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ELEMENTARY GEOMETRY
PRACTICAL AND THEORETICAL
CAMBRIDGE UNIVERSITY PRESS
C. F. CLAY, Manager
LONDON : Fetter Lane, E.G. 4
NEW YORK : G. P. PUTNAM'S SONS
BOMBAY, CALCUTTA, MADRAS: MACMILLAN AND CO., Ltd.
TORONTO : J. M. DENT AND SONS, Ltd.
TOKYO: THE MARUZEN-KABUSHIKI-KAISHA
All rights rciifvcd
ELEMENTARY GEOMETRY
PRACTICAL AND THEORETICAL
BY
C. GODFREY, M.V.O., M.A.
HHAD MASTER OF THE ROYAL NAVAL COLLEGE, OSBORNE,
FORMERLY SENIOR MATHEMATICAL MASTER AT WINCHESTER COLLEGE
AND
A. W. SIDDONS, M.A.
LATE FELLOW OF JESUS COLLEGE, CAMBRIDGE;
' ASSISTANT MASTER AT HARROW SCHOOL.
THIRD EDITION
CAMBRIDGE
AT THE University Press
1919
First Edition, Septenibtr 1903.
Reprinted October 1903, November 1903,
1904 {three times), 1905 {three times),
1906, 1907, 1908 (twice).
Second Edition, 1909.
Reprinted 1910, 19 11
Third Edition 1912.
Reprinted 1913, '915, 1916, 1918,
1919
PREFACE.
THE aim of the authors of the present work has been to
produce a book which will help to make Greometry an
attractive subject to the average British boy or girl.
The new schedule of geometry recently adopted by Cambridge
has been taken as a basis of operations. These regulations will
affect candidates for the Previous Examination after March 1904.
It has been found easy to follow this schedule closely and
at the same time to have regard to the reformed schedules of
\ arious other examinations, such as Oxford and Cambridge Locals,
Oxford Responsions, together with the examinations of the Uni-
versity of London, and the Civil Service Commissioners. The
reports of the British Association and of the Mathematical
Association have been very helpfuL
The book opens with a course of experimental work;
great pains have been taken to make the exercises perfectly
explicit and free from ambiguity. The beginner is taught to
use instruments, to measure accurately lines and angles (this
will in future be regarded as an indispensable part of geo-
metrical work), to construct and recognize the simpler plane
and solid figures, to solve problems by drawing to scale. At the
same time he is led to discover many geometrical truths which
are proved later; he should be encouraged to put into words
and make notes of any such discoveries. There is much in this
part which will be useful revision work for more advanced pupils.
n PREFACE
Then follows the dourse of Theoretical Geometry, which
is divided into four 'books.' The experimental inetho<l is still
prominent, in the shape of exercises leading up to propositions.
The sequence of theorems is Euclidean in form, but greatly
simplified by the omission of non-essentials, and by the use of
hypothetical constructions. There is reason to hope that it is
now possible to adopt a sequence (not differing very greatly
from that of Euclid) which will be generally accepted for some
time to come.
The treatment of problem.S Ls practical, though proofs arc
given ; for this part of the subject the present work is designed
to fulfil the purposes of a book on geometrical drawing.
Among the exercises, some are experimental and lead
up to future propositions, some are graphical and numerical
illustrations of known propositions, some are 'riders' of the
ordinary type*. In a great number of the earlier exercises the
figures are given. There is a collection of exercises on plotting
loci and envelopes; a subject which is found interesting,
and introduces the learner to other curves than the circle and
straight line.
Book I. deals with the subject-matter of Euclid I. 1 — 34;
angles at a point, parallels, angles of polygons, the triangle,
the parallelogram, sub-division of straight lines, the earliest
constructions and loci.
Book 11. treats of area. The notion of area is enforced
by a large number of exercises to be worked on squared paper,
the use of coordinates being explained incidentally. Euclid's
second book appears in a new garb as geometrical illustrations
of algebraical identities.
* We are indebted to the kindness of Mr R. Levett and of Messrs Swan
Sounenschein and Co. for permission to use a few of the riders from The
Elements of Plane Geometry issned under the auspices of the A. I. G. T.
PREFACE 711
Book IIL — ^the circle; relieved of a great number of useless
propositions. In addition to the topics usually treated, there
are sections on the mensuratioii of the circle, a knowledge
of which is generally assumed in works on solid geometry.
Book IV. — similarity. Here again much of Euclid VL
is omitted, as not really illustrating the subject of similar figures.
Euclid's definition of proportion has gone, and is replaced by an
easy algebraic treatment applicable (as is now permitted) to
commensurable magnitudes only.
On the whole, the authors believe that with two-thirds of
the number of theorems^ more ground is covered than by Euclid
L~IV. and VL
Keferences have generally been given in the proof of
propositions; it is not supposed however that pupils will be
required to quote references. Their presence in a book can be
justified only on the ground that they may help a reader to
foUow the argument.
The authors desire to express their gratitude to many friends,
whose criticisms have been both salutary and encouraging.
C. G.
CAMBsmGE, August, 1903. A. W S.
An appendix on the pentagon group of constructions is now
added.
July, 1906.
Revision papers have been added at the end of the book.
For permission to print certain items we are indebted to the
courtesy of H.M. Stationery Office, the Oxford Local Examina-
tions Delegacy, the Cambridge Local Examinations Syndicate,
the Joint Board, the LTniversity of London and the Board of
Management of the Common Entrance Examination.
C. G.
• December, 1916. A. W. S.
PKEFACE TO SECOND EDITION.
IN this edition the first four theorems of Book II. (areas of
parallelogram and triangle) have been i-ewritten and com-
pressed into three theorems, the enunciations now following the
arrangement of the Cambridge Syllabus. The proofs of III. 6
and 7 have also been rewritten.
In the first edition references were given, as a rule, in the
proofs of theorems; but in some cases an easy step was left
to the reader, by the insertion of (why ?). This is now deleted
from theorems, and the reference is given in all such cases.
An additional set of exercises on drawing to scale has been
inserted.
A very full table of contents now appears : this, in fact, was
added in an earlier reprint.
Other minor changes have been made (e.g. new figure for I. 3,
II. 7, lY. 1).
For the convenience of users of the first edition, it has been
arranged that there is no change in the numbering of pages or
exercises.
C. G.
A. W. S.
April, 1909.
PKEFACE TO THIRD EDITION.
IN this edition no changes have been made in the numbering of
pages or of exercises. The most important change is that
exercises of a theoretical character (riders) have been marked
thus fEx. 326, and exercises intended for discussion in class are
distinguished thus HEx. 30.
In order to economise time some of the drawing exercises in
the later part of the book have been slightly changed so that they
now require only a description of the method of performing the
construction instead of requiring that it shall actually be per-
formed.
C. G.
A. W. S.
December, 1911.
CONTENTS.
• PART I. EXPERIMENTAL GEOMETRY.
PAGE
Straight lines 3
Angles 7
R^ular polygons 17
Pattern drawing 19
Triangles 20
Pyramids — the Tetrahedron 26
Triangles (continued) 28
ParaUek and perpendiculars 35
Parallelogram, rectangle, square, rhombus .... 38
Cube, cuboid, prism, wedge 42
Drawing to scale 46
Heights and distances 48
How to copy a given rectilinear figure 50
Symmetry 61
Points, lines, svufaces, solids ..;.... 66
PART II. THEORETICAL GEOMETRY.
Book I.
Distinction between verification and proof .... 63
Angles at a point 64
Theorem 1. If a straight line stands on another straight
line, the sum of the two angles so formed is equal to
two right angles 65
Cor. If any number of straight hnes meet at a point,
the siun of aU the angles made by consecutive lines
is equal to four right angles 65
G. 6. b
jt CONTENTS — BOOK 1
PAQft
Theorem 2. If the sum of two adjacent angles is equal to
two right angles, the exterior arms of the angles are in
the same straight line 67
Theorem 3. If two straight lines intersect, the vertically
opposite angles are equal 69
Parallel straight unes 70
A fii-st treatment of parallels (for boi^inners) . . . 70a
Theorem 4. When a straight line cuts two other straight
lines, if
(1) a pair of alternate angles are equal,
or (2) a pair of corresponding angles are equal,
or (3) a pair of interior angles on the same side of
the cutting line are together equal to two
right angles,
then the two straight lines are parallel ... 71
Cor. If each of two straight lines is perpendicular to
a third straight line, the two straight lines are
parallel to one another 73
Playfair's Axiom 74
Theorem 5. If a straight line cuts two parallel straight
lines,
(1) alternate angles are equal,
(2) corresponding angles are equal,
(3) the interior angles on the same side of the
cutting line are together equal to two right
angles 74
A theorem and its converse 76
Theorem 6. Straight lines which are parallel to the same
straight line are parallel to one another ... 77
f Theorem 7. If straight lines are drawn from a point
parallel to the arms of an angle, the angle between
those straight lines is equal or supplementary to the
given angle 78
+ See note d- vii.
CONTENTS — BOOK I XI
PAGE
Angles of a tbiangle, a polycwn ..... 80
Theorem 8. The siim of the angles' of a triangle is equal
to two right angles 80
Cor. 1. If one side of a triangle is produced, the ex-
terior angle so formed is equal to the sum of the
two interior opposite angles 80
CoR. 2. If one side of a triangle is produced, the ex-
terior angle so formed is greater than either of the
interior opposite angles 80
Cor. 3. Any two angles of a triangle are together less
than two right angles 81
Cor. 4. Every triangle has at least two of its angles acute 81
CoR. 5. If two triangles have two angles of the one
equal to two angles of the other, each to each, then
the third angles are also equal 81
CoR. 6. The sum of the angles of a quadrilateral fs
equal to four right angles 81
Classification op triangles 81
Theorem 9. If the sides- of a convex polygon ai'e pro-
duced in order, the sum of the angles so formed is
equal to four right angles 83
CoR. The sum of the interior angles of any convex
polygon together with four right angles is equal to
twice as many right angles as the polygon has sides 83
Congruent triangles 85
Method of superposition 85
Theorem 10. If two triangles have two sides of the one
equal to two sides of the other, each to each, and also
the angles contained by those sides eqiial, the triangles
are congruent 86
Theorem 11. If two triangles have two angles of the one
equal to two angles of the other, each to each, and also
one side of the one equal to the corresponding side of
the other, the triangles are congruent .... 90
b 2
XU CJONTENTS — BOOK I
PAGE
Theorem 12. If two sides of a triangle are equal, the
angles opposite to these sides are equal . ■. 93
Theorem 13. If two angle.s of a triangle are equal, the
sides opposite to these angles are equal ... 96
Theorem 14. If two triangles have the three sides of the '
one equal to the three sides of the other, each to each,
the triangles are congruent 98
Theorem 15. If two right-angled triangles have their
hypotenuses equal, and one side of the one equal to
one side of the other, the triangles are congruent . 101
Constructions 103
Use of straight edge and of compasses .... 103
To construct a triangle having its sides equal to three
given straight hnes 104
Through a point O in a straight hne OX to draw a straight
line OY so that L XOY may be equal to a given angle
BAG 105
To bisect a given angle 107
To draw the perpendicular bisector of a given straight hne 109
To bisect a given straight line 109
To draw a straight line perpendicular to a given straight
line AB from a given point P in AB . . . Ill
To draw a straight line perpendicular to a given straight
line AB from a given point P outside AB . . 112
Construction of triangles from given data — the ambiguous
case 113
Miscellaneous exercises 115
Inequalities 119
Theorem 16. If two sides of a triangle are unequal, the
greater side has the greater angle opposite to it . 120
Theorem 17. If two angles of a triangle are unequal, the
greater angle has the greater side opposite to it . 122
t Theorem 18. Any two sides of a triangle are together
greater than the third side 124
CONTENTS — BOOK I Xlll
PAGE
+ Theorem 19. If two triangles have two sides of the one
equal to two sides of the other, each to each, and the
included angles unequal, the triangle which has the
greater included angle has the greater third side . 126
t Theorem 20. If two triangles have two sides of the one
equal to two sides of the other, each to each, and the
third sides unequal, the triangle which has the greater
third side has the greater included angle . . . 128
Theorem 21. Of all the straight lines that can be drawn
to a given straight line from a given point outside it,
the perpendicular is the shortest .... 130
Miscellaneous exercises 131
Parallelograms 139
Theorem 22. (1) The opposite angles of a parallelogram
are equal . 133
(2) The opposite sides of a parallelogram are equal 133
(3) Each diagonal bisects the parallelogram . . 133
(4) The diagonals of a parallelogram bisect one another 134
Cor. 1. If two straight lines are parallel, all jK>ints on
either line are equidistant from the other . . 134
Cor. 2. If a parallelogram has one of its angles a right
angle, aU its angles must be right angles . . . 134
Cor. 3. If one pair of adjacent sides of a parallelogram
are equal, all its sides are equal .... 134
Rectangle, square, rhombus, trapezium .... 135
t Theorem 23. (1) A quadrilateral is a parallelogram if
both pairs of opposite angles are equal . . 136
(2) A quadrilateral is a parallelogram if one pair of
opposite sides are equal and parallel . . . 136
(3) A quadrilateral is a parallelogram if both pairs of
opposite sides are equal 136
(4) A quadrilateral is a parallelogram if its diagonals
bisect one another 136
Cor. If equal perpendiculars are erected on the same
XIV CONTENTS — BOOK I
PA.UE
side of a straight line, the straight line joining
their extremities is parallel to the given line . 136
Construction. Through a given point to draw a straight
line parallel to a given straight line . . ; . i:i8
Construction. To draw a straight line parallel to a given
straight line and at a given distance from it . . 139
Subdivision op a straight line 140
Theorem 24. If there are three or more parallel straight
lines, and the intercepts made by them on any straight
line that cuts them are equal, then the corresponding
intercepts on any other straight line that cuts them
are also equal 140
Construction. To divide a given straight line into a
given nimiber of equal parts 142
ljl>CI 143
Theorem 25. The locus of a point which is equidistant
fix>m two fixed points is the perpendicular bisector of
the straight line joinifig the two fixed points . . 146
Theorem 26. The locus of a point which is equidistant
from two intersecting straight lines consists of the
pair of straight lines which bisect the angles between
the two given lines 147
Intersection of loci ........ 148
Construction of triangles, etc. by means of loci . . 151
Co-ordinates • . 152
Miscellaneous exercises 154
Book II. Area.
Area by counting squares — sqiiared paper 159
Rectangle 159
Right-angled triangle 163
Any rectilinear figure may be divided into rectangles and
right-angled triangles 163
Cionnlinear figures 166
CONTENTS — BOOK II XV
PAGB
Area of parallelogram 167
Theorem 1. Parallelograms on the same base and between
the same parallels (or, of the same altitude) are equi-
valent 169
Cor. 1. Parallelograms on equal bases and of the same
altitude are equivalent 169
Cor. 2. The area of a parallelogram is measured by the
product of the base and the altitude . . . 169
Area op trl/lngle 172
Theorem 2. Triangles on the same base and between
the same parallels (or, of the same altitude) are
equivalent . . 173
Cor. 1. Triangles on equal bases and of the same
altitude are equivalent 173
Cor. 2. The area of a triangle is measured by half the
product of the base and the altitude . . . 173
Area op polygon 178
Construction. To construct a triangle equivalent to a
given polygon . . 178
Land surveyors' method of finding area of polygon . . 180
Theorem 3. Equivalent triangles which have equal bases
in the same straight line, and on the same side of it,
are between the same parallels 183
Cor. 1. Equivalent triangles on the same or equal
bases are of the same altitude . . . . 183
Cor. 2, Equivalent triangles on the same base and on
the same side of it are between the same parallels . 183
+Theorem 4. If a triangle and a parallelogram stand on
the same base and between the same parallels, the
area of the triangle is half that of the parallelogram . 185
Miscellaneous exeroises on area 186
xvi CONTENTS — BOOK II
PAGE
The theorem op Pythagoras 187
Theorem 5. In a right-angled triangle, the square on the
hypotenuse is equal to the sum of the squares on the
sides containing the right angle ...'.. 190
Note on "error per cent." 198
Applications of Pythagoras' theorem .... 193
Square-roots found graphically 196
t Theorem 6. If a triangle is such that the square on one
side is equal to the sum of the squares on the other
two sides, then the angle contained by these two sides
is a right angle 199
Illustrations of algebraical identities by means op
geometrical figures 201
(A) {a+b)k=a^+bk 202
(B) {a+b)(c+cr)=ac+bc+ad+bd .... 203
(C) (a+bf=a^ + b^ + 2ah 204
(D) (a-6)2=a2-i-62-2aft 206
(E) a2-b^={a + b){a-b) 207
Exercises on the translation of verbal statements into
algebraical form 208
Projections 210
Extension op Pythagoras* theorem , . , . . 211
Theorem 7. In an obtuse-angled triangle, the square on
the side opposite to the obtuse angle is equal to the
sum of the squares on the sides containing the obtuse
angle pl^l8 twice the rectangle contained by one of
those sides and the projection on it of the other . 212
Theorem 8. In any triangle, the square on the side
opposite to an acute angle is equal to the sum of
the squares on the sides containing that acxite angle
minus twice the rectangle contained by one of those
sides and the projection on it of the other . . 213
Exercises — Apollonius' theorem 214
CONTENTS — BOOK III XVll
Book III. The Circtb.
I'AGE
Section I. Pbeliminaky 217
Definition of circle; centre, radius, circumference . . 217
Equal circles 217
Point and circle 218
Straight line and circle — chord, diameter, tangent . . 218
Arcs— minor and major; semicircle 218
S^ment, sector 219
Section II. Chord and centre . . . . . . 219
Symmetry of the circle 219
Theorem 1. A straight line, drawn from the centre of a
circle to bisect a chord which is not a diameter, is at
right angles to the chord.
Conversely, the perpendicular to a chord from the centre
bisects the chord 220
Cor. a straight line drawn through the mid-point of a
chord of a circle at right angles to the chord will, if
produced, pass through the centre of the circle . 221
Construction. To find the centre of a given circle . 222
Construction. To complete a circle of which an arc is
given 222
Theorem 2. There is one circle, and one only, which
passes through three given points not in a straight line 223
Cor. 1. Two circles cannot intersect in more than two
points 223
Cor. 2. The i)erpendicular bisectors of AB, BC, and
CA meet in a point 223
Construction. To circumscribe a circle about a given
tria.ngle 224
Section III. Arcs, anciles, chords 226
Theorem 3. In equal circles (or, in the same circle)
(1) if two arcs subtend equal angles at the centres,
they are equal.
(2) Conversdy, if two arcs are equal, they subtend
equal angles at the c^itres .... 226
XVlll CONTENTS — ^BOOK III
I'AGK
Theorem 4. In equal circles (or, in the same circle)
(1) if two cho«ls are equal, they cut off equal arcs.
(2) Conversely, if two arcs are equal, the chords of
the arcs ai-e equal 228
Construction. To place in a circle a chord of given
length 229
Construction. To inscribe a regulai* hexagon in a circle 230
Circumference of circle 231
Theorem 5. In equal circles (or, in the same circle)
(1) equal chords are equidistant from the centres.
(2) Conversely, chords that are equidistant from the
centres are equal 235
Lengths of chords 236
Section IV. The tangent 238
Theorem 6. The tangent at any point of a circle and the
radiiis through the point are perpendicular to one
another 239
Cor. a straight line drawn through the point of con-
tact of a tangent at right angles to the tangent will,
if produced, pass through the centre of the circle . 239
Construction. To draw the tangent to a circle at a given
point on the circle . . . . . . . 240
Construction. To inscribe a circle in a given triangle . 243
The escribed circles of a triangle 244
Section V. Contact of circles . . . . . . 245
Theorem 7. If two circles touch, the point of contact lies
in the straight line through the centres . . . 246
Construction of circles to satisfy given conditions . . 247
Section VI. Angle properties 250
Reflex angles 251
Theorem 8. The angle which an arc of a circle subtends
at the centre is double that which it subtends at any
point on the remaining part of the circumference , 252
CONTENTS — BOOK III XIX
PAGE
Theorem 9. Angles in the same segment of a circle are
equal 254
Theorem 10. The angle in a majoi* segment is acute ; the
angle in a semicircle is a right angle ; and the angle
in a minor segment is obtuse 255
Theorem 11. If the line joining two points subtends
equal angles at two other points on the same side
of it, the four points lie on a circle .... 257
Theorem 12. The opposite angles of any quadrilateral
inscribed in a circle are supplementary . . . 258
Theorem 13. If a pair of opposite angles of a quadri-
lateral are supplementary, its vertices are concyclic . 260
Section VII. Construction of tangents .... 261
To draw tangents to a given circle ABC from a given point
T outside the circle 262
Common tangents to two circles 263
To construct an exterior common tangent to two unequal
circles 264
To construct an interior common tangent to two circles . 266
Section VIII. Constructions depending on angle pro-
perties 268
The locus of pomts (on one side of a given straight line) at
which the line subtends a constant angle is an arc of
a circle, the given line being the chord of the arc . 268
To construct a triangle with given base, given altitude,
and given vertical angle 269
To inscribe in a given circle a triangle with given angles 270
To circumscribe about a given circle a triangle with given
angles 271
Section IX. "Alternate segment" 272
Theorem 14. If a straight line touch a circle, and from
the point of contact a chord be drawn, the angles
which this chord makes with the tangent are equal
to the angles in the alternate segments . . . 272
XX CONTENTS — BOOK III
PAQK
Construction. On a given straight line AB to construct ^
a segment of a circle to contain a given angle X . 274
Construction. In a given circle to inscribe a triangle
equiangular to a given triangle XYZ . . . 276
Tangent as limit of chord 276
Miscellaneous exercises on Sections VI., VIII., IX. . 277
Section X. Arcs and angles at the circumference 278
Regular polygons 280
Theorem 15. If the circumference of a circle be divided
into n equal arcs, (1) the points of division are the
vertices of a regular n-gon inscribed in the circle ;
(2) if tangents be drawn to the circle at these points,
these tangents are the sides of a regular n-gon circum-
scribed about the circle 281
Section XI. Area op circle 284
Area of sector of circle 287
Area of segment of circle 287
Section XIL Further examples op loci .... 288
Envelopes 293
Miscellaneous exercises 295
Book IV. Similarity.
Ratio and proportion 301
Internal and external division 304
Proportional division op straight lines .... 305
Theorem I. If a straight line HK drawn parallel to the
base BC of a triangle ABO cuts AB, AG in H, K
respectively, then tt^^-i-^ 306
* •" AB AC
CONTENTS — BOOK IV XXI
PAGE
Cor. 1. If a straight line is drawn parallel to one side
of a triangle, the other two sides are divided pro-
portionally 307
Cor. 2. If two straight lines are cut by a series of
parallel straight lines, the intercepts on the one
have to one another the same ratios as the corre-
sponding intercepts on the other .... 308
Construction. To find the fourth proportional to three
given straight lines 309
Theorem 2. If H, K are points in the sides AB, AC of a
AH AK
triangle ABC, such that ^^ = ^ , then HK is imrallel
to BC -311
AB AC
Cor. 1. If Tr;= ^5 then HK and BC are parallel . 311
AH AK
Cor. 2. If a straight line divides the sides of a triangle
proportionally, it is parallel to the base of the
triangle . . . 311
Similar triangles 313
Theorem 3. If two triangles are equiangular, their corre-
sponding sides are proportional 314
t Construction. On a given straight line to construct a
figure similar to a given rectiUnear figure. (First
method) 317
BC CA AB
Theorem 4, If, in two triangles ABC, DEF, pp = ^^ = — ,
then the triangles are equiangular .... 319
The diagonal scale 320
Theorem 5. If two triangles have one angle of the one
equal to one angle of the other and the sides about
these equal angles proportional, the triangles are
similar '. 321
Proportional compasses 322
XXU CONTENTS — BOOK IV
I'AOE
Areas op similar tkianglks ....... 323
Theorem 6. The ratio of the areas of similar triangles is
equal to the ratio of the squares on corresponding
sides 324
Rectangle properties 326
Theorem 7 (i). If AB, CD, two chords of a circle, intersect
at a point P inside the circle, then PA . PB = PC . PD . 328
Theorem 7 (ii). If AB, CD, two chords of a circle, in-
tersect at a point P outside the circle, then PA. PB= '
PC.PD 329
Construction. To find the mean proportional between
'two given straight lines 332
Construction. To describe a square equivalent to a given
rectilinear figure 333
Theorem 8 (i). The internal bisector of an angle of a triangle
divides the opposite side internally in the ratio of the
sides containing the angle .3.34
Theorem 8 (ii). The external bisector of an angle of a triangle
divides the opposite side externally in the ratio of the sides
containing the angle 335
tSlMILAR POLYGONS 337
t Theorem 9. If the straight lines joining a jwint to the
vertices of a given polygon are divided (all internally
or all externally) in the same ratio the jxjints of
division are the vertices of a similar polygon . . 337
+ Construction. On a given straight line to constnict a
figure similar to a given rectilinear figure. (Second
method) 339
t Theorem 10. If a polygon is divided into triangles by
lines joining a point to its vertices, any similar
polygon can be divided into corresponding similar
triangles 340
Contents — book iv xxm
PAGE
Cor. If in two similar figures whose sides are in the
ratio 1 : ^, Oj, O2 corresi)Oiid to X^, X2, then O^Og :
XiX2=l:yfc 342
t Theorem 11. The ratio of the areas of similar jwlygons
is equal to the ratio of the squares on c<jrresj)onding
sides 343
t Construction. To construct a figure equivalent to a
given figure A and similar to another figure B. . 344
MiscelIjAneous exercises 345
Appendix I.
Euclid 11. 14 . '. 353
Eviclid in. 35, 36 354
tAppendix II. The Pentagon.
Construction. To divide a given straight line into two
parts such that the square on the greater part may be
equal to the rectangle contained by the whole line and
the smaller part ........ 356
Extreme and mean ratio 358
Construction. To construct an isosceles triangle such
that each of the base angles is twice the vertical angle 359
Construction. To describe a regular pentagon . . 361
To prove that sin 18°=^-—^ 362
Kevision Papers 363
List of Definitions 391
PABT 1.
BXPEEIMBNTAL GBOMBTET.
Q. S.
INSTRUMENTS.
The following instrnments will be required : —
A hard pencil (HH).
A ruler about 6" loug (or more) graduated in inches and tenths of an
inch and also in cm. and mm.
A set square (60°) ; its longest side should be at least 6" loug.
A semi-circular protractor.
A pair of compasses (with a hard pencil point).
The pencil should have a chisel-point.
The compass pencil may have a chisel-point, or may be sharpened in the
ordinary way.
In testing the equality of two lengths or in transferring lengths, compasses
should always be used.
Exercises distinguished by a paragraph sign thus: ^Ex. 27,
are intended for discussion in class.
Exercises of a theoretical character {riders) are marked with
a dagger thus: fEx. 323.
EXPEEIMENTAL GEOMETRY.
Straight Lines.
In stating the length of a line, remember to give the unit ; the following
abbreviations may be used: — in. for inch; cm. for centimetre; mm. for
millimetre.
In Ex. 1-163, all lengths measured in inches are to be given to the
nearest tenth of an inch, all lengths measured in centimetres to the nearest
millimetre.
Always give your answers in decimals.
Ex. 1. Measure the lengths AB, CD, EF, GH in fig. 1
(i) in inches,
(ii) in centimetres.
X X
E F
X X
G H
fig. 1.
Ex. 2. Measure in inches and centimetres the lengths of
the edges of your wooden blocks.
1—2
EXPERIMENTAL GEOMETRY
Ex. 3. Measure in inches the lengths AB, BC, CD in fig. 2 ;
arrange your results in tabular form and add them together.
B
C
fig.
2.
AB
=
in.
BC
=
in.
CD
=
in.
AB + BC + CD= in.
Check by measuring AD.
Ex. 4. Repeat Ex. 3, using centimetres instead of inches.
Ex. 5. Repeat Ex. 3, for fig. 3, (i) using centimetres,
(ii) using inches.
XX X X
A 8 C D
fig. 3.
Ex. 6. Measure in centimetres the lengths AB, BC in
fig. 4, and find their difference ; arrange your results in tabular
form.
-I 1 1_
C
fig. 4.
AB= cm.
BC= cm.
AB — BC = cm.
Check by measuring AC.
Ex. 7. Repeat Ex. 6, using inches instead of centimetres.
STRAIGHT LINES 5
Ex. 8. Repeat Ex. 6, for fig. 5, (i) using inches, (ii) using
centimetres.
X * XX
A C B
fig. 6.
Ex. 9. Measure in inches, and also in centimetres, the
length of the paper you are using.
Tour ruler is probably too short to measure directly ; divide the length
into two (or more) parts by making a pencil mark on the edge, and add these
lengths together.
Ex, 10. Measure the breadth of your paper in inches and
also in centimetres.
Ex. 11. Draw a straight line about 6 in. long and cut oflf
a part AB = 2 in., a part BC = 1*5 in., and a part CD =1*8 in.;
find the length of. AD by adding these lengths ; check by measur-
ing AD. [Make a table as in Ex. 3.]
Ex. 12. Repeat Ex. 11, with
(i) AB = 2-7 cm., BC = 9-6 cm.,
(ii) AB ^ 5-2 cm., BC = 3*9 cm.,
(iii) AB = -7 in., BC = 2-6 in.,
(iv) AB = -8 cm., BC = '5 cm.,
(v) AB = 1-8 in., BC = 2-9 in.,
Ex. 13. A man walks 3*2 miles due north and then 1*5 miles
due south, how far is he from his starting point 1 Draw a plan
(1 mile being represented by 1 inch) and find the distance by
measurement.
Ex. 14. A man walks 5*4 miles due west and then 8-2 miles
due east, how far is he from his starting point? (Represent
1 mile by 1 centimetre.)
Ex. 15. A man walks 7*3 miles due south, then 12*7 miles
due north, then 1-1 miles due south, how far is he from his
starting point? (Represent 1 mile by 1 centimetre.)
CD =
1-3 cm.
CD =
2-8 cm.
CD =
2-4 in.
CD =
2-4 cm.
CD =
•6 in.
6
EXPERIMENTAL GEOMETRY
fig. 6.
Ex. 16. Draw a straight line, guess its middle point and
mark it by a short cross-line ; test your guess by measuring the
two parts.
Ex. 17. Repeat Ex. 16, three or four times with lines of
various lengths. Show by a table how far you are wrong.
Ex. 18. Draw a straight line of 10-6 cm. ; bisect it by calcu-
lating the length of half the line and measuring off that length
from one end of the line, then measure the remaining part.
When told to draw a line of some given length, you should draw a line
a little too long and cut off a part equal to
the given length as in fig. 6. You should -<
also write the length of the line against it,
being careful to state the unit.
Ex. 19. Draw a straight line 3 '2 in. long, bisect it as
in Ex. 18.
Ex. 20. Draw a straight line 2*7 in. long, bisect it as
in Ex. 18.
Ex. 21. Draw straight lines of the following lengths, bisect
each of them : (i) 7-6 cm., (ii) 10*5 cm., (iii) 4*1 in., (iv) -9 in.,
(v) 5-8 cm., (vi) 11-3 cm.
A good practical method of bisecting a straight line (AB) is
as follows : — measure off with dividers equal lengths (AC, BD)
from each end of the line (these lengths should be very nearly
half the length of the line) and bisect the remaining portion
(CD) by eye.
c D
H Hfl H-
A E B
fig. 7.
Ex. 22. Draw three or four straight lines and bisect them
with your dividers (as explained above) ; verify by measuring
each part of the line (remember to write its length against each
part)
Ex. 23. Open your dividers 1 cm., apply them to the inch
scale and so find the number of inches in 1 centimetre.
STRAIGHT LINES — ANGLES 7
Ex. 24. Pind the number of inches in 10, cm. as in Ex. 23 ;
hence express 1 cm. in inches. Arrange your results in tabular
form.
Ex. 25. Eind the number of centimetres in 5 in. as in Ex. 23;
lience find the number of centimetres in 1 inch.
h
Ex 26. Guess the lengths of the lines in fig.
inches, (ii) in centimetres ; verify by measurement,
table thus : —
8 (i) in
Make a
Line
Guessed
Measured
a
h
Angles.
If you hold one arm of your dividers firm and turn the other
about the hinge, the two arms
may be said to form an angle.
In the same way if two
straight lines OA, OB are drawn
from a point O, they are said
fig. 9.
8
EXPERIMENTAL GEOMETRY
to form an angle at o. o is called the vertex of the angle,
and OA, OB its arms.
fig. 10.
An angle may be denoted by three letters ; thus we speak of
the angle AOB, the middle letter denoting the vertex of the
angle and the outside letters denoting points on its arms.
If there is only one angle at a point O, we call it the
angle O.
Sometimes an angle is denoted by a small
letter placed in it; thus in the figure we have two 16
ani^les a and h.
L is the abbreviation for angle.
fig. 11.
Two angles AOB, CXD (see figs. 10 and 12), are said to be
equal when they can be made to fit on one another exactly (i.e.
when they are such that, if CXD be cut q
out and placed so that X is on O and XC
along OA, then XD is along OB). It is
important to notice that it is not necessary
for the arins of the one angle to be equal
to those of the other, in fact the size of an
angle does not depend on the lengths of its
arms.
HEx. 27. Draw an angle on your paper and open your
dividers to the same angle.
ANGLES
HEx. 28. Which is the greater angle in fig. 13? Test by
making on tracing paper an angle equal to one of the angles
and fitting the trace on the other. O
fig. 13.
liEx. 29. Name the angle at O in fig. 14 in as many different
■ways as you can.
HEx. 30. Take a piece of paper and fold it, you will get
something like fig. 15, fold
it again so that the edge OB
fits on the edge OAj now
open the paper; you have
four angles made by the
creases, as in fig. 16; they
are all equal for when folded
they fitted on one another.
Such angles are called right
angles. An angle less than
a right angle is called an
acute angle. An angle
greater than a right angle
is called an obtuse angle.
HEx. 31. Make a right
angle BOG as in Ex. 30, cut
it out and fold so that OB
falls on OC. Does the crease
(OE) bisect z. BOO? (Le. are
L. s BOE, EOC equal?) What
fraction of a right angle is each of the /. s BOE, EOC?
10
EXPERIMENTAL GEOMETRY
HEx. 32. If the ^ BOE of Ex. 31 were bisected by folding,
what fraction of a right angle would be obtained 1
If a right angle is divided into 90 equal angles, each of these
angles is called a degree.
25° is the abbreviation for '* 25 degrees," ,
fig. 17.
Fig. 17 represents a protractor; if each graduation on the
edge were joined to C, we should get a set of angles at C each
of which would be an angle of one degree.
HEx. 33. What fractions of a right angle are the angles
between the hands of a clock at the following times : — (i) 3.0,
(ii) 1.0, (iii) 10.0, (iv) 5.0, (v) 8.0? State in each case whether
the angle is acute, right, or obtuse.
I^Ex. 34. Find the number of degrees in each of the angles
in Ex. 33. [Use the residts of that Ex.]
ANGLES
1.
Z
^ u,
/
^
r -^
^
\ \
^
^
/■/
^^
\\
//
^
\^^^
V'^
^\
///
' e^
^hN^
I^\:
l/^
. ^\n^
w <
~~~~~
^^^--^
vr/byS^'
^
^
> E
X /
7/
\ \\
V \
^^^^
.^".
//
//
\ \
\\
^^
//
/
(P
\ 1
'V
"^ ^
-o^ ^
' ^
(P a>
C9
« S
t^
01
C3
fig.
18.
<<^
11
^Ex. 35. Fig. 18 shows the points of the compass ; what are
the angles between (i) N and E, (ii) W and S W, (iii) "W and
WN W, (iv) E and E by S, (v) NE and NN W, (vi) S W and SE 1
To measure an angle, place the protractor so that its centre
C is at the vertex of the angle and its base, CX, along one arm
of the angle ; then note under which graduation the other arm
passes ; thus in fig. 17, the angle — 48°.
In using a protractor such as that in fig. 17, care must be taken to choose
the right set of numbers ; e.g. if the one arm of the angle to be measured
lies along CX, the set of numbers to be used is obviously the one in which
the numbers increase as the line turns round C from CX towards CX'.
You should also check your measurement by noticing whether the
angle is acute or obtuse.
12
EXPERIJIENTAL GEOMETRY
When joa measure an angle in a figure that yoa have drawn /
(or make an angle to a given measure), always indicate in your /
figure the number of degrees, as in fig. 19. / SI
Ex, 36. Cut out of paper a right angle, bisect it
by folding, and measure the two angles thus formed.
fig. 19.
Ex. 37. Measure the angles of your set square (i) directly,
(ii) by making a copy on paper and measuring the copy.
It is difficult to draw a straight line right to the comer of a set square ;
it is better to draw the lines to within half a centimetre of the comer and
afterwards produce them (i.e. prolong them) with the ruler till they meet.
Ex. 38. Measure the angles of your models — this may be
done either directly, or more accurately by copying the angles
and measuring the copy.
Ex. 39. Measure z.s AOB, BOC in fig. 20; add; and check
your result by measuring l AOC. (Arrange in tabular form.)
fig. 20.
Ex. 40. Measure z.s AOC, COD, AOD in fig. 20. Check
your results.
ANGLES 13
Ex. 41. Mea.sure ^s AOB, BOD, AOD in fig. 20. Check
your results.
fig. 21.
Ex. 42. Repeat the last three exercises for fig. 21.
14 EXPERIMENTAL GEOMETRY
Ex. 43. Draw a circle (radius about 2*5 in.), cut off equal
parts from its circumference (this can be done ^ °
by stepping off with compasses or dividers).
Join OA, OB, OF. Measure l s AOB, AOF. Is
L AOF = 5 times ^AOB?
To make an angle to a given measure.
Suppose that you have a line AB and that at the fig- 22.
point A you wish to make an angle of 73°. Place the protractor
so that its centre is at A and its base along AB, mark the 73°
graduation with your dividers (only a small prick should be
made), and join this point to A. (Remember to write 73° in
the angle.)
Ex. 44. Make a copy of the smallest angle of your set
square and bisect it as follows : — measure the angle with your
protractor, calculate the number of degrees in half the angle,
mark off this number (as explained above) and join to the
vertex. Verify by measuring each half. (This will be referred
to as the method of bisecting an angle hy meoms of the pro-
tractor.)
Ex. 45. IVIake angles of 20°, 35°, 64°, 130°, 157°, 176° (let
them point in different directions). State whether each one is
acute, right, or obtuse.
Ex. 46. Make the following angles and bisect each by
means of the protractor, 24°, 78°, 152°, 65°, 111°. (Let them
point in different directions.)
HEx. 47. Draw an acute angle AOB; produce b
AO to 0 J what kind of angle is BOO ? {freehand) \
HEx. 48. Draw an obtuse angle BOC; produce ^
CO to A; what kfbd of angle is AOB? {freehand) gg 23.
^Ex. 49. Make /l AOB = 42°; produce AO to 0. By how
much is z. AOB less than a right angle? By how much is
L. BOC greater than a right angle 1
ANGLES 15
HEx. 50. (i) Make ^AOB = 65°; produce AO to C; measure
L BOC ; what is the sum of z. s AOB, BOC?
(ii) Repeat (i) with z. AOB = 77°.
(iii) Repeat (i) with z. AOB = 123\
Compare the results of (i), (ii), (iii) ; how many right angles
are there in each sum ?
HEx. 51. If, in fig. 23, l AOB = 57°, what is l. BOC % Check
by drawing and measuring.
UEx. 52. (i) If, in fig. 23, l. BOC = 137°, what is l AOB ?
(ii) „ „ Z.BOC= 93" „ „ Z.AOB?
(iii) „ „ ^AOB= 5° „ „ ^BOC?
liEx. 53. Draw a straight line OB; on opposite sides of OB
make the two angles AOB = 42°, BOC =129°. What is their
sum ? Is AOC a straight line ?
UEx. 54. Repeat Ex. 53, with
(i) /.AOB = 42°, /.BOC = 138°.
(ii) /.AOB = 90°, /.BOC =90°.
(iii) /.AOB = 73°, /.BOC = 113°.
(iv) /.AOB = 113°, /.BOC = 76°.
llEx. 55. What connection must there be between the two
angles in the last Ex. in order that AOC may be straight?
f Ex. 56. Make an /. AOB = 36° ; produce AC
to C ; make /. COD = 36° ; calculate l BOC ; is
BOD a straight line in your figure ? Give a reason,
UEx. 57. From a point O in a straight line
AB, draw two lines OC, CD (see fig. 25) ; measure
the three angles ; what is their sum 1
UEx. 58. Repeat Ex. 57, with AOB drawn in ^
a different direction.
16
EXPERIMENTAL GEOMETRY
^Ex. 59. Draw fig. 26 making l. BOC = 67° and l. B'O'd' = 29°.
What ia the sum of the four angles ?
A o/ B
a' d\. b'
o'
fig. 26.
fig. 27.
UEx. 60. Draw fig. 27 making l. BOC = 67° and l BOD = 29°.
What is the sum of the four angles at O ? Give a reason.
UEx. 61. From a point O in a straight line
AB, draw straight lines OC, OD, OE, OF, OG as in
fig. 28. Measure the angles AOG, COD, <fec. What b
is their sum ?
UEx. 62. From a point O, draw a set of straight
lines as in fig. 29, measure the angles so formed.
What is their sum? How many right angles is
the sum equal to 1
fig. 29.
Ex. 63. From a point O, draw a set of straight
lines as in fig. 29. Guess the size of the angles so formed ;
verify by measurement. Make a table thus : —
r
Angle
Guessed
Measured
a
h
c
45°
27°
47°
153°
ANGLES — REGULAR POLYGONS 17
Ex, 64. Draw two straight lines as in fig. 30; g
measure all the anglea
Ex. 65. Make aAOB = 47°; produce AO to
C and BO to D ; measure all the angles.
Ex, 66. Repeat Ex, 65 with l AOB = 166°,
HEx, 67. In fig. 30, if z. AOB = 73°, what are the remaining
angles ? Verify by drawing.
HEx, 68. (i) In fig. 30, if l AOD = 132°, what are the re-
maining angles ?
(ii) In fig, 30, if L COD = 58°, what are the remaining
angles ?
(iii) In fig, 30, if L BOG = 97°, what are the remaining
angles ?
Regular Polygons.
Ex. 69. Describe a circle of radius 5 cm.; at its centre O
draw two lines at right angles to cut the circle A
at A, B, C, D. Join AB, BC, CD, DA. Measure
each of these lines and each of the angles ABC,
BCD, CD A, DAB.
A square has all its sides equal and all
its angles right angles.
Ex, 70. Describe a circle of radius 5 cm.; at its centre
i.e. —J- ) ;
join the points where the arms cut the circle;
the figure you obtain is a hexagon (6-gon), and
it is said to be inscribed in the circle. What do
you notice about its sides and angles ?
a. s.
