards = 307". 
 
 3. For scale, 600 yards = 5". Show tens of yards. Vii. 7. 
 
 Take D for vertex of equivalent triangle, and base in 
 AB. Then area = ^ x 290 x 580 = 84,000 square yards. 
 
 4. II. 17. 
 
 5. 250 miles = 4'72". 5 primary divisions,- 5 secondary 
 
 divisions. 
 
 6. IX. 6. 7. See Paper III. 4. Side of square = 1-84". 
 
 VI. 
 
 1 
 
 1. 10,000 yards = 7-36". R. F.=^^ 
 
 2. The diagonals intersect in the centre of the circle to be 
 
 described, and, with the vertices of the triangles, give 
 the points of the octagon on the circle. 
 
 3. VI. 7. 4. 50 feet = 6 -88". 27 feet 8 inches = 3-25". 
 
 5. IX. 9. The line joining the centre of the triangle to its 
 
 vertices cuts the circles in the points of contact of the 
 fourth circle. 
 
 6. III. 4. Note. X. 11. Side =1-30". 7. 11. 7. 
 
 VII. 
 
 1. 15 miles =5'22". R. F. = j-^^V^^. 
 
 7 miles 3 furlongs = 2 -5 7". 
 2.1.6,2,3. I. Ex. 11. EF= 2-70". 
 
 3. 200 paces = 6 -06". 
 
 4. V. 1, 3. a;=l-23", 2/ = 4-13", z=2'03". 
 
 5. VII. 2. For scale, 1000 yards -= 5" 
 
 ^a= 800 yards, AD = 680, BC.= 1080, BD = 760. 
 
 6. II. 10. Sides = 1-97", 1-53", and 1-00". 
 
 7. IV. 2. Side of square = 1-53". 
 
142 ANSWERS TO PAPERS. 
 
 VIIL 
 
 1. 500 yards = 7 -41". E. F.=^f^ 127 yards =l-88'', 
 
 2. For scale 400 yards = 5". Show distances of 10 yards. 
 
 AF= 100. AG=^ 360. 
 Reduce to an equivalent triangle with, vertex at D, and 
 base in AB. 
 Area = | x 300 x 480 = 72,000 square yards. 
 
 3. (1) IX. 2. (2) IX. 3. 
 
 4. 50 chains = 7-5". R F. = 5^. 5. II. 7. 
 6. IX. Principle II. ; IX. 16; radius = 1-5 6". 
 
 IX. 
 
 1. 2000 yards = 6 -82". R. F.=yTyi^. 2. 11. 8, CoB. 
 
 3. For scale, 600 yards = 5 -45". For angles, see i. 9. 
 
 Area = |- X 160 x 650 = 52,000 square yards. 
 
 4. IX. Principle 11. ; ix. 7. 
 
 5. 10 metres = 5 -88"; 10 primary divisions: no secondary 
 
 divisions: 10 parallels. 6 metres 8 decimetres = 4". 
 
 6. II. 1 ; IV. 3 ; side of square = 1 -42". 
 
 X. 
 
 1. , 4000 yards = 5". Show hundreds of yards. 
 A walks 7500, B 9700 yards. 
 
 2. VII. 5. Side of square = 1-92"; 
 IV. 4. Side of triangle = 1 -99", 
 
 3. 40 feet= 7-27". R. F. =t^. 23 feet 5 inches = 4-26". 
 
 4. First draw the triangle required, and then the first circle. 
 
 Then IX. 15; radius = 1 •54". 
 
 5. 50 feet = 4 -56". R. F. = ^i^. 
 
 6. Draw two diameters at right angles, and produce them. 
 
 Then ix. 17 Coe.: radius = 1 •81". 
 
ANSWERS TO PAPERS. 143 
 
 XI. 
 
 1. 200 feet = 7 -27". 2. v. 6. ^(7=4-73". 
 
 3. ix 15. Eadius = 2-36". 
 
 4. 250 chains = 4 -98". R. F. = .^^i^. 137 chains = 2 -73". 
 5.x. 11, or 8. Eadius=l-63". 6. iv. 2. Side of square 
 
 = 2-21". 
 
 XII. 
 
 1. 30 miles = 4- 13". ^^=^^6^71- 
 17 miles 3 furlongs =2-39". 
 
 2. (i.)v. 1. AB=l-W. (ii.)v. 3. 5C=1-91". 
 
 3. II. 6. Sides = 3-15", 2-45". 4. 40 Irish mUes = 5-8r. 
 
 5. IX. 16. Radius =-78". 
 
 6. IV. 2 Cob. Side of square = 1-36". 
 
 Additional Scales. (Pages 132, 133, 134.) 
 
 1. ^5=2-06jR.F.=^,j^. 2. 5-84; j,,!^. 
 
 CI»= 2-38 ;R.F. = ^,,i^. 
 3. 4"; 6-06"; TT^. 4. 3-48; ^Aj. 
 
 5- 6-82" J T^. 6. 701 ; 3^^,^,. 
 
 7. 1 hour =5". 
 
 8. Base =205 yards. Altitude= 160 yards. Area=20,500 
 
 square yards. For scale 400 yards =5'71". 
 
 9. ^P=536 yards. For scale 1000 yards=6-67". 
 10. lEDC=132% i.DCB=57°,CI)=2i-3ieet 
 
 Printed by T. and A. Cohsiable, Printers to Her Majesty 
 at the Edinburgh University Press.