IS
EXPERIMENTAL GEOMETRY
A figure bounded by equal straight lines, which has all its
angles equal, is called a regular polygon.
A figure of 3 sides is called a triangle (A).
„ „ 4 „ „ „ quadrilateral (4-gon).
5 „ „ „ pentagon (5-gon).
„ „ 6 „ „ „ hexagon (6-gon).
„ „ 7 „ „ „ heptagon (7-gon).
„ „ 8 „ „ „ octagon (8-gon).
The comers of a triangle or polygon are called its verticea
The perimeter of a figure is the sum of its sides.
Ex. 71. What is the perimeter of a regular 6-gon, each of
whose sides is 2*7 in. long?
Ex. 72. In a circle of radius 5 cm. make a regular pentagon
(5-gon) as in Ex. 70 ; the angles you make at the centre must all
be equal and there will be five of them ; what is each angle 1
Ex. 73. Calculate the angle at the centre for each of the
following regular polygons; inscribe each in a circle of radius
5 cm.
(i) 8-gon, (ii) 9-gon, (iii) triangle, (iv) 10-gon, (v) 1 6-gon.
Ex, 74. Make a table of the results of Ex. 73.
1 REGUiiAR Polygons
1 1
Number
of sides
Angle at
centre
Length
of side
Perimeter
3
120°
4
90°
5
.
REGULAR POLYGONS — PATTERN DRAWING
19
Ex. 75. Explain in your own words a simple construction
for a regular hexagon depending on the fact you discovered in
Ex. 70, that each side of the hexagon was equal to the radius
of the circle.
Pattern Drawing
Ex. 76. Copy fig. 33, taking 5 cim. for
the radius of the lai^e circle. The dotted
lines are at right angles to one another.
How will you find the centres of the small
circles ?
If you describe only part of a circle, the
curve you make is called an arc of the
circle.
Ex. 77. Copy fig. 34, taking 5 cin. for
the radius of the circle. The six points
on the circle are the vertices of a regular
hexagon (see Ex. 75); each of these points
is the centre of one of the arcs.
Ex. 78. Copy fig. 35, taking 5 cm. for
the radius of the circle. The centres of
the arcs are the midpoints of the sides of
a square inscribed in the circle.
Ex. 79. Copy fig. 36, taking 5 cm. for
the radius of the circle. The angles between
the dotted lines are equal ; what size is each
of these angles? The centres of the arcs
are the midpoints of the dotted lines.
fig.
34.
^^"f \\^\
\ ^■ \
/ / \
1 \ \
/ /
\ \
'/~K
/_\\
iWl^
r^
\ \
y 1
\ \
]/ /
\^
Ji^
fig. 35.
20
EXPERIMENTAL GEOMETRY
fig.
38.
^
^
>
~'\-'^
A^
V-^ IS-
J^
~- ]/
V- r~
^^
\::::i^
^^
fig. 39.
Ex, 80. Copy fig. 37, taking 5 cm. for
the radius of the circle. Where are the
centres of the arcs ?
A straight line drawn through the centre
of a circle to meet the circumference both
ways is called a diameter.
The two parts into which a diameter
divides a circle are called semicircles.
Ex. 81. Copy fig. 38, taking AD = 9 cm.
AD is a diameter of a circle and is divided
into three equal parts at B and C ; semi-
circles are described on AB, AC, CD, BD as
diameters.
Ex. 82. Draw a figure showing the
points of the compass. See fig, 18.
Ex. 83. Copy fig. 39, taking 5 cm. for
the radius of the large circle. The radius
of the small circle is half that of the large
circle ; the centres of the arcs are the
vertices of the regular hexagon.
Ex. 84. Copy fig. 40. The points of the
star'iire the vertices of a regular pentagon.
fig. 40.
Triangles.
Ex. 85. Draw a triangle (each side being at least 2'5 in.
long). Measure all its angles ; find the sum of its angles.
PATTERN DRAWING — TRIANGLES 21
Ex. 86. Repeat Ex. 85 three or four times with triangles of
different shapes.
When told to construct a figure to given measurements, first make a
rough sketch of the figure on a small scale and write the given measurements
on the sketch.
Ex. 87. Make an angle ABC = 74°; cut off from its arms
BC = 3'2 in., BA = 2-8 in.; join AC. Measure the remaining side
and angles of the triangle ABC.
In all cases where triangles or quadrilaterals are to be constructed to
given measurements, measure the remaining sides (in inches if the given
sides are measured in inches, in centimetres if the given sides are measured
in centimetres) ; also measure the angles, and find their sum.
Ex. 88. Construct triangles to the following measurements: —
(i) z. ABC = 80°, AB = 2-2in., BC = 2-9in-
§(ii) z.B = 28°, AB = 7-3cm., BC=12-lcra.
(iii) ^A = 42°, AB = 3-7in., CA = 3-7in.
§(iv) ^B = 126°, AB = 6-lcm., BC = 6-lcm.
(v) ^C = 90°, BC = 3-9in., CA = 2-8in,
(vi) BC = 6-7cm., z.C = 48°, CA = 9-0cm.
(vii) AB=4-7in., BC = 2-9in., ^B = 32°.
§(viii) CA = 2-6in., AB = 3'3in., ^A = 162°.
(ix) /.C = 79°, CA = 4-7cm., ■BC = 6-lcm.
(x) AB=4-6cm., CA = 8-7cm., ^A=:58°.
Ex. 89. Draw a straight line AB 9 cm. long, at A make an
angle BAC = 60°, at B make an angle ABC = 40°, produce AC, BC
to cut at C. Measure the remaining sides and angle of the
triangle ABC. What is the sum of the three angles 1
Ex. 90. Construct triangles to the following measurements : —
(In case the construction is impossible with the given measurements, try
to explain why it is impossible.)
(i) AB = 8-3cm., z.A = 45°, ^B = 72°.
§(ii) AB=3-9in., ^A = 39°, z.B = 39°.
§ These will be enough exercises of this typo unless much practice is needed.
98 |;XFEBIMENTAL GEOMETRY
(iii) ^B = 90°, Bp = 7-2om., ^C = 42°.
§(iv) z.C = 116°, CA=l-8in., z.A=78°.
(v) A A = 60% ^C = 60°, AQ = 6-6cm.
(vi) /.B = 33% aC = 113°, 80=6-9 cm.
(vii) ^A = 73°, ^8 = 24°, A8 = 3-2in.
(viii) CA = 9-2 cm., z. C = 31°, z. A = 59'.
§(ix) A8=2-8in., ^A = 50°, z.8=130°.
(x) AB = 12-lcm., z.A = 27°, LB = 37°.
Ex. 91. Construct triangles to the following measurements: —
(i) BC = 10-8cm., z.A=90°, ^ C = 60°.
(ii) CA = 9-0cm., /.C = 48°, lB = 57°.
Ex. 92. Construct quadrilaterals A8CD to the following
measurements : —
(Here it is especially important that, before beginning the constmction,
a rough sketch should be made showing the given parts. ^ ^^^^
Note that the letters must be taken in order round the
quadrilateral; e.g. the quadrilateral in fig. 41 is called
ABCD and not ABDC.)
D C
fig. 41.
(i) AB=6-3cm., ^8 = 82% BC=8'2cm., lO = 90%
CD =7-7 cm.
(ii) A8 = 3-4in., BC = 2-2in., AD = 2-9 in., lA = 68%
z_B = 86°.
(iii) ^8=116°, BC = l-4in., z.C = 99°, CD = 1-9 in.,
2.D = 92°.
(iv) ^A = 67°, ^.8=113°, z.D = 46°, A8 = 6-3cm.,
AD = 8*6 cm.
(v) z. 8 = 122°, z.C = 130°, z.D = 130°, BC = CD = l-6in.
(vi) AD = 3-0 in., /l.D = 118°, z.DAC = 27°, Z.BAC = 35°,
A8 = 2-4in.
(vii) AC = 5-6 cm., z.8AC = 58°, ^DAC = 69°, z.8CA = 68°,
A. DOA = 69°.
(viii) A8 = l'9in., BD = l-7in., CD = 2-0in., iLABD = 118°,
^ BDC = 23°.
§ These will be enough exercises of this type unless much practice is needed.
TRIANGLES 23
(Lx) AP = CD = 5-8 cm., AD ==4-7 cm., L^-=12°,
z.BDC = 46°.
(x) ABi = 6-3Qm., eD,-5-4cm., ^ BAG = 64°, /.ACO=59°,
z.D=76°.
(xi) AB = 6'2cm., AG = 6-8 cm., AD=5'6cm., z.BAG=106°,
^ BAD = 122°.
(xii) z.ABD= /.ADB=50°, /. C = 68°, BC=2-3in.,
CD = 3-0 in.
(xiii) AC = irOcm., AB := 5*9 cm., BD = 7*4 cm.,
^ BAG = 22°, zlABD = 68°.
HEx. 93. Take a point O on your paper and mark a number
of points each of which is 2 in. from O. [To do this most easily,
open your dividers 2 in., place one point at O, and mark points
with the other.] The pattern you obtain is a circle; all the
points 2 in. from O are on this circle.
HEx. 94. How does ^ gardeuer mark out a circular bed ?
Ex. 95. Draw a figure to represent the area commanded
by a gun which can fire a distance of 5 miles in any direction.
(Represent 1 mUe by 1 cm.)
Ex. 96. Two forts are situated 7 miles apart; the guns in
each have a range of 5 miles ; draw a figure showing the area
in which an enemy is exposed to the fire of both forts. (Repre-
sent 1 mile by 1 cm.)
Ex. 97. A circular grass plot 70 feet in radius is watered
by a man standing at a fixed point on the edge with a hose
which can throw water a distance of 90 feet ; show the ar^a that
can be w^.tered. (Represent 10 feet by 1 cm.)
What is the distance between the two points on the edge
of the grass which the water can only just reach 1
I^Ex. 98. Mark two points A, B, 3 in. apart.
(i) On what curve do all the points lie which are 2*7 in.
from A?
24 EXPERIMENTAL GEOMETRY
(ii) On what curve do all the points lie which are 2 2 in.
from B?
(iii) la there a point which is 2*7 in. from A and also
2-2 in. from B ?
(iv) Is there more than one such point ?
Ex. 99. A and B are two points 7 '4 cm. apart; find, as in
Ex. 98, a point which is 5*7 cm. from
A and 3 "6 cm. from B. ,>if>>. - '^-^^
Ex. 100. Repeat Ex. 99, without ^.'''' %
drawing the whole circles. See fig. 42, A ' i,> 7-4 cm -v B
Ex.101, (i) Construct a triangle, ^' yi^b^
the lengths of whose sides are 12*1 cm., ~ .„
8-2 cm., 6-1 cm. See Ex. 100.
(ii) In how many points do your construction circles
intersect?
(iii) How many triangles can you construct with their
sides of the given lengths? Are these triangles congruent
(ie. could they be made to fit on one another exactly) ?
Ex. 102. Construct triangles to the following measurements : —
(It IB best to draw the longest side first.)
§(i) BC = 8-9 cm., CA = 8-3 cm., AB = 6*7 cm.
(ii) BC = 6-9 cm., CA = 11-4 cm., AB = 6-8 cm.
§(iii) BC = 6-3 cm., CA = 8*3 cm., AB = 2-5 cm.
(iv) BC = 3-9 in., CA = 2-5 in., AB = 2-5 in.
(v) BC = 3-2 in., CA = 3*2 in., AB = 1-8 in.
(vi) BC = 6-6 cm., CA = 6-6 cm., AB = 9-3 cm.
(vii) BC = 6-9 cm., CA = 6-9 cm., AB = 6-9 cm.
(viii) BC = 6-5 cm., CA = 9-6 cm., AB = 7-2 cm.
§(ix) BC = 2-1 in., CA = M in., AB = 3-2 in.
§(x) BC = 4-1 in-, CA = 4-1 in., AB = 4*1 in.
I These will be enough exercises of this type unless much practice is needed.
TRIANGLES 25
A triangle which has two of its sides equal is called an
isosceles triangle (to-os equal, o-kcXos a leg).
A triangle which has all its sides equal is called an equi-
lateral triangle (aequus equal, lattis a side).
A triangle which has no two of its sides equal is called a
scalene triangle (a-Ka\.r]v6<: lame or uneven).
HEx. 103. Which of the triangles in Ex. 102 are isosceles, and
which are equilateral 1
HEx. 104. Make a triangle of strips
of cardboard, its sides being 4 in., 5 in.,
6 in. long.
To do this, cut out strips about ^ in.
longer than the given lengths, pierce holes ^*
at the given distances apart and hinge the strips together by means of string,
or gut with knots, or by means of " eyes " such as a shoemaker uses.
Can the shape of the triangle be altered without bending or
straining the sides ?
HEx. 105. Make a quadrilateral of strips of cardboard, its
sides being 3 in., 3*5 in., 4*5 in., 6 in. long.
Can its shape be altered without bending or straining ?
Could it be made rigid by a strip joining two opposite
comers 1
The straight line joining opposite comers of a quadrilateral
is called a diagonaL
HEx. 106. Repeat Ex. 105 with a pentagon each of whose sides
is 3 in. long. How many additional strips must be put in to
make the frame- work rigid 1
Ex. 107. Construct quadrilaterals ABCD to the following
measurements : —
(i) AB = 2-3in., BC = 2-lin., CD = 3-3 in., DA =1-5 in.,
BD = 3-4 in.
(ii) AB = CD = 6-4cm., BC = DA = 3'7 cm., BD = 5'7cm.
(iii) AB = AD=l-9in., CB = CD = 2-9in., BD = 2-5in.
/
26
EXPERIMENTAL GEOMETRY
(iv) AB = BO = CD = DA = 5-1 cm., AC = 9-2 cm.
(v) AB = 3-8m., BC = l-7in., CD = 1-0 in., DA = 4-9 in.,
(vi) AB = 5'3cm., BC=6-3cm., CD = 6*7 cm., iLB = 70°,
AC =48°.
(vii) AB=2-7cm., BC=7-5cm., AD = 8-4 cm., z.C = 98°,
L DBC = 28°.
(viii) BC = CD = 2-4in., BD = l-9in., z. ABD = Z. ADB = 67°.
(ix) AB = 9-3cm., BC=DA = 6-7cm., ^A = lir, ^ B = 28°.
Ex. 108. Construct pentagons ABCDE to the following
measurements : —
(i) AB = 2-0in., BC=2-2in., CD = 1-7 in., DE==2'2in.,
EA=2-5in., ^B = lir, ^C = 149°.
(ii) AB=l-7in., BC-l'Oin., CD = 2-2 in., DE = 3-4in.,
EA = 0-5in., ^A = 126°, aB = 137°.
(iii) AB = 5cm., BC = 3-7cm., CD = 3-6 cm., DE = 4*3cm.,
EA = 3*8 cm. , AC = 6*4 cm., AD = 6*7 cm.
(iv) AB=BC = CD=DE = EA = 5-0cm., AC = BE =8-1 cm.
Pyramids. — The Tetrahedron.
Figs. 44, 45 represent a tetrahedron, i.e. a solid bounded
by four faces (tct/w- four-, ISpa a seat, a base).
fig. 44.
fig. 45.
THE TETRAHEDRON
27
HEx. 109. Make a tetrahedron of thin cardboard (or thick
paper) ; fig. 46 represents what you will have to
cut out (this will be referred to as the net of
the tetrahedron) ; each of the small triangles is
equilateral (their sides should be 4 in. long) ;
the paper is to be creased (not cut) along the
dotted lines, and the edges fastened with stamp- fig. 46.
edging.
HEx. 110. How many corners has a tetrahedron?
lIEx. 111. How many edges meet at each comer 1
^Ex. 112. What is the total number of edges ?
HEx. 113. Can you explain why the total number of edges
is not equal to the number of comers multiplied by the number
of edges at each corner ?
UEx. 114. What is the greatest number of faces you can see
at one time 1
Ex, 115. Make sketches of your model in three or four
different positions.
Figs. 47, 48 represent a square pyramid (ie. a pyramid on
a square base).
fig. 47.
?. 48.
^
28
EXPERIMENTAL GEOMETRY
HEx. 118. Make a square pyramiil
(fig. 49 represents its net) ; make each
side of the square 2 in. long and the
equal sides of each triangle 2*5 in. long.
II Ex. 117. How many comers has a
square pyramid?
HEx. 118. How many edges ?
fig. 49.
ITEx. 119. What is the greatest number of faces you can see
at one time 1
Ex. 120. Make sketches of your model in three or four
different positions.
Ex. 121. Draw the net of a regular hexagonal pyramid, and
make a rough sketch of the solid figure.
Triangles (contimied).
^Ex. 122. What is the sum of the angles of a triangle 1
HEx. 123. Out out a paper triangle;
mark its angles; tear off the comers and
fit them together with their vertices at
one point, as in fig. 50.
What relation between the angles of a
triangle is suggested by this experiment ?
HEx. 124. Cut out a paper quadrilateral and proceed as in
Ex. 123.
^Ex. 125. If two angles of a triangle are 54°, 76°, what is the
third angle ?
*OEx. 126. If two angles of a triangle are 27°, 117°, what is
the third angle ?
TRIANGLES 29
HEx. 127. If two angles of a triangle are 23°, 31°, what is the
third angle ?
HEx. 128. If two angles of a triangle are 65°, 132°, what is the
third angle?
lIEx. 129. If the angles of a triangle are all equal, what is the
number of degrees in each 1
HEx. 130. If one angle of a triangle is 36°, and the other two
angles are equal, find the other two angles.
HEx. 131. Repeat Ex. 1 30 with the given angle (i) 90°, (ii) 1 32°,
(iii) 108°.
^:^
HEx. 132. In fig. 51, triangle ABC has Z. A = 90°,
AD is drawn perpendicular to BC. If z. B = 27°,
find the angles marked x, y, z.
& i iff 8 D C
HEx. 133. Repeat Ex. 132 with (i) z. B = 54°, ^8- ^l-
(ii) /.B = 33°, (iii) ^B-45°.
HEx. 134. A triangle ABC has ^ A = 75°, z. B = 36° ; if AD is
drawn perpendicular to BC, find each angle in the figure.
HEx. 135. Would it be possible to have triangles with angles
of (i) 90°, 60°, 30°, (ii) 77°, 84°, 20°, (iii) 59°, 60°, 61°,
(iv) 135°, 22°, 22°, (v) 73°, 73°, 33°, (vi) 54°, 54°, 72°?
HEx. 136. (i) Give two sets of angles which would do for the
angles of a triangle.
(ii) Give two sets which would not do.
Ex. 137. Construct a triangle ABC, having z.A=76°, L B=54°,
BC = 2-8 in. What is z. C ?
FiZst find z. C by calculation, then constract the triangle as though
BC, L B and z. C were given.
Measure Z.A; this will be a means of testing the accuracy of your
drawing.
30 EXPERIMENTAL GEOMETRY
Ex. 138. Construct triangles to the following measurements : —
(i) BC = 8-0cm., LA =^71", iLB = 46°.
§(ii) AB = 7-3cm., ^B=^C = 57°.
(iii) LB = IW, ^C = 33°, AC = 9-4 cm.
§{iv) lC= L/K = 60% AB=2'7in.
(v) AB=4-3cm., z.A = 57°, z.C=33°.
§(vi) BC=Min., z.A=14°, lC = 52''.
HEx. 139. Draw a quadrilateral ABCD ; join AC.
(i) What is the sum of the angles of A ABC ?
(ii) „ „ „ „ „ A ADC?
(iii) „ „ „ „ „ the quadrilateral ?
ITEx. 140. If three of the angles of a quadrilateral are 110°,
60°, 80°, what is the fourth angle 1
UEx. 141. Repeat Ex. 140 with angles of (i) 75°, 105°, 76°,
(ii) 90°, 90°, 90°, (iii) 123°, 79°, 35°.
flEx. 142. If two angles of a quadrilateral are 117° and 56°,
and the other two angles are equal, what are the other two
angles ?
liEx. 143. If the four angles of a quadrilateral are all equal,
what is the number of degrees in each ?
^Ex. 144. Draw a pentagon ABCDE freehand ; join AC and AD.
What is the sum of the angles of the pentagon ?
UEx. 145. If the five angles of a pentagon are all equal^ what
is the niunber of degrees in each 1
Ex. 146. Construct a triangle ABC having BC == 6 in., CA = 5 in.,
AB = 4 in.
^ Construct a triangle A'B'c' having B'C' = 6 cm., C'A' = 5 cm,
A'B' = 4cm.
f
Measure and compare the angles of the two triangles.
S These will be enough exercises of this type unless much practice is needed.
TRIANGLES 31
Ex. 147. Construct a triangle ABC having BC = 4 in., lB = 90°,
z. C = 30°.
Construct a triangle A'B'C' having B'C' = 2 in., /. B' = 90°,
z.C'=30°.
Measure and compare the sides of the two triangles.
Ex. 148. Construct a triangle ABC having BC=9 cm., /. B = 18°,
^C = 35°.
Construct a triangle A'B'C' having B'C' = 6 cm., z.B' = 18°,
^ C' = 35°.
Measure and compare the sides of the two triangles.
Ex. 149. Draw any triangle. Without using a graduated
ruler, draw three straight lines respectively double the lengths of
the sides of the triangle ; with these three lines as sides construct
a triangle. Compare the angles of the two triangles.
HEx. 150. How many triangles of different sizes can you make
which have their angles 30°, 60° and 90° 1
Figures which are of the same shape (even though of different
sizes) are called similar figures.
HEx. 151. Which of the following pairs of figures are of
necessity similar : — (i) two circles, (ii) two right-angled triangles,
(iii) two isosceles triangles, (iv) two equilateral triatigles, (v) two
squares, (vi) two rectangles, (vii) two right-angled isosceles
triangles, (viii) two regular hexagons, (ix) two spheres, (x) two
cubes ?
HEx. 152. What is a triangle called which has two of its sides
equal ? What do you know about the angles of such a triangle 1
HEx. 153. What is a triangle called which has all its sides
equal ? What do you know about the angles of such a triangle 1
Ex. 154. Sketch a right-angled triangle (i.e. a triangle which
has one of its angles a right angle).
What kind of angles are the other two 1 Give a reason.
32
EXPERIMENTAL GEOMETRY
HEx. 155. Try to make a triangle on a base of 1*5 in. having
the angles at the ends of the base each right angles.
UEx. 156. Draw an obtuse-angled triangle freehand (i.e. a
triangle which has one of its angles obtuse).
What kind of angles are the other two ? Give a reason.
HEx- 157. Try to make a triangle on a base of 2 in. having
angles of 120°, G0° at the ends of the base.
How could you have foretold the result of your experiment ?
^Ex. 158. Sketch a tiiangle which is neither right-angled nor
obtuse-angled. "What do you note about its angles? What
would you call such a triangle?
HEx. 159. Can you draw a right-angled isosceles triangle?
What will its other angles be ?
ITElx. 160. Can you draw an obtuse-angled isosceles triangle ?
^Ex. 161. Can you draw an isosceles triangle with the equal
angles obtuse ?
HEk. 162. Which of the following combinations of angles are
possible for a triangle ?
(i) Right, acute, acute. (ii) Right, acute, obtuse,
(iii) Acute, acute, acute, (iv) Obtuse, obtuse, acute,
(v) Right, right, acute. (vi) Acute, acute, obtuse.
Ex. 163. Make a table showing in column A whether the
triangles in fig. 52, are acute-, right-, or obtuse-angled, and in
column B whether they are equilateral, isosceles, or scalene.
Triangle
numbered
A
B
1
2
3
TRIANGLES
33
So far you have only measured to one place of decimals in
inches or centimetres, but you will often need to measure more
fig. 52.
accurately. To do this you must imagine each tenth of an inch
(or centimetre) divided again into 10 equal parts.
The line AB is more than 1*2 in.
and less than 1"3 in.; "T ~ J"
if its length is almost exactly half-
fig. 53.
Avay between these measurements you will say it is 1-25 in.;
if it is a little more than half-way you will say it is 1'26 in.;
if it is about a third of the way you will say it is 1 '23 in.;
if it is about two-thirds of the way you will say it is 1*27 in.
and so on.
With a little practice you ought to get this figure nearly
accurate.
In the same way you can measure angles to within less than
a degree.
o. s. 3
34 EXPERIMENTAL QEOMETRY
HEx. 164. (i) What fraction of an inch does a figure in the
second place of decimals repi-esent ?
(ii) What fraction of an inch is '03 ?
Ex. 165. Construct triangles to the following measurements :
(All lengths should be measured to 2 decimal places and all angles to
within one-fifth of a degree*.)
(i) BC=3-18in., AB = 3-18in., ^B = 33•5^
(ii) BC = 2-39in., CA = 2-44in., aC = 63-5°.
(iii) AB = 2-82in., AC - 2-77 in., z.A = 137°.
(iv) AB = 3-00ia, lA=6V, ^B = 59°.
(v) BC- 3-52 in., z.B = 25°, ^C-23°.
(vi) AC =10-65 cm., ^ A = 54-5% z.C = 36°.
(vii) BC = 6-40cm., CA==9-05cm., AB = 7-63cm.
(viii) BC = 7-69cm., CA = 9'30 cm., ,AB = 5-30cm.
(ix) BC = 4 -53 in., CA = 2-68 in., ' AB = 2-02 in.
(x) AB=2-71in., z.B = 55-5°, ^C = 67-5°.
(xi) ^ A - 24°, /. C = 47 -5°, BC = 3-04 cm.
(xii) ^A = 133°, BC = 10-73 cm., ^ B = 23-5°.
(xiii) ^C=90°, BC =1-00 in., CA=2-00in.
(xiv) BC=^4-09cm., CA = 3-31cm., AB = 7'54cm.
(xv) ^ A = 90-5°, ^ B = 78°, BC = 3-54 in.
(xvi) AB = 2-99in., ^B = 127-5°, z.C = 53-5°.
(xvii) AB = 2-92in., ^B-59°, AC = 2-39 in.
(xviii) ^ B = 33-5°, BC = 2 -61 in., CA = 1 -54 in.
(xix) CB = 2-16in., CA=2-64in., /.B = 64-5°.
(xx) Z- A = 24°, AB - 7-76 cm., BC ^ 2-87 cm.
" This can be done with a well graduated protractor of 2-inch radius,
with a smaller protractor it is difficult.
TRIANGLES — PARALLELS 35
Parallels and Perpendiculars.
HEx. 166. Give instances of parallel straight lines {e.g. the
flooring boards of a room, the edges of your paper).
UEx. 167. Draw with your ruler two straight lines
as nearly parallel as you can judge ; draw a straight
line cutting them as in fig. 54 ; measure the angles
marked. These are called corresponding angles.
Are they equal 1
UEx. 168. Repeat Ex. 167 two or three times drawing the
cutting line in different directions.
HEx. 169. Draw two straight lines which are not parallel
and proceed as in Ex. 167. Are the angles equal?
HEx. 170. Draw a straight line AB (see fig. 55).
In AB take a point C; through C draw CD making
2. BCD = 90° (use your set square); through A ^
draw AE making lBAE = 90°. Are AE and CD B
parallel? fig- 55.
HEx. 171. In the figure you obtained in the last Ex. draw two
more straight lines at right angles to CD ; measure the part of
each of these three straight lines cut off between AE and CD;
are these parts equal ?
"Would these three parts be equal if the lines aU made
different angles with CD ?
^Ex. 172. Repeat Ex. 170 with z. BCD = /. BAE ^- 60° (use
your set square) ; draw three straight lines at right angles to CD ;
measure the parts cut off between AE and CD.
^lEx. 173. Repeat Ex. 170 with z. BCD =z. BAE = 30° (use
your set square) ; measure as in Ex. 172.
3—2
36
EXPERIMENTAL GEOMETRY
In the course of Ex. 166-173, you should have observed
the following properties of parallel straight lines : —
(i) they do not meet however far they are produced in
either direction.
(ii) if a straight line cuts them, corresponding angles are
equal.
(iii) parallel straight lines are everywhere equidistant.
To draw a parallel to a given line QR through a given
point P by means of a set square and a straight edge.
It is important that the straight edge should not be bevelled (if it is
bevelled the set square will slip over it) ; in the figures below a ruler with
an unbevelled edge is represented, but the base of the protractor or the edge
of another set square will do equally welL
Place a set square so that one of its edges lies along the
given line QR (as at (i)) ; hold
it in that position and place the
straight (unbevelled) edge in
contact with it ; now hold the
straight edge firmly and slide
the set square along it. The
edge which originally lay along
QR wiU always be parallel to
QR. Slide the set square till
this edge passes through P (as fig, 5(5,
at (ii)), hold it firmly and rule the line.
This method of drawing parallels suggests an explanation of
the term corresponding angles.
Ex. 174. Draw a straight line QR and mark a point P;
through P draw a parallel to QR.
Ex. 175. Repeat Ex. 174 several times using the different
edges of the set square. (See fig. 57, and Ex. 170.)
PARALLELS AND PERPENDICULARS
37
Ex. 176. Near the middle of your paper draw an equilateral
triangle with its sides 1 in. long ; through each vertex draw a
line parallel to the opposite side.
If the angle between two straight lines is a right angle the
straight lines are said to be at right angles to one another or
perpendicular to one another.
To draw through a given point P a straight line
perpendicular to a given straight line qr.
The difficulty of drawing a line right to the comer of a set
square can be overcome as follows : —
Place a set square so that one of the edges containing
the right angle lies along
the given line QR (as at (i));
place the straight edge in
contact with the side op-
posite the right angle; now
hold the straight edge firmly
and slide the set square
along it; the edge which
lay along QR will always be
parallel to QR and the other edge containing the right angle will
always be perpendicular to QR. SHde the set square till this
other edge passes through P ; then draw the perpendicular.
Ex. 177. Through a given point in a straight line draw a
perpendicular to that line.
Ex. 178. Draw an acute-angled triangle; from each vertex
draw a perpendicular to the opposite side.
Ex. 179. Repeat Ex. 178 with an obtuse-angled triangle.
(You will find it necessary to produce two of the sides.)
Ex. 180. Describe a circle, take any two points A, B upon it,
join AB; from the centre draw a perpendicular to AB; measure
the two parts of AB.
fig. 67.
38 EXPERIMENTAL GEOMETRY
Ex. 181. DraAV an acute-angled triangle ; from the middle
point of each side draw a straight line at right angles trt that,
side.
Kx. 182. Hej)eat Ex. 181 with an obtuse-angled triangle.
Parallklogram, Rectangle, Square, Rhombus.
Ex. 183. Make an angle ABC = 65°, cut off BA = 2-2 in.,
EC = 1*8 in. 5 through A draw AD parallel to EC, through C draw
CD parallel to BA.
A four-sided figure with its opposite sides parallel is called a
parallelogranL
Ex. 184. Make a parallelogram two of whose adjacent sides
(i.e. sides next to one another) are 6*3 cm. and 5*1 cm., the angle
between them being 34°.
Measure the other sides and angles.
Ex. 185. Repeat Ex. 184 with the following measurements:
10-4 cm., 2-6 cm., 116°.
Ex. 186. Repeat Ex. 184 with the following measurements:
10-4 cm., 2-6 cm., 64°.
HEx. 187. Draw a parallelogram two of whose sides are 3*7 in.,
and 0*8 in., and One of whose angles is 168°.
Are its opposite sides and angles equal 1
It will be proved later on tliat the opposite sides and angles
of a parallelogram are always equal.
liEx. 188. Construct a quadrilateral AECD having AB = CD =
4-7 cm., AD = BC = 7'2 cm., and ^ A — 85°. Is it a parallelogram 1
HEx. 189. Make a parallelogram of strips of cardboard, one
pair of sides being 5 in. long and the other pair 3 in.
PARALLELOGRAMS 39
HEx. 190. Open one of the acute angles of the framework yon
Jiave just made until it is a right angle ; examine the other angles.
A parallelogram which has one of its angles a right angle is
called a rectangle.
Ex. 191. Draw a rectangle having sides =7'3cm. and 3*7 cm.
Measure all its angles.
Ex. 192. Draw a parallelogram having sides = 9-2 cm. and
4-3 cm., and one angle = 125°. Draw its diagonals, and measure
their parts.
Ex. 193. Repeat the last Ex. Avith the following measure-
ments, 8 "6 cm., 6*8 cm., 68° ; test any facts you noted in that Ex.
Ex. 194. Draw a parallelogram and measure the angles
between its diagonals ; are any of them equal ? Give a reason.
Ex. 195. Draw a rectangle having sides =3*5 in. and 2-3 in.
Measure its diagonals.
Ex. 196. Repeat the last Ex. with the following measure-
ments, (i) 8-6 cm., 11-2 cm., (ii) 14'3 cm., 2*8 cm.
A rectangle which has two adjacent sides equal is called a
square.
Ex. 197. Draw a square having one side = 5*6 cm. Measure
all its sides and angles.
Ex. 198. Draw a square having each side = 3-2 in. Measure
its diagonals and the angles between them.
HEx. 199. Explain how you would test by folding whether
a pocket handkerchief is square.
HEx. 200. Make a paper square by folding.
A parallelogram which has two adjacent sides equal is called
a rhombus.
40 EXPERIMENTAL GEOMETllY
Ex. 201. Draw a rhombus having one side = 2-2 in. and one
angle = 54°. Measure the sides, angles, diagonals, and the angles
between the diagonals.
Ex. 202. Repeat Ex. 201 making one side = 6-8 cm. and one
angle = 105°.
In the course of Ex. 183-202, you should have observed
the following properties : — *
(i) The opposite sides and angles of a parallelogram
are equal.
(ii) The diagonals of a parallelogram bisect one another.
The above properties, (i) and (ii), must be true for a rectangle,
square, and rhombus, since these are particular cases of a
parallelogram (i.e. special kinds of parallelogram).
(iii) All the angles of a rectangle are right angles.
(iv) The diagonals of a rectangle are equal.
(iii) and (iv) must be true for a square, since a square is a
particular case of a rectangle.
(v) The diagonals of a square intersect at right angles.
(vi) The diagonals of a rhombus intersect at right angles.
Since a square may be regarded as a particular case of a
rhombus, (v) might have been deduced from (vi).
Ex. 203. Copy the table given below ; indicate for which
figures the given properties are always true by inserting the
words " yes " or " no " in the corresponding spaces.
PARALLELOGRAMS
41
•S to
a O
.2 s
m
u
2 IB
a <i>
-*3 CO
a
<
"3
OQ
a
ft
i
s
1
--a
Parallelogram
Rectangle
Square
Rhombus
A square inch is a square whose sides are one inch long.
' Ex. 204. Draw a square inch, and measure its diagonals.
Ex. 205. Draw a square ABCD having its sides 3 in. long;
divide AB and EC into inches and through the points of division
draw parallels to the sides of the square. Into how many square
inches is ABCD divided?
Ex. 206. Repeat Ex. 205 with a square 5 in. long.
Ex. 207. Draw a square having its sides 6 cm. long ; divide
it up into square centimetres ; how many are there ?
Ex. 208. Describe a rectangle ABCD having AB=5 in.,
BC = 4 in. ; divide it up into square inches ; how many are there 1
Ex. 209. Describe a rectangle 6 cm. by 3 cm.; divide it up
into square centimetres ; how many are there 1
42
EXPERIMENTAL GEOMETRY
CuBB, Cuboid, and Prism.
Figs. 58, 59 represent a cube (i.e. a .solid l)OU)ule<i by six
equal squaras).
fig. 58.
JL-.
fig. 69.
HEx. 210. Make a cube of thin cardboard; its net is given in
fig. 60 (see Ex. 109); each edge should
be 2 in. long.
HEx. 211. How many corners has a
cube?
UEx. 212.
cube?
How many edges has a
HEx. 213. How many edges meet at
each comer?
Hg. 60.
UEx. 214. Is the number of edges equal to the number of
comers multiplied by the number of edges which me-et at each
comer ? Give a reason.
UEx. 215. How many edges has each face ?
UEx. 216. Is the number of edges equal to the number of faces
multiplied by the number of edges belonging to each face ? Give
a i-eason.
CUBE AND CUBOID
43
flEx. 217. Is tlie number of angles equal to the number of
faces multiplied by the numljer of angles belonging to each facte?
Give a reason.
HEx. 218. What is the greatest nuriiljer of faces, edges, and
comers you can see at one time 1
Ex. 219. Make sketches of a cube from three different points
of view.
Figs. 61, 62 represent a cuboid or rectangular block (i.e.
a solid like a brick).
fig. 61. fig. 62.
HEx. 220. How does a cuboid differ from a cube ?
fig. GJ.
11 Ex. 221. Make a cuboid of thin caitlboard ; its net is given
in fig. 63 ; it should measure 3 in. by 1-9 in. by 1*3 in.
HEx. 222. Choose one edge of the cuboid ; how many other
edges are equal to this edge?
44 EXPERIMENTAL GEOMETRY
Kgs, 64, 65 represent a regular three-sided prism.
fig. G4. fig, 6:.
HEx. 223. What sort of figures are the ends of the prism in
fig. 64 ? What are the sides ?
HEx. 224. Make a regular three-sided prism ; its net is given
in fig. 66 ; the short lines should
each be 2 in. long, and the long
ones 3*5 in.
11 Ex. 225. How many edges has
a three-sided prism? How many
faces? How many corners ?
HEx. 226. What is the greatest
number of faces, edges, and comers
you can see at one time ?
ITEx. 227. Make sketches of your
model from three difierent points of view.
UEx. 228. Draw the net of a three-sided prism whose ends are
triangles with sides 3 in., 3 in., 1 in. and whose length is 1*5 in.
Such a prism is often called a wedge.
Figs. 67, 68 represent a regular hexagonal prism.
fig. 66.
fig. 67.
fig. 68.
PRISMS — DRAWING TO SCALE 45
HEx. 229. What sort of figures are the ends of the prism in
fig. 67 ? What are the sides ?
HEx. 230. Draw its net.
HEx. 231. What is the number of edges, faces, and comers t
HEx. 232. What is the greatest number of edges, faces, and
corners you can see at one time ?
HEx. 233. Make sketches of a regular hexagonal prism from
three different points of view.
Drawing to Scale.
When drawing a map, or plan, to scale you should always begin
by making a rough sketch showing the given dimensions, and then
work from the sketch.
The bearing of a place A from a second place B is the point
of the compass towards which a man at B would be facing if he
were looking in the direction of A.
By "N. 10° W." or " 10° W. of K" is meant the direction in
which you would be looking if you first faced due north and then
turned through an angle of 10° towards the west.
Ex. 234. A is 2-5 miles W. of B, and C is 4*5 miles S. of A.
What is the distance from B to C? Wh^t is the bearing of B
from C, and of C from B ? (Scale 1 mOe to 1 inch.)
Ex. 235. G is 7-5 miles S. of H, and 10 miles W. of K.
What is the distance and bearing of K from H ? (Scale 1 mile
to 1 cm.)
Ex. 236. X is 17-5 mUes KW. of Y, Y is 23 mUes N.E. of Z.
What is the distance and bearing of X from Z ? (Scale 10 miles
to 1 inch.)
Ex. 237. P is 64 miles W. of Q, R is due N. of Q; if PR is
72 miles, what is QR 1 What is the bearing of P from R 1 (Scale
10 mUes to 1 cm.)
46 EXPERIMENTAL GEOMETRY
Ex. 238. Draw a plan of a room 30 ft. by 22 ft. ; find the
distances between opposite corners. (Scale 2 ft. to 1 cm.)
Ex. 239. Exeter is 4& miles W. of Dorcliester, and Barnstaple
is 35 miles N.W. of Exeter. What is the distance and bearing
of Barnstaple from Dorchester? (Scale 10 miles to 1 in.)
Ex. 240. Rugby is 44 miles N. of Oxford, and Reading is
24 miles S. 30° E. of Oxford. Find the distance from Rugby to
Reading. (Scale 10 miles to 1 in.)
Ex. 241. Southampton is 72 miles S. 53° W. of London,
Gloucester is 75° W. of N. from London, and •19"' AV. of N. from
Southampton. Find the distance between Southampton and
Gloucester. (Scale 10 miles to 1 cm.)
In the following exercises, use any suitable scale; always state what
scale yon use.
Ex. 242. Draw a plan of a rectangular field 380 yards by
270 yards. AVTiat is the distance between the opposite corners ?
Ex. 243. The legs of a pair of compasses are 10 cm. long.
I open them to an angle of 35°. What is the distance between
the compass points ?
Ex. 244. Two blockhouses are known to be 1000 yards apart,
and one of them is due E. of the other. A party of the enemy
are observed by one blockhouse in a N.W. direction, and at the
same time by the other in a N.E. direction. How far are the
enemy from each blockhouse 1
Ex. 245. A and B are two buoys 800 yards apart, B due N.
of A. A vessel passes close to B, and steering due E., observes
that after 5 minutes the bearing of A is 57* W. of S. ¥ind the
distance the vessel has moved.
Ex. 246. Stafford is 27 miles from Derby and the same
distance from Shrewsbury, and the three towns are in a straight
line. Birmingham is 40 miles from Shrewsbury and 36 from
Derby. How far is Stafford from Birmingham?
DRAWING TO SCAIJE 47
Ex. 247. A buoy is moored by a cable 55 feet long ; at low
tide the distance between the extreme positions the buoy can
occupy is 100 feet. What Avill be the distance between the
extreme positions when the water is 24 feet higher?
Ex. 248. Two ships sail from a port, one due N. at 15 miles
an hour, the other E.N.E. ; at the end of half an hour they are in
line with a lighthouse which is 11 miles due E, of the port.
At what rate does the second ship sail?
Ex. 249. A donkey is tethered to a point 20 feet from a
long straight hedge ; he can reach a distance of 35 feet from the
point to which he is tethered. How much of the hedge can he
nibble?
Ex. 250. A is a lighthouse. B and C are two ships 3 "5 miles
apart. B is due north of A, C due east of B, and C north-
east of A. Find the distance of both ships from the light-
house.
Ex. 251. A man standing on the bank of a river sees a tree
on the far bank in a direction 20° W. of N. He walks 200 yards
along the bank and finds that its direction is now N.E. If the
river flows east and west, find its breadth.
Ex. 252. A ferry-boat is moored by a rope 30 yards long to
a point in the middle of a river. The rope is kept taut by the
current. What angle does it turn through as the boat crosses
the river, whose width is 30 yards 1
Ex. 253. The case of a grandfather clock is 16 inches wide ;
the pendulum is hung in the middle of the case and its length is
39 inches. Assuming that the end of the pendulum swings to
within 3 inches of each side of the case, find the angle through
which it swings.
Ex. 254. Brixham is 4*6 miles N.E of Dartmouth, Torquay
is 4 miles N. of Brixham, Totnes is 7 '4 miles S. 75° W. of
Torquay; what is the distance and bearing of Totnes from
Dartmouth ?
48 EXPERIMENTAL GEOMETRY
Ex. 255. From Q go 9 miles W. to H, from H go 12 miles K
to A, from A go 17 miles W, to R. What is the distance from Q
to Rl
Ex. 256. A is 12 miles K of H, D is 24 miles S. of H, O is
due W. of A and OH is 42 ; find OD and OA,
Ex. 257. XT = 19 miles, MX = 1 1 miles, MT = 17-5 miles; how
far is M from the line XT 1
Heights and Distancbs*.
If a man who is looking at a tower through a telescope holds
the telescope horizontally, and then raises (or " elevates") the end
of it till he is looking at the top of the tower, the angle he has
turned the telescope through is called the angle of elevation of
the top of the tower.
If a man standing on the edge of a cliff looks through a
horizontal telescope and then lowers (or " depresses ") the end of
it till he is looking straight at a boat, the angle he has turned
the telescope through is called the angle of depression of the
boat.
Bcmember that the angle of elevation and the angle of depression are
always angles at the observer's eye.
If O is an observer and A and P two points (see fig. 14), the
angle AOP is said to be the angle subtended at O by AP.
Ex. 258. In fig. 51, name the angles subtended (i) by BD at
A, (ii) by AD at B, (iii) by AC at B.
Ex. 259. A vertical flagstaff 50 feet liigh stands on a hori-
zontal plane. Find the angles of elevation of the top and middle
point of the flagstaff from a point on the horizontal plane 1 5 feet
from the foot of the flagstaff.
Ex. 260. The angle of elevation of the top of the spire of
Salisbury Cathedral at a point 1410 feet from its base was found
to be 16°. What is the height of the spire?
* For further exercises on heights and distances see p. 59.
HEIGHTS AND DISTANCES 49
Ex. 261. A torpedo lx)at passes at a distance of 100 yards
from a fort the guns of which are 100 feet above sea-level; to
what angle should the guns be depressed so that they may point
straight at the torpedo boat ?
Ex. 262. From the top of Snowdon the Menai Bridge can be
seen, the angle of depression being 4°. The height of Snowdon
is 3560 feet. How far away is the Menai Bridge?
Ex. 263. From a point A the top of a church tower is just
visible over the roof of a house 50 feet high. If the distance
from A to the foot of the tower is known to be 160 yards, and
from A to the foot of the house 60 yards, find in feet the height
of the tower, and the angle of elevation of its top as seen from A.
Ex. 264. A flagstaff stands on the top of a tower. At a
distance of 40 feet from the base of the tower, the angle of
elevation of the top of the tower is found to 'be 23|°, and the
flagstaff subtends an angle of 25|°, Find the length of the
flagstaff and the height of the tower.
Ex. 265. At two points on opposite sides of a poplar the
angles of elevation of its top are 39° and 48°. If the distance
between the points is 1 50 feet, what is the height of the tree ?
Ex. 266. From the top of a mast 80 feet high the angle of
depression of a buoy is 24°. From the deck it is 5|°. Find the
distance of the buoy from the ship.
Ex. 267. At a window 15 feet from the ground a flagstaff
subtends an angle of 43° ; if the angle of depression of its foot is
11°, find its height.
Ex. 268. A man observes the angle of elevation of the top of
a spire to be 23°; he walks 40 yards towards it and then finds
the angle to be 29°. What is the height of the spire 1
Ex. 269. An observer in a balloon, one mile high, observes
the angle of depression of a church to be 35°. After ascending
vertically for 20 minutes, he observes the angle of depression to
be now 55^°. Find the rate of ascent in miles per hour.
a s. 4
60 EXPERIMENTAL GEOMEl'RY
Kx. 270. An observer finds that the lino joining two forts A
and B subtends a right angle at a point C ; from C he walks 100
yards towards B and finds that AB now subtends an angle of 107°;
find the distance of A from the two points of observation,
Ex. 271. A man on the top of a hill sees a level road in the
valley running straight away from him. lie notices two con-
secutive mile-stones on the road, and finds their angles of depression
to be 30° and 13° respectively. Find the height of the hill (i) as
a decimal of a mile, (ii) in feet.
How TO COPY A GIVEN RBCTILINEAR FIGURE.
A rectilinear figure is a figure made up of straight lines.
An exact copy of a given rectilinear figure may be made in
various ways.
Isi method. Suppose that it is required to copy a pentagon
ABCDE (as in fig. 69). First copy side AB; then /.ABC; then
side BC ; then l BCD ; etc. You will not find it necessary to
copy all the sides and angles.
Ex. 372. Draw a good-sized quadrilateral; copy it by Method L If you
have tracing paper, make the copy on this ; then see if it fits the original,
Ex. 373. Repeat Ex. 272, -with an (irregular) pentagon.
2nd method. A simpler way is to prick holes through the
different vertices of the given figure on to a sheet of paper
below ; then join the holes on the second sheet by means of
straight lines.
Srd inetJiod. Place a sheet of tracing paper over the given
figure, and mark on the tracing paper the positions of the
different vertices. Then join up with straight lines.
Uh metJwd — by intersecting arcs.
To copy ABCDE by this method (see fig. 69). Make A'b' = AB.
With centre A' and radius equal to AC describe an arc of
a circle.
With centre B' and radius equal to BC describe an arc of a
circle.
Let these arcs intersect at C'. Then C' is the copy of C.
COPYINa FIGURES — SYMMETRY
51
Similarly, fix D' by means of the distances A'D' and B'D' ; fix
E* by "means of the distances A'e' and B'E'.
D
fig. 69.
The five vertices A'B'C'D'E' are now fixed, and the copy may
be completed by joining up.
In Ex. 274 — 276 the copies should be made on tracing paper if possible ;
the copies can then be fitted on to the originals.
Ex. 274. Draw, and copy (i) a quadrilateral, (ii) a pentagon, by the
method of intersecting arcs. If tracing paper is not used, the copy may
be checked by comparing its angles vnth. those of the original.
Ex. 276. By intersecting arcs, copy figs. 45 and 48.
Ex. 276. By intersecting arcs, copy the part of fig. 52 which consists
of triangles 1, 2, 3, 4, 6, 6.
Symmetry,
HEx. 277. Eold a piece of paper once; cut the folded sheet
into any pattern you please;
then open it out (see fig. 70).
The figure you obtain is
said to be. symmetrical about
the line of folding. This line
is called an axis of symmetry.
fig. 70.
4—2
62
EXPERIMENTAL GEOMETRY
^iEx. 278. Make sketches of the syiuinetrical figures pro-
(iuced when the folded sheet is cut into the following shapes.
((Jive names if possible.)
(i) a rt. l'^ A with its shortest side along the crease,
(ii) an isosceles A with its base along the crease,
(iii) a scalene A with its longest side along the crease,
(iv) an obtuse z. •* A with its shortest side along the crease,
(v) a semi-circle with its diameter along the crease,
(vi) a rectangle with one side along the creasa
(vii) a parallelogram with one side along the crease,
HEx. 279. Which of the following figures possess an axis
of symmetry t (You may find that in some cases there is more
than one axis.) In each case make a sketch showing the axis
(or axes), if there is symmetry, (i) isosceles A , (ii) equilateral A ,
(iii) square, (iv) rectangle, (v) parallelogram, (vi) rhombus,
(vii) regular 5-gon, (viii) regular 6-gon, (ix) circle, (x) a semi-
circle, (xi) a figure consisting of 2 unequal circles, (xii) a figure
consisting of 2 equal circles.
UEx. 280. Fold a piece of paper twice (as in Ex. 30), so that
the two creases are at right z. s ;
cut the folded sheet into any shape,
being careful to cut away all of
the original edge of the paper. On
opening the paper you will find
that you have made a figure with
two axes of symmetry at right
angles. fig. 71.
^Ebc. 281. Cut out a paper parallelogram (be careful not
to make it a rhombus). Fold it about a diagonal ; do the two
halves fit?
You will notice that the parallelogram has no axis of
symmetry. Yet it certainly has symmetry of some kind.
The nature of this symmetry will be made clear by the follow-
ing exercise.
SYMMETRY 53
^Ex. 282. Draw a parallelogram. Through O, the intersection
of the diagonals, draw a number of straight lines, meeting the
boundary of the parallelogram.
Suppose that one of these lines meets the boundary in P and
P'. Notice that PP' is bisected at O. This is the case for each
of the lines. In fact, every straight line drawn through O to
meet the boundary in two points is bisected at O.
The parallelogram is therefore said to be symmetrical
about the point O. O is called the centre of symmetry.
HEx. 283. Which of the figures in Ex. 279 are symmetrical
about a centre 1
HEx. 284. Fasten a sheet of paper to the desk (or to a drawing
board), and on it draw a parallelogram. Drive a pin through
the centre of the parallelogram into the desk. With a knife,
cut out the parallelogram. When it is cut free from the sheet
of paper, turn it round the pin and see if you can bring it into
a position where it exactly fits the hole from which it was cut ;
what angle must it be turned through to fit in this manner 1
HEx. 285. Has figure 71 (2) central symmetry ?
HEx. 286. Describe the symmetry of the following capital
letters : —
A, C, H, I, O, S, X, Z.
Solids may have symmetry. The human body is more or less
symmetrical about a plane. Consider the reflexion in a mirror
ot the interior of a room. The objects in the room together
with their reflexions form a symmetrical whole ; the surface of
the mirror is the plane of symmetry.
HEx. 287. Give 4 instances of solids possessing planes of
symmetry.
HEx. 288. Fold a sheet of paper once. Prick a number of
lioles through the double paper, forming any pattern. On opening
the paper you will find that the pin-holes have marked out a
symmetrical figure.
54 EXPERIMENTAL GEOMETRY
Join corresponding points as in fig. 72. Notice that when
p
N
P'
fig. 72.
the figure was folded NP' fitted on to NP. This shows that
NP' = NP.
The line joining any pair of corresponding points,
in a figure which is symmetrical about an axis, is bisected
by and perpendicular to the axis of symmetry.
^Ex. 289. If a point P lies on the axis of symmetry, where is
the corresponding point P' 1
^Ex. 290. Draw freehand any curve (such as APB in fig. 73) ;
and rule a straight line XY. Mark a number of points on the
A
P
/
y
/'
\
\
\
\
V
B
X
A'
"•-
N
■■•
"■"
._
''
/
> '"
B'
z. 73.
POINTS, LINES, SURFACES, SOLIDS 56
curve; draw perpendiculars to the line (e.g. PN); produce to
an equal distance below the line (e.g. NP'= PN). Draw a curve,
freehand, through the points thus obtained.
HEx- 291. What points would you describe as "corresponding"
in the case of a figure with a centre, but no axis of symmetry ?
UEx, 292. By a method similar to that of Ex. 290 construct
a ciurve symmetrical about a centre.
Points, Lines, Surfaces, Solids.
This shcndd be taken viva voce; the definitions a/re not intended
to he learnt.
In Ex, 109, 116, 210, 221, 224 you have made some solids.
The term does not refer to the stuff of which the solids are made,
but to the space occupied — geometry deals with size and shape,
and not with material, colour, hardness, temperature, &c.
Any body, such as a brick, a sheet of cardboard or paper,
a planet, a drop of water, the water of a lake, the air inside a
football, the flapae of a candle, a smoke-ring, is called a solid in
the geometrical sense of the word.
UEx. 293. Has a brick any length ? Has it any breadth ? Has
it any thickness 1
A solid is bounded by one or more surfaces.
UEx. 294. Which of the solids mentioned above is bounded by
one surface only?
UEx. 295. A bottle is filled partly with water and partly with
oil ; the water and oil do not mix ; the boundary between them
is neither water nor oil, it is not a body but a surface. Has it
any thickness 1
UEx, 296. Consider the boundary between the water of a calm
lake and the air. Is it water or is it air 1 Has it any thickness 1
Has it any length ? Has it any breadth ?
56 EXPERIMENTAL GEOMETRY
IfEx, 297. Suppose the end of the lake is formed by a wall
built up out of the water ; what would you call the boundary
which separates the wall from the air and water ? Has it any
thickness t Has it any length 1 Has it any breadth 1
A surface has length and breadth, but no thickness.
ITEx. 298. Part of the surface of the wall is wet and part dry ;
is the boundary between these two parts wet or dry? Has it
any thickness 1 Has it any length ? Has it any breadth 1
This boundary is really the intersection (or cutting place) of
the air-water surface and the wall surface.
The intersection of two surfaces is a line. A Jine has length
but no breadth or thickness.
We cannot represent a line on paper except by a mark of some breadth ;
but, in order that a mark may be a good representation of a line, it ehoald
be made as narrow as possible.
HEx. 299. Take a model of a cube ; what are its edges 1 Have
they any length, breadth, or thickness 1
^iEx. 300. If you painted part of your paper black, would the
boundary between the black and the white have -any width?
HEx. 301. If part of the wall in Ex. 297 were painted red and
the rest painted black, would the boundary between the two
parts be red or black?
HEx. 302. Suppose that the red and black paint were continued
below the water as well as above, the line bounding the red and
black would be partly wet and partly dry; has the boundary
between the wet and dry parts of this line any length ?
The intersection of two lines is a point. A point has neither
length, breadth, nor thickness, but it has position.
We cannot represent a point on paper except by a mark of some bvac ;
the best way to mark a point is to draw two fine lines through the point.
We have now considered in turn a solid, a surface, a Hue,
and a point. We can also consider them in the reverse order.
POINTS, LINES, SURFACES, SOLIDS 57
A point has position but no magnitude.
If a point moves, its path is a line (it Ls said to generate
a line).
A pencil point when moved over a sheet of paper leaves a streak behind,
showing the hne it has generated (of coarse it is not really a line because it
has some thickness).
If a line moves, as a rule it generates a surface.
A piece of chalk when laid flat on the blackboard and moved sideways
leaves a whitened surface behind it. Consider what would have happened
if it had moved along its length.
If a surface moves, as a rule 'it generates a solid.
The rising surface of water in a dock generates a (geometrical) soUd.
HEx. 303. Does a flat piece of paper moved along a flat desk
generate a solid ?
A straight line cannot be defined satisfactorily in a simple
way; the idea of a straight line however is familiar to everyone.
lIEx. 304. How can you roughly test the straightness of (i) a
billiard cue, (ii) a railway tunnel, (iii) a metal tube 1
HEx. 305. How does a gardener obtain a straight line ?
fig. 74.
liEx. 306. Test whether the two thick lines in fig. 74 aro
straight.
68 EXPERIMENTAL QEOMETTRT
Make a careful tracing of one line ; move the tracing along and Bee if it
can be made to fit on the line everywhere else ; turn the tracing over and
try again. If it is impossible to find a position in which they do not fit
on one Einotber, then the line must be straight.
The above assumes that the paper is plane.
UEx. 307. Test the straightness of the lines in fig. 74 by
means of a stretched thread.
Ex. 304-7 lead us to a concliision which may be stated in
various ways as follows : —
(i) Two straight lines cannot enclose a space.
(ii) Two straight lines cannot intersect in more than one
point.
(iii) If two straight lines have two points in common, they
must coincide.
(iv) One straight line, and one only, can he drawn through
ttoo given points.
(v) Ttoo points determine a straight line.
A surface which is such that the straight line joining every
pair of points in it lies wholly in the surface is called a plane
surface, or, briefly, a plane.
^Ex. 308. Push a straight knitting needle through an apple ;
does the straight line joining the two points where the needle
cuts the surface lie wholly in the surface of the apple ?
HEx. 309. Test whether the surface of your desk is plana
Place a straight edge {e.g. the edge of your ruler or set square) against
the surface and see if the straight edge touches the surface all along its
length ; if it does so in all positions, the surface is plane.
UEx. 310. Is the lid of your instrument box plane?
UEx. 311. Is the glass of your watch plane?
U Ex. 312. Are the faces of your cuboid plane ?
STRAIGHT LINE AND PLANE 59
HEx. 313. Could you find two points on the surface of a garden
roller such that the straight line joining them lies wholly in the
surface ? Is the surface plane ?
Parallel straight lines are defined to be straight lines in the
same plane which do not meet however far they are produced in
either direction.
HEx. 314. Can you explain why the words in italics are
necessary ?
UEx. 315. A five-barred gate is half-open ; there is one of the
gate-posts which the line of the top bar does not meet ; is the
top bar parallel to this post?
UEx. 316. Give instances of pairs of straight lines which are
not parallel but do not meet however far they are produced.
f Ex. 317. Would a set of telegraph poles along the side of
a straight road be parallel to one another? Would they be
parallel if the road were crooked?
UEx. 318. Are the upright edges of a box parallel ?
Heights and Distances (Continued from p. 50).
Ex. 318a. The shadow of a tree is 30 feet long when the
sun's altitude is 59° ; find, by drawing, the height of the tree,
taking a scale of 1 inch to 10 feet.
Ex. 3186. A telegraph pole standing upright on level ground
is 23 '6 feet high and is partly supported by a wire attached
to the top of the pole at one end and fixed to the ground at
the other so that its inclination to the pole is 54° 22'.
Find the length of the wire.
Ex. 318 c. The angle of elevation of the top of a tower
on level ground is read off on a theodolite. Find the height of
the tower from the following data : '
Reading of theodolite = 15°.
-Height of theodolite telescope above ground = 3' 6".
Distance of theodolite from foot of tower = 372 yards.
60 EXPERIMENTAL GEOMETRY
Ex. 318 d. From a ship at sea the top of Aconcagua has an
angle of elevation of 18". The ship moves out to sea a distance
of 5 nautical miles further away from the mountain. The angle
of elevation of the top of the mountain is now 13". Find the
height of Aconcagua above sea level in feet. (1 nautical
mile = 6080 ft.)
Ex. 318 «. Two observations are made to find the height
of a certain monument. From the first station the angle of
elevation of the top is found to be 32° and from the second
station, which is 27 yards from the first and exactly between
it and the foot of the monument, the angle of elevation is 43°.
If the telescope of the theodolite with which these observations
are made is 3 feet above the ground, what is the height of the
monument in feet ?
Ex. 318/ Wishing to find the height of a cliff I fix two
marks A and B on the same level in line with the foot of the cliff.
From A the angle of elevation of the top of the cliff is 37° and
from B the angle of elevation is 23° 30'. If A and B are 120 feet
apart, calculate the height of the cliff.
Ex. 318^. From a point on a battleship 30 ft. above the
water, a Torpedo Boat Destroyer is observed steaming away
in a straight line. The angle of depression of the bow is
observed to be 11°, and that of the stern to be 21*. Find the
length of the T. B. D.
Ex. 318 A. From the top of a mast 70 feet high, two buoys
are observed due N. at angles of depression 57° and 37° j find
the distance between the buoys to the nearest foot.
Ex. 318t. The angles of depression of two boats in a line
with the foot of a cliff are 25° 16' and 38° 39' as observed by a
man at the top of the eliff. If the man is 250 feet above sea-
level, find how far apart the boats are.
Ex. 318 ^. A torpedo boat is steering N. 14° E., and from
the torpedo boat a lighthouse is observed lying due N. If the
HEIGHTS AND DISTANCES 61
speed of the vessel is 15 knots and it passes the lighthouse
40 minutes after the time of observation, find the clearance
between the vessel and the lighthouse, and its distance from
the lighthouse at the first observation.
Ex. 318^. A landmark bears N. 32" W. from a ship. After
the ship has sailed 7 "2 miles N. 22° E. the landmark is observed
to bear N. 71° W. How far is it then from the ship 1
Ex. 318 m. The position of an inaccessible point C is required.
From A and B, the ends of a base line 200 yards long, the
following bearings are taken :
-, . fBearing of B is N". 70° 30' E.
^^'^"^M „ „CisK30°20'E.
FromB „ „ CisK 59°40'W.
Find the distances of C from A and B.
Ex. 318n. A ship observes a light bearing N. 52° E. at a
distance of 5 miles. She then steams due S. 6 miles, and again
observes the light What does she find the bearing and distance
of the light to be at the second observation 1
Ex. 318 o. An Admiral signals to his cruiser squadron
(bearing N. 40° W. 50 miles from him) to meet him at a place
N. 50° K, 70 miles from his present position. Find bearing and
distance of the meeting place from the cruisers.
Ex. 318^0. A is 1 mile due W. of B.
From A, C bears N. 28° W. and D bears N. 33° E.
From B, C bears N. 34° W. and D bears N. 9° W.
Find the distance and bearing of D from C.
Ex. 318 g. It is required to find the distance between Stokes
Bay Pier and a buoy from the following readings :
Bearing of Stokes Bay Pier from Ryde Pier, N. 9° E.
„ „ Buoy from Ryde Pier N. 36° W.
„ „ Buoy from Stokes Bay Pier S. 79° W.
Known distance from Stokes Bay Pier to Ryde Pier, 2*29 m.
02 EXPERIMENTAL GEOMETRY
Ex. 318r. A lies 7 miles N. 32° W. of B ; C is 5 miles
S. 67° E. of A. Find the distance and bearing of C from B,
Ex. 318s. Two rocks A, B are seven miles apart, oiie being
due East of the other. How many miles from each of them
is a ship from which it is observed that A bears S. 24° W. and
B bears S. 35° K ?
Ex. ZlSt. From a ship at sea the following observations are
made: Dover bears N. 16° E., and Boulogne S. 81° E. From
the chart it is found that Dover is 26 miles N. 24° W. of
Boulogne. Find the distance of the ship from Dover.
Ex. 318u. O and P are points on a straight stretch of shore.
P is 4'5 miles N. 74° E. of O. From O a ship at sea bears S. 58° E.,
and from P the ship bears S. 32° W. Find the distance of the
ship from P, and also its distance from the nearest point of the
shore.
Ex. 318v. A ship steaming due E. at 9*15 knots through the
Straits of Gibraltar observes that a point on the Rock bears
N. 35° E. ; 40 minutes later the same point bears due N. ; how
far is she from the point at the second observation ?
Ex. 318 w. A ship is observed to be 3 miles N. 28° E. from
a coast-guard station, and to be steaming N. 72° W. After
15 minutes the ship bears N. 36° W. At what rate is she
steaming 1
Ex. 318a;. C and D are inaccessible objects. A and B are
points 100 yards apart, B due East of A.
From A the bearing of C is due North.
„ A „ „ D is K 46" E.
„ B „ „ C is N. 63° W.
„ B „ „ D is N. 10° W.
Find (i) distance of C from B,
(ii) distance of D from B,
(iii) distance of C from D.
PAET II.
THEOEETICAL GEOMETBY.
BOOK I.
We are now going to prove theoretically that certain geo-
metrical statements are always true.
By using instruments we have been led to assume that certain statements
are true. For instance, by measuring the angles of a large number of
isosceles triangles we were led to assume that two angles of such a triangle
are always equal ; we now need something more than this, we must prove
that this is true for every isosceles triangle whether it is possible to measure
its angles or not.
Theoretical proof has two advantages over verification hy
measurement,
(i) Measurement is at best only approximate.
(ii) It is impossible to measure every case.
In theoretical geometry, we must never assume that things
are equal because they look equal or because our instruments lead
us to suppose them equal, and we must never make a statement
unless we have a sound reason for it. The reasons which we
use will in some cases depend on facts which we sliall have
already proved in the co"urse of our theoretical work, in some
cases on the definitions, in others on self-evident truths (called
axioms).
It is impossible to state here all the axioms we shall employ,
but we may give two examples.
64 BOOK I
Things which are eqiial to the same thing are equal to oiia
another.
[John is the same height as James, and William is the same height
as James ; therefore John is the same height as William.]
Ifeqvnh he added to equals the sums are equal.
[If two hoys each have five shillings and are each presented with another
shilling, the amounts which they then have mnst he equal.]
Angles at a point.
Points, lines, surfaces, etc. The formal definitions are
given later; for the present, the general ideas obtained from
the introduction are sufficiently definite. (See pp. 55 — 58.)
Dep. When two straight lines are drawn from a point
they are said to form, or contain, an angle. The point is called
the vertex of the angle, and the straight lines are called the
arms of the angle.
The size of an angle does not depend on the lengths of its arms.
(See Ex. 27, 28.)
Dep. When three straight lines are drawn from a point,
if one of them is regarded as lying between the other two, the
angles which this line makes with the other two are called
adjacent angles (e.g. l' a and h in fig. 11),
Dep. When one straight line stands on another straight
line and makes the adjacent angles equal, each of the angles
is called a right angle; and the two straight lines are said
to be at right angles or perpendicular to one another.
We shall assume that all right angles are equal.
HEx. 319. How would you test the accuracy of the right angle of your
set square ?
Dep. An angle less than a right angle is said to be acute.
Dep. An angle greater than a right angle is said to be
obtuse.
ANGLES AT A POINT 66
Revise Ex. 47—50, 57, 58.
ITEx. 320. A, B, C, D are four towns in order on a straight road; a
man walks from A to B and then on from B to D; another man walks
from A to C and then on from C to D ; have they walked the same distance?
ITEx. 321. If in fig. 75 a straight line OP revolves about O from the
position OB to the position OA, and then on to the position OC; and if
another straight line OQ revolves about O from the position OB to the
position OD and then on to OC ; will OP and OCl have turned through
the same angle?
Theorem 1.
If a straight line stands on another straight line, the
sum of the two angles so formed is equal to two right
angles.
.' D
^
COS
fig. 75.
Data The st. line AO meets the st. line BC at O.
To prove that l BOA + L AOC = 2 rt. /. s.
Construction Draw OD to represent the line through O per-
pendicular to BC,
Proof L BOA = z. BOD + Z. DO A,
L AOC = L DOC - L DOA,
.'. L BOA + L AOC := L BOD + L DOC
= 2 rt. z. s. Constr.
Q. E. D.
Cor. If any number of straight lines meet at a point,
the sum of all the angles made by consecutive lines is
equal to four right angles.
Q. S. R
66
BOOK I
Revise Ex. 51, 52.
TtSz. 3aa. If two straight lines AOB, COD intersect at O and Z.AOC is
a right angle, prove that the other angles at O are right angles.
ftEx. 323. If A ABC has /-ABC = Z. AC B, prove that
the exterior angles formed by producing the base both
ways tnust be equal to one another (i.e. prove that
^ABD = ^ACE). (See fig. 70.)
When angles or lines are given or made
equal it is well to indicate the fact in your
figure by putting the same mark in each.
tEx. 324. In A ABC, Z.ABC = /.ACB and AB
and AC are produced to X and Y, prove that
iLCBX = /.BCY. (See fig. 77.)
Def. If a straight line or angle is
divided into two equal parts it is said to be
bisected.
Ex. 325. If in fig. 75, /.BOA = 110°, and OP is drawn bisecting ^BOA
and OQ bisecting Z. AOC ; what are /.' POA, AOQ ? What is their snm ?
tEx. 326. Three straight lines OA, OB, OC are
drawn from a point O ; OP is drawn bisecting L. BOA,
and OQ bisecting /.AOC. Prove that Q
/.POQ:=KBOG.
tEx. 327.
If a straight line stands on another
fig. 78.
straight line, prove that the bisectors of the two
adjacent angles so formed are at right angles to one another.
Ex. 325, 326.)
Ex. 328. Prove the corollary to Theorem 1. See Ex. 59—62.
(See
Def. When the sum of two angles is equal to two right
angles, each is called the supplement of the other, or is said
to be supplementary to the other.
IfEx. 329. Name the supplements of /.ABC and /.BOY in fig. 77.
Name the supplements of /.AOB, /.COD, and /.AOC in fig. 75.
Ex. 330. State Theorem 1, introducing the term "supplementary."
Ex. 331. In fig. 75, show how to obtain another supplement of /.AOB,
t£x. 332. If two angles are equal, their supplements are equaL
ANGLES AT A POINT 67
Bevise Ex. 53—55.
Theoeem 2.
[CoKYERSE OP Theorem 1.]
If the sum of two adjacent angles is equal to two
right angles, the exterior arms of tJie angles are in the
same straight line.
M
fig. 79.
Data The svun of the adjacent L s BOA, AOO = 2 rt. Z.&
To prove that BO C is a straight line.
Construction Produce BO to D.
Proof Since AO meets the st. line BD at O,
.". /. BOA + z. A0D = 2rt. Z-S. 1.1.
But I. BOA + L AOO = 2 rt. ^ s, Data
.'. L BOA + L AOD = L BOA 4- L. AOC,
/. ^AOD= Z.AOC,
.'. OC coincides with OD.
Now BOD is a st. line, C(mstr.
.'. BOG is a st. line.
Q. E. D.
5—2
68 BOOK I
tEz. 888. From a point A in a straight line A B, straight lines AC and AD
are drawn at right angles to AB on opposite sides of it ; prove that CAD is
a straight line.
o
^Ex. 834. From a point O in a straight line AOC, OB ^
and OD are dravm on opposite sides of AC so that ^j r*^— — ^
^ AOB = Z.COD ; prove that BOD is a straight line. /
tEx. 336. Three straight lines OB, OA, OC are drawn
from a point (see fig. 78), OP bisects Z.BOA, Oa bisects ^8* 80.
Z. AOC ; prove that, if L. POQ is a right angle, BOC is a straight line.
tEx. 336. Two straight lines XOX', YOY' intersect at right angles ; OP
bisects ^XOY, OQ bisects Z.X'OY'. Is POQ a straight line? [Find the
sum of Z.» POY, YOX', X'OQ.]
Eevise Ex, 64—66.
lIEx. 337. If a straight line rotates about its middle point, do the two
parts of the straight line turn through equal angles ?
II the line rotates about any other point, are the angles equal? ,
HEx. 338. A BCD are four points in order on a straight
line; if AC=BD then AB=CD.
IFEx. 339. If two straight lines AOB, COD intersect at
O (see fig. 81) what is the sum of Z.» AOB, BOC?
What is the sum of ^» BOC, COD ?
Def. The opposite angles made by two in-
tersecting straight lines are called vertically
opposite angles {vertically opposite because they
have the same vertex).
HEiZ. 340. Kame two pairs of vertically opposite angles in fig. 81.
angles at a point 69
Theorem 3.
If two straight lines intersect, the vertically opposite
angles are equal.
c
fig. 82.
Data The two st lines AOB, COD intersect at O.
To prove that /. AOD = vert. opp. a. BOC,
L AOC = vert. opp. l BOD.
Proof Since st. line OD stands on st. line AB,
.'. -L AOD + L DOB = 2 rt. z. s, L 1.
and since st. line OB stands on st. line CD,
.". z.DOB + z.BOC=: 2 rt. ^s, 1.1.
.*. L AOD + L. DOB = L DOB + L BOC,
.". Z.AOD = A BOC.
Sim^y L AOC = Z. BOD. Q. K D.
Revise Ex. 67, 68.
tEx. 341. Write oat in full the proof that L AOC= z BOD in i. 3.
^Ex. 342. Draw a triangle and produce every side both ways ; number all
the angles in the figure, using the same numbers for angles that are equal.
tEx. 343. In fig. 83, prove that
(i) if Z. 5 = z. /, then I. c= l-f.
(U) if Z.c=Z./, then Z.d=Z.e.
(iii) if Z. d + Z./= 2 rt. W, then Lh=l.f.
(iv) if Z.gf=Z.c, then Z. <i=Z.A.
(v) \i L.h=.L. a, then i.e = Ld.
(vi) M L a— L e, then Lh=Lg.
(vii) if L c = ^f, then L d+ Lf =2rt. L\
70 BOOK I
tEx. 344. If two Btraight lines ACC, BOD intersect at O and OX bisectfl
Z.AOB, then XO produced bisects Z.COD.
tEx. 846. Tb« bls«etors of a pair of verticaUy opposite angles are
in one and tlie same straight line.
Parallel Straight Lines.
Def. Parallel straight lines are straight lines in the same
plane, which do not meet however far they are produced in either
direction.
Dep. In the figure two straight
lines are cut by a third straight line;
7. * c and / are called alternate angles,
L ' h and /corresponding angles (some-
times L. * h and f are spoken of as " an
exterior angle and the interior opposite
angle on the same side of the cutting
line").
f^Ex. 346. Name another pair of alternate angles in fig. 83.
^Ex. 347. Name another pair of corresponding angles.
HEx. 348. What are the names of the following pairs:
(i) c,f, (ii) 6,/, (iii) h, d, (iv) o, d, (v) c, g, (vi) e,f, (vii) e, a, (viii) c, d?
tEx. 349. Prove that if a straight Une cuts two other straight lines and
makes a pair of alternate angles equal, then a pair of corresponding angles
are equal.
[That is, in fig. 83, prove that if Lc=Lf, then L b=Lf.'\
tEx. 360. In fig. 83, prove that, if ^c=Lf, then Z.d+/./=2rt. /.«.
State this formally as in Ex. 349. (Z.' d and / are interior angles on the
same side of the cutting line.)
Bevise Ex. 167.
HEx. 361. Draw two parallel straight lines apd a line cutting them;
measure a pair of alternate angles.
ITEx. 362. Take a strip of paper about two inches wide with parallel sides,
cut it across as in fig. 84 ; measure the angles so formed with your protractor,
noting which are equal, and test whether the two pieces can be made to
coincide (i.e. fit on one another exactly).
PARALLEL STRAIGHT LINES 70 a
A First Treatment of Parallels (for Beginners).
The strict treatment of parallels given on pages 71, 72
may be found difficult for beginners. The following treat-
ment, based upon the equality of corresponding angles, is
recommended as more suitable for a first reading of theo-
retical geometry; it must not however be regarded as a
satisfactory proof.
fig. 83 (a). fig. 83 (6).
On page 36 was explained the set-square method of drawing
through P a parallel to QR, fig. 83 (a). Figure 83 (b) shews
the lines with the set-square removed.
It will be seen at once that the corresponding angles PSV,
RVX were covered by the same angle of the set-square, and must
be equal. Thus, the actual method of drawing parallel lines
suggests that
When a straight line cuts two other straight lines,
if a pair of corresponding angles are equal, then the
two straight lines are parallel.
From this it is easy to deduce
Theorem A,
When a straight line cuts two other straight lines, if
(2) a pair of alternate angles are equal,
or (3) a pair of interior angles on the same side of
tiie cutting line are together equal to two
right angles (supplementary),
then the two straight lines are parallel.
706 BOOK I
(2) Data The st. line AB cuts the two st. lines TP, QR forming
the z. s a, 6, c, c?, (3 ;
/. 6 = alternate c c.
To prove that TP, QR are parallel
Froof ^c = vert. opp. Le.
But Lh = Lc. DcUa
.'. Lh= Le,
and these are corresponding angles,
.'. TP, QR are parallel.
(3) Data Lh + ^d=2 rt. ls.
To prove thai TP, QR are parallel.
Le-\- Ld=2 Tt. L^ l1.
But z.6 + ^o?=2 rt. /.s. DeOa
.'. Le + i.d=^b + Ld.
.'. Le = Lh,
and these are corresponding angles,
.'. TP, QR are parallel. Q. e. d.
After this point the class may return to the ordinary
treatment at the middle of page 73 ; and deal with the
converse theorem. But it is probably a mistake to lay
any stress, in a first reading, upon the difficulties connected
with the parallel theorem and its converse.
The above presentation is easily seen to be open to objec-
tion; in fact we have virtually assumed Th. 4 (2), But
no harm is likely to result from adopting this treatment
of parallels with beginners, so long as it is clearly under-
stood to be provisional
PARALLEL STRAIGHT LINES
71
Theorem 4.*
(1) When a straight line cuts two other straight lines,
if a pair of alternate angles are equal, then the two
straight lines are parallel.
fig. 84.
(1) Data The st line EF cuts the two st. lines AB, CD at E, F,
forming the z. s a, 6, c, c? ; and z. a = alternate L d.
To prove thai
Proof
AB, CD are parallel.
I. 1.
Ll.
Daia
La + Lb = 2 rt. I.S,
LC + Ld=2rt. L8,
. La+ Lb= LC + Ld.
But La^ Ld.
.'. Lb = LC.
Take up the part AEFC, call it A'E'f'C'; and, turning it
round in its own plane, apply it to the part DFEB so that E'
falls on F and E'a' along FD.
': La- Ld, Data
:. E'F' falls along FE,
E'f' ^ FE (being the same line),
.". F' falls on E,
again •/ Lc = Lb, Proved
.'. F'C' falls along EB.
The proof of this theorem should be omitted at a first reading.
and
72 BOOK I
Now if EB and FD meet when produced towards Band D,
F'C' and E'A' must also meet when produced towards C' and A',
ie. FC and EA must also meet when produced towards C and A.
.'. if AB, CD meet when produced in one direction, they
will alsp meet when produced in the other direction ; but
this is impossible, for two st. lines cannot enclose a space.
.'. AB, CD cannot meet however far they are produced
in either direction.
.*. AB and CD are parallel.
Q. E. D.
When a straight line cuts two other straight lines, if
(2) a pair of corresponding angles are equal,
or (3) a pair of interior angles on the same side of
the cutting line are together equal to two
right angles,
then the two straight lines are parallel.
(2) Data The st. line GH cuts the two st. lines AB, CD forming
the z. s a, 6, c, d, e.
Le = corresp. i. d.
To prove thai AB, CD are paralleL
PARALLEL STRAIGHT LINES 73
Proof Le = vert. opp. L.a. , I. 3.
But Le = Ld, Data
.'. La~Ld,
and these are alternate angles,
.". AB, CD are parallel by (1).
(3) Baia z.6 + Ac? = 2i-t. z.s.
To 'prove that AB, CD are parallel.
Proof z. 6 + z. a = 2 rt. /. s. I. 1.
But Z.6 + ^c?= 2rt. iLS. Bata
.*. Lh-^ La = Lh-v Ld^
:. La-=Ld,
and these are alternate angles,
.'. AB, CD are parallel. by (1).
Q. K. D.
Cor. If each of two straight lines is perpendicular to
a third straight line, the two straight lines are parallel to
one another.
tEx. 353. Prove the corollary.
tEx. 364. Prove that the straight lines in fig. 83 would be parallel
(i) ii La = Lh, or (ii) if Lh+Lh = 2ri. L.\
Def. a plane figure bounded by three straight lines is
called a triangle.
Def. a plane figure bounded by four straight lines is called
a quadrilateral
Def. The straight lines "which join opposite corners of a
quadrilateral are called its diagonals.
Def. a quadrilateral with its opposite sides parallel is called
a parallelogram.
tEx. 355. ABCD is a quadrilateral, its diagonal AC is drawn; prove
that, if Z.BAC = /.ACD and /.DAC = Z.ACB, ABCD is a parallelogram.
74 BOOK I
Playfair's Axiom. Through a given point one straight line,
and one only, can be drawn parallel to a given straight line.
V. Theorem 5.
[CONVEESK OF THEOREM 4.]
If a straight line cuts two parallel straight lines,
(1) alternate angles are equal,
(2) corresponding ajigles are equal,
(3) the interior angles on the same side of the
cutting line are together equal to two right angles.
fig. 86.
Data AB cuts the parallel st. lines CD, EF at G, H.
To prove that (1) ii.CGH = alt. z. GHF,
(2) i. AGD= corresp. L GHF,
(3) ^DGH + z.GHF = 2rtz.s.
(1) Construction If z. CGH is not equal to iL GHF,
suppose GP drawn so that L PGH = L GHF.
Proof V ^PGH = alt. Z.QHF,
.*. PG is II to EF. I. 4.
.*. the two straight lines PG, CG which pass through the
point G are both || to EF.
But this is impossible. Playfair's Axiom
PARALLEL STRAIGHT LINES 75
.*. Z. CGH cannot be unequal toz. GHF,
.'. Z.CGH= ^GHF.
(2) Since, by (1), aCGH= ^GHF
and z. OGH = vert. opp. l AGD,
.'. ^AGD = ^GHF.
(3) Since GH stands on CD^
.'. aDGH + ^CGH==2 rt. Z.S, 1.1.
and, by (1), ^CGH = ^GHF,
.*. A DGH + Z.GHF = 2 rt ^ s. Q. E. D
Ex. 366. Copy fig. 86, omitting the line PG. If Z. AGD = 72°, find all
the angles in the figure, giving your reasons ; make a table.
+Ex. 367. Prove case (2) of Theorem 5 from first principles [i.e. without
assmning case (1)].
tEx. 868. Prove case (3) of Theorem 5 from first principles [Le. without
assuming cases (1) or (2)].
tEx. 369. In fig. 87 there are two pairs
of parallel lines ; prove that the following
pairs of angles are equal : — (i) 6, l, (ii) /, "k,
(iii) m, «, (iv) /, A, (v) r, I, (vi) s, h, (vii) s,g,
(viii) «, k, (ix) «, a, (x) g, I.
[State your reasons carefully.
e.g. WX, YZ are || and ST cuts them,
.'. Lq = Lf (corresponding angles).]
Ex. 860. What do you know about the sums of (i) /.» /, g, (ii) Z." /, Z,
(iii) Z.' m, 71, in fig. 87? Give your reasons.
Ex. 861. Draw a parallelogram ABCD, join AC, and
produce BC to E; what pairs of angles in the figure are
equal ? Give your reasons.
tEx. 362. A triangle ABC has Z.B = Z.C, and DE is
drawn parallel to BC ; prove that Z-ADE = Z.AED.
tEx. 363. Xf a straigbt line is perpendicular to one
off two parallel straight lines, it is also perpendicular
to the other.
76 . BOOK I
iEx. 364. The opposite angles of a parallelogram are eqaaL [See
Ex. 360.]
1 Ex. 865. What is the sum of the angles of a parallelogram?
Hence find the sum of the angles of a triangle.
tEx. 366. If one angle of a parallelogram is a right angle, prove that all
its angles must be right angles.
Note on a Theorem and its Converse.
The enunciatiou of a theorem can generally be divided into
two parts (1) the data or hypothesis, (2) the conclusion.
If data and conclusion are iiiterchanged a second theorem
is obtained which is called the converse of the first theorem.
For example, we proved
in I. 4, that, if La= Ld (data), then AB, CD are || (conclusion) ;
in I. 6, that^ if AB, CD are || (data), then La = i.d (conclusion).
The data of I. 4 is the conclusion of I. 5, and the conclusion
of I. 4 is the data of I. 5; so that i. 5 is the converse of i. 4
(and I. 4 is the converse of i. 5).
It must not be assumed that the converses of all true theorems
are true; e.g. "if two angles are vertically opposite, they are
equal" is a true theorem, but its converse "if two angles are
equal, they are vertically opposite " is not a true theorem.
1[Ex. 367. State the converses of the following: are they true?
(i) If two sides of a triangle are equal, then two angles of the triangle
are equal.
(ii) If a triangle has one of its angles a right angle, two of its angles
are acute.
(iii) London Bridge is a stone bridge.
(iv) A nigger is a man with woolly hair.
PARALLEL STRAIGHT LINES 77
Theorem 6.
Straight lines which are parallel to the same straight
line are parallel to one another.
fig. 89.
DcUa AB, CD are each || to XY.
To prove that AB is || to CD.
Construction Draw a st, line cutting AB, CD, XY and forming
with them corresponding Lsp, q, z respectively.
Proof ': AB is II to XY,
.'. z.^ = corresp, Lz. I, 5.
Again -.-CD is || to XY,
.'. Lq = corresp. /.a, L 5.
:.Lp = Lq.
. Now these are corresponding angles,
.'. AB is II to CD. L 4.
(^ E. D.
• Ex. 868. Prove i. 6 by means of Playfair's Axiom.
[Suppose AB and CD to meet.]
UEx. 369. Are the theorems true which you obtain (i) by substituting
"perpendicular" for "parallel" in i. 6, (ii) by substituting "equal" for
" parallel " in i. 6 ?
78 BOOK 1
Theorem 7.+
If straight lines are drawn f^om a point parallel to the
arms of an angle, the angle between those straight lines
is equal or supplementary to the given angle.
Data BAC is an angle.
From O, OX is drawn |] to AB and in the same sense* as AB,
and OY is drawn 1| to AC and in the same sense as AC ;
XO, YD are produced to Z, W respectively.
To prove that z.XOY= iL ZOW = z. BAC,
L YOZ = L WOX = supplement of l BAC.
* A straight line may be generated by the motion of a point, and the point
may move in either of two opposite directions or senses ; thus, in fig. 30,
the line AB may be generated by a point moving from A to B or from B to
A, and the Une OX by a point moving from O to X or from X to O. If a
point moves from A to B and another from O to X we say that they move
in the same sense, or AB and OX have tlie same sense ; but if the one moves
from A to B and the other from X to O they move in opposite senses,
or AB and XO have opposite senses.
PARALLEL STRAIGHT LINES 79
Proof Let WY cut AB at P,
then L XOY = corresp. L BPY, I. 5.
and u SAC = corresp. l. BPY, I. 5
.•. L XOY = L. BAG.
But A ZOW = vert. opp. L XOY,
.*. AZ0W= Z.BAC.
Again z. YOZ = L XOW = supplement of z. XOY
= supplement of L BAG.
Q. E. D.
tEz. 370. If straigbt lines are drawn tsova. a point perpendicular
to the arms of an svngle, the angle between those straight lines is eqnal
or supplementary to the given angle.
(Take BAG as the given angle, tlirough A draw straight lines parallel to
the given perpendiculars ; first prove that the angle between these lines is
equal or supplementary to L BAG.)
O. S,
80 BOOK I
Theorem 8.
The sum of the angles of a triangle is equal to two
right angles.
Data ABC is a triangle.
To prove iliat /. A + z. B + z. ACB == 2 rt. z. s.
Construction Produce BC to D.
Through C draw CE 1| to BA.
Proof Since AC cuts the ||s AB, CE,
.*. /. A = alt. z. ACE.
And since BC cuts the lis AB, CE,
.'. z. B = corresp. z. ECD,
.'. z. A + Z. B = z. ACE + z. ECD.
Add ^ ACB to each side,
.*. ^A+Z.B + Z.ACB = Z.ACB+Z.ACE + z. ECD
= 2 rt. A 8 (for BCD is a st. line),
.'. sum of z. 8 of A ABC = 2 rt. z. s.
Q. E. D.
Cor. 1. If one side of a triangle is produced, the
exterior angle so formed is equal to the sum of the two
interior opposite angles. (Proof as above.)
CoR. 2. If one side of a triangle is produced, the exterior
angle so formed is greater than either of the interior opposite
angles.
ANGLES OF A TRIANGLE 81
CoE. 3. Any two angles of a triangle are together less than
two right angles.
Cor. 4. Every triangle has at least two of its angles acute.
Cor. 5. If two triangles have two angles of the one
equal to two angles of the other, each to each, then the
third angles are also equal.
CoR. 6. The sum of the angles of a quadrilateral is
equal to four right angles. (Draw a diagonal.)
Revise Ex. 127—136.
+Ex. 371. Write out the full proof of Cor. 1.
tEx. 372. Prove i. 8 by drawing through A a straight line PAQ parallel
to BC.
+Ex. 373. In a triangle ABC, Z.A=Z.B; prove that if BC is produced
to D, Z.DCA = 2^B.
tEx. 374. Prove Cor. 6.
tEx. 376. What is the sum of the angles of a pentagon ?
[Join one vertex to the two opposite vertices.]
Ex. 376. If, in fig. 91, LA=56° and Z.ACD = 100°, find all the other
angles.
Ex. 377. In a quadrilateral ABCD, Z.A=77°, Z.B=88°, Z.C=99°;
find L D.
Ex.378. In a quadrilateral ABCD, ^A=37°, Z.B=lll°,and Z.C = Z.D;
find L.C and Z.D.
Ex. 379. If the exterior angles formed by producing the base of a
Iriangle both ways are 105° and 112°, find all the angles of the triangle.
+Ex. 380. If one angle of a triangle is a right angle, the other two
angles must be acute.
fEx. 381. If one angle of a triangle is obtuse, the other two angles
mast be acute.
Def. a triangle which has one of its angles an obtuse angle
is called an obtuse-angled triangle.
Def. a triangle which has one of its angles a right angle
is called a right-angled triangle.
The side opposite the right angle is called the hypotenuse.
6—2
82
BOOK I
Def. a triangle which has all its angles acute angles is
called an acute-angled triangle.
In Ex. 378 — 9, we have seen that •Terjr triangle xmtut bava at least
two of It* angles acute.
Def. a triangle which has two of its sides equal is called
an isosceles triangle.
Def. a triangle which has all its sides equal is called an
equilateral triangle.
Def. a triangle which has no two of its sides equal is called
a scalene triangle.
Def. a triangle which has all its angles equal is said to be
equiangular.
Revise Ex. 159—163.
HEx. 882. If two of the angles of a triangle are 67° and 79°, what is
the third angle? What are the exterior angles, formed
by producing the sides in order (see fig. 93)? What is
their sum?
lIEx. 383. Produce the sides of a Square* in order
(see fig. 92) ; what is the sum of the exterior angles?
ITEz. 384. In fig. 93 the sides of a triangle are produced
in order; what are the following sums: (i) ^a+^x,
(u) Lb + Ly, (iii) Lc + Lz, (iv) /.a + Lb+Lc?
Hence find £.x + Z,y + L.z.
%Ex, 38ft. Which angles are equal to the following sums:
(i) Lb + ^e, (ii) l.c+La, (iii) La+Lh?
Hence find ^x+Ly + ^z.
Wx. 886. If a yacht sails from A round the pentagon
BCDEF back to A, what angles does it turn through at
B,C, D.E, F?
When it gets back to A, it has headed towards every
point of the compass ; what then is the sum of the angles
through which it has turned ?
HEx. 887. Draw a figure to show which angles a yacht
turns through in sailing round a triangular course.
What is the sum of these angles?
* In Ex. 383-^689, the following properties of a square may be assumed:
^) all its sides are equal and (ii) all its angles are right angles.
ANGLES OF A POLYGON jk 83
Dep. a plane figure bounded by straight lines is called
a polygon.
Theorem 9.
If the sides of a convex polygon are produced in order,
the sum of the angles so formed is equal to four right
angles.
T-
fig. 95.
Data ABODE is a convex polygon; its sides are produced in
order and form the exterior angles, w, v, x, y, z.
To prove that Lw+ lv+ lx+ Ly + /.z = 4rt, ^s.
Comstniction Through any point O draw OP, OQ, OR, OS, OT
II to and in the same sense as EA, AB, BC, CD, DE respectively.
Proof Since OP, OQ are respectively 1| to and in the same sense
as EA, AB,
.'. Z.w=Z.POQ, 1.7.
Sim'y/Lz? = /.QOR,
LX=L ROS,
Ly = LSOT,
LZ = LT0P,
.'. ^fV + Lv + Lx + Ly + Lz = simi of /. s at O
= 4 rt. z. s. I. 1 Cor.
Q. E. D.
Cor. The sum of the interior angles of any convex
polygon together with four right angles is equal to twice
as many right angles as the polygon has sides.
84 ]| BOOK I
Ex. 888. Three of the exterior angles of a quadrilateral are 79°, 117°,
65°; find the other exterior angle and all the interior angles.
tEx. 880. Prove the corollary for a pentagon
(i) by considering the sum of the exterior and interior angles at each
corner, and the sum of all the exterior angles ;
(ii) by joining a point O inside the pentagon to each comer, and
considering the sums of the angles of the triangles so formed and the sum
of the angles at the point O.
Def. a polygon which has all its sides equal and all its
angles equal is called a regular polygon.
IFEx. 890. What is the size of each exterior angle of a regular octagon
(8-gon) ? Hence find the size of each interior angle.
Ex. 891. What are the exterior angles of regular polygons of 12, 10,
5, 3 sides ?
Hence find the interior angles of these polygons.
Ex. 892. The exterior angle of a regular polygon is 60°, how many
sides has the polygon ?
Ex. 398. How many sides have the regular polygons whose exterior
angles are (i) 10°, (u) 1°, (iii) 2^°?
Ex. 894. Is it possible to have regular polygons whose exterior angles
are (i) 15°, (ii) 7°, (iii) 11°, (iv) 6°, (v) 5°, (vi) 4°?
UEx. 896. Is it possible to have regular polygons whose exterior angles
are obtuse ?
Ex. 896. Is it possible to have regular polygons whose interior angles
are (i) 108°, (ii) 120°, (iii) 130°, (iv) 144°, (v) 60°? (Think of the exterior
angles.)
In the cases which are possible, find the number of sides.
Ex. 897. Make a table showing the exterior and interior angles of
regular polygons of 3, 4, 5 ... 10 sides.
Draw a graph showing horizontally the number of sides and vertically
the number of degrees in the angles.
Ex. 898. Construct a regular pentagon having each side 2 in. long.
(Calculate its angles, draw AB = 2 in., at B make ^ABC=the angle uf
the regular pentagon, cut off BC = 2 in., Ac, &o.)
Ex. 899. Construct a regular octagon having each side 2 in. long.
Ex. 400. Construct a regular 12-gon having each side 1 -5 in. long.
CONGRUENT TRIANGLES 85
Congruent Triangles.
If two figures when applied to one another can be made
to coincide (i.e. fit exactly) they must be equal in all respects.
This method of testing equality is known as the method
of superposition,
TEx. 401. How did you test the equality of two angles ? (See Ex. 28.)
HEx. 402. How would you test whether two cricket bats were of the
same length ?
Pigures which are equal in all respects are said to be
congruent
The sign = is used to denote that figures are congruent.
ITEx. 403. Draw a triangle DEF having DE = 3in., DF = 2in., Z.D = 26°;
on tracing paper draw a triangle ABC having AB = 3 in., AC = 2in., Z.A = 30°.
Apply aABC to aDEF so that A falls on D; put a pin through these
two points ; turn A ABC round until AB falls along DE.
B falls on E. Why is this ?
Does AC fall along DF?
(Keep the A ABC for the next Ex.)
ITEx. 404. Draw a triangle DEF having DE = 3in., DF = 2in., ^D = 30°.
Apply A ABC (made in the last Ex.) to a DEF so that A falls on D; put
a pin through these two points ; turn a ABC round until AB falls along DE.
B falls on E. Why is this?
AC falls along DF. Why is this ?
C falls on F. Why is this ?
Do the triangles coincide altogether ?
86 BOOK I
Theorem 10.
If two triangles have two sides of the one equal to two
sides of the other, each to each, and also the angles
contained by those sides equal, the triangles are con-
gruent.
Data ABC, DEF are two triangles which have AB — DE, AC — DF,
and included z. BAC = included l EDF.
To prove that A ABC = A DEF,
Proof Apply AABCto ADEFso that A falls on D, and AB falls
along DE.
V AB = DE,
.'. B falls on E.
Again *.• l BAC = L EDF,
.'. AC falls along DF.
And •.•AC = DF,
.'. C falls on F,
.'. A ABC coincides with A DEF,
.'. AABC= A DEF.
Q. E. D.
N.B. It must be carefully noted that the congruence of the triangles
cannot be inferred unless the equal angles are the angles included (or con-
tained) by the sides which are given equal.
Ex. 405. Make a list of all the equal sides and angles in A" ABC and
DEF of I. 10. Say which were given equal and which were proved equal.
CONGRUENT TRIANGLES
8T
tEx. 406. Draw two triangles PQR, XYZ and mark GIR = XY, RP=l: YZ,
and Z.Gl=^Z. Would this theorem prove the triangles congruent ? Give
two reasons.
iEx. 407. A BCD is a square, E is -the mid-point
of AB; equal lengths AP and BQ are out off from AD
and BC. Join E P and EQ. Prove that A A E P = A B EGL
Write down all the pairs of lines and angles in these
triangles which you have proved equal.
t£x. 408. ABCD is a square, E is the mid-point of AB ;
join CE and DE. Prove that A AED = a BEC.
Write down all the pairs of lines and angles in these
triangles which you have proved equaL
tEx. 409. PQRS is a quadrilateral in which
PQ=SR, ^Q=Z.R, and O is the mid-point of QR.
Prove that OP = OS.
[You must first join OP and OS, and mark in your
figure all the parts that are given equal ; you will then
see that you want to prove that a OGlP= a ORS.] ^8" ^^*
tEx. 410. ABCD is a square; E, F, G are the mid-points of AB, BC,
CD respectively. Join EF and FG Mid prove them equal.
[Which are the two triangles that you must prove equal ?]
tEx. 411. ABC, DEF are two triangles which are equal in all respects ;
X is the mid-point of BC, Y is the mid-point of EF. Prove that AX = DY,
and/.AXB=z.DYE.
[You will of course have to join AX and DY.] t,^ - -^^
+Ex. 412. The equal sides QP, RP of an isosceles
triangle PQR are produced to S, T so that PS = PT; prove
thatTQ=SR.
fig. 100.
+Ex. 413. D is the mid -point of the side BC of a a ABC,
AD is produced to E so that DE = AD. Prove that AB = EC
and that AB, EC are parallel.
[First prove a pair of triangles congruent.]
88
BOOK I
tEx. 414. Show that the distanoe
between G and H, the opposite cor-
ners of a house, can be found as
follows. At a point P set up a post ;
step off HP and an equal distance
PN, taking care to keep in a straight
line with the post and the comer H ;
step off GP and an equal distanoe
PM, M being in the same straight
line as G and P. Measure M N ; this
must be equal to GH.
Draw a ground plan and prove
that MN = GH.
M
fig. 102.
+Ex. 415. W is the mid-point of a straight line YZ,
WX is drawn at right angles to YZ. Prove that
XY = XZ.
[A line which is a side of each of two triangles is
said to be eonunon to the two triangles.]
+Ex. 416. The bisector of the angle between the
equal sides of an isosceles triangle is perpendicular to
the base.
[Let XYZ be an isosceles triangle, having XY=XZ ; let XW bisect Z.YXZ
and let it meet YZ at W ; prove ilXWY= Z.XWZ. See fig. 103.]
+Ex. 417. XYZ is an isosceles triangle having
XY = XZ; prove that ^Y = ^Z.
[Draw XW the bisector of /.YXZ.]
1 Ex. 418. OA, OB, OC are three radii of a circle. If
iLAOB = ^COB, prove that BO bisects AC.
tEx. 410. In fig. 105, AB and DC are equal and
parallel ; prove that AD = BC.
[Join BD. Since AB is parallel to DC .-. ^? = ^? .]
tEx. 420. The equal sides AB, AC of an isosceles
triangle ABC are produced to X and Y respectively ;
BX is made equal to CY (see fig. 77). If Z.CBX = Z.BCY, prove that
CX = BY.
[By the side of your figure make sketches of the triangles BOX, CBY]
CONGRUENT TRIANGLES 89
tEx. 421. XY is a straight line, XP and YQ are drawn at right angles
to X Y and X P is made equal to YQ. Prove that L PYX = Z. QX Y.
+Ex. 422. A BCD is a quadrilateral in which AB = CD, AD = BC and
Z. A= Z.C ; prove that A BCD is a parallelogram.
[Join BD.]
tEx. 423. If the diagonals of a quadrilateral bisect one another it must
be a parallelogram.
HEx. 424. In two A» ABC, DEF, /.A=Z.D, Z.B = iLE; prove that
UEx. 425. Draw a triangle DEF having EF = 3-7in., Z.E = 35°, Z.F = 64°;
on tracing paper draw a triangle ABC having BC=3*7in,, ^B = 35°,
iCC = 64°. Apply A ABC to A DEF so that B falls on E, and BC falls
along EF. Do the two triangles ooiuoide?
90 BOOK I
Theorem 11.
If two triangles have two angles of the one equal to
two angles of the other, each to each, and also one side
of the one equal to the corresponding side of the other,
the triangles are congruent.
fig. 106.
Data ABC, DEF are two triangles which have BC = EF and two
angles of the one equal to the two corresponding angles of
the other.
To prove thai, A ABC = A DEF.
Proof Since two angles of A ABC are respectively equal to two
angles of A DEF,
.'. the third angle of A ABC = the third angle of A DEF,
I. 8, C(yr. 6.
.*. Z. A = A D, /. B = Z. E, and lC = L?.
Apply A ABC to A DEF so that B falls on E, and BC falls
along EF.
•.• BC = EF,
.*. C falls on F.
Now z. B = /. E,
.'. BA falls along ED,
.*. A fells somewhere along ED or ED produced.
Again z. C = ^ F,
.'. CA falls along FD,
.'. A falls somewhere along FD or FD produced,
.*. A falls on D,
.'. A ABC coincides with A DEF,
.". A ABC = A DEF. Q. K. D.
CONGRUENT TRIANGLES 91
Ex. 436. Make a list of all the equal sides and angles in A' ABC,
DEF of I. 11.
tEx. 427. Draw two A» GHK, XYZ, and mark GH = XY, iLH = z.Y,
and Z. K = /. X ; are the triangles congruent ?
+Ez. 428. ABCD is a square, E is the mid-point of AB; at E make
Z.AEP=60° and ^BEQ=60°; let EP, EQ cut AD, BC at P and Q re-
spectively. Prove that AP= BQ. (See fig. 97.)
tEx. 429. In a aXYZ, Z.Y = Z.Z; XW is drawn so that /.X is
bisected; prove that XY=XZ. (See fig. 103.)
tEx. 430. If the bisector of an angle of a triangle cuts the opposite side
at right angles, the triangle must be isosceles.
[Let XYZ be a triangle; and let XW, the bisector of Z.X, cut YZ at right
angles at W ; prove that X Y = X Z . See fig. 103.]
tEx, 431. ABC, DEF are two triangles which are equal in all respects ;
AP, DQ are drawn perpendicular to BC, EF respectively. Prove that
AP=DQ.
tEx. 432. A ABC= A DEF. AG, DH are the bisectors of Z.A, Z.D and
meet the opposite sides in G, H. Prove that AG = DH.
tEx. 433. The following method
may be used to find the breadth
of a river. Choose a place where
the river is straight, note some con-
spicuous object T (e.g. a tree) on
the edge of the other bank; from
a point O opposite T measure a
distance OS along the bank; put
a stick in the ground at S ; walk on ^q
to a point P such that SP = OS; g j^q™
from P walk at right angles to the
river till you are in the same straight line as S and T. PQ is equal to the
breadth of the river. Prove this.
tEx. 434. The perpendiculars drawn to the arms of
an angle from any point on the bisector of the angle are
equal to one another.
tEx. 435. ABCD is a parallelogram, prove that AB= CD,
[Join AC and use i. 5.]
fig. 108.
92 BOOK I
tEx. 4S6. If the diagonal PR of a qnadrilateral PQRS bisects the angles
at P and R , prove that the quadrilateral has two pairs of ^
equal sides.
+Ex. 487. A triangle XYZ has LS = lZ\ prove that
the perpendiculars from the mid-point of YZ to XY and y^
XZ are equal to one another.
fig. 109.
tEx. 488. A triangle ABC has /.B=^C; prove that
the perpendiculars from B and C on the opposite sidet*
are equal to one another.
+Ex. 489. A triangle ABC has AB = AC ; prove that g
the perpendiculars from B and C on the opposite sides
are equal to one another. fig. na
tEx. 440. The diagonal AC of a quadrilateral ABCD bisects the angle
Aandz.ABC = Z.ADC; doe8BC = CD?
HEx. 441. Draw two or three isosceles triangles ; measure their angles.
congruent triangles 93
Theorem 12.
If two sides of a triangle are equal, the angles opposite
to these sides are equal.
A
B D
fig. 111.
Daia ABC is a triangle which has AB = AC.
To prove that z. C = z. B.
Construction Draw AD to represent the bisector of L BAC.
Let it cut BC at D.
Proof In the As ABD, ACD
( AB = AC, DcOa
AD is common,
L BAD = L CAD (included L s), Constr.
.*. AABD= A ACD, I. 10.
.'. /L B = /. C.
Q. K. D.
The phrase " the sides " of an isosceles triangle is often used to mean the
equal sides, " the base " to mean the other side, " the vertex " to mean the
point at which the equal sides meet, and ' ' the vertical angle " to mean the
angle at the vertex.
Ex. 442. State the converse of this theorem.
Ex. 443. In a triangle XYZ, XY = XZ ; find the angles of the triangle in
the foUowing cases: (i) Z.Y = 74°, (ii) Z.X=36°, (iii) Z.X = 142°, (iv) Z.Y = 13°,
(v) Z.Z=97°, (vi) Z.Z=45°.
tEx. 444. Each base angle of an isosceles triangle must be acute.
Ex. 445. Find the angles of an isosceles triangle in which each of the
base angles is half of the vertical angle.
. Ex. 446. Find the angles of an isosceles triangle in which each of the
base angles ia double of the vertical angle.
94 BOOK I
fEx. 447. Prove tliat a triangle whleb la equilateral ia alao eqm*
angnlar. (See definition, p. 82.)
[If PQR is an equilateral triangle, •.•QP = GlR.-.z.? = z.?.]
Ex.448. In a triangle ABC, AB = 9-2cm., iLC = 82°, AC=9-2cm. ;
AS, AC are produced to D, E respectively. Find all the angles in the figure.
tEz. 449. ABC is an isosceles triangle; the equal sides AB, AC are
produced to X, Y respectively. Prove that Z.XBC = iLYCB.
State the converse of this theorem.
tEx. 450. EDA, FDA are two isosceles triangles on opposite sides of
the same base DA ; prove that /.EDF= Z.EAF. See fig. 123.
' tEx. 451. EDA, FDA are two isosceles triangles on the same side of
the same base DA; prove that ^EDF = Z.EAF.
tEx. 463. Through the vertex P of an isosceles triangle PQR a straight
line XPY is drawn parallel to QR ; prove that ^QPX = ^RPY,
tEx. 468. From the mid-point O of a straight line AB /\\
a straight line OC is drawn; if OC = OA, ^ACB ia a / \^-^
right angle. A O B
tEx. 464. In fig. 113, A ABC ia isosceles and BP = CQ;
prove that Z.APCl=Z-AQP.
[First prove AP = AQ.]
tEx. 466. The perpendicular from the vertex of an isos- ^^—p
celes triangle to the base bisects the base. fig. 113.
tEx, 466. The perpendiculars to the equal sides of an isosceles triangle
from the mid-point of the base are equal. (See fig. 109.)
tEx. 467. The perpendiculars from the ends of the base of an isosceles
triangle to the opposite sides are equaL (See fig. 110.)
tEx. 468. The straight lines joining the mid-point
of the base of an isosceles triangle to the mid-points
of the sides are equal.
tEx. 460. If A, B are the mid-points of the equal
sides XY, XZ of an isosceles triangle, prove that
AZ = BY.
^L,
fig. 115.
CONGRUENT TRIANGLES
95
tEx. 460. The bisectors of the base angles of an
isosceles triangle are equal.
tEx. 461. At the ends of the base BC of an isosceles
triangle ABC, perpendiculars are drawn to the base to
meet the equal sides produced; prove that these per-
pendiculars are equal.
tEx. 462. XYZ is an isosceles triangle (XY = XZ),
the bisectors of Z. X and L Z meet at O ; prove that
OY bisects Z.Y.
tEx. 463. The angle between a diagonal and a side
of a square is 45°.
^^
tEx. 464. OA, OB are radii of a circle, AO is
produced to P; prove that ;LB0P = 2z.BAP. p
fig. 119.
tEx. 466. In fig. 119, prove that the perpendicular from O to AB
bisects AB.
tEx. 466. If a four-sided figure has all its sides equal, its opposite
angles are equal.
[Draw a diagonal.]
tEx. 467. Draw a line BC, at B and C make equal angles CBA, BCA
so as to form a triangle ABC. Measure AB and AC.
TEx. 468. Repeat Ex. 467 two or three times with other lines and angles.
O. 8 7
9G
BOOK I
Theorem 13.
[Converse of Theorbm 12.]
If two angles of a triangle are equal, the sides opposite
to these angles are equal.
Proof
Data ABC is a triangle which has /l B = /L C.
To prove that AC = AS.
Construction Draw AD to represent the bisector of L BAC
Let it cut BC at D.
In the As ABD, ACD,
r Z. B = Z. c,
\ L BAD = L CAD,
( AD is common,
.*. A ABD = A ACD,
.'. AB=AC.
Q. E. D.
Data
Constr.
1.11.
tEz. 469. Prove that If a triangle PQR is eqnianeular, it must
also be equilateral.
[z.Q=z.R, .•. side ?= side?.]
tEx. 470. The sides AB, AC of a triangle are produced to XY; prove
that, if Z.XBC = Z.YCB, a ABC is isosceles. (See fig. 77.)
tEx. 471. A straight line drawn parallel to the base of
an isosceles triangle to cat the equal sides forms another
isosceles triangle.
fig. 121.
CONGRUENT TRIANGLES 9T
tEx. 472. XYZ is an isosceles triangle ; the bisectors of the equal angles
(Y, Z) meet at O ; prove that A OYZ is also isosceles. (See fig. 118.)
tEx. 47S. Prom Gl and R, the extremities of the base of aa isosceles
triangle PQR, perpendiculars are drawn to the opposite sides. If these
perpendiculars intersect at X, prove that XGl=XR.
tEx. 474, XYZ is an isosceles triangle (XYsaXZ), the bisectors of /.Y
and Z.Z meet at O; prove that OX bisects Z.X.
tEx. 476. If through any point in the bisector of an n^
angle a line is drawn parallel to either of the arma of the
angle, the triangle thus formed is isosceles.
tEx. 476, ABCD is a quadrilateral in which AB=sAD, „ .
and Z.B=^D; prove that CB=CD. ^'
[Draw a diagonal.]
tEx. 477. In the base BC of a triangle ABC, points P, d are taken
Buchthat Z.BAP=/.CAGl; if AP=AQ, prove A ABC is isosceles.
tEx. 478. In a quadrilateral ABCD, L* A, B are equal and obtuse, and
AB is parallel to CD ; prove that AD= BC.
[Produce DA, CB till they meet.]
tEx. 470. If the L* G, H of a triangle FGH are each double of /.F,
and if the bisector of /.Q meets FH in K, prove that FK=GKs»GH,
ITEx. 479 a. If one side of a triangle is double another, is the angle
opposite the former double the angle opposite the latter?
In order to answer this question, take the following instances :
(1) Consider a triangle whose angles are 45°, 45°, 90°.
(2) Consider a triangle whose angles are 30°, 60°, 90°.
(3) Draw aABC in which AB = 8-2cms., BC = 4-lcms., CA = 6cms.
Measure the angles. Is C double A ?
(4) Draw a ABC in which A = 82°, B = 41°, BC = 3". Measure the
remaining sides. Is BC double CA?
(5) Prove that in a ABC whose angles are 30°, 60°, 90°, the longest
side AB is double the shortest BC.
[Make z CAD = 30°
and produce BC to meet AD in D.
How many degrees in / D ?
What kind of a triangle is ABD ?]
98
BOOK I
Theorem 14.
If two triangles have the three sides of the one equal
to the three sides of the other, each to each, the triangles
are congruent.
Data ABC, DEF are two triangles which have BC = EF, CA = FD,
and AB = DE.
To prove tluxi A ABC = A DEF.
Proof Apply A ABC to A DEF so that B falls on E and BC falls
along EF but so that A and D are on opposite sides of EF j
let A' be the point on which A falls. Join DA'.
Since BC = EF, C will fall on F.
Cask i When DA' cuU EF.
^
fig. 128.
In A EDA', ED = EA' (i.e. BA),
.'. z. EA'D = A EDA'. I. 12.
In A FDA', FD = FA' (i.e. CA),
.'. aFA'D = ^FDA', I. 12.
, L EA'D + L FA'D - L EDA' + L FDA',
ie. /LEA'F = Z.EDF,
i.e. L BAC = Z. EDF,
in As ABC, DEF,
L AB = DE, Data
\ AC = DF, Data
( L BAC = L EDF (included L s). Proved
.•. AABC= A DEF. i. 10.
CONGRUENT TRIANGLES
Cask 1L When DA' passes through one end of EF, say-?.
A D
Cash hi.
fig. 124.
In A EDA', ED = EA' (i.e. BA),
.'. L EA'D = L EDA',
Le. L BAC = L EDF,
,*. as in Case I. A ABC = ADEF.
When DA' does not cut EF.
A
fig. 125.
As in Case l. l EA'D = z. EDA',
and L FA'd = z. FDA',
L EA'D - ^ FA'D = L EDA' - z. FDA',
i.e. L EA'F = I. EDF,
i.e. Z- BAC = z. EDF,
;. as in Case l. A ABC = ADEF.
I. 12.
Q. K. D.
100 BOOK I
Ex. 480. State the converse of this theorem. Is it true?
+Ex. 481. If, in a quadrilateral ABCD, AB = AD, CB = CD, prove that
AC bisects Z.A and Z.C.
tEz. 482. PQ and RS are two equal chords of a
circle whose centre is O. Prove that Z.POQ=^ROS.
(A chord of a circle is a straight line joining anj
two points on the circle.)
tEx. 483. AS is a chord of a circle whose centre is O;
C is the mid-point of the chord AB. Show that OC is
perpendicular to AB.
tEx. 484. If the opposite sides of a quadrilateral are equal, it is a parallel-
ogram.
[Draw a diagonal, and use i. 4.]
+Ex. 486. Equal lengths AB, AC are cut off from the
arms of an angle BAC ; on BC a triangle BCD is drawn
having BD = CD. Show that AD bisects ^BAC.
tEx. 486. The bisectors of the equal angles Y, Z of
an isosceles triangle XYZ meet at O. Prove that XO
bisects ^X.
tEx. 487. EDA, FDA are two isosceles triangles on opposite sides of the
same base DA; prove that EF bisects DA at right angles.
[First prove A DEF= a AEF ; see fig. 123.]
tEx. 488. EDA, FDA are two isosceles triangles on the same base DA and
on the same side of it ; prove that EF produced bisects DA at right angles.
tEx. 489. In a quadrilateral ABCD, AD = BC and the diagonals AC, BD
are equal; prove that ^ADC=Z.BCD.
Also prove that, if AC, BD intersect at O, aOCD is isosceles.
tEx. 490. Two drcles intersect at X, Y ; prove tbat XY is bisected at
right angles by the straight line joining the centres of the two circles.
[Join the centres of the circles to X and Y.]
CONGRUENT TRIANGLES
101
Theorem 15.
If two right-angled triangles have their hypotenuses .
equal, and one side of the one equal to one side of the
other, the triangles are congruent.
Data ABC, DEF are two triangles which have z. s C, F right L s,
AB = DE, and AC = DF.
To prove thai
AABC= A DEF.
Proof Apply A DEF to A ABC so that D falls on A
AC, but so that E and B are on opposite sides
be the point on which E falls.
Since DF = AC, F will fall on C.
Since z. s ACB, ACE' (i.e. DFE) are two rt.
BCE' is a sfc. line.
.'. ABE' is a A.
In this A, AB = AE' (ie. DE)
.-. ^ E' = ^ B.
Now in the As ABC, AE'C,
f^B = ^E',
\ L ACB = L ACE',
( AB = AE',
.*. AABC= A AE'C,
.-. AABC= A DEF.
and DF along
of AC ; let E'
Z.S, DatQ,
1.2.
Data
1.12.
Proved
Data
Data
I. 11.
Q. E. D.
102 BOOK I
tEz. 401. In fig. 97, given that E is the mid-point of AB and EP = EQ,
prove that aAEP= aBEQ.
tEz. 40a. Ina triangle XYZ, XY = XZ, and XW is drawn at right angles
to YZ : prove that a XYW= a XZW. (Use i. 16.)
+Ex. 493. Perpendiculars are drawn from a point P to two straight'lineP
XA, XB which intersect at a point X ; prove that, if the perpendiculars are
equal, PX bisects Z.AXB. (See fig. 108.)
fEz. 494. AB is a chord of a circle whose centre is O. Show that the
perpendicular from O on AB bisects AB.
tEz. 495. The perpendiculars from the centre of a circle on two equal
chords of the circle are equal to one another. (See fig. 126 ; use Ez. 494.)
tEz. 496. In fig. 129, PM, QN are drawn perpen-
dicular to the diameter AOB, O being the centre ot the
circle ; show that, if PM = QN, then / POM = L GION.
i Ex. 497. If the perpendiculars from the mid-point
of the base of a triangle to the other two sides are equal, n 1 20
the triangle is isosceles. (See fig. 109.)
tEz. 498. If the perpendiculars from two corners of a triangls to the
opposite sides are equal, the triangle is isosceles. (See fig. 110.)
tEz. 499. From the vertices A, X of two triangles ABC, XYZ, lines
AD, XW are drawn perpendicular to BC, YZ respectively. If AD = XW,
AB = XY, and AC = XZ, prove that the triangles ABC, XYZ are congruent,
provided they are both acute-angled, or both obtuse-angled.
tEz. 600. With the same notation as in Ex. 499, prove that, if AD = XW,
<\B = XY, and BC = YZ. the triangles are congruent.
CONSTRUCTIONS 103
Constructions.
Hitherto we have constructed our figures with the help of
graduated instruments. We shall now make certain construc-
tions with the aid of nothing but a straight edge (not graduated)
and a pair of compasses.
We shall use the straight edge
(i) for drawing the straight line passing through any two
given points,
(ii) for producing any straight line already drawn,
We shall use the compasses
(i) for describing circles with any given point as centre
and radius equal to any given straight line,
(ii) for the transference of distances; i.e. for cutting off
from one straight line a part equal to another straight line,
[(ii) is really included in (i).]
By means of theorems which we have already proved, we
shall show that our constructions are accurate.
In the exercises, when you are asked to construct a figure,
you should always explain your construction in words. You need
not give a proof unless you are directed to do so.
In the earlier constructions the figures are shown with
given lines — thick,
construction lines — fine,
required lines — of medium thickness,
lines needed only for the proof — broken.
In making constructions, only the necessary parts of construotiou circles
sliould be drawn even though "the circle" is spoken of.
Bevise Ex. 98—102,
104 BOOK I
To construct a triangle having its sides equal to three
given straight lines.
Q
fig. 130.
Let X, Y, Z be the three given straight lines.
Construction Draw a straight line PGL
From PQ cut off" a part PR = X.
With centre P and radius = Y describe a circle.
With centre R and radius = Z describe a circle.
Let the circles intersect at S.
Join PS, RS.
Then PRS is the required triangle.
Note. It is best to draw the longest Une first.
It should be observed that the construction is impossible if
one of the given straight lines is greater than the sum of the
other two. (Why?)
Ex. 601. Draw a large* triangle and construct a congruent triangle.
Ex. SOS. Construct a triangle having its sides equal to the lines b, d, h
of fig. 8.
Ex. 603. Draw a straight line (about 3 in. long) ; on it describe an
equilateral triangle. Measure its angles.
Ex. 604. Construct an isosceles triangle of base 5 cm. and sides 10 cm.
Measure the vertical angle.
t£x. 605. Draw an angle ABC ; complete the parallelogram of which AB,
BC are adjacent sides. [On AC construct aACD= a CAB.] Give proof.
Ex. 606. Make an angle of 60° (without protractor or set square).
Ex. 607. Make an angle of 120° (without protractor or set square).
Eevise Ex. 274—276.
* Constructions should always be made on a large scale; an error of
•5 nun. is less important in a large figure than in a small one. In this case
let the shortest side be at least 3 in. long.
CONSTRUCTIONS 105
Through a point o in a straight line OX to draw
a straight line OY so that Z.XOY may be equal to a
given angle bac.
Canstruction With centre A and any radius describe a circle
cutting AB, AC at D, E respectively.
With centre O and the same radius describe a circle PY
cutting OX at P.
With centre P and radius = DE describe a circle cutting
the circle PY at Y.
Join OY.
Then l XOY = L BAC.
Proof Join DE and PY.
In the A» OPY, ADE,
fOP = AD, Constr.
OY = AE,
[pY = DE,
/. APOY= AADE, I. U
.-. L POY = L DAE,
Le. L XOY = L BAC.
106 BOOK I
[The protractor must not be used in Ex. 608 — 518.]
Ex. 508. Draw an acute angle and oonstmot an equal angle*.
Ex. 509. Draw an obtuse angle and make a copy of it.
Ex. SIO. Draw an acute angle ABC ; at C make an angle BCD = ^ ABC.
Let BA, CD intersect at O. Measure OB, OC.
Ex All. Draw a triangle ABC ; at a point O make a copy of its angles
in the manner of fig. 60.
Ex. 01 a. Bepeat Ex. 511 for a quadrilateral.
Ex. 013. Draw two straight lines and an angle. Construct a triangle
having two sides and the included angle equal respectively to these lines
and angle.
Ex. 014. Construct a triangle ABC having given BC, Lti and Z.C.
Ex. 010. Construct a triangle ABC having given BC, ^A and LB,
Ex. 016. Draw a straight line EF and mark a point G (about 2 in. from
the line) ; through G draw a line parallel to EF.
[Draw any line through G cutting EF at H ; make ^HGC = ^QHF;
z^ fig. 86.]
Ex. 017. Bepeat Ex. 516, using corresponding instead of alternate
angles.
Ex. 018. Draw a large polygon and make a copy of it, using the first
method described on p. 50.
Bevise «• Synunetry " pp. 51 — 55.
Ex. 019. Cut out an angle of paper; bisect it by folding as in Ex. 31.
* It is convenient to draw the angle on tracing paper so as to compare
it with the angle made equal to it.
CONSTRUCTIONS
To bisect a given angle.
307
Let BAG be the given angle.
Construction From AB, AC cut off equal lengths AD, AE.
With centres D and E and any convenient radius de-
scribe equal circles intersecting at F.
Join AF.
Then AF bisects l BAG.
Proof Join DF and EF.
In the A* ADF, AEF,
{AD^AE,
DF^EF,
AF is common.
.•. AADF = AAEF,
.". AF bisects l BAC.
Constr.
I. 14.
"Ant/ convenient radius." If it is found that the equal
circles do not intersect, the radius chosen is not convenient, for
the construction breaks down ; it is necessary to take a larger
radius so that the circles may intersect.
i08 BOOK I
f^Ex. sao. If fig. 132 were folded about AF, what points wonld coincide?
What lines?
ITEx. 631. Make two equal angles and bisect them ; in one case join the
vertex to the nearer point at which the equal circles intersect, in the other
to the further point.
Which gives the better result ?
ITEx. 022. Is there any case in which one point of intersection wonld
coincide with the vertex of the angle ?
Ex. 523. Draw an acnte angle and bisect it. Check hj measurement.
Ex. 624. Draw an obtuse angle and bisect it. Check by measurement.
Ex. 626. Quadrisect a given angle (i.e. divide it into four equal parts).
Ex. 626. Draw an angle of 87° and bisect it (1) by means of the pro-
tractor, (2) as explained above.
Do the results agree ? (This will test the accuracy of your protractor.)
Ex. 627. Construct angles of 15°, 30° and 150° (without protractor).
Ex. 628. Draw a large triangle and bisect each of its angles.
Ex. 620. Construct an if^osceles triangle, bisect its vertical angle and
measure the parts into which the base is divided.
Ex. 630. Draw a triangle whose sides are 5 cm., 10 cm., 12 cm. Bisect
the greatest angle and measure the parts into which the opposite side is
divided.
HEx. 631. Draw a straight line AB on tracing paper ; fold it so that A
falls on B ; measure the parts into which AB is divided by the crease and the
angles the crease makes with AB.
Bevise Ex, 288—290.
CONSTRUCTIONS
109
To draw the perpendicular bisector of a given straight
line.
To bisect a given straight line.
O
A<- )B
fig. 133.
Let AB be the given straight line.
Construction With centres A and B and any convenient radius
describe equal circles intersecting at C and D.
Join CD and let it cut AB at E.
Then CD is the perpendicular bisector of AB, and E is the
mid-point of AB.
Proof Join AC, AD, BC, BD.
In the A* ACD, BCD,
rAC = BC, Constr.
AD = BD,
(CD is common,
.•. AACDs A BCD, I. 14.
.•. z.ACD=- z. BCD.
In the A" ACE, BCE,
[AC = BC, Constr.
■^ CE is common,
i z. ACE = z. BCE, Proved
:. AACE = A BCE, I, 10.
.-. AE=BE,
and L. ' CEA, CEB are equal and are therefore rt. l ', Def.
.'. CD bisectvS AB at right angles.
110 BOOK I
^Ex. 692. Demribe the symmetry cf fig. IBS.
Ex. 633. Draw a straight line and bisect it.
Ex. 634. Quadrisect a given straight line.
HEx. 636. Draw a straight line AB and its perpendicular bisector CO.
Take any point P in CD and measure PA and PB. Take three other points
on CD and measure their distances from A and B.
Ex. 636. Draw a large acute-angled triangle ; draw the perpendicular
bisectors of its three sides.
Ex. 637. Bepeat Ex. 536 for (i) a right-angled triangle, (ii) an obtuse-
angled triangle.
Ex. 638. Draw any chord of a circle and its perpendicular bisector.
Dep. The straight line joining a vertex of a triangle to the
mid-point of th6 opposite side is called a median.
Ex. 630. Draw a large triangle; and draw its three medians. Are the
angles bisected?
IfEx. 640. Call one of the short edges of your paper AB ; construct its
perpendicular bisector by folding. Fold the paper again so that the new
crease may pass through A, and B may fall on the old crease ; mark the
point C on which B falls and join CA, CB. "What kind of triangle is ABC?
Ex. 641. Draw a large obtuse angle (v€ ry nearly 180°) and bisect it.
CONSTRUCTIONS 111
To draw a straight line perpendicular to a given
straight line ab from a given point P in ab.
^
A »C p D^
fig. 134.
Construction From PA, PB cut oflf equal lengths PC, PD.
With centres C and D and any convenient radius de-
scribe equal circles intersecting at E.
Join PE.
Then PE is x to AB.
Proof Join CE, DE.
In the A" CPE, DPE,
PC=PD, Constr.
■ CE= DE (radii of equal 0*),
PE is common.
.*. A CPE = ADPE, I. 14.
/. L EPC = L EPD,
.'. PE is ± to AB. Def.
[The protractor and set square must not be used in Ex. 542 — 555.]
Ex. 542. Draw a straight line, and a straight line at right angles to it.
Test with set square.
Ex. 543. Draw an isosceles triangle; at the ends of the base erect
perpendiculars and produce the sides to meet them (see fig. 117). Measure
all the hues in the figure.
Ex. 544. Construct angles of 45° and 75°.
Ex. 545. Draw a chord AB of a circle, at A and B erect perpendiculars
to cut the circle at P and Q respectively. Measure AP, BQ.
+Ex. 546. Make an angle AXB ; from XA, XB cut off equal lengths XM,
XN ; from M, N draw MP, NP at right angles to XA, XB respectively;
join PX. Prove that PX bisects Z.AXB. Check by measurement.
[See fig, 108.]
O. & 8
112
BOOK I
To draw a straight line perpendicular to a given
straight line ab from a given point P outside ab.
^^^JX
^
fig. 136.
Construction With centre P and any convenient radius describe
a circle cutting AB at X and Y.
With centres X and Y and any convenient radius de-
scribe equal circles intersecting at Q.
Join PQ cutting AB at Z.
Then PZ is J. to AB.
Proof Join XP, XQ, YP, Ya
In the A« PQX, PQY,
!PX = PY (radii of a 0),
QX = QY (radii of equal Q%
PQ is common.
.*. A PQX = A PQY, L 14.
.'. Z.XPQ=^YPa
We can now prove that
A PXZ = A PYZ, (give the three reasons)
.•. L PZX = i. PZY,
,'. PZ is ± to AB,
£z. 647. Draw a large acute-augled triangle ; from each vertex draw a
perpendicular to the opposite side.
Ex. 648. Bepeat Ex. 547 with a right-angled triangle.
Ex. 649. Bepeat Ex. 547, with an obtuse-angled triangle. [You will
have to produce two of the sides.]
CONSTRUCTIONS 113
Ex. 660. Draw an acute angle and bisect it ; from any point on the
bisector drop perpendiculars on the arms of the angle ; measure the perpen-
diculars.
Ex. 651. Bepeat Ex. 550 for an obtuse angle.
Ex. 552. From the centre of a circle drop a perpendicular on a chord
of the circle.
Ex. 553. Cut out of paper an acute-angled triangle ; by folding con-
struct the perpendiculars from each vertex to the opposite side.
Ex. 554. Gut out a paper triangle ABC (Z.* B and C being acute) ; by
folding construct AD perpendicular to BC. Again fold so that A, B and 0
all fall on D.
Ex. 655. Gut out of paper an equilateral triangle ABC (see Ex. 540).
Construct two of the perpendiculars from the vertices to the opposite sides ;
let them intersect at O. Fold so that A falls on O, and then so that B and
C fall on O. What is the resulting figure ?
Construction of Triangles from given data.
We have seen how to construct triangles having given
(i) the three sides (Ex. 99-102, and p. 104) ;
(ii) two sides and the included angle (Ex. 87, 88, 513) ;
(iii) one side and two angles (Ex. 89, 90, 514, 515).
L 14, 10, 11 prove that if a set of triangles were constructed
from the same data, such as those given above, they would all be
congruent.
In Ex. 146-150, we saw that, given the angles, it is possible
to construct an unlimited number of different triangles.
If two angles of a triangle are given, the third angle is known ; hence the
three angles do not constitute more than two data.
We have still to consider the case in which two sides are
given and an angle not included by these sides.
8—2
114
BOOK I
ITEx. 556. C!onstruct a triangle ABC having given BC = 2*4in.,
CA = l-8in., and z. B = 32^
First make BC = 2-4 in. and l. CBD = 32^
A must lie somewhere on BD, and must be 1*8 in, from C.
Where do all the points lie which are 1 'S in. from C ?
How many points are there which are on BD and also I'S in.
from C?
We see that it is possible to construct two unequal triangles
which satisfy the given conditions. This case is therefore called
the ambiguous case.
Ex. 657. Construct triangles to the following data : —
(i) BC = 8-7cm., CA=5-3cm., /.B = 29
(ii) BC = 7-3cm., CA = 90cm., z.A=63=
(iii) AB = 3-9in., AC = 2-6in.,
AB = 2-2in., BC = 3-7in.,
AC =5-3 cm., BC = 10cin.,
AC = 1-6 in., BC = 4-7m.,
^C=68°
^A=90°
Z.B=32°
^B = 26°.
(iv)
(V)
(vi)
tEx. 558. Prove (theoretically) that the two triangles obtained in
Ex. 657 (iv) are congruent.
We may summarise the cases of congruence of triangles as
follows : —
Data
Gcmclusiim
Theorem,
3 sides
AU the triangles are congruent
1.14
2 sides and included
angle
All the triangles are congruent
1. 10
2 sides and an angle
not included
Two triangles are generally
possible (ambiguous case)
Ex. 867
1 side and 2 angles
All the triangles are congruent
1.11
3 angles
All the triangles have the same
shape, but not necessarily the
same size
—
EXERCISES ON I. 1 — 15 115
MISCELLANEOUS EXERCISES.
Constructions.
Ex. 559. Construct angles of (i) 135°; (ii) 105°; (iii) 22J° (withont pro-
tractor or set square).
Ex. 560. Show how to describe an isosceles triangle on a given straight
line, having each of its equal sides double the base.
Are the base angles double the vertical angle ?
Ex. 561. Describe a circle and on it take three points A, B, C; join
BC, CA, AB. Bisect angle BAG and draw the perpendicular bisector of BC.
Produce the two bisectors to meet,
Ex. 562. Having given two angles of a triangle, construct the third
angle (without protractor).
Ex. 563. Draw an isosceles triangle ABC; on the side AB describe an
isosceles triangle having its angles equal to the angles of the triangle ABC
(without protractor).
Ex. 564. Show how to describe a right-angled triangle having given its
hypotenuse and one acute angle.
Ex. 565. Construct a triangle ABC having AB = 3 in., BC=5 in., and
the median to BC = 2-5 in. Measure CA.
Ex. 566. Construct a triangle ABC having given AB = 10 cm., AC =8 cm.,
and the perpendicular from A to BC = 7"5 cm. Measure BC. Is there any
ambiguity ?
[First draw the line of the base, and the perpendicular.]
Ex. 567. Construct a triangle ABC having given AB = 11'5 cm.,
BC = 4'5 cm., and the perpendicular from A to BC = 8-5 cm. Measure AC.
Is there any ambiguity?
Ex. 568. Show how to construct a quadrilateral having given its sides
and one of its angles.
Ex. 569. Four of the sides, taken in order, of an equiangular hexagon
are 1, 3, 3, 2 inches respectively: construct the hexagon and measure the
remaining sides.
[What are the angles of an equiangular hexagon ?]
+Ex. 570. Show how to construct an isosceles triangle having given the
base and the perpendicular from the vertex to the base. Give a proof.
[See Ex. 466.]
116 BOOK I
+Ex. 671. A, B are two points on opposite sides of a straight line CD;
in CD find a point P such that Z.APC= ^ BPD. Give a proof.
+Ex. 672. A, B are two points on tbe same side of a straiRht line CD;
in CD find a point P snch that z. APC = z. BPD. Give a proof.
[From A draw AN perpendicular to CD and produce it to A' so that
NA'— NA; if P is any point in CD, /.» APN and A'PN can be proved equal;
in fact, A and A' are symmetrical points with regard to CD.]
Ex. 678. Show how to construct an isosceles triangle on a given base,
having given the sum of the vertical angle and one of the base angles.
Ex. 674. Construct a triangle, having one angle four times each of the
other two. Find the ratio of the longest side to the shortest.
[First calculate the angles.]
+Ex. 676. Show how to construct an isosceles triangle on a given base,
having its vertical angle equal to a given angle. Give a proof.
+Ex. 676. Show how to construct an equilateral triangle with a given
line as median. Give a proof.
fEx. 677. Through one vertex of a given triangle draw a straight line
cutting the opposite side, so that the perpendiculars upon the line from the
other two vertices may be equal. Give a proof.
[See Ex. G70.]
+Ex. 678. From a given point, outside a given straight line, draw a line
making with the given line an angle equal to a given angle. (Without
protractor.) Give a proof.
[Use parallels. ]
+Ex. 670. Through a given point P draw a straight line to cut off equal
parts from the arms of a given angle XOY. Give a proof.
[Use parallels.]
Ex. 680. Draw a triangle ABC in which Z.B is less than /.C. Show
how to find a x>oint P in AB such that PB = PC.
+Ex. 681. In the equal sides AB, AC of an isosceles triangle ABC show
how to find points X, Y such that BX = XY = YC. Give a proof.
Theorems.
Ex. 683. How many diagonals can be drawn through one vertex of
(i) a quadrilateral, (ii) a hexagon, (iii) a n-gon?
Ex. 683. How many different diagonals can be drawn in (i) a quadri-
lateral, (ii) a hexagon, (iii) a n-gon?
EXERCISES ON I. 1 — 15 117
tEx. 684. The bisectors of tti« fonr angles formed by two inter-
secting straiglit lines are two straight lines at rigbt angles to one
anotber.
tEx. 585. If the bisector of an exterior angle of a triangle is parallel to
one side, the triangle is isosceles.
tEx. 686. The internal bisectors of two angles of a triangle can never be
at right angles to one another.
tEx. 687. AB, CD are two parallel straight lines drawn in the same sense,
and P is any point between them. Prove that ^BPD = Z.ABP + /.CDP.
tEx. 588. ABC is an isosceles triangle (AB = AC). A straight line is
drawn at right angles to the base and cuts the sides or sides produced in D
and E. Prove that a ADE is isosceles.
tEx. 680. From the extremities of the base of an isosceles triangle
straight lines are drawn perpendicular to the opposite sides ; show that the
angles which they make with the base are each equal to half the vertical
angle.
tEx. 690. The medians of an equilateral triangle are equal.
tEx. 691. The bisector of the angle A of a triangle ABC meets BC in D,
and BC is produced to E. Prove that ^-ABC + A.ACE=:2z.ADC.
tEx. 692. From a point O in a straight line XY, eqnal straight lines
OP, OQ are drawn on opposite sides of XY so that ^.YOP=^YOQ. Prove
that aPXY= aQXY.
tEx. 698. The sides AB, AC of a triangle are bisected in D, E; and BE,
CD are produced to F, G, so that EF = BE and DG =CD. Prove that FAG
is a straight line.
tEx. 594. If the straight lines bisecting the angles at the base of an
isosceles triangle be produced to meet, show that they contain an angle
equal to an exterior angle at the base of the triangle.
tEx. 695. The bisectors of the angles B, C of a triangle ABC intersect
at I; prove that Z.BIC=90° + JZ.A.
tEx. 696. XYZ is an isosceles right-angled triangle (XY = XZ); YR
bisects ^.Y and meets XZ at R; RN is drawn perpendicular to YZ. Prove
that RN = XR.
tEx. 697. The perpendiculars from the vertices to the opposite sides of
an equilateral triangle are equal to one another.
118 BOOK I
tEx. 698. If two of the bisectors of the angles of a triangle meet at a
point I the perpendicnlars from I to the sides are all equal.
fjix. 699. The perpendicular bisectors of two sides of a triangle meet
at a point which is equidistant from the vertices of the triangle.
tEx. 600. In the equal sides PQ, PR of an isosceles triangle PQR points
X, Y are taken equidistant from P; QY, RX intersect at Z. Prove that
A" ZQR, ZXY are isosceles.
tEx. 601. ABC is a triangle right-angled at A; AD is drawn perpen-
dicular to BC. Prove that the angles of the triangles ABC, DBA are respec-
tively equal.
tEx. 602. From a point O in a straight line XOX' two equal straight
lines OP, OQ are drawn so that A POQ is a right angle. PM and QN are
drawn perpendicular to XX'. Prove that PM = ON.
tEx. 603. If points P, Q, R are taken in the sides AB, BC, CA of a:\
equilateral triangle such that AP=BQ = CR, prove that PQR is equi-
lateral.
tEx. 604. ABC is an equilateral triangle; DBC is an isosceles triangle
on the same base BC and on the same side of it, and Z BDC=Jz BAG,
Prove that AD = BC.
Ex, 606. How many sides has the polygon, the sum of whose interior
angles is three times the sum of its exterior angles ?
[What is the sum of all the exterior and interior angles ? What is the
sum of an exterior angle and the corresponding interior angle ?]
tEx. 606. If two isosceles triangles have equal vertical angles and if the
perpendiculars from the vertices to the bases are equal, the triangles are
congruent.
tEx. e07. If, in two quadrilaterals ABCD, PQRS,
AB=PQ, BC=QR, CD = RS, /B=zQ, and /C=2R,
the quadrilaterals are congruent.
Prove this (i) by superposition (see i. 10 and 11) ;
(ii) by joining BD and QS and proving triangles congruent.
tEx. 608. K two quadrilaterals have the sides of the one equal re-
spectively to the sides of the other taken in order, and have also one angle of
the one equal to the corresponding angle of the other, the quadrilaterals are
congruent.
[Draw a diagonal of each quadrilateral, and prove triangles congruent.]
tEx. e09. If points X, Y, Z are taken in the sides BC, CA, AB of an
equilateral tritmgle, such that Z BAX= / CBY= l ACZ, prove that, unless
AX, BY, CZ pass through one point, they form another equilateral triangle.
EXERCISES ON I. 1 — 15 119
+Ex. 610. If points X, Y, Z are taken in the sides BC, CA, AB of any
triangle, such that / BAX= / CBY = ACZ, prove that, unless AX, BY, CZ
pass through one point, they form a triangle whose angles are equal to the
angles of the triangle ABC.
tEx. 611. If AA', BB', CC are diameters of a circle, prove
aABCs aA'B'C.
tEx. 612. On the sides AB, BC of a triangle ABC, squares ABFG,
BCED are described (on the opposite sides to the triangle); prove that
aABD= aFBC.
tEx. 613. On the sides of any triangle ABC, equilateral triangles BCD,
CAE, ABF are described (aU pointing outwards); prove that AD, BE, CF are
all equaL
tEx. 614. The side BO of a triangle a6c is produced to D; / ACB is
bisected by the straight line CE which cuts AB at E. A straight line
is drawn through E parallel to BC, cutting AC at F and the bisector of
I ACD at G. Prove that EF= FG.
tEx. 615. ABC, DBC are two congruent triangles on opposite sides of
the same base BC; prove that either AD is bisected at right angles by BC,
or AD and BC bisect one another.
tEx. 616. In a triangle ABC, the bisector of the angle A and the perpen-
dicular bisector of BC intersect at a point D ; from D, DX, DY are drawn
perpendicular to the sides AB, AC produced if necessary. Prove that
AX = AY and BX = CY.
[Join BD, CD.]
Inequalities.*
ITEx. 617. Draw a scalene triangle, measure its sides and arrange them
in order of magnitude. Under each side in your table write the opposite
angle and its measure, thus: —
Sides I AC =5-8 in.
Angles I ^ B =
BC=4-3in.
Z.A=
AB=3-2in.
Z.C=
Are the angles now in order of magnitude ?
ITEx. 618. In fig. 136, AD = AC ; if Z.A=:88°, find Z.ADC and Z.ACD.
What is the sum of Z. B and /L DCB ?
* This section, pp. 119 — 132, may be omitted at a first reading.
120 BOOK I
The sign > means " is greater than."
The sign < means "is leas than."
Tliese signs are easily distinguished if it is borne in mind that the greater
quantity is placed at the greater end of the sign.
Theorem 16.
If two sides of a triangle are unequal, the greater side
has the greater angle opposite to it. .
?. 136.
Daia ABC is a triangle in which AB > AC.
To -prove thnt L ACB > ^ B.
Construction From AB, the greater side, cut oflFAD = AC.
Join CD.
Proof In AACD, AD=AC,
.•. ^ACD = Z.ADC. I. 12.
But since the side BD of the A DBC is produced to A,
.'. ext. L ADC > int. opp. L B, I, 8, Cor. 2.
.'. L ACD > Z. B.
But /. ACB > its part L ACD,
.*. Z- ACB>/. B.
Q. E. D.
INEQUALITIES 121
HEx. 619. In a A ABC, BC = 7 cm., CA = 6-7 cm., AB = 7-5 cm. ; which is
the greatest angle of the triangle? Which is the least angle? Verify hy
drawing.
ITEx. eao. If one side of a triangle is known to be the greatest side, the
angle opposite that side must be the greatest angle. (Notice that i. 16 only
compares two angles ; here we are comparing three.)
tEx. 631. The angles at the ends of the greatest side of a triangle are
acute.
+Ex. 632. In a parallelogram ABCD, AB > AD ; prove that
Z.ADB > Z.BDC.
[What angle is equal to Z. BDC ?]
tEx. 638. In a quadrilateral ABCD, AB is the
shortest side and CD is the longest side ; prove that
/-B > LD, and /.A > Z.C. ^
D o
[Draw a diagonal.] g_ j^g^
fEx. 634. Assuming that the diagonals of a parallelogram ABCD bisect
one another, prove that, if BD > AC, then Z. DAB is obtuse.
[Let the diagonals intersect at O, then OB > OA and OD > OA ; what
follows ?]
A
TEx. 635. Prove Theorem 16 by means of the
following construction: — from AB cut off AD=AC, ^
bisect Z.BAC by AE, join DE.
B E Q
fig. 138.
122 BOOK I
Theorem 17.
[CoNVERSB OP Theorem 16.]
If two angles of a triangle are unequal, the greater
angle has the greater side opposite to it.
fig. 139.
Data ABC is a triangle in which /. C > z. B.
To prove thai A B > AC.
Proof
Either (i) AB>AC,
or (ii) AB=AC,
or (iii) AB<AC.
If, as in (iii), AB < AC,
then z. C < z. B,
L 16.
which is impossible.
Data
If, as in (ii), AB = AC,
then z. C = z. B,
I. 12.
which is impossible.
Data
.'. AB must be > AC.
y. K D.
Note. The method of proof adopted in the above theorem
is called reductio ad absurdum.
ITEx. 626. In a A ABC, Z.A = 68° and Z.B = 28°. Which is the greatest
side of the triangle? Which is the shortest side?
INEQUALITIES 123
Ex. 627. Bepeat Ex. 626 with ^ B = 34°, Z. C = 73°.
Ex. 638. Draw accurately a triangle whose sides measure 5 cm., 7 cm.,
9 cm. ; guess the number of degrees in each angle, and verify your guesses by
measurement.
tEx. 629. In a right-angled triangle, the hypotenuse is the longest
side.
tEx. 630. The side opposite the obtuse angle of an obtuse-angled
triangle is the greatest side.
tEx. 631. If one angle of a triangle is known to be the greatest angle,
the side opposite to it must be the greatest side.
tEx. 632. If ON is drawn perpendicular to a straight line AB, and O is
joined to a point P in AB, prove that ON <0P.
Ex. 633. The side BA of a triangle ABC is produced to E so that
AE = AC; if / BAG =86° and /ACB=52°, find all the angles in the
figure.
Ex. 634. In the last Ex. prove that BE> BC.
tEx. 635. AD is drawn perpendicular to BO the opposite side of a
triangle ABC; prove that AB> BD and AC^CD.
Hence show that AB + AC > BC.
[There wUl be two cases.]
tEx. 636. The bisectors of the angles B, 0 of a triangle ABC intersect
at O. Prove that, if AB>AC, OB>OC.
tEx. 637. If the perpendiculars from B, 0 to the opposite sides of the
triangle ABC intersect at a point X inside the triangle, and if AB>AC,
prove that XB>XC.
tEx. 638. The sides AB, AC of a triangle are produced, and the bisectors
of the external angles at B, 0 intersect at E. Prove that, if AB>AC,
EB<EC.
tEx. 639. A straight line cuts the equal sides AB, AC of an isosceles
triangle ABC in X, Y and cuts the base BC produced towards 0. Prove
that AY > AX.
tEx. 640. Prove that the straight line joining the vertex of an isosceles
triangle to any point in the base produced is greater than either of the equal
sides.
tEx. 641. Prove that the straight line joining the vertex of an isosceles
triangle to any point in the base is less than either of the equal sides of the
triangle.
124
BOOK I
Theorem 18l +
Any two sides of a triangle are together greater than
the third sida.
8"
Q
fig. 140.
Data
ABO is a tn'an^e.
To prove ttuxt
(1) BA + AO^BO,
(2) CB + BA:^CA,
(3) AO + CB>Aa
'
(1) Construction
Produce BA to D.
Prom AD cut off AE » AC.
Join CE.
Proof
In the A AEG, AEaAO)
/. /.ACE = /.AEC,
Conatr.
I. 12.
Bat I. BCE > its part /.ACE,
.*. L BCE > i. AEG,
A in the AEBC, l BCE >z. BEG,
.*. BE>BC
Le. BA + AE>BC,
.*. BA + AC > BC,
(2) Sim'y CB + BA>CA,
(3) and AC + CB>Aa
1.17.
Cotistr,
Q. ■. D.
INEQUALITIES 125
tEx. 642. Prove this theorem by drawing AD the bisector of /.A, and
applying i. 17 to the two triangles thus formed.
tEx. 643. The difference between any two sides of a triangle is less than
the third side.
Prove this (i) by means of the same construction as in fig. 136.
(ii) by means of the result of i. 18.
ITEx. 644. Why would it be impossible to form a triangle with three rods
whose lengths are 7 in., 4 in., and 2 in. ?
ITEx. 646. If you had four rods of lengths 2 in., 3 in., 4 in., and 6 in.
with which sets of three of these would it be possible to form triangles ?
tEx. 646. S is a point inside a triangle PQR such that PS = PGl; the
bisector of Z.QPS cuts QR at T. Prove that GIT = TS.
Hence from aSTR prove that RQ > RS.
tEx. 647. Any three sides of a quadrilateral are together greater than
the fourth side.
[Draw a diagonal.]
1 Ex. 648. If D is any point in the side AC of a triangle ABC, prove that
BA+AC > BD + DC.
tEx. 649. If O is any point inside a triangle ABC, prove that
BA + AC > BO + OC.
[Produce BO to cut AC]
A
tEx. 660. Any chord of a circle which does not pass through the centre
is less than a diameter.
[Join the ends of the chord to the centre.]
tEx. 661. In fig. 141, O is the centre of the circle
and POA is a straight line ; prove that PA s» PB.
[Join OB.]
tEx. 662. In fig. 141, prove that PC < PB.
Tie a piece of elastic to the ends of the arms of your dividers so as tu
form a triangle ; notice that the more the dividers are opened the more the
elastic is stretched; or, in other words, the greater the angle between the
sides of the triangle the greater the base.
126
BOOK I
Theorem 19.* +
If two triangles have two sides of the one equal to two
sides of the other, each to each, and the included angles
unequal, the triangle which has the greater included
angle has the greater third side.
A
E
fig. 142.
Data ABC, DEF are two triangles which have AB = DE, and
AC= DF but L BAC> L EDF.
To prove tJiat BC > EF.
Proof Apply AABCto ADEFso that A falls on D and AB falls
along DE; then B falls on E (for AB = DE),
Since l BAG > ^ EDF,
.'. AC falls outside L EDF.
Let C' be the point on which C falls.
If EFC' is a st. line, EC' > EF,
i.e. BC ^ EF.
Case ii. If EFC' is not a st. line.
Construction Draw DG to represent the bisector of L FDC';
let DG cut EC' at G.
Proof
Cask i.
In As DGF, DGC',
DF = DC' (i.e. AC),
Data
DG is common.
A FDG = z. C'DG (included L s),
Constr.
.*. the triangles are congruent,
I. 10.
.". GF = GC'.
Now, in A EFG, EG + GF > EF.
I. 18.
But GF = GC',
Proved
.-. EG + GC' > EF,
i.e. EC' > EF,
i.e. BC > EF.
Q. E. D.
This proposition may be omitted.
INEQUALITIES 127
Ex. 653. Draw a figure for i. 19 in which AC, DF are greater than
AB, DE. Does the proof hold for this figure?
tEx. 664. A, B, C, D, are four points on a circle whose centre is O, such
that Z.AOB > ^COD ; prove that chord AB > chord CD.
Also state the converse. Is it true ?
tEx. 656. If, in fig. 155, a point P' is taken not in the straight line PN,
prove that P'A, P'B must be unequal.
[Join P'N.]
+ Ex. 656. In a quadrilateral ABCD, AD=BC and /1.ADC > Z.BCD;
prove that AC > BD.
tEx. 667. Equal lengths YS, ZT are cut off from the sides YX, ZX of a
triangle XYZ ; prove that, if XY > XZ, YT > ZS.
tEx. 658. The sides XY, XZ of a triangle XYZ are produced to S, T so
that YS = ZT ; prove that, if XY > XZ, ZS > YT.
G. S.
128
BOOK I
Theorem 20.* t
[COKVERSK OF THEOREM 19.]
If two triangles have two sides of the one equal to two
sides of the other, each to each, and the third sides
unequal, the triangle which has the greater third side
has the greater included angle.
fig. 143.
Data ABG, DEF are two triangles which have AB = DE. and
AG = DF but BC> EF.
To prove that
Proof
^BAC>z.EDF.
Either (i) z.BAG>/.EDF,
or (ii) L BAG = L EDF,
or (iii) L. BAG < L EDF.
If, as in (iii), l BAG < z. EDF,
then BG < EF,
I. 19.
which is impossible.
Data
If, as in (ii), l BAG = L EDF,
then BG = EF,
I. 10.
which is impossible.
Data
.*. L BAG must be > z. EDF.
Q. E. D.
* This proposition may be omitted.
INEQUALITIES 129
tEx. 6S9. In a triangle ABC, AB>AC; D is the mid-point of BC.
ProTe that L ADC is acute.
tEx. 660. P is any point in the median AD of a triangle ABC; prove
that, if AB > AC, PB > PC. (Use Ex. 659.)
tEx. 661. Equal lengths YS, ZT are cut off from the sides YX, ZX of
a triangle XYZ ; prove that, if YT>ZS, XY>XZ.
tEx. 662. State and prove the converse of Ex. 658.
tEx. 663. In a circle A BCD whose centre is O, the chord AB> the
chord CD ; prove that /.AOB> Z.COD.
tEx. 664. In a quadrilateral ABCD, AD = BC, but AC > BD ; prove that
Z.ADC>Z.BCD.
tEx. 665. In a quadrilateral ABCD, AD = BC, but AB<CD; prove that
Z.DAC>Z. ACB.
tEx. 666. In a quadrilateral ABCD, AD = BC, and Z.ADC>Z.BCD;
prove that Z.ABC>Z.BAD.
HEx. 667. Draw a straight line AB, and draw ON perpendicular to AB
(see fig. 144); from O draw six or fwven straight lines to meet AB. Measure
all these lines.
9—2
180
BOOK 1
Theorem 21,
Of all the straight lines that can be drawn to a given
straight line from a given point outside it, the per-
pendicular is the shortest.
A N p B
fig. 144.
Data AB is a straight line and O a point outside it ; ON is drawn
i. to AB meeting it at N.
To "prove thai ON < any other st. line that can be drawn from
O to AB.
Construction Draw any other st. line from O to meet AB at P.
Fnx/ In the AONP,
z-N + ^P<2rt.A8, I. 8, Cor. 3.
and LH = lrt. L,
.'. LP <l rt. L,
.'. /.P<Z.N,
.•. ON < OP, I. 17.
Sim'' ON may be proved less than any other st. line
drawn from O to meet AB.
.". ON is the shortest of all such lines.
Q. E. D.
Note. Since the perpandicular is the shortest line that can
be drawn from a given point to a given line, it is called the
distance of the point from the line.
INEQUALITIES 131
tEz. 668. In fig. 144, prove that OB > OP.
Ex. 669. Is it possible, ia fig. 144, to draw from O to AB a straight line
equal to OP?
tEx. 670. The extremities of a given straight line are equidistant from
any straight line drawn through its middle point.
tEx. 671. If the bisectors of two angles of a triangle are produced to
meet, their point of intersection is equally distant from the three sides of
the triangle.
Miscellaneous Exercises.
Ex. 673. How many triangles can be formed, two of whose sides are
S in. and 4 in. long and the third side an exact number of inches?
tEx. 673. ABC, APQC, are a triangle and a convex quadrilateral on the
same base AC, P and Q. being inside the triangle; prove that the perimeter
of the triangle is greater than that of the quadrilateral.
[Produce AP, PQ to meet BC and use i. 18.]
tEx. 674. O is a point inside a triangle ABC ; prove that A BOC > /. BAC.
[Produce BO to cut AC]
tEx. 676. The sum of a median of a triangle and half the side bisected
is greater than half the sum of the other two sides.
tEx. 676. Two sides of a triangle are together greater than twice the
median drawn through their point of intersection.
[Use the construction and figure of Ex. 413.]
tEx. 677. O is a point inside a quadrilateral ABCD ; prove that
OA+OB+OC+OD
cannot be less than AC + BD.
tEx. 678. The sum of the distances of any point O from the vertices of
a triangle ABC is greater than half the perimeter of the triangle.
[The perimeter of a figure is the sum of its sides. Apply i. 18 to
A" OBC, OCA, OAB in turn and add up the results.]
tEx. 679. The sum of the distances from the vertices of a triangle of
any point within the triangle is less than the perimeter of the triangle.
[Apply Ex. 649 three times.]
Would this be true for a point outside the triangle?
132 BOOK I
tEx. eao. The Bnm of the diagotialB of a quadrilateral is greater than
half its perimeter.
tEx. 681. The stun of the diagonals of a quadrilateral is less than its
perimeter.
tEx. 683. The sum of the medians of a triangle is less than its peri-
meter.
[Use Ex. 676.]
i Ex. 683. The sum of the distances of any point from the angular points
of a polygon is greater than half its perimeter.
tEx. 684. In a triangle A6C, 0 is the mid-point of BC; if AD<BD,
A ABC must be obtuse-angled.
tEx. 685i Find the position of P within a quadrilateral ABCD, for
which PA + PB+PC + PD is least. Give a proof.
[See Ex. 677.]
tEx. 686. ABC, DBC are two triangles on the same base BC, and AD
is parallel to BC. If the triangle ABC is isosceles its perimeter is less than
that of the triangle DBG.
[Produce BA to E so that AE=AB. Join DE.]
tEz; dlB^. P is anj point in the median AD Of a triangle ABC * prove
that, if AB>AC, PB>PG.
tEx. 688. In a quadrilateral ABCD, / BCA ^ z DAC ; prove thait
/ADB5> zDBC.
tEx. 689. O is a point within an equilateral triangle ABO ; if
/ GAB > ^ GAG, / OCB > / GBG.
Paballeloorams.
Dep. a quadrilateral with its opposite sides {)arallel ia
called a parallelogram.
Bevise Ex. IBS— 203.
parallelograms 133
Theorem 22.
(1) The opposite angles of a paraUelogram are equal.
B C
fig. 145.
Dala ABCD is a parallelogram. .
To 'prove that lIK = lC, /.B^/lD.
Proof Since AD and BC are ||, and AB meets them,
.•. /. A + /L B == 2 rt. z. s. I. 5.
Sim'y /. B + /. C = 2 rt. ^ s. 1. 5.
.*. Z.A+z-B = Z.B4-Z.C,
.'. Z.A-Z.C.
Sim^ L B ='/L D.
Q. K D.
(2) The opposite sides of a parallelogram are equal.
(3) Each diagonal bisects the parallelogram.
fig. 146.
Data ABCD is a parallelogram, and BD one of its diagonals.
To prove tJiat AB = CD, AD = CB, and that BD bisects the paral-
lelogram.
Proof Since AD is || to BC and BD meets them,
Z. ADB = alt. Z.CBD. L 5.
Since AB is || to CD and BD meets them,
L ABD = alt. L CDB. I. 5.
134 BOOK I
,". in As ABD, CDB,
I Z.ADB = iLCBD,
( BD is common,
.'. AABD = ACDB, I. 11.
.•. AB^^CD, AD = CB.
And since A ABD = ACDB,
BD bisects the parallelogram.
Sim'y AC bisects the parallelogram. Q. E. D.
(4) The diagonals of a parallelogram bisect one
another.
fig. 147.
Data A BCD is a parallelogram; its diagonals AC, BD intersect
at O.
To -prove thai OA = OC and CD = OB.
Proof Since AD is || to BC and BD cuts them,
.'. Z-ADO = ^CBO,
.*. in As OAD, OCB
I L ADO = L CBO,
\ L AOD = vert. opp. u COB,
( AD - CB, L 22 (2).
.*. the As are congruent, I. 11.
.'. OA = OC and OD - OB. Q. E. D.
CoK. 1. If two straight lines are parallel, all points
on either line are equidistant from the other.
Cor. 2. If a parallelogram has one of its angles a
right angle, all its angles must be right angles.
Cor. 3. If one pair of adjacent sides of a parallelo-
gram are equal, all its sides are. equal.
PARALLELOGRAMS 185
tEx. 690. Prove Cor. 1. (See note to i. 21.)
tEx. 601. Prove Cor. 2.
tEx. 692. Prove Cor. 3.
Dep. a parallelogram which has one of its angles a right
angle is called a rectangle.
Cor. 2 proves that all the angles of a rectangle are right angles.
Def. a rectangle which has two adjacent sides equal is
called a square.
Cor. 3 proves that all the sides of a square are equal to one another.
Again, since a square is a rectangle, all its angles are right angles.
Def. a parallelogram which has two adjacent s'des equal is
called a rhombus.
Cor. 3 proves that all the sides of a rhombns are equal to one another.
Revise p. 40 and Ex. 203.
Def. a quadrilateral which has only one pair of sides
parallel is called a trapezium.
Def. a trapezium in which the sides which are not parallel
are equal to one another is called an isosceles trapezium..
tEx. 693. Draw an isosceles triangle ABC and a line parallel to the
base cutting the sides in D, E; prove that DECB is an isosceles trapezium.
HEx. 694. In fig. 195, what lines are equal to (i) PQ, (ii) QR ? Give a
reason.
TEx. 696. In fig. 199, what are the lengths of SV, VT, ST, Z Y, RV ?
Ex. 696. Draw a parallelogram ABOD ; from AB, AD cut off equal
lengths AX, AY ; through X, Y draw parallels to the sides. Indicate in your
figure what lines and angles are equal. (Freehand)
tEx. 697. In fig. 167, ABOD is a parallelogram and PBCQ is a rectangle ;
prove that A BPA= aCQD.
tEx. 698. The bisectors of two adjacent angles of a parallelogram are at
right angles to one another.
tEx. 699. The bisectors of two opposite angles of a parallelogram are
parallel.
tEx. 700. Any straight Une drawn through O, in fig. 147, and terminated
by the sides of the parallelogram is bisected at O.
tEx. 701. ABOD is an isosceles trapezium (AD = BC); prove that
/C=/D.
[Through B draw a parallel to AD.]
tEx. 70a. If in Ex. 701 E, F are the mid-pointa of AB, CD, then EFifl
jierpendicular to AB. [Join AF, BF.]
136 BOOK I
Theorem 23.+
[Converses op Theorem 22.]
(1) A quadrilateral is a parallelogram if both pairs of
opposite angles are equal.
fig. 148.
Data ABCD is a quadrilateral in which
jlA= lC= lx (say) and ^B= ^D= Ly (say).
To -prove that ABCD is a parallelogram.
Prcof The sum of the angles of a quadrilateral is equal to
4 rt. /. 8, L 9, Cor.
.'. 2 z. as + 2 z. y = 4 rt. z. 8,
.'. z. « + z. ^ = 2 rt. z. 8,
.'. z. A + z. B = 2 rt. z. 8,
.•. AD is II to BC. L 4.
Also z- A + z. D = 2 rt. z. 8,
.". AB is II to DC, I. 4.
.*. ABCD is a ||<>8™m q ^ P
(2) A quadrilateral is a parallelogram if one pair of
opposite sides are equal and parallel.
(Draw a diagonal and prove the two triangles congruent.)
(3) A quadrilateral is a parallelogram if both pairs
of opposite sides are equal.
(Draw a diagonal and prove the two triangles congruent.)
(4) A quadrilateral is a parallelogram if its diagonals
bisect one another.
(Prove two opposite triangles congruent.)
Cor. If equal perpendiculars are erected on the same
side of a straight line, the straight line joining their
extremities is parallel to the given line.
PARALLELOGRAMS 137
tEx. 708. Prove i. 23 (2).
tEx. 704. Prove i. 23 (3).
tEx. 705. Prove i. 23 (4).
tEx. 706. Prove the Corollary.
tEx. 707. The straight line joining the mid-points of two opposite sides
of a parallelogram is parallel to the other two sides.
tEx. 70a. ABCD is a parallelogram; AB, CD are bisected at X, Y re-
spectively; prove that BXDY is a parallelogram.
tEx. 709. If the diagonals of a quadrilateral are equal and bisect one
another at right angles, the quadrilateral must be a square.
tEx. 710. Two straight lines bisect one another at right angles; prove
that they are the diagonals of a rhombus.
tEx. 711. If the diagonals of a parallelogram are equal, it must be a
rectangle.
tEx. 712. Au equilateral four-sided figure with one of its angles a right
angle must be a square.
lEx. 713. In a quadrilateral ABCD, Z-A = iLB and LC = l.D; prove
that ABCD is an isosceles trapezium. In what case would it be a parallelo-
gram?
Eevise Ex. 616, 517.
138 BOOK I
Through a given point to draw a straight line parallel
to a given straight line.
rL a
%
B P IQ C
fig. 149.
Let A be the given point and BC the given straight line.
Construction In BC take any point P and cut off any length PQ.
With centre A and radius PQ. describe a circle.
"With centre P and radius AQ describe a cirde.
Let the circles intersect at R.
Join AR.
Then AR is || to BC.
Proof Join AQ and PR.
In the quadrilateral ARPQ
(AR = QP,
tAQ=RP,
.-. ARPQ is a \f^. L 23 (3).
.'. AR is II to BC.
The set square method of dr&wing parallels is the most practical (see
p. 36).
Ex. 714. Show how to construct, without using set square, a parallelogram
having given two adjacent sides and the angle between them,
Ex. 715. Show how to construct a square on a given straight line.
Ex. 7ie. Show how to construct a rectangle on a given f-traight line,
having each of its shorter sides equal to half the given line.
tEx. 717. Show how to construct a rhombus on a given straight line,
having one of its angles = 60° (without protractor or set square). Give a proof.
Ex. 718. Construct a parallelogram having two sides and a diagonal
equal to 5 cm. , 12 cm., 13 cm, respectively. Measure the other diagonal.
Ex. 719. Construct a rectangle having one side of 2*5 in, and a diagonal
of 4 in. Measure the sides.
Ex. 7 SO. Construct a parallelogram with diagonals of 3 in. and 5 in.
intersecting at an angle of 53°. Measure the shortest side.
CONSTRUCTION OF PARALLELS 139
Ex. 731. Construct a rectangle with a diagonal of 7 cm., the angle
between the diagonals being 120°. Measure the shortest side.
Ex. 723. Construct a rhombus with diagonals of 4 in. and 2 in.
Measure the side.
Ex. 733. Construct a square whose diagonal is 3 in. long. Measure its
side.
Ex. 734. Construct an isosceles trapezium whose /\ \
sides are 4 in., 3 in., 1*5 in., 1-5 in. Measure its acute j \ \
angles.
^ fig. 150.
To draw a straight line parallel to a given straight
line and at a given distance from it.
9-^ — . ^ -— Q
fig. 151.
Let AB be the given straight line and '5 in. the given distance.
Construction In AB take any two points 0, D, as far apart as
possible.
"With C, D as centres and radius of "5 in. describe
two circles.
With at ruler draw a common tangent PQ to the two circles.
Then PQ is parallel to AB.
Proof This must be postponed, as it depends on a theorem in
Book III;
Ex. 735. On a base 3 in. long construct a parallelogram of height 1*2 in.
with an angle of 55°. Measure the other side.
Ex. 736. Construct a rhombus whose side is 7*3 cm,, the distance
between a pair of opposite sides being 5*6 cm. Measure its acute angle.
HEx. 737. Draw a straight line and cut off from it two equal parts AC,
CE; through A, 0, E draw three parallel straight lines and draw a line
cutting them at B, D, F ; measure BD, DF. (See fig. 152.)
liEx. 738. Draw a straight line and mark off equal parts PR, RQ ; join
P, Q, and R to a point O ; draw a straight line (not parallel to PQ) to out
^P, OQ, OR at2», q, r ; ia pr=qr^
140 BOOK I
Theorem 24.
If there are three or more parallel straight lines, and
the intercepts made by them on any one straight line that
cuts them are equal, then the corresponding intercepts
on any other straight line that cuts them are also equal.
■4
1 \
.
H /\
! \f
/
ag-
152.
Data The parallels AB, CD, EF are cut by the straight lines
ACE, BDF, and the intercepts AC, CE are equaL
To prove that the corresponding intercepts BD, DF are equal
Construction Through B draw BH |1 to ACE to meet CD at H.
Through D draw DK 1| to ACE to meet EF at K,
Proof [AsBHD, DKF must be proved congruent],
AH is a ll**™", .'. AC = BH, I, 22,
CK is a ll"*""", .-.CE^DK.
But AC = CE, Data
:. BH = DK.
Kow CD is II to EF,
.'. L BDH = corresp. L DFK. I. 6.
Again BH, DK are || (each |1 to ACE),
.'. L DBH = corresp. L FDK, I. 6.
/. in As BHD, DKF
I LBDH=lDFK,
\ L DBH = L FDK,
( BH = DK,
.'. the As are congruent,
.•. BD = DF. I. 11.
Q. B. D.
SUBDIVISION OF A STRAIGHT LINE 141
*^Ex. 720. In fig. 152, if AB, CD are parallel and AC = CE and BD = DF,
prove that EF is parallel to CD.
[Use reductio ad absurdum.]
tEx. 730. The straight line drawn through the mid-point of one side of
a triangle parallel to the base bisects the other side.
[Let A, B coincide in fig. 152.]
tEx. 731. Tbe straigbt line jbinlng tbe mid-points of the sides of a
triangle is parallel to the base.
[Prove this (i) by reductio ad absurdum ;
(ii) directly, with the following construction : —
Let ABC be the triangle ; D, E the mid-points of AB, AC. Produce DE
to F so that EF = DE. Join CF.]
tEx. 732. The straight line joining the mid-points of the sides of a
triangle is equal to half the base.
[Join the mid-point of the base to the mid-point of one of the sides.]
tEx. 733. The straight lines jofaing the mid-points of the sides of a
triangle divide it into four congruent triangles.
tEx. 734. Given the three mid-points of the sides of a triangle, construct
the triangle. Give a proof.
tEx. 735. If AD = ^AB and AE=JAC, prove that DE is parallel to BC
and equal to a quarter of BC.
ilZx. 736. If the mid-points of the adjacent sides of a quadrilateral are
joined, Iha figure thus formed is a parallelogram.
[Draw a diaconal of the quadrilateral.]
tEx. 737. The straight lines joining the mid-points of opposite sides of
a quadiilateral bisect one another.
Ex. 733. Draw a straight Une 4 in. long ; divide it into seven equal parts
hy calculating the length of one part and stepping off with dividers.
142 BOOK I
To divide a given straight line into five equal parts.
-c
Let AB be the given straight line.
Construction Through A draw AC making any angle with AB.
From AC cut off any part AD.
From DC cut off parts DE, EF, FG, GH, equal to AD, so
that AH is five times AD.
Join BH.
Through D, E, F, G draw st. lines || to BH.
Then AB is divided into 5 equal parts.
Proof AD = DE = ,.., Constr.
and Dd, Ee, ..., HB are all parallel. Constr.
.-. Ad=de= ..., I. 24.
.'. AB is divided into 5 equal parts.
The graduated rider must not be tised in the constructions
qf Ex. 739—747.
SUBDIVISION OF A STRAIGHT LINE
143
Ex. 739. Dlvld« a given stxalght line AB into five equal parts by
means of tbe following construction : —
As in fig. 153, draw AC and out off equal
parts AD, D£, EF, FG. GH ; through B draw
BK parallel to HA and cut off from it BP, PQ,
QR, RS, ST each equal to AD. Join OP,
FQ, .... These lines divide AB into five
equal parts.
Give a proof. ' fig. 154,
Ex. 740. Trisect a given straight line by eye; check by making the
construction.
Ex. 741. Divide a straight line of 10 cm. into six equal parts; measure
the parts. Give a proof.
Ex. 742. From a given straight Une cut off a part equal to ^ of tbe
whole line.
Ex. 743.
Ex. 744.
Ex. 745.
Divide a straight line decimally (i.e. into ten equal parts).
Construct a line equal to (i) 1^, (ii) 1-2 of a given line.
Divide a straight line of 13*3 cm. in the ratio of 3 : 4.
[Divide the straight hne (AB) into seven (i.e 3 + 4) equal parts ; if D is the
third point of division from A, AD contains three parts and DB contains four
AD 3
DB~4"^
parts,
Ex. 746. Divide a straight line in the ratio of 5 : 3.
Ex. 747. Divide a straight line 10 om. long so that the ratio of the two
parts may be 4.
Loci.
Mark two points A and B, 2 inches apart. Mark a point
3 inches from A and also 3 inches from B : then a point 4 inches
from A and B.
In a similar way mark about 10 points equidistant from A
and B ; some above and some below AB.
Notice what pattern this set of points seems to form. Draw
a line passing through all of them.
Find a point on AB equidistant from A and B ; this belongs to
the set of points.
The pattern formed by all possible points equidistant from
two fixed points A and B is called the locus of points equidistant
from A and B,
G. s. XO
144 BOOK I
ITEz. 748. What is the loons of points at a distance of 1 inch from a fixed
point O?
IF Ex. 740. Draw a straight line right across your paper. Construct the
locus of points distant 1 inch from this Une. *
(Do this either hy marking a number of such points; or, if you can,
without actually marking the points. Remember that the distance is
reckoned perpendicular to the line.)
ITEx. 760. A bicyclist is riding straight along a level road. What is the
locus of the hub of the back wheel?
VEx. 751. What is the locus of the tip of the hand of a clocks
II Ex. 762. What is the locus of a man's hand as he works the handle of
a common pump?
H Ex. 768. A stone is thrown into stiU water and causes a ripple to spread
outwards. What is the locus of the points which the ripple reaches after
one second?
HEx. 764. Sound travels about 1100 feet in a second. A gun is fired;
what is the locus of all the people who hear the sound 1 second later.
ITEx. 766. A round ruler rolls down a sloping plank; what is the locus
of the centre of one of the ends of the ruler ?
11 Ex. 766. A man walks along a straight road, so that he is always equi-
distant from the two sides of the road. What is his locus?
HEx. 767. A runner runs round a circular racing-track, always keeping
one yard from the inner edge. What is his locus?
HEx. 768. Two coins are placed on a table with their edges in contact.
One of them is held firm, and the other roUs round the circumference of the
fixed coin. What is the locus of the centre of the moving coin ? Would the
locus be the same if there were slipping at the point of contact?
'^Ex. 769. What is the locus of a door-handle as the door opens?
ITEx. 760. What is the locus of a clock-weight as the clock runs down?
ITEx. 761. Slide your set-square round on your paper, so that the right
angle always remains at a fixed point. What are the loci of the other two
vertices?
The above exercises suggest the foUoMring alternative defini-
tion of a locus.
Def. If a point moves so as to satisfy certain conditions the
path traced out by the point is called its locus.
LOCI 145
Ex. 763. A man stands on the middle rung of a ladder against a wall.
The ladder slips down ; find the locns of the man's feet.
(Do this by drawing two straight lines at right angles to represent the
wall and the ground; take a length of, say, 4 inches to represent the
ladder; draw a considerable number of different positions of the ladder as it
slips down; and mark the middle points. This is called plotting a locus.
The exercise is done more easily by drawing the ladder (the line of'
4 inches) on transparent tracing-paper ; then bring the ends of the ladder
on to the two lines of the paper below ; and prick through the middle point.)
ITEx. 763. Draw two unlimited lines, intersecting near the middle of your
paper at an angle of 60°. By eye, mark a point equidistant from the two
lines. Mark a number of such points, say 20, in various positions. The
pattern formed should be two straight lines. How are these lines related
to the original lines? How are they related to one another?
HEx. 764. (On squared paper.) Draw a pair of lines at right angles
(OX, OY); plot a series of points each of which is twice as far from OX as
fromOY. What is the locus? (Keep your figure for the next Ex.)
Ex. 766. Using the figure of Ex. 764, plot the locus of points 3 times as
far from OX as from OY; also the locus of points ^ as far from OX as
from OY.
Ex. 7,66. (On squared paper.) Plot the locus of a point which moves
so that the sum of its distances from two lines at right angles is always
4 inches.
Ex. 767. (On squared paper.) Plot the locus of a point which moves so
that the difierence of its distances from two lines at right angles is always
1 inch.
Ex. 768. Draw a line, and mark a point O 2 inches distant from the line.
Let P be a point moving along the line. Experimentally, plot the locus of
the mid-point of OP.
Ex. 769. A point O is 3 cm. from the centre of a circle of radius 5 cm.
Plot the locus of the mid-point of OP, when P moves round the cfrcum-
ference of the oirde.
10—2
146
BOOK I
Theorem 25.
The locus of a point which is equidistant from two
fixed points is the perpendicular bisector of the straight
line joining the two fixed points.
Constimction
Proof
N B
fig. 155.
Data P is any one position of a point which is always equi
distant from two fixed points A and B.
To prove that P lies on the perpendicular bisector of AB.
Join AB ; let N be the middle point of AB.
Join NP.
In the As ANP, BNP,
/AP = BP,
J AN = BN,
[PN is common,
.'. the triangles are congruent,
.'. z.ANP = z.BNP,
.'. PN is ± to AB,
.'. P lies on the perpendicular bisector of AB.
Sim'y it may be shown that any other point equidistant
from A and B lies on the perpendicular bisector of AB.
Q. K D.
Note, It will be noticed that N is a point on the locus.
tEx. 770. Prove that any point on the perpendicular bisector of a line
AB is equidistant from A, B.
Data
Constr.
I. 14.
LOOI
147
Theorem 26.
The locus of a point which is equidistant from two
intersecting straight lines consists of the pair of straight
lines which bisect the angles between the two given lines.
fig. 156.
DcUa AOA', bob' are two intersecting straight lines; P is any
one position (in l AOB, say) of a point which is always
equidistant from AOA', BOB'.
To prove that P lies on one of the bisectors of the angles formed
by AOA', BOB'.
Construction Draw PM, PN x to AA', BB' respectively.
Join OP.
Troof In the rt. z.d As POM, PON,
' /. s M and N are rt. l s, Ccmstr.
OP is common,
PM = PN, Lata
.'. the triangles are congruent, i. 15.
.•. z.POM= Z.PON,
.'. P lies on the bisector of L AOB (or l A'OB').
Sim^, if P be taken in L AOB' or L A'OB, it may be shown
that the point lies on the bisector of L AOB' (or z. A'OB).
Q. E. D.
148 BOOK I
fEx. 771. Prove that any point on the bisector of an augle is equidistant
from the arms of that angle.
tEx. 778. Prove formally that the locus of points at a distance of 1 incli
from a given line, on one side of it, is a parallel line. (Take two such
points, and show that the line joining them is parallel to the given line.)
tEx. 778. O is a fixed point. P moves along a fixed line; Q is in OP
produced, and PQ=OP. Prove that the locus of Q, is a parallel line.
Intebsection op Loci.
Draw two unlimited straight lines AOA', BOB', intersecting
at an angle of 45°. It is required to find a point (or points)
distant 1 inch from each line.
First draw the locus of points distant 1 inch from AOA' ; this
consists of a pair of lines parallel to AOA' and distant 1 inch
from it. The points we are in search of must certainly lie some-
where upon this locus.
Next draw the locus of points distant 1 inch from BOS'. The
required points must lie upon this locus also.
The two loci will be found to intersect in four points. These
are the points required.
Measure the distance from O of these points.
Ex. 774. Draw two unlimited straight lines intersecting at an angle of
80°. Find a point (or points) distant 2 cm. from the one line and 4 cm. from
the other.
Ex. 775. Draw an unlimited straight line and mark a point O 2 inches
from the line. Find a point (or points) 3 inches from O and 3 inches from
the line. (What is the locus of points 3 inches from O ? What is the locus
of points 3 inches from the line ? Draw these loci.) Measure the distance
between the two points found.
Ex. 776. In Ex. 775 find two points distant 4 inches from O and from
the line. Measure the distance between them.
Ex. 777. In Ex. 775 find as many points as you can distant 1 inch from
both point and line.
Ex. 778. Qiven two points A, B 3 inches apart, find a point (or points)
distant 4 inches from A and 5 inches from B.
LOCI 149
Ex. 779. Make an angle of 45°; on one of the arms mark a point A
3 inches from the vertex of the angle. Find a point (or points) equidistant
from the arms of the angle, and 2 inches from A. Measure distance
between the two })oints found.
Ex. 780. Draw a circle of radius 5 cm. and mark a point A 7 cm. from
centre of circle. Find two points on the circle 3 cm. from A, and measure
the distance between them.
Ex. 781. Construct a quadrilateral ABCD, having AB = 6cm.,
BC = 13cm., CD = 10cm., 2ABC = 70°, / BCD =60°.
On diagonal BD (produced if necessary), find a point
(1) equidistant from A and 0,
(2) equidistant from AB and AD,
(3) equidistant from AB and DC.
In each case measure the equal distances.
Ex. 782. Find two points on the base of an equilateral triangle (side
3 inches) distant 2*7 Inches from the vertex. Measure distance between
them.
Ex. 788. Find a point on the base of an equilateral triangle (side 10 cm.)
which is 4 cm. from one side. Measure the two parts into which it divides
the base.
Ex. 784. On the side AB of an isosceles triangle ABC (base BCi=2 ins.,
Z A=36°), find a point P equidistant from the base and the other side AC.
Measure AP, and the equal distances.
tEx. 785. In Ex. 784 2>rove that AP=CP-CB.
Ex. 786. Find a point on the base of a scalene triangle equidistant from
the two sides. Is this the middle point of the base ?
Ex. 787. Draw a circle of radius 2 Ins.; a diameter; and a parallel line
at a distance of 3 ins. Find a point (or points) in the circle equidistant from
the two lines. Measure distance between these points.
Ex. 788. Draw a circle, a diameter AB, and a chord AC through A.
Find a point P on the circle equidistant from AB and AC. Measure PB
and PC.
Ex. 789. In Ex. 788, find a point on the oirole equidistant from AB and
CA« produced.
160 BOOK I
Ex. 7©0. Draw A ABC having AB = 2-8 ins., AC ==4-6 ins., BC = 4'6ins.
Find a. point (or points) equidistant from AB and AC, and 1 inch from BC.
Measure distance between points.
Ex. 701. Using the triangle of Ex. 790, find a point (or points) equi-
distant from AB and AC, and also equidistant from B and C. Test the
equidistance by measurement.
Ex. 793. In triangle of Ex. 790, find a point (or points) 2 inches from
A, and equidistant from B, C. Measure the distance between them.
Ex. 793. Draw a triangle ABC ; find a point O which is equidistant from
B, C ; and also equidistant from C, A. Test by drawing circle with centre O
to pass through A, B, C.
Ex. 794. Two lines XOX', YOY' intersect at O, making an angle of 25°.
A lies on OX, and OA=7om. Through A is drawn AB parallel to YOY'.
Find a point (or points) equidistant from XOX' and YOY'; and also
equidistant &om AB and YOY'. Draw the equal distances and measure
them.
Ez. 796. Draw a triangle ABC. Inside the triangle find a point P which
is equidistant from AB and BC; and also equidistant from BC and CA.
From P draw perpendiculars to the three sides ; with P as centre and one of
the perpendiculars as radius draw a circle.
Ex. 796. A river with straight banks is crossed, slantwise, by a straight
weir. Draw a figure representing the position of a boat which finds itself at
the same distance from the weir and the two banks.
tEx. 797. P is a moving point on a fixed line AB; O is a fixed point out-
side the line. P is joined to O, and PO is produced to Q so that OQ= PO.
Prove that the locus of Q is a line parallel to AB. (See Ex. 772.)
Ex. 798. Use the locus of Ex. 797 to solve the following problem. O is
a point in the angle formed by two lines AB, AC.
Through O draw a line, terminated by AB, AC,
and bisected at O. ( C
Ex. 799. Draw a figure like fig. 157, making
radius of circle 2 ins., 00 = 3 ins., ON = 5 ins.
Through O draw a line (or lines), terminated by —
AB and the circle, and bisected at O. (See A N B
Ex. 797.) fig. 157.
Ex. 800. A town X is 2 miles from a straight railway ; but the two
stations nearest to X are each 3 miles from X. Find the distance between
the two stations.
LOCI 151
Construction op Triangles, etc. by means op Loci.
In Exs. 801 — 811 accurate figures need not be drawn unless
technical skill is required.
Ex. 801. Construct A ABC, given
(i) base BC = 14 cm., height = 9 cm., Z.B = 65°. Measure AB.
(ii) AB = 59 mm., AC = 88 mm., height AD = 49 mm. (Draw height
first.) Measure base BC.
(iii) BC = 4 in., Z-B = 80°, median CN = 4 in. Measure BA,
(iv) base BC = 12 cm., height AD = 4 cm., median AL = 5 cm. Measure
AB, AC.
Ex. 802. Construct a right-angled triangle, given
(i) longest side =10 cm., another side = 5 cm. Measure the smallest
angle.
(ii) side opposite rigM angle=4 in., another Bide=3 inches. Measure
the third side. 4
Ex. 808. Oonstruot a right-angled triangle ABC, given ^ A =90°,
AB = 7 cm., distance of A from BC = 2-5 cm. Measure the smallest angle.
Ex. 804. Construct an isosceles triangle having each of the equal sides
twice the height. Measure the vertical angle.
Ex. 805. Construct a triangle, given height = 2 in., angles at the ex-
tremities of the base =40° and 60°. Find length of base.
Ex. 806. Construct an isosceles triangle, given the height and the angle
at the vertex (without protractor).
Ex. 807. Construct a parallelogram ABCD, given
AB = 12cm., AD = 10 cm., distance between AB, DC = 8 cm.
Measure the acute angle.
Ex, 808. Construct a rhombus, given that the distance between the
parallel sides is half the length of a side. Measure the acute angle.
Ex. 809. Construct a quadrilateral ABCD, given diagonal AC = 9 cm.,
diagonal BD = 10cm., distances of B, D from AC 5 om. and 4 om. re-
spectively, side AB = 7 cm. Measure CD.
Ex. 810. Construct a trapezium ABCD, given base AB = 10 cm.,
height=4 cm., AD = 4-5 cm., BC = 4*2 cm. Measure angles A and B.
(There are 4 cases.)
Ex. 811. Construct a trapezium ABCD, given base AB=3'5 in.,
height =1*7 in., diagonals AC, BD = 2'5, 3*5 ins. respectively. Measure CD.
162
BOOK I
CO-ORDINATKS.
Take a piefce of squared paper; near the middle draw two
straight lines intersecting at right angles (XOX, YOY in fig. 158).
These will be called axes ; the point O where they intersect will
be called the origin.
- — H
"
Y
f
'
B
h
"
A
0
p
-X
-
H
_c
)_
-
b
Xi
R
s
6
^F
c
._
_
__
_Y.
L
_
_.
fig. 158.
In order to arrive at the point A, starting from the origin O,
one may travel 3 divisions along towards X4-, to the right, and
then 4 divisions wpwards. Accordingly the point A is fixed
by the two numbers (3, 4). These two numbers are called the
co-ordinates of the point A.
Ex. 812. Mark on a sheet of squared paper
(i) the pomts (3, 6), (3, 10), (8, 10), (8, 6).
(ii) the points (1, 2), (2, 4), (3, 6), (4, 8), (5, 10).
(iii) the points (4, 3), (4, 2), (4, 1), (4, 0), (4, - 1), (4, -2).
(iv) the pomts (6, 6), (4, 6), (2. 6). (0, 6), ( - 2, 6).
CO-ORDINATES IBS
To reach B (fig. 158) from O, one may travel 3 divisions along
towards X— to the left, and then 4 divisions upwards. Accordingly
the point B is fixed by the co-ordinates (- 3, 4),
To reach C from O, go 3 divisions^ along to the right, then
4 divisions downwourds. C is therefore (3, — 4).
N.B. To the right is reckoned 4 ; to the left, - .
Upwards is reckoned + ; downwards - .
To get from O to E, it is not necessary to travel alcmg at all;
the journey is simply 4 divisions upwards. Accordingly, E is
the point (0, 4),
Ex. 813. Write down the co-ordinates of the following points in fig. 158 :
D, F, G, H, O, P, Q, R, S.
Ex. 814. Plot (i.e. mark on squared paper) the following points : (5, 0),
(4, 3), (3, 4), (0, 5), (-3, 4), (-4, 3), (-5, 0), (-4, -3), (-3, -4), (0, -5),
(3, -4), (4, -3), (5, 0).
Ex. 816. Plot the points: (8, 16), (6, 9), (4, 4), (2, 1), (0, 0), (-2, 1),
(-4, 4), (-6, 9), (-8, 16).
Ex. 816. Plot the points: (0, 0), (2, 0), (-2, 0), (0, 13), (1, -10), (8, 6),
(_8, - 6), (-3, -5). (The constellation of Orion.)
Ex. 817. Plot the points: (-12, -2), (-8, 0), (-4,0), (0,0), (3,-2),
(7, 0), (5, 4). (The Great Bear.)
Ex. 8 18. (Inch paper.) Find the co-ordinates of two points each of which
is 3 inches from (0, 0) and (2, 2).
Ex. 819. (Inch paper.) Find the co-ordinates of all the points which
are 2 inches from the origin and 1 inch from the a;-axis (XOX).
Ex. 820. (Inch paper.) Find the co-ordinates of all the points which are
equidistant from the two axes and 3 inches from the origin.
Ex. 821. (Inch paper.) Find the co-ordinates of a point which is
equidistant from
(i) (2, -1), (1,8), (-2,0),
(ii) (2.3), (2, -1), (-2,-1).
(iii) (2.3), (2, -1), (-2, -2).
Ex. 822. (Inch paper.) Find the co-ordinates of a point inside the
triangle given in Ex. 821 (i), and equidistant from its three sides.
Ex. 823. Repeat Ex. 822 for the triangles given in Ex. 821 (ii) and (iu).
154 BOOK I
MISCELLANEOUS EXERCISES.
Constructions.
Ex. 834. A ship is sailing due N. at 8 miles an hour. At 8 o'clock a
lighthonse is observed to be N .E. and after 90 minutes it is observed to bear
7i° S. of E. How far is the ship from the lighthouse at the second obser-
vation, and at what time (to the nearest minute) was the ship nearest to the
lighthouse?
Ex. 830. Is it possible to make a pavement consisting of equal equi-
lateral triangles?
Is it possible to do so with equal regular figures of (i) 4, (ii) 5, (iii) 6,
(iv) 7 sides?
Ex. 836. A triangle ABC hasZ.B = 60°, BC=8om.; what is the least
possible size for the side CA? What is the greatest possible size for /.C?
fEx. 837. Draw a triangle ABC and show how to find points P, Q in
AB, AC such that PQ is parallel to the base BC and =^BC. Give a proof.
[Trisect the base and draw a parallel to one of the sides. ]
fEx. 838. In OX, OY show how to find points A, B suoh that
Z.OAB = 3z.OBA. Give a proof.
[Make an angle equal to the sum of these angles.]
tEx. 839. A and B are two fixed points in two unlimited parallel straight
lines* show how to find points P and Q. in these lines such that APBQ is a
rhombus. Give a proof.
fEx. 880. Prove the following construction for bisecting the angle
BAC : — With centre A describe two circles, one cutting AB, AC in D, E,
and the other cutting them in F, G respectively; join DO, EF, intersecting
in H ; join AH.
fEx. 831. A, B are two points on opposite sides of a straight line CD;
show how to find a point P in CD so that Z. APC = L BPC. Give a proof.
fEx. 833. Show how to construct a rhombus PQRS having its diagonal
PR in a given straight line and its sides PQ, QR, RS passing through three
given points L, M, N respectively. Give a proof.
fEx. 833. A and B are two given points on the same side of a straight
line CD ; show how to find the point in CD the difference of whose distances
from A and B is greatest.
Also show how to find the point for which the difierence is least.
EXERCISES ON BOOK I 166
+Ex. 884. A and B are two points on the same side of a straight line
CD ; show how to find the point P in CD for which AP+ PB is least. Give
a proof.
tEx. 835. Show how to describe a rhombus having two of its sides along
the sides AB, AC of a given triangle ABC and one vertex in the base of the
triangle. Give a proof.
fEx. 886. Show how to draw a straight line equal and parallel to a given
straight line and having its ends on two given straight lines. Give a proof.
t^x. 887. To trisect a given angle.
Much time was devoted to this famous problem by the Greeks and
the geometers of the Middle Ages ; it has now been shown that it
is impossible with only the aid of a pair of compasses and a straight
edge (ungraduated).
In fig. 160, DE=the radius of the oirole; prove that L BDE= |Z.ABC.
fig. 159.
Fig. 159 shows a simple form of trisector; the instrument is opened until
the angle between the rods corresponding to BA and BC can be made to
coincide with the given angle ; then the angle between the long rods (corre-
sponding to D) is one-third of the given angle.
With a ruler, marked on its edge in two places, and a pair of compasses,
it is possible to trisect an angle as follows : —
Let A BC be the angle. With B as centre and radius = the distance between
the two marks describe a circle cutting BC at C ; place the ruler so that its
edge passes through C and has one mark on AB produced, the other on the
circle (this must be done by trial, a pin stuck through the paper at C will
help); rule the line DEC, then L^=\L ABC.
156 BOOK I
Theorem&
Ex. 838. The gable end of a boose is in the form of a pentagon, of
which the three angles at the ridge and eaves are equal to each other: show
that each of these angles is equal to twice the angle of an equilateral
triangle.
fEx. 830. If on the sides of an equilateral triangle three other equi-
lateral triangles are described, show that the complete figure thus formed
will be (1) a triangle, (ii) equilateral.
tEz. 840. Two isosceles triangles are on the same base: prove that the
straight line joining their vertices bisects the base at right angles.
tEx. 841. Two triangles ABC, DCB stand on the same base BC and
on the same side of it; prove that AD is parallel to BC if AB=DC and
AC = DB.
tEx. 843. In the diagonal AC of a parallelogram ABCD points P, Q
are taken such that AP=CQ; prove that BPDQ is a parallelogram.
tEx. 843. ABCD, ABXY are two parallelograms on the same base and
on the same side of it. Prove that CDYX is a parallelogram.
tEx. 844. The diagonal AC of a parallelogram ABCD is produced to E,
so that CE=CA; through E, EF is drawn parallel to CB to meet DC pro-
duced in F. Prove that ABFC is a parallelogram.
tEx. 846. E, F, G, H are points in the sides AB, BC, CD, DA re-
spectively of a parallelogram ABCD, such that AH = CF and AE=CG:
show that EFGH is a parallelogram.
tEx. 846. C is the mid-point of AB ; from A, B, C perpendiculars
AX, BY, CZ are drawn to a given straight Une. Prove that, if A and B
are both on the same side of the line, AX-i-BY=2CZ.
What relation is there between AX, BY, CZ when A and B are on
opposite sides of the line?
tEx. 847. If the bisectors of the base angles of an isosceles triangle
ABC meet the opi)osite sides in E and F, EF is parallel to the base of
the triangle.
t£x. 848. In a quadrilateral ABCD, AB = CD and /B=/C; prove
that AD is parallel to BC.
'tEx. 840. Prove that the diagonals of an isosceles trapezium are equal.
EXERCISES ON BOOK I 157
tEx. 850. ABCD is a quadrilateral, such that Z A= / B and / C= / D;
prove that AD = BC.
tEx. 851. The figure formed by joining the mid-points of the sides of a
rectangle is a rhombus.
tEx. 862. The medians BE, CF of a triangle ABC intersect at G;
GB, GC are bisected at H, K resi)ectively. Prove that HKEF is a parallelo-
gram. Hence prove that G is a point of trisection of BE and CF.
■^Ex. 853. The diagonal AC of a parallelogram ABCD is produced to E,
BO that CE = CA; through E and B, EF, BF are drawn parallel to CB, AC
respectively. Prove that ABFC is a parallelogram.
tEx. 854. T, V are the mid-points of the opposite sides PGi, RS of a
parallelogram PQRS. Prove that ST, QV trisect PR.
tEx. 855. Any straight line drawn from the vertex to the base of a
triangle is bisected by the line joining the mid-points of the sides.
tEx. 856. The sides AB, AC of a triangle ABC are produced to X, Y
respectively, so that BX = CY = BC; BY, CX intersect at Z. Prove that
Z.BZX-fiZ.BAC = 90°.
tEx. 857. ABCD is a parallelogram and AD = 2AB; AB is produced
both ways to E, F so that EA=AB=BF. Prove that CE, DF intersect at
right angles.
tEx. 858. In a triangle whose angles are 90°, 60°, 30° the longest side is
double the shortest.
[Complete an equilateral triangle.]
tEx. 859. In a right-angled triangle, the distance of the vertex from the
mid-point of the hypotenuse is equal to half the hypotenuse.
[Join the mid-point of the hypotenuse to the mid-point of one of the
sides.]
tEx. 860. Given in position the right angle of a right-angled triangle
and the length of the hypotenuse, find the locus of the mid-point of the
hypotenuse. (See Ex. 859.)
tEx. 861. ABCD is a square; from A Unes are drawn to the mid-
points of BC, CD; from C lines are drawn to the mid-points of DA, AB.
Prove that these lines enclose a rhombus.
168 BOOK I
tEx. 863. ABC is an equilateral triangle and D is any point in AB; on
the side of AD remote from C an equilateral triangle ADE is described;
prove that BE=CD.
tEx. 863. In a triangle ABC, BE and CF are drawn to cut the opposite
sides in E and F; prove that BE and CF cannot bisect one another.
tEx. 864. If P be any point in the external bisector of the anglQ A
of a triangle ABC, AB + AC<PB + PC.
tEx. 866. ABC is an acute-angled triangle, whose least side is BC.
With B as centre, and BC as radius, a circle is drawn cutting AB, AC at D,
E respectively. Show that, if AD = DE, /.ABC = 2L BAC.
tEx. 866. ABC is an isosceles triangle (AB=AC); a straight line is
drawn cutting AB, BC, and AC produced in D, E, F respectivdy. Prove
that, if DE=EF, BD=CF.
tEx. 867. If two triangles have two sides of the one equal to two sides
of the other, each to each, and the angles opposite to two equal sides equal,
the angles opposite the other equal sides are either equal or supplementary;
»nd in the former case the triangles are congruent.
tEx. 867 a. A quadrilateral ABCD, that has AB = AD and BC = DC, is
called a kite. Use Th. i. 25 to prove that the diagonals of a kite are at
right angles.
tEx. 867 b. If two circles cut at P, Q, use i. 25 to prove that the line
joining their centres bisects PQ at right angles.
BOOK IL
Abea.
■~
—
~
Y
f
■"
"■
"
■^ *
B
h
A
Q
p
X
_
H
n
0
Xf
R
s
D
F
c
_j
y
z.
__
fig. 161.
Area of rectangle. Count the squares in the rectangle
ABDC (fig. 161). They are 48 in number. We say, then, that the
area of ABDC is 48 squares of the paper.
Ex. 868. In each of the following exercises plot the points mentioned,
join them up in the order given, and find the number of squares in the
area.
(i) (1, 16). (9, 16). (9, 1). (1, 1).
(ii) (-6, 2), (2, 2), (2. -13), (-6, -13),
(iu) (0. 0), (8, 0), (8, -15), (0. -15).
(iv) (10, 20), (-10, 20), (-10. -20), (10, -20).
So far, we have taken the unit of length to be one division
of the paper, and the unit of area to be one square of the paper.
G. s. 11
160 BOOK n
If we wish to use the inch for unit of length, we shall need
paper ruled in squares 1 inch each way. On inch paper there are
generally finer lines at distances of ^ inch. The paper will
show larger squares and smaller squares ; the larger sqxiares
1 incli each way, and therefore of area 1 sq. inch; the smaller
squares ^ inch each way. Paper ruled like this will be referred
to as inch paper.
Ex. 869. On inch paper, draw a square inoh. (Use the lines of the
paper to gnide your drawing.)
Ex. 870. On inch paper draw reotan^es whose areas, in square inches,
are 6, 9, 16, 4, 2, 2^, 1.
Ex. 871. Draw two rectangles of different shape so that the area of each
shall be 12 sq. inches. See whether the two rectangles have the same
perimeter (the perimeter is the sum of the sides).
Ex. 87a. Count the number of small squares in one square inch. What
fraction of a square inch is each of these small squares ? What decimal ?
Ex. 873. Mark out a square containing 25 of these small squares. What
decimal of a square inch is this square ? What fraction ?
Ex. 874. Mark out a square containing 64 small squares. What decimal
of a square inch is this ?
Ex. 876. On inch paper, draw the rectangle whose comers are (2, 15),
(7, 15), (7, 2), (2, 2). (Take the side of a small square for unit of length.)
How many hundredths of a square inch are contained in this rectangle?
How many square inches ? (Always express your answer in decimals.)
Ex. 876. Bepeat Ex. 875, taking, instead of the points there mentioned,
the following :—
(i) (-1, 10), (14, 10), (14, -10), (-1, - 10).
(u) (0, 0), (0, 12), (11, 12), (11, 0).
(ui) (-3, 7), (14, 7), (14, -3), (-3, -3).
You will probably have noticed that the most convenient
way of counting the number of squares in a rectangle is as
follows : — count how many squares there are in one row, and
multiply by the number of rows. Or, we may say : count the
number of divisions in the length, and multiply by the number
of divisions in the breadth. Use this plan in the following
exercises:
AHEA — SQUARED PAPER 161
Ex. 877. How many squares are contained in a rectangle drawn on
squared paper, the length being 30 divisions and the breadth 20 ?
Ex. 878. On inch paper draw a rectangle 55 tenths in length and
33 tenths in breadth. How many hundredths of a square inch are there
in the area ? How many square inches?
Ex. 879. Bepeat Ex. 878 with the following numbers for length and
breadth respectively:
(i) 40, 25, (ii) 125, 80, (iii) 23, 17, (iv) 125, 8.
Hitherto we have dealt only with rectangles whose dimensions
are expressed by whole numbers. We will now see whether the
same rule will hold for rectangles whose dimensions are not
expressed by whole numbers.
On inch paper draw a rectangle 5-3 inches long and 4*7 inches
broad. Coiint the number of tenths of an inch in the length
and breadth. Ilence find the number of hundredths of a square
inch in the area. Reduce this to square inches; the result
should be 24-91 sq. inches. Now multiply together the numbers
of inches in the length and breadth: 5*3 x 4*7. The result
is again 24*91.
Why are these two results the same 1 The reason is as
follows : —
53 X 47 53 47
X
= 5-3 X 4-7.
100 10 10
We may now state the rule for the area of any rectangle : —
To find the nmnber of square units in the area of
a rectangle, multiply together the numbers of units in
the length and breadth of the rectangle.
Ex. 880. What is the corresponding rule for calculating the area of
a square ?
Ex. 881. Find the area of a rectangle,
(i) 16-7 ins. by 14-3 ins.
(ii) 10 mm. by 10 mm., in square mm. and also in sq. cm.
(iii) 21-6 cm. by 14-5 cm. , in sq. cm. and also in sq. mm.
(iv) 7 kilometres 423 metres by 1 km. 275 m., in sq. km. and also
in sq. m.
(v) a incAes by b inches,
(vi) X cm. by 2x cm.
11— 'J
162
BOOK n
Ex. 883. Find the area of a sqnare whose side is (i) 70 yards, (ii) 69 yds.
Say in each case whether the square is greater or less than an acre.
Ex. 883. Find the areas of squares of side (i) 2 inches, (ii) 1 foot (in sq.
ins.), (iii) 1 yd. (in sq, ins.), (iv) a cm., (v) 2x ins.
IFEz. 884. Draw a figure to show that if the side ofone square is 8 times
the side of another square, the area of the one square is 9 times the area
of the other, (Freehand)
Ex. 885. Find (i) in sq. ins., (u) in sq. cm., the area of the rectangle
which encloses the print on this page. Hence find the number of sq. cm. in
a sq. inch (to 1 place of decimals).
vr
V9»
^
•5"
,7//
4
10
Ol
.
*_
V
•6^
^
1.2"
i"
fig. 162.
Ex. 886. Make freehand sketches of the given figures (fig, 162), In
each case find the area.
Ex. 887. Find the oiOcier dimension of a rectangle, given
(i) area = 140 sq. ft., one dimension =35 ft.
(il) area=l sq. ft., one dimension =6 ins.
(iii) area = 304 sq. yds., one dimen8ion=5J yds.
(iv) area=l acre ( = 4840 sq. yds.), one dimension =22 yds.
(v) area = 2^^ sq. ins., one dimension =x ins.
Ex. 888. How many bricks 9 in. by 4 in. are required to cover a fioor
34 ft, long by 17 ft, wide ?
AREA — SQUARED PAPER
163
Area of right-angled triangle. By drawing a diagonal of
a rectangle we divide the rectangle into two equal right-angled
triangles. Hence the area of a right-angled triangle may be
found by regarding it as half a certain rectangle.
Ex. 889. Find the number of squares contained by a triangle whose
corners are
(i) (0, 0), (0, 2), (6, 0). (Complete the rectangle.)
(ii) (2, 5), (17, 5), (17, 10).
(iii) (5, -5), (-5, -5), (-5. 5).
(iv) (5, -5), (-5, -5), (5, 5).
Ex. 890. Find the areas of right-angled triangles in which the sides con-
taining the right angle are (i) 2", 3", (ii) 6'5 cm., 4-4 cm., (iii) 4-32", 3-71",
(iv) 112 mm., 45 mm. (in sq. mm. and also in sq. cm.).
Area of any rectilinear figure (on squared paper).
With the aid of rectangles and right-angled triangles we can
find the area of any figure contained by straight lines (i.e. any
rectilinear figure). This way is especially convenient when
one side of the figure runs along a line of the squared paper.
fig. 163.
Mg. 163 shows how a 4-sided figure may be divided up into
rectangles and right-angled triangles ; the number inside each
rectangle and triangle indicates the number of squares it contains;
and the complete area is 199^ or 199'5 squares.
164 BOOK II
Ex. B91. Measoxe the size of the small BqnareB in fig. 168 ; hence find
the area of the 4-sided figure in eq. inches.
Ex. aoa. Find the area (in squares of yonr paper) of each of the following
figures by dividing up the figures into rectangles and right-angled triangles :
(i) (2, 1), (11, 1), (8, 6), (2, G).
(u) (1, 2), (1, 10), (6, 13), (6, 2).
(iu) (5, 0), (8, 4), (-5, 4), (-6, 0).
(iv) (0, 6), (-8, 2), (-3, -2), (0, -8).
(V) (0, 0). (1, 4), (6, 0).
(vi) (1, 4), (C, 3), (1, -3).
(vii) (-4, -3), (-3, 3), (5, 6), (10, -8).
(viu) (3, 5), (-3, 2), (-5, -3), (3, -7).
(ix) (3, 0), (0, 6), (-3, 0), (0, -6).
(X) (2, 5), (5, 2), (5, -2), (2, -6), (-2, -5), (-6, -2), (-5, 2), (-2, 5).
Are all the sides of this figure equal ?
(xi) (3, 4), (4, 3), (4, -3), (3, -4), (-3, -4), (-4, -3), (-4, 3), (-3, 4).
(xii) (5, 0), (4, 3), (3, 4), (0, 5). (-3, 4), (-4, 3), (-6, 0), (-4, -3),
(-3, -4), (0, -5), (3. -4), (4, -3).
Ex. 803. Draw the three following figures on the same axes ; find the
area and perimeter of each.
(i) (1, 1), (1, 6), (6, 6), (6, 1).
(ii) (1, 1), (4, 5), (9, 5). (6. 1).
(iii) (1. 1), (6. 4). (10, 4), (6, 1).
(This exercise shows that two figures may have the same perimeter and
different areas.)
Ex. 894. Draw the two following figures on the same axes ; find the area
and perimeter of each.
(i) (0. 0), (7, 0), (9, 6), (2, 6).
fii) (0. 0), (7, 0), (3, 6), (-4, 6).
(This exercise shows that two figures may have the same area and
different perimeters.)
Ex. 806. Find the area of
(i) (1, 0), (1, 8), (4, 14), (2, 14), (0, 10), (-2, 14), (-4, 14), (-1, 8), (-1, 0).
(u) (5, 7), (-4, 7), (-6, 6), (1. 6), (-5, -7), (5, -7), (6, -5), (-1, -6).
AREA — SQUARED PAPER
165
If there is no side of the figure which coincides with a hne
of the paper (ABCD in fig. 164), it is generally convenient to draw
lines outside the figure, parallel to the axes, thus making up a
rectangle (PQRS) ; the area required can then be found by sub-
tracting a certain number of right-angled triangles from the
rectangle.
-
p
-
^t"^
I
-
-
-
-
-
-
-
-
Q
)
/
s
^
J
/
sj
s
li
i-
—
-
il
/
s
s
/
^
>s
J
^J
S
R
/
/
D
/
i
f
>
/
.
f
J
D
6
f
^
^
/
to
1
.8
L_
^
*i«
/
'
*
1
1
,
^
>«
^
f\
"
_
S
,
-\
-1
-
-
-
-(
-
^-
fig. 164.
Thus in fig. 164
ABCD = PQRS - AQB - BRC - CSD - DPA
= 221-25-20-18-35
= 123,
Ex. 896. Find the areas of the following figures s —
(i) (1, 1), (16, 5), (9, U).
(ii) (6, 3), (12, 9), (3, 11). >=-;>
(iii) (10, -20), (20, -24), (12, 4).
(iv) (0,0), (9,-l),(7,6)#(2, 5).
(v) (1, 0), (6, 1), (5, 6), (0, 5).
(vi) (3, 0), (7, 3), (4, 7), (0, 4).
(vii) (4, 0), (10, 4), (6, 10), (0, 6).
(viii) (5, 0), (0, 5), (-5, 0), (0, -6).
166
BOOK II
Area of a curvilinear figure. This cannot be found exactly
by the method of counting squares : tlie approximate vahie
however is easily calculated as follows.
fig. 165.
To find the area of the fig. ACBA, notice that the curved
boundary ACB cuts through various squares ; in counting squares
we have to decide what is to be done with these broken squares.
The following rule gives a useful approach to the true value : —
Jf the broken square is more than half a complete square, cou/nt 1 /
if less tham. half a square, Qownt 0.
Counting up the squares in ACBA on this system, we find
that the area is 72 squares. As each of the above squares is
j^ sq. inch, the area is -72 sq. inches.
Ex. 897. On inch paper draw a circle of radius 1 inch; find its area
as above, and reduce to square inches. (The counting can be shortened
in various ways ; e.g. by dividing the circle into 4 quarters by radii.)
Ex. 898. Find the area of circles of radii 2, and 3 inches. Calculate,
to 2 places, how many times each of these circles contains the 1-inch
circle of Ex. 897.
Ex. 899. Plot the graph V = 6 - -^ , and find the area contained between
o
the curve and the ^-axis.
AREA OF PARALLELOGRAM 167
Def. Any side of a parallelogram may be taken as the
base. The perpendicular distance between
the base and the opposite (parallel) side ^/-v 7*^
is called the height, or altitude. /" j^^--,. /
Thus in fig. 166 if BC be taken as base, MN g^- ^ q
(which may be drawn from any point of the
base) is the height (or altitude). If AB be taken fig. 166.
as base, G H is the height.
II Ex. 900. In fig. 166 what is the height if CD be taken as base? if AD
be taken ?
Ex. 901. Prove that the altitudes of a rhombus are equal.
Area of parallelograiu. Take a sheet of paper (a rectangle)
and call the corners P, B, C, Q; BC being
one of the longer sides (fig. 167). Mark
a point A on the side PQ. Join BA, and
cut (or tear) off the right-angled triangle
PBA. You now have two pieces of paper;
you will find that you can fit them
together to make a parallelogram (A BCD ~ fig. 167.
in % 167).
Notice (i) that the rectangle you had at first and the paral-
lelogram you have now made, a,re composed of the same paper,
and thefrefore have the sa/me area.
(ii) thait the rectangle and the parallelogram are on the
same base BC, and both lie between the same pair of parallel
lines BC and PAQD. Or, we may say that they have the same
height.
ITEx. 002. Make a paper parallelogram with sides of 6 and 4 ins. and an
angle of 60°. Cut the parallelogram into two pieces which you can fit
together to make up a rectangle. Find its area.
Ex. 80a. Repeat Ex. 902 with sides of 12 and 6 cm. and angle of 60°.
168 BOOK II
Ex. 0O4. Draw a parallelogram ABCD, having AB= 13 om., BC = 16om.,
angle B = 70° ; on the same base draw a rectangle of equal area ; find the area.
Measure the two altitudes of the parallelogram and calculate the products
BC . MN and AB . GH (see fig. 166).
Ex. 90S. On base 2 inches draw a parallelogram of angle 50° and height
4 inches. On the same base construct a rectangle of the same area; and
find the area. Also calculate the products BC.MN and AB.GH as in
Ex. 904.
Ex. 0O6. Repeat Ex. 905 with the same base and height, but with
angle of 76°.
Dep. Figures which are equal in area are said to be
equivalent
Notice that congruent figures are necessarily equivalent ; but that equiva-
lent figures are not necessarily congruent.
HEx. 807, Give the sides of a pair of equivalent rectangles, which are not
congruent.
AREA OF PARALLELOGRAM 169
Theorem 1.
Parallelograms on the same base and between the
same parallels (or, of the same altitude) are equivalent.
Data ABCD, PBCQ are \\°«'^°"^ on the same base BC, and between
the same parallels BC, PD.
To prove that ABCD and PBCQ are equivalent.
Proof In the As PBA, QCD,
L BAP = corresp. L CDQ ( •.• BA, CD are ||), I. 5.
L BPA = corresp. l CQD (•/ BP, CQ are ||), I. 5.
BA = CD (opp. sides of ll"*"^ ABCD), I. 22.
.*. the triangles are congruent. l. 11.
Now if A PBA is subtracted from figure PBCD, ||oeram ^q jg
left; and if A QCD is subtracted from figure PBCD, Ij"*""*" BQ
is left.
Hence the ||°8™°>8 are equivalent.
Q. E. D.
Cor. 1. Parallelograms on equal bases and of the same
altitude are equivalent.
(For they can be so placed as to be on the same base and
between the same parallels.)
Cor. 2. The area of a parallelogram is measured by
the product of the base and the altitude.
(For the y^""^™ is equivalent to a rectangle on the same base
and of the same altitude, whose area = base x altitude.)
170 BOOK II
Ex. 908. Find the area of a parallelogram of sides 2 ins. and 3 ins. and
of angle 30°.
Ex. 9O0. Draw a rectangle on base 12 cm. and of altitude 10 cm. ; on the
same base construct an equivalent parallelogram of angle 60°; and measure
its longer diagonal.
Ex. OlO. Show how to construct a parallelogram equivalent to a given
rectangle, on the same base and having one of its angles equal to a given
angle (without using protractor).
Ex. 911. Draw a rectangle of base 4 ins., and height 8 ins. : on the
same base make an equivalent parallelogram with a pair of sides of 5 ins.
Measure the angle between the base and the shorter diagonal.
Ex. 012. Show how to construct on the same base as a given rectangle
an equivalent parallelogram having its other side equal to a given straight
line (without using scale). Is this always possible ?
Ex. 013. Draw a rectangle whose base is double its height; on the same
base construct an equivalent rhombus and measure its acute angle.
Ex. 014. Transform a rectangle of base 4-53 cm. and height 2*97 cm.
into an equivalent parallelogram having a diagonal of 8'45 cm. Measure the
angle between the base and that diagonal.
Ex. 015. Transform a parallelogram of sides 2 and 1 ins. and angle 80°
into an equivalent parallelogram of sides 2 and 2*5 ins. Measure acute angle
of the latter.
Ex. 016. Transform a parallelogram of sides 8-3 and 12-4 cm. and angle
12° into an equivalent rhombus of sides 8*3 cm. Measure angle of rhombus.
Ex. 017. Repeat Ex. 916, making side of rhombus 12-4 cm.
Ex. 018. Transform a parallelogram of base 2-34 ins., height 2*56 ins.
and angle 67° into an equivalent parallelogram on the same base with angle
60°. Measure the other side of the latter.
Ex. 010. Transform a given parallelogram into an equivalent parallelo-
gram with one of itf> angles = a given angle (without using protractor).
Ex. 02O. Make parallelogram ABCD, with AB=2-5 ins., AD = 3ins.,
angle A = 60°, Transform this into an equivalent parallelogram with sides
of 2 ins. , and 4 ins.; measure acute angle of the latter.
(First, keeping the same base AB, make equivalent parallelogram ABEF
having AE=4 ins. Next, taking AE for iMise, oonstmct an equivalent
parallelogram with sides 2 and 4 ins.)
AREA OF PARALLELOGRAM 171
Ex. 921. Show how to make a parallelogram equivalent to a given
rectangle, having its sides equal to two given lines. Is this always possible?
Ex. 922. Construct a parallelogiam of sides 9 and 8 cm. and angle 20° ;
make an equivalent rhombus of side 6 cm. and measure its angle.
Ex. 923. Bepeat Ex. 922, with angle 30° mstead of angle 20°.
Ex. 924. What is the locus of the intersection of the diagonals of a
parallelogram whose base is fixed and area constant ?
In calculating the area of a parallelogram by means of
IL 1 (area = base x height), you will notice
that the product may be formed in two
different ways; e.g. in fig. 169 we may uV!"
take either BC.MN or AB.GH; these / ""
two products should be equal, being both [_
equal to the area. In practice it will B N C
be found that the two results do not ^' ^^^'
generally agree exactly; (what is the
reason for this ?). The difference however should not be greater
than 1 or 2 per cent. In order to get the best possible result
for the area, calculate both products and take the average.
Ex. 926. Find the area of each of the following parallelograms, taking
the average of two results as explained above.
(i) Sides 3-6 and 4-5 ins., angle of 70°.
(ii) Sides 12-7 and 14-5 cm., angle of 120°.
(iii) Sides 10 and 6 cm., angle of 30° (in this case one of the altitudes
will fall partly outside the parallelogram ; produce a side).
(iv) Sides 5-53 and 1-61 ins., angle of 160°.
(v) Diagonals 3*7 and 2-2 ins., angle hetween diagonals 55°.
(vi) Equal diagonals of 3-2 ins. , angle between diagonals 150°.
(vii) Sides 6-6 and 8-8 cm., a diagonal of 11 cm.
Ex. 926. Find the area of a rhombus of side 2 inches and angle 30°.
Ex. 927. Find (correct to jfg inch) the height of a rectangle whose aiea
is 10 sq. ins. and whose base =3*16 ins.
Ex. 928. Draw a parallelogram of area 24 sq. cm., base 6 cm. and angle
75°. Measure the other sides.
172 BOOK II
Ex. 939. Draw a parallelogram of area 12 sq. ins., sides of 4 and 8'5 inB.
Measure its acute angle.
Ex. 980. Draw a rhombus of area 24 sq. om. and side 6 cm. Measure
its acute angle.
Ex. 981. Draw a parallelogram of area 15 sq. ins., base 5^ins. and
diagonal 4 ins. Measure the acute angle.
Area of Triangle.
Dep. Any side of a triangle may be t«iken as base. The
line drawn perpendicular to the base from the opposite vertex is
called the height, or altitude.
There will be three different altitudes according to the side which is
taken as base.
ITEx. 932. Draw an acute-angled triangle and draw the three altitudes.
(Freehand.)
HEx. 933. Eepeat Ex, 932 for a right-angled triangle. {Freehand.)
^Ex. 934. Bepeat Ex. 932 for an obtuse-angled triangle. {Freehand.)
IfEx. 936. In what case are two of the altitudes of a triangle equal ?
^Ex. 936. In what case are all three altitudes equal ?
^Ex. 937. In what case do some of the altitudes fall outside the triangle?
HEx. 938. By making rough sketches, try whether you can find a triangle
(1) in which one (and only one) altitude falls outside, (2) ini which all three
altitudes fall outside.
ahea of triangle 173
Theorem 2.
Triangles on the same base and between the same
parallels (or, of the same altitude) are equivalent.
Data ABC, PBC are As on the same base BC, and between the
same parallels BC, PA.
To prove that ABC, PBC are equivalent.
Construction Complete the ||08ramB abCD, PBCQ by drawing
CD, CQ II to BA, BP respectively, to meet PA (produced if
necessary) in D, GL
Then A ABC = 1 1|<^*'» ABCD, i. 22 (3).
and A PBC = | f^"^ PBCQ. I. 22 (3).
But ||<«ram8 ABCD, PBCQ are equivalent, being on the
same base and between the same parallels. li. 1.
.•. A ABC = A PBC.
Q. E. D.
Cor. 1. Triangles on equal bases and of the same altitude
are equivalent.
(For they can be so placed as to be on the same base and
between the same parallels.)
Cor. 2. The area of a triangle is measured by half the
product of the base and the altitude.
tEx. 939. Prove Cor. 2.
tEx. 940. Prove that, in general, the area of a triangle is less than half
the product of two of its sides.
tEx. 94X. Prove that the area of a right-angled triangle is half the pro-
duct of the sides which contain the right angle.
174 BOOK II
Since any one of the three sides may be taken for base, there
are three different ways of forming the product of a base and the
corresponding altitude. Thus the area may be calculated in
three different ways ; and of course, theoretically, the result
is the same in each case. Practically, none of the measurements
will be quite exact, and the results will generally differ slightly.
To get the best possible value for the area take the average of
the three results.
Ex. 943. Find, to three significant figures, the areas of the following
triangles, taking the average of three results in each case :
(i) sides 3, 4, 4-5 ins.
(ii) sides 6, 8, 9 cm.
(iii) sides 3, 4, 5 ins.
(iv) sides 6, 8, 10 cm.
(v) sides 2, 3, 4-5 in&
(vi) sides 4, 7, 10 cm.
(vii) sides 3, 4 ins., included ^120°.
(viii) BC=7-2cm., Z.B=20°, Z;C=40°.
Ex. 943. Make a copy of your set-square and find its area (i) in sql
inches, (ii) in sq. cm.
Ex. 944. (On inch paper.) The vertices of a triangle are the points
(2, 0), ( - 1, 2), ( - 2, - 2). Find the area (i) by measuring sides and altitudes,
(ii) as on p. 165.
Ex. 946. ' (On inch paper.) Kepeat Ex. 944 with the following vertices :
(i) (-1,2), (0,-1), (2, -2).
(ii) (-2, -2), (1.1), (3,0).
AREA OF TRIANGLE 17;")
Ex. 946. Find the area of an equilateral triangle of Bide (i) 1 inch,
(ii) 2 inches. Find the ratio of the greater area to the smaller.
Ex. 947. Find the surface (i.e. the sum of the areas of all the faces) :
(i) of the tetrahedron in Ex. 109.
(ii) of the square pyramid in Ex. 116.
(iii) of the cube in Ex. 210.
(iv) of the cuboid in Ex. 221.
(t) of the 3-sided prism in Ex. 224.
Ex. 948. Find the combined area of the walls and roof of the house in
fig. 102 ; take width of house = 8 yds., depth (front to back) = 4 yds., height
of front wall = 6 yds., height of roof-ridge above ground =7^ yds. Neglect
doors and windows.
Ex. 949. Find the area (i) in sq. inches, (ii) in sq. cm., of the triangle
whose vertices are ACD in fig. 20.
tEx. 950. Prove that the area of a rhombus is half the product of its
diagonals.
tEx. 951. D is the mid-point of the base BC of a triangle ABC;
prove that triangles ABD, ACD are equivalent.
tEx. 952. ABCD is a parallelogram; P, Q the mid-points of AB, AD.
Prove that a APQ=i of ABCD. (Join PD, BD.)
tEx. 953. The base BC of A ABC is divided at D so that BD^^BC;
prove that aABP = ^aABC.
tEx. 954. The base BC of a ABC is divided at D so that BD = ?BC;
prove that aABD=:|a ACD.
tEx. 955. The ratio of the areas of triangles of the same height is
equal to the ratio of their bases.
tEx. 956. The ratio of the areas of triangles on the same base is equal
to the ratio of their heights.
G. s. 12
176 BOOK II
tEx. 9S1'. ABCD is a quadrilateral and the diagonal
AC bisects the diagonal BD. Prove that AC divides the
quadrilateral into equivalent triangles (fig. 171).
"I Ex. 068. E is the mid-point of the diagonal AC of
% quadrilateral ABCD. Prove that the quadrilaterals
ABED, CBED are equivalent.
tEx. 959. E is a point on the median AD of a ABC; prove that
aABE=aACE.
tEx. 960. D is a point on the base BC of A ABC ; E is the mid-point
of AD ; prove that A EBC = ^ a ABC.
lEx. 961. Divide a triangle into 4 equivalent triangles. (FreehaTid)
Ex. .902. The base of a triangle is a fixed line of length 3 inches, and
the vertex moves so that the area of the triangle is always 6 sq. ins. What
is the altitude ? What is the locus of the vertex ?
tEx. 968. Prove that the locus of the vertex of a triangle of fixed base
and constant area is a pair of straight lines parallel to the base.
Ex. 964. Draw a scalene triangle, and transform it into an equivalent
isosceles triangle on the same base. (Keep the base fixed ; where must the
vertex be in order that the triangle may be isosceles ? Where must the vertex
be in order that the triangle may be equivalent to given triangle?) {Freehand.)
Ex. 966. Show how to transform a given triangle
(i) into an equivalent right-angled triangle.
(11) into an equivalent triangle on the same base, having one side of
2 inches. Is this always possible ?
(iii) into an equivalent triangle with an angle of 60°.
(iv) into an equivalent triangle having one angle = a given angle
(without protractor).
(v) into an equivalent right-angled triangle with one of the sides
about the right angle equal to 5 cm. (First make one side 5 cm. ; then
take this as base and make the triangle right-angled.)
(vi) into an equivalent isosceles triangle with base equal to a given
line.
AREA OF TRIANGLE 177
Ex. 966. Transform an equilateral triangle of uiie 3 ins. into an equi-
valent triangle with a side of 4 ins., and an angle of 60° adjacent to that
side. Measure the other side adjacent to the 60° angle.
Ex. 967. Transform a given triangle into an equivalent triangle with its
vertex (i) on a given line, (ii) one inch from a given line, (Hi) one inch from
a given point, (iv) equidistant from two given intersecting lines.
^Ex. 968. Transform a given quadrilateral ABCD into an equivalent
quadrilateral ABCD', so that the three vertices A, B, C may be unchanged,
and z BAD' = 170°.
H Ex. 969. Repeat Ex. 968, making / BAD' = 180°. What kind of figure
is produced?
P A
tEx. 970. In fig. 172 PA is paraUel to BC.
Prove that a POB = a AOC.
B C
fig. 172.
tEx. 971. A line parallel to the base BC of a ABC cuts the sides AB, AC
in D, E respectively. Prove that aABE= aACD.
tEx. 973. F is any point on the base BC of A ABC : E is
the mid-point of BC. ED is drawn parallel to AF. Prove
that aDFC = JaABC. (JoinAE.) Fig. 173. B
fig. 173.
tEx. 973. Draw a line through a given point of a side of a triangle to
bisect the area of the triangle. (See Ex. 972.) Verify your construction by
measuring and calculating areas.
12—2
178
BOOK II
Area of any rectilinear figure. Thin may be detorniined
in various ways.
Method L By dividing up the figure into triangles.
Method II. Perhaps the most convenient method is tlutt of
constructing a single triangle equivalent to the given figure, as
follows :
To construct a triangle equivalent to a given quadri-
lateral ABCD.
Construction Join CA. Through D draw
DD' II CA, meeting BA produced in D'.
Join CD'.
Then A BCD' = quadrilateral ABCD.
A ACD' = A ACD. (Why 1)
Add to each AACB.
.-. A BCD' = quadrilateral ABCD.
Proof
In a similar way a pentagon may be reduced, first to an
equivalent quadrilateral and then to an q
equivalent triangle: and so for figures of >i^^*^C
more sides. The area of the triangle can
then be found as already explained. A
convenient method of dealing with the E A B C
pentagon is shown in fig. 175. ^8- 175.
tEx. 974. Explain the construction of fig. 175 and prove that
A C'DE' = figure ABCDE.
tEx. 976. Given a quadrilateral ABCD, construct an equivalent triangle
on base AB having Z.A in common vnth the quadrilateraL (Freehand)
Ex. 976. Construct a triangle whose area is equal to the sum of the areas
of two given triangles. (First transform one triangle till it has a side equal
to a side of the other triangle; then fit the triangles together to form a
quadrilateral, and consider how to reduce the sum to a single triangle.)
AREA OF POLYGON
179
Ex. 077. Construct a triangle equivalent to the difference of two given
triangles.
Ex. 878. Find the area of a quadrilateral A BCD, when
(i) DA = lin., Z.A = 100°, AB=2-3in3., Z.B = 64°, BC = l-5ins,
(ii) AB = 5-7 cm., BC=5-2cm., CD = l-7 cm., DA=3-9 cm., /.A=76°.
Ex. 979. Find the area of a pentagon ABODE, given AB=6-5 cm.,
BC=2-4cm., CD = DE=4cm., EA=2-5cm., Z.A=80°, ii.B = 133°.
Ex. 980. Find the area of a regular hexagon inscribed in a circle of
radius 2 ins.
Ex. 981. Find the area of a regular pentagon of side 6 cm.
Ex. 983. Find the areas of the 4-gon8 and 5-gonc in Ex. 107 (i), (ii),
108 (i), (ii).
Ex. 988. Find the area of a trapezlnin
ABCD (fig. 176), given AB = 3in8., height=2ins.,
/.A =70°, Z.B = 50°. (Divide into 2 as, and
notice that their heights DE, BF are equal.)
Ex. 984. Find the area of a trapezium
ABCD, given
(i) AB = 7'5cm., height=4cm., AD = 5cm., BC=:4-3cm., Z.Aobtuse,
L B acute.
(ii) AB=3-6 ins., CD=2-6 ins., height=l-3 ins., ^A = 60°
(iii) Same dimensions as in (ii) except that Z.A = 80°.
(iv) AB = 5 cm., AD=4 cm., BD=5 cm., Z-DBC = Z.BDC.
+Ex. 985. In fig. 177 E is the mid-point of
BC, PQ is II to AD. Prove that trapezium
ABCD = ||ogram APQD.
tEx. 986. Prove tliat the area of a trape-
zium is equal to half the product of the
height and the sum of the two parallel
sides (see Ex. 985).
IfEx. 987. Cut out of paper two congruent trapezia, and fit them together
to make up a parallelogram. Hence prove Ex. 986.
180
BOOK n
Method III. This method is used by Icund-surveyora and
depends on the following principle. It is required to find the
area of the field ABCDEFG (fig. 178). The field is treated as a
polygon, the sides of the polygon being chosen so that the small
irregularities may roughly compensate one another. The longest
diagonal AE is chosen as base-line. In AE points L, M, N, P
are determined, namely the points where the perpendiculars from
the corners meet AE. The field is thus divided up into right-
angled triangles, trapezia and rectangles, whose areas can be
calculated as soon as the necessary measurements have been made.
The surveyor now measures with a chain the difierent distances
along the base-line, AL, AM, AN, AP, AE; also the distances to
the difierent comers, right and left of .the base-line*, namely,
LB, MC, MG, NF, PD. These measurements are recorded in the
Field-Book in the following form : —
Yards.
ToE
600
240
460
360
50
240
300
120
200
100
From
A
go North
* The distances at right angles to the base-line are called oBaetm; in
practice they are never allowed to exceed a few yards, on account of the
difficulty of determining accurately the feet of the perpendiculars.
AREA OF POLYGON
181
This record is to be read wpwcurds. In the middle column
are set down the distances from A of the different points on the
base-line ; on the right and left are set down the offsets as they
occur; e.g. L is 100 yards North of A, and B is 200 yds. to the
left of L ; and so on.
Ex- 988. On inch paper draw a plan of the field represented in fig. 178
from the measurements given (scale, 1 inch to represent 100 yards) ;
calculate its area in square yards.
Ex. 989. Give the coordinates of the comers of the field in fig. 178,
taking AE as axis of y and A as origin.
Ex. 990. Draw a plan and find the area of the field in the following
survey: — (Preehand)
Yards
To D
400
340
50
70
300
90
200
30
100
50
From
A
go North
1
In practice, distances are measured with a chain of 100
links. The length of the surveyors' chain is the same as the
length of a cricket-pitch, namely 22 yards. A square whose side
is 1 chain has area 22^ or 484 sq. yards. Now an acre contains
4840 sq. yards ; hence 10 sq. chains = 1 acre.
182
BOOK II
Ex. 091- Draw plans and find the area (in acres) of tho fields whose
dimensions are recorded below : {Freehand)
(i)
Links
ToB
800
600
400
500
300
100
200
From
A
go East
(ii)
Links
To B
1100
400
1000
800
800
400
600
1
From
A
go S.E.
(iii)
Links
To B
800
150
700
100
500
200
400
350
300
300
100
200
From
1
A
go N.W,
Ex. 993. Draw a plan of a field whose comers are represented by the
points ABCDO in fig. 20; choose the longest diagonal as base-line and draw
offsets ; enter measurements as for Field-Book (taking 1 inch to rspresent
100 yards) ; find the area of the field in square yards.
Also find the area by constructing a single equivalent triangle.
area of triangle 183
Theorem 3.
Equivalent triangles which have equal bases in the
same straight line, and are on the same side of it, are
between the same parallels.
A B D E
fig. 179.
Data ABC, DEF are equivalent triangles on equal bases AB, DE,
these being in a straight line, and C and F being on the
same side of AE.
To prove tJuit CF is parallel to AE.
ConstrVjCtion Join CF.
If possible, draw a line CG || to AE, distinct from CF,
meeting FD (produced if necessary) in G. Join EG.
Proof Since AB = DE, and CG is || to AE,
.'. AABC = ADEG. II. 2.
But AABC = ADEF, Data
.•. ADEF = ADEG,
.'. F coincides with G, and CF with CG,
.'. CF is II to AE.
Q. E. D.
CoR. 1. Equivalent triangles on the same or equal bases are
of the same altitude.
CoR. 2. Equivalent triangles on the same base and on the
same side of it are between the same parallels.
fEx. 993 a. Give another proof of Cor.l.
184 BOOK It
Ex. 999. What is the oonverse of the above Theorem? P
tEx. 994. D Earethemid-polntaof tli«BldeaAB,AC
of a 1riaxi«l« ABC; prove tbat DE U parallel to BC.
(Join DC, EB.) Q 8
%. 180.
tEx. 995. In fig. 180 aPXQ= aRXS; prove that PR ^
is parallel to QS. o/\6
tEx. 996. In fig. 181 aAEB= aADC; prove that DE b q
is parallel to BC. « 28i
ABEA OF TBIANGLE
185
Theorem 4. t
If a triangle and a parallelogram stand on the same
base and between the same parallels, the area of the
triangle is half that of the parallelogram.
fig. 182.
Data A EBC and H"*™"* ABCD stand on the same base BC and
between the same parallels BC, AE.
To prove that A EBC = HI**™" ABCD.
Join BD.
Construction
Proof
Since AE is |1 to BC,
.•. AEBC^ADBC,
and ADBC = |1|°«'*°'ABCD,
.-. AEBC = i|l°«''""ABCD.
<J. E. D.
]86 BOOK 11
fEx. 997. Construct a rectangle equal to a given triangle. Give a proof.
fEx. 998. F, E are the mid-points of the sides AD, BC of a parallelogram
ABCD; P is any point in FE, Prove that A APB = i ABCD.
tEx. 999. P, GL are any points upon adjacent sides AB, BC of a parallelo-
gram ABCD; prove that aCDP=aADGI.
tEx. XOOO. AB, CD are parallel sides of a trapezium ABCD; E is the
mid-point of AD; prove that aBEC= ^trapezium. (Through E draw line
parallel to BC.)
fEx. lOOl. O is a point inside a parallelogram ABCD; prove that
AOAB+ AOCD = i ABCD.
Miscellaneous Exercises on Area.
Ex. lOOa. Find the area of a triangle whose sides are
(i) y=2x+2, y=^, y=2-x.
(ii) y=2x + 2, y = 2-x, y=0.
(iii) x=0, y=l- g, y=x-l.
Ex. 1003. The area of a parallelogram of angle 30° is half the area
of a rectangle with the same sides.
+Ex. 1004 *. O is any point on the diagonal
BD of a parallelogram ABCD. EOF, GOH
are parallel to AB, BC respectively. Prove that
parallelogram AC = parallelogram CO. ** HO
fig. 183.
tEx. 1005. Any straight line drawn through the centre of a parallelo-
gram (i.e. through the intersection of the diagonals) bisects the parallelogram.
Ex. 1006. Show how to divide a parallelogram into three equal
parallelograms.
Ex. 1007. Show how to bisect a parallelogram by a straight line drawn
perpendicular to a side.
* This exercise appears in old books on Geometry as a proposition, and
was used by Euclid in the proof of later propositions. It was enunciated as
follows: "The complements of the parallelograms which are about the
diagonal of any parallelogram are equaL"
EXERCISES ON AREA
187
Q A
B E
r
^N-P
OH
fig.
184.
0
tEx. 1008. E is any point on the diagonal AC of a parallelogram
ABCD. Prove that aABE=aADE.
tEx. 1009. Produce the median BO of a triangle ABC to E, making
DE=DB. Prove that A EBC= A ABC.
tEx. lOlO. P, Q are the mid-points of the eidee
BC, AD of the trapezium ABCD; EPF, GQH are
diawn perpendicular to the base. Prove that
trapezium = rectangle GF. (See fig. 184.)
i£x. lOll. L, M are the mid-points of the parallel sides AB, CD of
a trapezium ABCD. Prove that LM bisects the trapezium.
tEx. lOia. In Ex. 1011 O is the mid-point of Ll\4; prove that any line
through O which cuts AB, CD (not produced) bisects the trapezium.
I Ex. lOlS. Prove that the area of the parallelogram formed by joining
the mid-points of the sides of any quadrilateral ABCD (see Ex. 736) is half
the area of the quadrilateral,
tEx. 1014. The medians BD, CE of a ABC intersect at Q ; prove that
quadrilateral ADGE= A BGC. (Add to each a certain triangle.)
The Theorem op Pytiiaqoras.
Fig. 185 represents an isosceles right-angled triangle with
squares described upon each of the
sides. The dotted lines divide up the
squares into right-angled triangles,
each of which is obviously equal to
the original triangle. This sub-division
shows that the square on the hypo-
tenuse of the above right-angled tri-
angle is equal to the sum of the squares
on the sides containing the right angle.
(A tiled pavement often shows this
fact very clearly.)
v
N
N.
fig. 185.
188
BOOK n
vx/
x><
X
fig.
186.
II Ex. lOl A. Construct a right-angled triangle with
sides of 3 cm. and 4 cm. containing the right angle.
Construct squares on these two sides, and upon
the hypotenuse. Measure the length of the hypo-
tenuse, and ascertain whether or no the square on
the hypotenuse is equal to the sum of the squares
on the sides containing the right angle. See fig. 186.
Ex. 1016. Bepeat Ex. 1015 taking 4*3 cm. and
6*5 cm. as the sides containing the right angle.
Ex. 1017. Draw a good-sized scalene right-angled triangle ABC, right-
angled at A. Measure the three sides and calculate the areas of the squares
upon them. Add together the areas of the two smaller squares, and arrange
your results like this —
AB=...cm., sq. on AB = ...sq. cm.,
AC = ...cm., sq. on AC=...sq. cm.,
sum of sqq. on AB, AC = ...sq. cm.,
BC=...cm., sq. on BC = ...sq. cm.
Ex. 1018. Bepeat Ex. 1017 with a different right-angled triangle.
Ex. 1019. Bepeat Ex. 1017 making ^A = 60° instead of 90°.
fig. 187.
THEOREM OF PYTHAGORAS
189
Ex. loao. In fig. 187 find (in squares of the paper) the area of the square
BD by first finding the area of the square AG and then deducting the four
triangles at the corners. Also calculate the areas of the squares on AB and
AC, and see whether these add up to the square on BC.
Ex. 1021. Repeat Ex. 1020 (drawing your own figure on squared paper)
with different numbers instead of 8 and 13.
The above exercises lead up to the fact that
"In a right-angled triangle the square described on
the hypotenuse is equal to the sum of the squares on the
other two sides."
^This famous theorem was discovered by Pythagoras (b.c. 570
— 500). Before proving it, the -pupil
may try the following experiment.
Ex. 1032. Draw (on paper or, better, on
thin cardboard) a right-angled triangle and
the squares on the tlaree sides (see fig. 188).
Choose one of the two smaller squares and
cut it up in the following manner. First
find the centre of the square by drawing
the diagonals. Then, through the centre,
make a cut across the square parallel to BC,
the hypotenuse, and a second cut perpen-
dicular to BC. It will be found that the
four pieces of this square together with the
other small square exactly make up the
square on the hypotenuse.
(Perigal's dissection.)
fig. 188.
The following exercises lead up to the method o/'proo/' adopted
for the theorem of Pythagoras.
tEx. 1023. On two of the sides AB, BC of any triangle ABC are described
squares ABFG, BCED (as in fig. 189) ; prove that triangles BCF, BDA are
congruent; and that CF = DA.
tEx. 1024. On the sides of any triangle ABC are described equilateral
triangles BCD, CAE, ABF , their Yertices pointing outwards. Prove that
AD = BE=CF.
190
BOOK n
Theokkm 6.
[Thk Throbbm of Pythagoras.]
In a right-angled triangle, the square on the hypo-
tenuse is equal to the sum of the squares on the sides
containing the right angle.
Data ABO is a triangle, right-angled at A.
The figures BE, CH, AF are squares described upon BC, CA,
AB respectively.
To prove tlwi sq. BE = 8q. CH + sq. AF.
Construction Through A draw AL || to BD (or CE).
Join OF, AD.
Proof f rt ^ CBD = rt. L FBA,
add to each L ABC,
.". ^ABD-- /.FBC.
Hence, in As ABD, FBC
/■ Z. ABD = /L FBC,
J AB = FB (sides of a square),
( BD = BC,
/. AABD= A FBC. I. 10.
AABD = AFBC^
THEOREM OF PYTHAGORAS
191
A FBC = Jsq. AF
AABD=Jrect. BL
/ Since each of the angles BAC, BAG is a
right angle
,". CAG is a st. line, L 2.
and this line is 1| to BF.
.'. A FBC and sq. AF are on the same base
BF, and between the same parallels BF, CG,
A FBC - I sq. AF.
II. 4.
Again AABD and rect. BL are on the
same base BD and between the same parallels
BD, AL,
.'. AABD =^ rect. BL. n. 4.
sq. AF = rect. BL
sq. OH = rect. CL
.*.sq. AF + sq. CH \
= sq. BE
/ But A FBC = AABD. Proved
.'. sq. AF = rect. BL.
In a similar way, by joining BK, AE, it
may be shown that
sq. CH = rect. CL.
Hence
^ sq. AF + sq. CH - rect. BL + rect. CL
=sq. BE.
Q. K D.
Q. 8.
13
192
BOOK n
An alternative method of proof is indicated below ; the pupil
should work out for himself the actual details of the proof.
The figs. AF, GE are two squares placed side by side.
Mark off AC = GK and join BC.
Then BAC is a rt. z.** A and AF,
GE are equal to the squares on the
sides containing the right angla
Produce GF to D so that FD = GK.
Join BD, DE, EC.
Prove that A» BAC, CKE, DHE,
BFD are congruent.
Prove that fig. CD is a square,
namely the square on the hj'potenuse
of ABAC.
From the figure AKEHFB subtract the triangles (1) and (2)
and fit them on to (3) and (4), thus making up the sq. CD.
fig. 190.
fig. 191.
Another proof oi the theorem is shown in fig. 191.
The triangles marked 1, 2, 3, 4, 5, 6, 7, 8 are all congruent
right-angled triangles. A is the square on the hypotenuse of one
of these triangles, B and C are the squares on the sides containing
the right angle. A little consideration will make it evident, that
A = B + C. «
THEOREM OF PYTHAGORAS 193
Ex. 1026. What is the side of a square whose area is 4 sq. in. ; 9 sq. in. ;
16 sq. cm. ; 17 sq. cm. ; 2 sq. in. ; 6 sq. mUes ; a^ sq. in. ; 6 sq. cm. ?
Ex. 1026. What is the square on the hjrpotenuse of a right-angled
triangle if the sides containing the right angle are 6 cm. and 8 cm. ? Hence
calculate the length of the hypotenuse. Verify by drawing.
Note on "error per cent." In cases where a result is
obtained both by calculation and by drawing, it will generally be
found that there is a slight disagreement. To see whether this
disagreement, or *' error," is serious, it is necessary to reduce it to
a percentage. Thus, the calculation in Ex. 1026 would be as
follows : —
sum of sqq. on sides = (6^ + 8^) sq, cm.
— (36 + 64) sq. cm.
= 100 sq. cm.,
.'. sq. on hypotenuse = 100 sq. cm.,
.'. hypotenuse = ^100 cm.
= 10 cm. (by calculation).
Suppose that we find hypotenuse = 9*95 cm. (by drawing),
error = 0-05 in 10
= 0-5 in 100
= 0-5 per cent.
N.B. (1) It is not necessary to calculate the "error
per cent." to more than one significant figure.
(2) Do not be satisfied until your error is less than
1 per cent.
Work the following exercises (i) ly calculation, (ii) hy drawing, in every
case making a rough estimate of the error ■per cent. Every calculation is to
be " to three significant figures."
Ex. 1027. Find the hypotenuse of a right-angled triangle when the
sides containing the right angle are
(i) 5 cm., 12 cm., (ii) 4*5 in., 6 in.,
(iii) 7-8 cm., 9-4 cm., (iv) 2-34 in., 4-65 in.
(v) 44 mUes, 5| mUes, (vi) 65 mm., 83-5 mm.
13—2
194 BOOK n
Ex. loaa. Find the remaining side and the area of a right-angled
triangle, given the hypotennae and one side, as follows: —
(i) hjp. =15 cm., Bide=12 om.; (ii) hyp.=6in., Bide=4in.;
(iii) hyp. =8 in., side =4 in.; (iv) hyp. = 160 mm., side =100 mm.;
(v) hyp. = 143 mm., side=71"6 mm.
Ex. 1039. A flag-staff 40 ft. high is held np by several 50 ft. ropes; each
rope is fastened at one end to the top of the flag-staff, and at the other end
to a peg in the ground. Find the distance between the peg and the foot of
the flag-staff.
Ex. X030. Find the diagonal of a rectangle whose sides are (i) 4 in. and
6 in., (ii) 9 cm. and 11 cm.
Ex. 1031. Find the remaining side and the area of a rectangle, given
(i) diagonal=10 cm., one Bide=7 cm.; (ii) diagonal=4*63 in., one side
= 3-47 in.
Ex. 1032. Find the diagonal of a square whose side is (i) 1 in. , (ii) 5 om.,
(iii) 6-72 cm.
Ex. 1033. Find the side and area of a square whose diagonal is (i) 2 in.,
(ii) 10 cm. , (iii) 14-14 cm.
Ex. 1034. Find the side of a rhombus whose diagonals are
(i) 16 cm., 12 cm.; (ii) 6 in., 4 in.
Ex. 1035. .Find the altitude of an isosceles triangle, given (i) base =4 in.,
side=5in., (ii) base=64mm., side=40mm.
Ex. 1036. Find the altitude of an equilateral triangle of side 10 cm.
Ex. 1037. In fig. 192, ABCD represents a square of g ^
side 3in. ; AE=AH = CF=CG = 1 in. Prove that EFGH
is a rectangle; find its perimeter and diagonal. **
Ex. 1038. Find how far a traveller is from his starting
point after the following journeys : — (i) first 10 miles N., then o' ' g 'o
8 miles E., (ii) first 8 miles E., then 10 miles N., (iii) 43 km. ^ jgn
S. W. and 32 km. S.E., (iv) 14 miles S. , 10 miles E. , 4 miles N.
(try to complete a right-angled triangle having the required line for hypo-
tenuse), (v) 4 miles E., 6 miles N., 3 miles E., 1 mile N.
Ex. 1039. (Inch paper.) If the coordinates of a point P are (1, 1) and
of Q. (2, 3), find the distance PQ. (PQ is the diagonal of a certain rectangle.)
THEOREM OF PYTHAGORAS 195
Ex. 1040. (Inch paper. ) In each of the following casea find the distance
between the pair of points whose coordinates are given : — (i) (2, 1) and (1, 3);
(ii) (0, 0) and (3, 1); (iii) (2, 0) and (0, 3); (iv) (-1,-1) and (2, 1);
(v) (-2, 2) and (1, -2); (vi) (0-4, 1-3) and (2-3, 0-4); (vii) (-0-9, 0-4) and
(1-6, -0-7).
Ex. 1041. Find the lengths of the sides of the triangle whose verticea
are (2, -2), (0, -3) and (-2, 1).
Ex. 1042. Newhaven is 90 nules N. of Havre, and 50 miles E. of Ports-
month. How far is it from Portsmonth to Havre?
Ex. 1043. St Albans is 32 miles N. of Leatherhead, and Leatherhead
is 52 miles from Oxford. Oxford is due W. of St Albans; how far is Oxford
from St Albans ?
Ex. 1044. A ship's head is pointed N., and it is steaming at 15 miles
per hour. At the same time it is being carried E. by a current at the
rate of 4 miles per hour. How far does it actually go in an hour, and in
what direction?
Ex. 1045. Two men are conversing across a street 30 feet wide from the
windows of their respective rooms. Their heads are 15 ft. and 30 ft. from
the level of the pavement. How far must their voices carry?
Ex. 1046. A man, standing on the top of a vertical cliS 700 ft. high,
estimates the distance from him of a boat out at sea to be 1500 ft. How far
is the boat from the foot of the cliff?
Ex. 1047. A ladder 60 ft. long is placed against a wall with its foot 20 ft.
from the foot of the waU. How high wiU the top of the ladder be ?
Ex. 1048. A field ABCD is right-angled at B and D. AB=400 yards,
AD =300 yards, the diagonal AC =500 yards. Find the area of the field.
Ex. 1049. Find the distance between the summits of two columns, 60
and 40 ft. high respectively, and 30 ft. apart.
Ex. 1050. An English battery (A) finds that a Boer gun is due N., at
a range of 4000 yards. A second English battery (B) arrives, and takes up
a pre-arranged position 1000 yards E. of A. A signals to B the range and
direction in which it finds the enemy's gun. Find the range and direction in
which B must fire.
Ex. 1051. What is the hypotenuse of a right-angled triangle whose
sides are a and & in.?
196 BC»OK II
Ex. lOsa. What is the remaining side of a right-angled triangle which
haa hypotenuse =x in. and one side =j/ in. ?
If fnrther practice is needed, the reader may solve, by calculation,
Ex. 234—239, 242, 244, 247, 249, 265, 266.
Ex. 1053. Given two squares of different sizes, show how to construct
a square equal to the sum of the two squares. (Will the side of the new
square be equal to the sum of the sides of the old squares ?)
Ex. 1064. C!onstruct a square equal to the sum of the squares BD, AG
in fig. 187, and measure the side of the resulting square in inches.
Ex. 1056. Given two squares of different sizes, show how to construct
a square equal to the difference of the two given squares.
Ex. 1056. CJonstruct a square equal to the difference of the squares
BD, AG in fig. 187, and measure the side of the resulting square in inches.
Ex. 1057. Draw three squares of different sizes and construct a square
equal to the sum of the three squares. (Begin by adding together twq of
(he squares and then adding in the third.)
Ex. 1058. Make a square to have twiioe the area of square BD in fig. 187.
Square-roots found graphically. The square on a side of
1 inch is 1 square inch. The square on a side of 2 inches is
4 square inches. What is the side of a square of 2 square
inches 1 Clearly ^2 inches. Such a square may be constructed
by adding together two 1 inch squares. If the side of the
resulting square be measured in inches and decimals of an inch,
we shall have an approximate niunerical value of ^2.
{The following exercUe$ are most easily done on inch paper.)
Ex. 1O50. Construct a square of area 2 sq. in^ Hence find ^2 to
two places of decimals; check by squaring.
Ex. 1060. Construct a square of area 6 sq. in. (by adding together
squares of area 1 and 4 sq. in.). Hence find ^^6; check.
Ex. 1061. As in Ex. 1060 find graphically ^10, ,^8, ^3, checking your
result in each case.
THEOREM OF PYTHAGORAS
197
In the preceding set of exercises a number of square roots
have been found graphically. There "were, however, gaps in the
series, e.g. ^3 did not appear. The square roots of all integers
may be found in succession by the following construction, which
is most easily performed on accurate inch paper.
A3 A4 A5 Ae A? A8A9
fig. 193.
Draw OX = 1 inch and draw at O a line of unlimited length
perpendicular to OX. Mark off OAj = OX = 1. Then A^X = J2.
Mark off OA2 = A^X = ^%
Then Apc" = O Aj' + OX^
=2+1
=3,
.-. A2X=V3.
Mark off OA3 = AjX = ^3,
OA4 = A3X,
OA5 = A4X, &c.
We now have
OAi^^l, OA2=^2, OA3-73, OA4=^4, OAb=^5, &c.,
and, by measurement, these square roots may be determined.
tEx. 1062. Prove i. 15 by means of Pythagoras' theorem.
"("Ex. 1063. AD is the altitude of a triangle ABC. Prove that
AB2-AC2=BD2-CD«.
198 BOOK II
Ex. 1064. In Ex. 1063 let AB = 3 in., AC =2 in., BC=3 in. Caloolate
BDi'-CD^. Hence find BD- CD.
[ BD3- CDa= (BD - CD) {BD + CD) = (BD - CD) BC. ]
Knowing BD - CD and BD + CD, you may now find BD and CD. Henoe
find AD. Hence find area of A ABC. Verify all your calculations by
drawing.
Ex. 1066. Repeat Ex. 1064, taking AB = 3in., AC=2in., BC=4in.
tEx. 1066. PQR IB a triangle, right-angled at GL On QR a point S is
taken. Prove that PS« + QR^ = PR2 + QS^.
tEx. 1067. ABC is a triangle, right-angled at A. On AB, AC respectively
points X, Y are taken. Prove that BY^ + CX2= XY^ + BC^.
tEx. 1068. The diagonals of a quadrilateral ABCD intersect at right
angles. Show that AB2+CD2=BC2+DA2.
tEx. 1069. O is a point inside a rectangle ABCD. Prove that
OA2 + OC2=OB2 + OD2.
(Draw perpendiculars from O to the sides of the rectangle.)
{The following S-dimensional exercises give further practice in the use
of Pythagoras^ Theorem.)
Ex. 1069a. The edges of a certain cuboid (rectangular block) are
8", 4k", 6" ; find the diagonals of the faces.
Ex. 1O60 b. A room is 18 ft. long, 14 ft. wide, 10 ft. high. Find the
diagonals of the walls. Find the diagonal of the floor.
Ex. 1069 e. Find the length of a string stretched across the room in
the preceding exercise, from one comer of the floor to the opposite corner of
the ceiling.
Ex. 1069 d. Find the diagonal of the face of a cubic decimetre. Also
find the diagonal of the cube.
Ex. 1069 e. Find the slant side of a cone of (i) height 6", base-radius
3"; (ii) height 4-6 cm., base-radius 7'5 cm.; (iii) height 55 mm., base-
diameter 46 mm.
Ex. 1069 f. Find the height of a cone of (i) slant side 10'', base-radius
4" ; (ii) slant side 5*8 m., base-diameter 11 m.
Ex. 1069 g. Find the base-radius of a cone of (i) slant side 7 ft., height
5 ft. ; (ii) slant side 11*3 cm., height 57 millimetres.
THEOREM OF PYTHAGORAS
199
Theorem 6.f
[Converse of Pythagoras' Theorem.}
If a triangle is such that the square on one side is
equal to the sum of the squares on the other two sides,
then the angle contained by these two sides is a right
angle.
Data The triangle ABC is such that BC^ = AB'' + AC''.
To prove that L BAC is a right angle.
Construction Coi
L EDF a rt.
Proof
istruct a A DEF, to have DE = AB,
DF = AC, and
L.
Since /. EDF is a right angle,
Constr.
.*. EF2=DE2+DF2
= AB2 + AC2
Gonstr.
= BC2,
Data
/. EF = BC.
Hence, in the As ABO, DEF,
AB=DE,
ConsPr.
•
AC=DF,
Constr.
,BC=EF,
Proved
.'. the triangles are congruent,
.'. Z.BAC = Z.EDF.
Now z. EDF is a right angle.
Constr.
/. /.BAC
) is a right angle.
Q. £. B.
200 BOOK II
Ex. 1070. Axe the triangles right-angled whose sides are
(i) 8, 17, 15; (ii) 12, 36, 34; (iii) 25-5, 25-7, 3-2;
(iv) 4n, 4n2_l, 4n2+l; (v) vi^ + n^, w?~n\ 2mn; (vi) o, 6, o + 6?
Ex. 1071. Bristol is 71 miles due W. of Beading; Beading is 55 miles
from Northampton; Northampton is 92 miles from Bristol. Ascertain whether
Northampton is due N. of Beading.
Ex. 1073. Ascertain (i) by measurement and calculation, (ii) by con-
structing the triangle, whether a right-angled triangle could be made having
for sides the lines d, h, k in fig. 8.
Ex. 1073. Ascertain, by considering the lengths of the sides, whether
the triangle of Ex. 821 (i) is right-angled.
Ex. 1074. Perform, and prove, the following construction for erecting
a perpendicular to a given straight line AB at its extremity A. Along AB
mark off AC = 3 imits. On AC as base construct a triangle ACD, having
AD =4, CD = 5. Then AD is perpendicular to AB. (Ancient Egyptian
method.)
ILLUSTRATIONS OF ALGEBRAICAL IDENTITIES 201
Illustration of Algebraical Identities by means op
Geometrical Figures.
It has been shown that the area of a rectangle 4 inches long
and 3 inches broad is 4 x 3 sq. inches.
In the same way the area of a rectangle a inches long and
b inches broad is ah sq. inches.
{Caution. Notice carefully the form of the above statement : —
area = 4x3 sq. inches. Never say, 4 inches x 3 inches; which
is nonsense. It is impossible to multiply by a length — such as
3 inches. The statement : — area of rectangle = length x breadth
is really a convenient but inaccurate way of abbreviating the
following statement : — the number of units of area in a rectangle is
equal to the product of the numbers of v/rdts of length in the length
and breadth of the rectangle^
ITEx. 1075. What is the area of a rectangle
(i) X cm. long, y cm. broad;
(ii) 1x cm. long, 2y cm. broad;
(iii) a cm. long, a cm. broad (a square)?
HEz. 1076. What is the area of a square whose side is x inches?
Ex. 1077. Write out the accurate form of the statement of which the
following is a convenient abbreviation : — area of square = square of its side.
HEx. 1078. Find an expression for the area of each of the following
rectangles (do not remove the brackets): —
(i) (a +6) inches long, ft inches broad;
(ii) (a + 6) cm. long, (c + d) cm. broad;
(iii) (a + 6) cm. long, (a -6) cm. broad.
HEx. 1079. What is the area of a square whose side is (a + &) inches? Is
the answer equal to (a^+ft^) gq. inches?
UEx. 1080. What is the area of a square whose side is (a -6) inches?
Is the answer equal to (a^-ft^) gq. inches?
HEx. 1081. Simplify the following expressions by removing brackets: —
(i) (a + fe)(c + d), (ii) (a + 6)2, (iii) (a -6)2,
(iv) (o+ 6)2+ (a -6)3, (y) (a + 6)2 -(o- 6)2-
202
BOOK II
(A) Geometrical illustration of the identity
(a + b) k = ok + bk,
ST V
Let PQ = a units of length, QR = 6 units of length.
Then PR = (a + b) units of length.
At P, Q, R erect equal perpendiculars PS, QT, RV ; the length
of each being k units of length.
Then STV is a straight line || to PQR I. 23, Cor.
and all the figures are rectangles.
Rect. PW = (a + b)k units of area.
Rect. PT= ak „ „ „
Rect. GIV= bk „ „ „
But rect. PV = rect. PT + rect. QV,
.*. (a + b) k^ak + bk.
tEx. loaa. In the above proof, why would it have been wrong to say,
PR=a6 units of length, instead of {a+b)2
Ex. 1083. Give geometrical illustrations of the following identities (i.e.
draw figures and give explanations) :
(i) {a+b+c)ksak+bk+ek,
(ii) {a-b)ksak-bkt
(iii) dbsba.
ILLUSTRATIONS OF ALGEBRAICAL IDENTITIES
203
(B) Geometrical illustration of the identity
(a+b)(c + d) = ac + bc + ad + bd.
X oY b Z
ad
bd
be
P aQ b R
fig. 196.
In the figure, all the angles are right angles and all the figures
rectangular.
Also PQ, QR, PS, SX are respectively a, b, c, d units of
length.
Then PR = (a + 6) units of length, PX = (c + d) units of length.
Rect. PZ ={a + b){c + d) units of area.
Rect, PT = ac „
Rect. QV = be „
Rect. SY = ad „
Rect, TZ = bd
But rect. PZ is the sxun of rectangles PT
QV, SY, TZ,
.*. (a + b) (c + d) = ac + be + ad + bd.
204
BOOK II
(C) Geometrical illustration of the identity
(a + by = a' + b' + 2ab.
X a Y frZ
ab
b*
ab
a Q6 R
fig. 197.
Let PQ = a units of length, QR = 6 units of length.
Then PR = (a + 6) units of length.
On PR construct the square PRZX.
From PX cut off PS = PQ = a units of length.
Through Q draw QTY |1 to PX.
Through S draw STV || to PR.
Then all the angles formed are right angles, and all the
figures rectangidar.
Also PT is a square. (Why 1)
'Again RZ = (a + b) units of length,
and RV = PS = a „ „ „
.'. VZ = 6 „ „ „
andYZ = QR = 6 „ „ „
.*. TZ is a square.
Sq. PZ = (a + by units of area.
Sq. PT= a' „ ,, „
Sq. TZ = b' „ „ „
Rect. sy= ab „ „ „
Rect. G>/—ab „ „ „
But sq. PZ = sq. PT + sq. TZ + rect. SY + rect. QV,
.-. {a + by=a'' + b^ + 2ab.
TZ is a square -
ILLUSTRATIONS OF ALGEBRAICAL IDENTITIES
205
Ex. 1084. State the above result In words.
Ex. 1085. Prove algebraically that
{a+b+c)^ = a^ + b^ + c^ + 2bc + 2ca + 2ab;
also give a geometrical illustration of the identity. (It will be enough to
draw a figure, and mark the lengths and areas.)
Ex. 1086. Illustrate the identities (i) {2x)^six^, (ii) (2a) (36)s6a6.
Numerical cases of identities may be illustrated on squared
paper. For instance, to illustrate the identity
{x+A){x + Q) = x'+10x + 24:.
6x
24
4}f
lr+-
fig. 198.
In fig. 198 AB = 6 units of length,
AD = 4 units of length,
BC = DE = a; units of length (any length).
The numbers inside the rectangles denote the areas. It is
now obvious how the figure illustrates the given identity.
Ex. 1087. By means of figures, illustrate the following identities: —
(i) {x + 5){x + 9)^x^+Ux+i5,
(ii) (2/ + 7)2 = 1/2 + 14?/ + 49,
(iii) 6(x+12)s5a; + 60,
(iv) 5(a;-12)s5a:-60(when.r>12),
(v) 5(12-a-)=60-5x(whena;<12),
(vi) a(b+lC) = a6 + 16a.
206
BOOK II
(D) Geometrical illustration of the identity
Z Y
a-b
a-b
(a-b)»
ab
a-b
W
o X
fig. 199.
Let PQ = a units of length.
From PQ cut off the length QR, containing h unita
Then PR = (a — h) units of length.
On PQ construct the square PQXW; its area is a* units of area.
On QR construct the square QRZY as in the figure.
The area of this square is 6' units of area.
Then the whole figure contains {a' + 6*) units of area.
From PW cut off PS = PR = (a - h) units of length.
Then SW = PW- PS =a-(a-6) units of length
~ '' » >j »
Through S draw ST || to PQ; produce ZR to meet ST in V.
All the figures so formed are rectangular.
Also figure SR is a square, and contains (a — b)' units of area
Rect. WT contains ab units of area.
Lastly, in rect. VY, side YZ = QR = 6 units of length,
and side YT = YQ + QT
= RQ + PS
= b + (a — b) units of length
— a )} » »
.'. Rect. VY contains ab units of area.
Now sq. SR = whole fig. - rect. WT — rect. VY,
.-. {a-by={a'' + b')-ab-ab
= a' + b^-2ab.
Ex. 1088. State the above resnlt In words.
ILLUSTRATIONS OF ALGEBRAICAL IDENTITIES 207
(E) Geometrical illustration of the identity
a' - 6^. = (a + b) (a - b).
S
T
a-b
Q
y/Mwm
m
fig. 20L
Let PQ = a units of length.
On PQ construct the square PS ; its area is a' units of area.
From PQl cut off the length PR, containing b units of length.
From PT cut off PX = PR; through X draw XY || to PQ.
Through R draw RZ |I to PT to meet XY in Z.
All the figures so formed are rectangular.
Also PZ is the square on PR ; its area is &* units of area.
If sq. PZ is subtracted from sq. PS, there remains the shaded
part of the figure.
The area of the shaded part is therefore (a^ - b-) units of area.
Now this part is composed of the rectangles XS and RY.
These rectangles have the same breadth, namely {a — b) units
of length. (Why?)
They might therefore be placed end to end, so as to form a
single rectangle (as shown above on the right).
The length of this single rectangle = TS+ ZR = (a + 5) units
of length, and the area of this rectangle ={a + b){a — b) units
of area,
:.a^-b''={a + b){a~b).
Ex. 1089. State the above result in words.
G. 8
U
208 BOOK II
Express each of the following theorems {Ex. 1090 — 1093) as an algebraical
identity ; prove the identity.
Ex. 1090. Xf there are two stralglit lines, one of which Is divided
into any number of parts (x, y, z say) while the other is of length a,
then the rectangle contained by the two straight lines Is equal to the
stun 6f the rectangles contained by the undivided straight line and
the several parts of the divided line. (Draw a figure.)
Ex. 1081. If a straight line is divided into any two parts (a; and y), the
square on the whole line is equal to the sum of the rectangles contained by
the whole line and each of the parts. (Draw a figure.)
Ex. 1092. If a straight line is divided into any two parts, the rectangle
contained by the whole line and one of the parts is equal to the square on
that part together with the rectangle contained by the two parts. (Draw
a figure.)
Ex. 1093. If a straight line is divided into any two parts, the
square on the whole line is equal to the sum of the squares on the
two parts together with twice the rectangle contained by the two parts.
(Draw a figure.)
Ex. 1094. What algebraical identity is suggested ^ p O 8
by fig. 202? (Take AO = OB = a, OP = 6.) ' ' ^
Ex. 1095. Express and prove algebraically: — If
a straight line is divided into any two parts, four
times the rectangle contained by the whole line and
one of the parts, together with the square on the other
part, is equal to the square on the straight line which
is made up of the whole line and the first part. gg^ 202.
Ex. 1096. Prove that the square on the difference of the sides of a
right-angled triangle, together with twice the rectangle contained by the
sides, is equal to the square on the hypotenuse. (Use Algebra.)
Ex. 1097. If a straight line AB (length 2a;) is bisected at O and also
divided unequally at a point P (distant y from O), what are the lengths of
the two unequal parts AP, PB ? Prove algebraically that the rectangle
contained by the unequal parts, together with the square on the line
between the points of section (OP), is equal to the square on half the
original line.
ILLUSTEATIOKS OF ALGEBRAICAL IDENTITIES 209
Ex. 1098. Show that in the above exercise AO is half the sum of
AP, PB ; and that OP is half the difference of AP, PB. (Most easily proved
by Algebra.)
Ex. 1099. If a straight line AB (length 2x) is bisected at O, and pro-
duced to any point P (OP=y) the rectangle contained by the whole line thus
produced and the part of it produced, together with the square on half the
original line, is equal to the square on the straight line made up of the half
and the part produced.
Ex. 1 lOO. If a straiglit line is divided into any two parts, the snm
of the squares on the whole line and on one of the parts is equal to
twice the rectangle contained by the whole and that part, together
with the square on the other part. (Draw figure.)
Ex. IIOI. If a straight line AB is bisected at O and also divided
unequally at a point P (as in Ex. 1097), the sum of the squares on the two
unequal parts is twice the sum of the squares on half the line and on the
line between the points of section (OP).
Ex. 1102, If a straight line is bisected and produced to any point
(as in Ex. 1099), the sum of the squares on the whole line thus
produced and on the part produced, is twice the sum of the squares on
half the original line, and on the line made up of the half and the part
produced.
Ex. 1103. Four points A, B, C, D are taken in order on a straight line;
prove algebraically that AB , CD + BC . AD = AC . BD. (Take AB = x,
BC = y, CD = 2.)
Verify numerically.
Ex. 1104. If a straight line is bisected and also divided unequally (as
in Ex. 1097) the squares on the two unequal parts are together equal to twice
the rectangle contained by these parts together with four times the square
on the line between the points of section.
14—2
210
BOOK II
Projections
Def. If from the extremities of a Hue AB perpendiculars
AM, BN are drawn to a straight line CD, then MN is called the
projection of AB upon CD (figs. 203, 204).
C M
fig. 203.
fig. 204
ITEx. 1105. In fig. 189 name the projection of AB upon DE ; of AE upon
BC ; of AC upon AL.
ITEx. 1106. In fig. 208 name the projection of AC upon BN ; of BC upon
NC.
UEx. 1107. (On squared paper.) What is the length of the projection
(i) upon the axis of x, (ii) upon the axis of y, of the straight lines whose
extremities are the points
(a) (2, 3) and (6, 6).
(6) (2, 4) and (6, 7).
(c) (0, 0) and (4, 3).
{d) (-1,-3) and (3, 0).
(c) (-5,0) and (-1, 3).
(/) (1, 1) and (5, 1).
(g) (0, -2) and (0,2).
tEx. 1 108. Prove tbat tbe projections on tbe same stxaigbt line of
equal and parallel straight lines are equal. (See fig. 205.)
tEx. 1 109. O is the mid-point of A B ; the pro-
jections of A, B, O upon any line are P, Q, T.
Prove that PT = QT.
fig. 205.
TEx. mo. Measure the projection of a line of
length 10 cm. when it makes with the line upon which it is projected the
foUowmg angles:— 15°, 30°, 45°, 60°, 76°, 90°. Draw a graph.
PROJECTIONS
211
HEx. 1111. In what case is the projection of a line equal to the line
itself?
HEx. 1112. In what case is the projection of a line zero?
Ex. 1113. Prove that, if the slope of a line is 60°, its projection is
half the line.
[Consider an equilateral triangle.]
Ex. 1114. A pedestrian first ascends at an angle of 12° for 2000 yards
and then descends at an angle of 9° for 1000 yards. How much higher is
he than when he started? What horizontal distance has he travelled
(i.e. what is the projection of his journey on the horizontal)?
Ex. 1115. The projections of a line of length { upon two lines at right
angles are x, y. Prove that x^ + y^=l\
HEx. 1116. How does the projection of a line of given length alter as the
slope of the line becomes more and more steep ?
Note. It may be necessary to
produce the line upon which we pro-
ject, e.g. if required to project AB upon
CD in fig. 206, we must produce CD.
D M
fig. 206.
Extension of Pytuagoras' Theorem,
BAG, BAOi, BACa (fig. 207) are triangles respectively right-
angled, acute-angled, and obtuse-angled
at A.
Also AG = ACi = ACg .
By I. 19 BGj < BG and BG2> BG.
Now BC'=CA2-)-AB^ .
.'. BGi^ = GjA^ + AB'* — some area,
and BGg^ = GjA^ + AB^ + some area,
The precise value of the quantity referred to as "some area"
is given in the two following theorems.
14-3
fig. 207.
212 BOOK n
Theorem 7.
Ill an obtuse-angled triangle, tlie square on tlie side
opposite to the obtuse angle is equal to the sum of the
squares on the sides containing the obtuse angle plus
twice the rectangle contained by one of those sides and
the projection on it of the other.
Data The A ABO has L BAG obtuse.
ON is the perpendicular from O upon BA (produced),
.'. AN is the projection of AC upon BA.
Let BO=a units, CA = 6 units, AB = c units, AN =/> units, ON— A units.
To prove that BC» = CA' + AB* + 2AB . AN,
Lo. that a* ^ 6" + c* + 2cp.
Proof Since A BNC is right-angled,
.'. BC" = BN* + NC^, Pythagoras
io. a'^ = {c+py + h^
= c' + 2cp-{-p^ + h\
But AANC is right-angled,
.'. p^ + h^ = h% Pythagoras
:. a' = c' + 2cp + h\
ie. BC = AB' + 2AB . AN + AC*.
Q. E, D.
EXTENSION OF PYTHAGORAS THEOREM
213
Theorem 8.
In any triangle, the square on the side opposite to an
acute angle is equal to the sum of the squares on the sides
containing that acute angle minus twice the rectangle
contained by one of those sides and the projection on it
of the other.
1 !W-tJ
N B
fig. 210.
Data The A ABC has l BAC acute
CN is the perpendicular from C upon AB (or AB produced),
.'. AN is the projection of AC upon AB.
Let BC- a units, CA=6uuits, AB = c units, AN- jt? units, CN=/t units.
To prove that BC^ = CA^ + AB* — 2AB . AN,
i.e. that a? — IP' ■\- c^ — 2cp.
Proof
Since A BNC is right-angled,
:'. BC* = BN''+NC^
le. in fig. 209, a' = {G-pf + h%
in fig. 210, a^=z(p^cY + h%
.'. in both figures,
a^ — c^ — 2cp + ]}' + h'.
But AANC is right-angled,
.-. p''-¥h^ = h\
.-. a' ^c'~ 2cp + h\
;.e. BC* == AB* - 2AB . AN -f AC^.
Pythagoras
Pythagoras
Q. E. C
214 BOOK II
tEx. 1117. Write out the proof of ii. 8 for the case in which z B is a right
ang1& What doea the theorem become?
Ex. 1118. Verify the truth of it 7, 8 by drawing and measurement.
Ex. 1119. What is the area of the rectangle referred to in the enuncia-
tion of n. 7, 8 for the following cases : —
(i) c = 5 cm., 6=4 cm., / BAG = 120° (by drawing) j
(ii) c=5cm., 6=4 cm., / B AC =00° (by drawing);
(iii) c=3in., 5=2 in., a=4 in. (by calculation; check by drawing);
(iv) c=3in., 6 = 2 in., fl=2in. ( „ „ )?
Ex. 1120. By comparing the square on one side with the sum of the
squares on the two other sides, determine whether triangles having the follow-
ing sides are acute-, obtuse-, or right-angled (check by drawing) : —
(i) 3, 4, 6; (ii) 3, 4, 8; (iu) 2, 3, 5; (iv) 2, 3, 4; (v) 12, 13, 5.
Ex. 1121. Given four sticks of lengths 2,3, 4, 5feet, how many triangles
can be made by using three sticks at a time? Find out whether each
triangle is swjute-, obtuse-, or right-angled.
Ex. liaa. Calculate BC when
AB=10cm., AC = 8cm„ /A=60°. (SeeEx. 1118.)
Ex. 1123. Calculate BC when
AB=10cm., AC=8cm., /A=120°.
Ex. 1124. Bristol is 26 miles E. of Cardiff ; Beading is 70 miles E. of
Bristol ; Naseby is due N. of Beading and 95 mUes from Bristol. Calcu-
late the distance from Cardiff to Kaseby, and check by measurement.
Ex. 1125. Brighton is 48 miles S. of London ; Hertford is 20 miles
N. of London ; Shoeburyness is due E. of London, and 64 miles from
Brighton. How far is it from Hertford? Verify graphically.
BeviseEx. 256.
I^Ex. 1126. Suppose that Z.A in fig. 208 becomes larger and larger til'
BAC is-a straight line. What does ii. 7 become in this case?
EXTENSION OF PYTHAGORAS' THEOREM 215
HEx. X127. Suppose that ^A in %. 209 becomes smaller and smaller till
C is on BA. What does u. 8 become in this case ?
tEx. 1128. In the trapezium ABCD (fig. 211), prove
that AC2 + BD2= AD2 + BC^ + 2AB . CD.
(Apply II. 9 to A" ACD and BCD.)
D p
tEx. 1129. D is a point on the base BC of an
' isosceles A ABC. Prove that AB« = AD2 + BD. CD.
(Let O be mid-point of BC, and suppose that D lies ^S- 211.
between B and O. Then
BD = BO-OD, CD = CO + OD = BO + OD.)
tEx. 1130. ABC is an isosceles a (AB = AC); BN is an altitude. Prove
that2AC.CN = BC2.
tEx. 1131. BE, CF are altitudes of an acute-angled A ABC. Prove that
AE.AC = AF.AB,
(Write down two dififerent expressions for BC^.)
tEx. 1132. In the figure of Ex. 1131, BC^^AB. FB + AC. EC.
tEx. 1133. Tlie sum of tlie squares on, the two sides of a triangle
ABC is equal to twice the sum of tlie squares on the median AD, and
half the base. (Apollonius' theorem.)
(Draw AN xto BC ; apply ii. 7, 8 to A» ABD, ACD.)
Ex. 1134. Use Apollonius' theorem to calculate the lengths of the three
medians in a triangle whose sides are 4, 6, 7.
Ex. 1135. Repeat Ex. 1134, with sides 4, 5, 7.
Ex. 1136. Calculate the base of a triangle whose sides are 8 cm, and
16 cm., and whose median is 12 cm. Verify graphically.
Revise Ex. 246.
tEx. 1137. The base BC of an isosceles A ABC is produced to D, so that
CD = BC; prove that AD2=AC24-2BC*.
216 BOOK II
tEx. 1138. A side PR of an iaosooles A PQR is prodaoed to 8 so that
RS = PR : prove that QS2=2<aR«+PR2.
tEx. 1130. The base AD of a triangle OAD is trisected in B, C. Prove
that OA2+20D2=30C2 + CCD2.
(Applj Apollonins' theorem to A' OAC, OBD ; then eliminate OB^.)
■^Ex. 1140. In the figure of Ex. 1139, OAa + OD2 = OBs + OC» + 4BC2.
iEx. 1141. A point moves so that the sum of the squares of its distances
&om two fixed points A, B remains constant ; prove that its locns is a circle,
having for centre the mid-point of AB.
tEx. 1142. The sum of the squares on the sides of a parallelogram is eqtial
to the sum of the squares on the diagonals.
tEx. 1143. In any quadrilateral the sum of the squares on the four sides
exceeds the sum of the squares on the diagonals by four times the square on
the straight line joining the mid-points of the diagonals.
(Let E, F be the mid-points of AC, BD ; apply Apollonius' theorem to
A* BAD, BCD and AFC.)
tEx. 1144. The sum of the squares on the diagonals of a quadrilateral is
equal to twice the sum of the squares on the lines joining the mid-points of
opposite sides. (See Ex. 736 and 1142.)
tEx. 1145. In a triangle, three times the sum of the squares on the sides
=:fonr times the sum of the squares on the medians.
HEx. 1146. "What does Apollonius' theorem become if the vertex moves
down (i) on to the base, (ii) on to the base produced?
BOOK m.
The Circijb.
Cone.
Sphere.
Section I. Preliminary.
Ctlindeb.
Def. a circle is a line, lying in a plane, such that all
points in the line are equidistant from a certain fixed point, called
the centre of the circle.
In view of what has been said already about loci we may give
the following alternative definition of a circle : —
Def. a circle is the locus of points in a plane that lie
at a fixed distance from a fixed point (the centre). The fixed
distance is called the radius of the circle.
The word " circle " has been defined above to mean a certain kind of
curved line. The term is, however, often used to indicate the part of the
plane inside this line. If any doubt exists as to the meaning, the line la
called the ciTcnznference of the circle.
Two circles are said to be equal if they have equal radii
If one of two equal circles is applied to the other so that the
centres coincide, then the circumferences also will coincide.
G. s. 11. 15
218
. BOOK UI
fig. 212.
r
O
;
\c
- T,
y
r-
>— H
Point and circle. A point may be
either outside a circle, on the circle or in-
side the circle. The point will lie outside
the circle if its distance from the centre >
the radius; it will lie on the circle if its
distance - the radius;' it will lie inside the
circle if the distance < the radius.
Straight line and circle. A straight line cannot cub a
circle in more than two pointa In fact, an unlimited straight
line may
(i) cut a circle in two points, e.g.
AB or CD in fig. 213. In this case the
part of the line which lies inside the
circle is called a chord of the circle.
(ii) The line may meet the
circle in one point only; thus EF
meets the circle in T. In this case
the line is said to touch the circle ;
it is called a tangent ; T is called the point of contact of the
tangent.
The tangent lies entirely outside the circle and has one point,
and one only, in common with the circle. It is obvious that
there is one and only one tangent which touches the circle at a
given point.
(iii) The line may lie entirely outside the circle, and have
no point in common with the circle, e.g. GH in fig. 213.
A chord may be said to be the straight line joining two points
on a circle. If the chord passes through the centre it is called a
diameter, e.g. aob in fig. 213.
The length of a diameter is twice the length of the radius;
all diameters are equal.
A chord divides the circumference into two parts called arcs.
If the arcs are unequal, the less is called the minor arc and the
greater the major arc.
fig. 213.
THE CIRCLE 219
Three letters are needed to name an arc completely ; e.g. in fig. 213, CTD
is a minor arc, CBD a major arc.
A diameter divides the circumference into two equal arcs,
each of which is called a semicircle.
It will be proved below that the two semicircles are equal.
The term "semicircle" like the term "circle" is used in two different
senses ; sometimes in the sense of an arc (as in the definition) ; sometimes
as the part of the plane bounded by a semi-circumference and the cor-
responding diameter.
A segment of a circle is the part of
the plane bounded by an arc and its chord
(fig. 214). A sector of a circle is the part of
the plane bounded by two radii, and the arc
which they intercept (fig. 214).
fig. 214.
IFEz. 1147. A circular hoop is cut into two pieces; what is each called?
IfEx. 1148. A penny is out into two pieces by a straight cut; what is the
shape of each piece?
ITEx. 1149. "What geometrical figure has the shape of an open fan?
TEx. 1150. A certain gun in a fort has a range of 5 miles, and can be
pointed in any direction from 15° E. of N. to 15° W. of N. What is the
shape of the area commanded by the gun?
Section II. Chord and Centke.
Symmetry of the circle. From what has been said about
symmetry (Ex. 277 onwards) it wiU be seen that the circle is
symmetrical about any diameter, and is also symmetrical about
the centre.
HEx. 1151. Draw a circle of about 3 in. radius; draw a set of parallel
chords (about 10) ; bisect each chord by eye. What is the locus of the mid-
points of the chords ? {Freehand.)
HEx- 1152. Draw a circle and a diameter. This is an axis of symmetry.
Mark four pairs of corresponding points. Is there any case in which a pair
of corresponding points coincide ? {Freehand.)
ITEx. 1153. What symmetry is possessed by (i) a sector, (ii) a segment,
(iii) an arc, of a circle ?
15—2
220 BOOK in
Theorem 1.
A straight line, drawn from the centre of a circle to
bisect a chord which is not a diameter, is at right angles
to the chord;
Conversely, the perpendicular to a chord from the
centre bisects the chord.
(1) Data OD is a straight line joining O, the centre of ©ABO,
to D, the mid-point of the chord AB.
To jrrove tliat OD is x to AB.
Constribction Join OA, OB.
Proof . In the As OAD, OBD
rOA- OB (radii),
J OD is common,
(ad=BD. Data
.'. the triangles are congruent, L 14.
.•. ^ODA= ^ODB,
,'. OD is ± to AB,
(2) Ck)NVEfiSK Theohem.
Data OD is a straight line drawn from O, the centre of ©ABC,
to meet the chord AB at right angles in D.
To prove tJiat AD = BD.
CHORD AND CENTRE 221
Construction Joia OA, OB.
Proof In the right-angled As OAD, OBD
1L s ODA, ODB are rt. /. s, Data
OA = OB (radii),
OD is common,
.■. the triangles are congruent, I. 15.
.•. AD = BD.
Q. E. D.
CoE. A straight line drawn through the midpoint of a
chord of a circle at right angles to the chord will, if produced,
pass through the centre of the circle.
(For only one perpendicular can be drawn to a given line at
a given point in it.)
222 BOOK III
To find the centre of a given circle.
I. 25.
fig. 216.
Construction Draw any two chords AB, CD (not parallel).
Draw EMF to bisect AB at right angles,
and GNH to bisect CD at right angles.
Let these straight lines meet at O.
Then O is the centre of the circle.
Proof Since EMF bisects chord AB at right angles
.'. the centre must lie somewhere on EMF.
Similarly the centre must lie somewhere on GNH.
Hence the centre is at O, the point of intersection of
EMF and GNH.
HEx. 1164. Why is it necessary that the chords AB, CD should not be
parallel?
To complete a circle of which an arc is given.
Find the centre of the circle as in the preceding construction.
Ex. 1156. With a fine-pointed pencil trace round part of the edge of a
penny, so as to obtain an arc of a circle. (Take care to keep the pencil
perpendicular to the paper.) Complete the circle by finding the centre.
Ex. 1156. By the method described in Ex. 1155, examine how far the
curved edge of your protractor differs from a true semicircle.
HEx. 1157. Describe five circles (in the same figure) to pass through two
given points A, B, 6 cm. apart. (The centre must be equidistant from A
and B ; what is the locus of points equidistant from A and B ?)
Ex. 1158. Describe a circle to pass through two given points A, B, 6cm.
apart, and to have a radius of 5 cm. Measure the distance of the centre from
AB.
HEx. 1159. What is the locus of the centres of circles which pass through
two given points?
to find the centre ^223
Theorem 2. *
There is one circle, and one only, which passes through
three given points not in a straight line.
fig. 217.
Data A, B, C are three points not in a straight line.
To prove that one circle, and one only, can be drawn to pass
through A, B and C.
Proof It is only necessary to show that there is one point (and
one only) equidistant from A, B, and C,
Now the locus of all points equidistant from A and B is
FE, the perpendicular bisector of AB ; I. 25.
and the locus of aU points equidistant from B and C is HG,
the perpendicular bisector of BC. I. 25.
These bisectors, not being parallel, wiU intersect.
Let the point of intersection be O.
The point O is equidistant from A and B; also from B
and C;
.*. O is equidistant from A, B and C;
and there is no other point equidistant from A, B and C.
Hence a circle with centre O and radius OA wiU pass
through A, B and C;
and there is no other circle passing through A, B and C.
Q. E. D.
CoR. 1. Two circles cannot intersect in more than two
points.
For if the two circles have three points in common, they
have the same centre and radius, and therefore coincide.
CoR. 2. The perpendicular bisectors of ab, bc, and CA
meet in a point.
224 BOOK III
HEx. 1 190. How vonld the proof of in. 2 fail if A, B, C were in a straight
line?
Ex. 1161. Prove Cor. 2. (Let two of the hisectors meet at a point O;
then prove that O lies on the third bisector.)
A
Def. If a circle passes through all the
vertices of a polygon, the circle is said to
be circuznscribed about the polygon ; and
the polygon is said to be inscribed in the
circle (fig. 218).
Def. If a circle touches all the sides
of a polygon, the circle is said to be in-
scribed in the polygon ; and the polygon
is said to be circumscribed about the
circle (fig. 219).
fig. 219.
To circumscribe a circle about a given triangle.
This is the same problem as that of describing a circle to pass
through three given points, namely the three vertices of the
triangle (see iii. 2).
Def. The centre of the circle circumscribed about a triangle
is called the circumcentre of the triangle.
Notice that, though the perpendicular bisectors of all three sides pass
through the circumcentre, yet it is not necessary to draw more than two of
these bisectors in order to find the centre.
Ex. 1163. (Inch paper.) Draw a circle to pass through the points
(0, 3), (2, 0), ( - 1, 0), and measure its radius.
Does this circle pass through
(i) (0, -3), (ii) (1, 3), (iii) (0, -I)?
Ex. 1163. (Inch paper.) Draw the circumcircle of the triangle whose
vertices are (0, 2), (4, 0), ( - 1, 0), and find its radius.
Does this circle pass through
(i) (0, -2), (u) (0, -3), (iii) (1-5,3)?
TO FIND THE CENTRE 225
Ex. 1164. (Inch paper.) Find the circumradius, and the coordinates
of the circumcentre of (0, 1), (3, 0), ( - 3, 0).
Ex. 1165. (Inch paper.) Find the circumradius, and the coordinates
of the circumcentre of each of the triangles in Ex. 821 (i), (ii), (iii).
Ex. 1165a. Find the circumradii of A' (i)— (vi) in Ex. 942.
Ex. 1166. Mark four points (at random) on plain paper, and find
out vrhether it is possible to draw a circle through all four.
Ex. 1167. (Inch paper.) Can a circle be drawn through the four points
(i) (2,0), (0,2), (-2, 0), (0, -2);
(ii) (2,0), (0,1), (-2, 0), (0, -1);
(iii) (2,0), (0, 2), (-2, 0), (0, 1)?
ITEx. 1168. Can a circle be circumscribed about a rectangle?
ITEx. 1169. Draw a parallelogram (not rectangular) and try if a circleoan
be circumscribed about it.
Ex. 1170. Draw an acute-angled scalene triangle ABC (no side to be less
than 3 inches). Draw the circumscribing circle. Find P, Q, R the middle
points of the sides. Draw the circle which passes through P, Q, R. Find
the ratio of the radius of the greater circle to that of the less ;
greater radius
smaller radius *
Ex. 1171. Bepeat Ex. 1170 with a right-angled triangle.
Ex 1172. Draw a scalene triangle, and on its three sides construct equi-
lateral triangles, pointing outwards. Draw the circumeircles of these equi-
lateral triangles; they should all pass through a certain point inside the
triangle.
Ex 1173. Draw four straight lines, such that each line meets the
three other lines. Four triangles are thus formed. Draw the circumeircles
of these triangles ; they should meet in a point.
tEx. 1174. If a chord cuts two concentric circles in A, B; C, D, then
AC= BD. (Draw perpendicular from centre on to chord.)
fEx. 1175. From a point O outside a circle two equal lines OP, OGl are
drawn to the circumference. Prove that the bisector of / POGl passes through,
the centre of the circle. (Join PQ.)
Ex 1176. O is a point 4 inches from the centre of a circle of radius
2 inches. Show how to construet with O as vertex an isosceles triangle
having for base a chord of the circle, and a vertical angle of 50°.
(Freehand)
Ex. 1177. If a polygon is such that the perpendicular bisectors of all
the sides meet in a point, a circle can be circumscribed round the polygon.
226 BOOK m
Section III*. Arcs, Angles, Chords
Theorem 3.
In equal circles (or. in the same circle)
(1) if two arcs subtend equal angles at the centre^
they are equal.
(2) Gorw&rsdy, if two arcs are equal, they subtend
equal angles at the centres.
(1) Data ABC, DEF are equal 0 s.
The arcs AGB, DHE subtend equal z.s APB, DQE at the
centres P, GU
To prove that arc AG B = arc DHE.
Proof Apply 0DEF to 0ABC, so that centre Gl may fall on
centre P.
Since the 0s are equal, the circumference of 0 DEF falls
on the circumference of 0 ABC.
Make 0DEF revolve about the centre till QD falls
along PA.
Then, since z. DQE = /. APB Data
QE falls along PB, and since the circumferences coincide^
D coincides with A, and E with B.
.'. arc DHE coincides with arc AGB.
.'. arc DHE = arc AGB.
(2) Converse Theorem.
Data arc AGB = arc DHE.
To prove that LS APB, DQE, subtended by these arcs at the
centres, are equal.
* This section may be omitted at first reading, with the exception of
Theorem 5 and the exercises which follow (pp. 235 — 237).
ABCS AND ANGLES 227
Proof Apply ©DEF to ©ABC, so that centre Q may fall on
centre P.
Since the ©s are equal, the circumference of ©DEF
falls on the circumference of ©ABC.
Make ©DEF revolve about the centre till D coincides
with A.
Then, since arc DHE =arc AGB Data
E coincides with B.
.'. QD coincides with PA, and QE with PB,
.'. /.DQE= Z.APB.
Q. E. D.
CoR. Equal angles at the centre determine equal sectors.
Note on the case op "the same circle."
The above proposition is proved for eqiud circles. To see
that it applies to arcs and angles in the same circle, let the arcs
AB, PQ of circle ABPQ (fig. 221 i.) subtend equal angles AGB, POQ
at the centre. To prove that arc AB = arc PGL
P ^Q
fig. 221 i. ii. iii.
Fig. i. may be regarded as consisting of the two circles
in figs, ii., iii. superposed. But these are equal © s,
.'. arc AB = arc PQ.
tEx. 1178. Show how to bisect a given arc of a circle. Give a proof.
tEx. 1179. P, A, B are points on a circle whose centre is O; PA = PB.
Prove that P is the mid-point of arc AB ; and that OP bisects AB.
+Ex. 1180. PQ, PR are a chord and a diameter meeting at a point P in
the circumference. Prove that the radius drawn parallel to PQ bisects the
arc QR.
tEx. 1181. P is ,1 point on the circumference equidistant from the radii
OA, OB. Prove that arc AP=arc BP.
228
BOOK m
Theorem 4.
In equal circles (or, in the same circle)
(1) if two chords are equal, they cut off equal arcs.
(2) Conversely, if two arcs are equal, the chords of
the arcs are equal.
H.
fig. 222.
(1) Data ABC, DEF are equal ©s ; their centres are P and Q.
Chord AB = chord DE.
To prove that arc AGB = arc DHE, and arc ACB = arc DFE.
Gonstructica Join PA, PB ; QD, QE.
Proof In the As APB, DQE
'AB=DE Daia
• AP =: DQ (radii of equal 0 s)
BP = EQ (radii of equal 0 s)
.*. the triangles are congruent, I. 14.
.•. L APB = L DQE,
.'. arc AGB = arc DHE, III. 3.
Again, whole circumference of 0ABC=whole circum-^
ference of 0 DEF.
.'. the remaining arc ACB = the remaining arc DFE.
(2) Converse Theorem.
Data
To prove that
Construction
arc AGB = arc DHE.
chord AB =: chord DE.
Join PA, PB; QD, QE.
ARCS AND CHORDS 229
Proof Since arc AGB = arc DHE, Data
:. L APB = L. DQE III. 3.
/. in the As APB, DQE
AP = DQ,
BP = EQ,
. Z. APB =:= Z. DQE.
/. the triangles are congruent, I. 10.
.'. chord AB = chord DE.
Q. E. D.
tEx. 1182. A quadrilateral ABCD is inscribed in a circle, and AB=CD.
Prove that AC = BD.
tEx. 1183. Prove the converse, in m. 4, by superposition. Also try to
prove the direct theorem by superposition, and point out where such a
proof fails.
To place in a circle a chord of given length.
Adjust the compasses to the given length. With a point
A on the circle as centre draw an arc cutting the circle in B.
Then AB will be the chord required.
Ex. 1184. Place in a circle, end to end, 6 chords each equal to the
radius.
Ex. 1185. Place in a circle, end to end, 12 chords each equal to \ the
radius.
Ex. 1 186. Draw a circle of radius 5 cm. Place in the circle a number of
chords of length 8 cm. Plot the locus of their middle points.
Ex. 1187. Show how to construct an isosceles triangle, given that the
base is 7 cm. and the radius of the circumscribing circle is 5 cm. (Which
will you draw first — the base, or the circle ?)
Ex. 1188. Construct a triangle, given BC=3in., Z.B= 30°, radios of
circumscribing circle =2 ins. Measure AC and L. A.
tEx. 1189. In a circle are placed, end to end, equal chords PQ, QR, RS,
ST. Prove that PR = QS=RT.
230
BOOK m
To inscribe a regular hexagon in a circle.
A
IB
fig. 223.
In the circle place a chord AB, equal to the i-adius.
Join A, B to O, the centre.
Then A CAB is equilateral,
.•". ^ AOB = 60°.
Place end to end in the circle 6 chords each equal to the
radiiis.
Each chord subtends 60° at the centre,
.'. the total angle subtended by the 6 chords is 360°.
In other words, the 6 chords form a closed hexagon inscribed
in the circle.
Since each side of the hexagon = the radius,
the hexagon is equilateral ;
and since each angle of the hexagon = 120°, the hexagon is
equiangular,
.'. the hexagon is regular.
tEz. 1 lOO. The side of an isosceles triangle of vertical angle 120° is equal
to the radius of the circumcircle.
Ex. 1191. Find the area of a regular 6-gon inscribed in a circle of
radios 2 in.
Bevise "Eegular polygons," Ex. 69 — 74.
Ex. 1192. Find the perimeter and area of a regular 8-gon inscribed in
a circle of radius 2 in.
REGULAR HEXAGON — CIRCUMFERENCE OF CIRCLE 231
Circumference of Circle.
Consider any circular object, such as a penny, a I'ouud tin,
a gaixlen-roller, a bucket, a running track. Measure the cir-
cumference and the diameter; how many times does the cir-
cumference contain the diameter? Work out your answer to
three significant figures
Methods of measuring the circtunference : —
(i) Put a small spot of ink on the edge of a penny; roll the penny
along a sheet of paper, and measure the distance between the ink spots left
on the paper.
(ii) Wrap a piece of paper tightly round a cylinder; prick through two
thicknesses of the paper; unroll the paper and measure the distance between
the pin-holes.
(in) Wrap cotton round a cylinder several times, say 10 times ; measure
the length of cotton used, and divide by 10.
In measuring the diameter, make sure that you are measuring the greatest
width.
XI - Ti. T ,, , , ,, ,■ ^ circumference . ,
Hix. 1193. Find the value of the quotient — =^ ■ for several
diameter
circular objects of different sizes, and take the average of your answers.
Theory shows that the value of this quotient (or ratio) is the
same for all circles; it has been worked out to 700 places of
decimals and begins thus
3-1416926535
For the sake of brevity this number is denoted by the Greek
letter tt ; a useful approximation for tt is ^.
232
BOOK in
The ratios of the perimeters (or circumferences) of regular
polygons to the diameters of their circumscribing circles are
shown in the following table : —
Table shotoirig the perimeters of regtdar polygons inscribed
in a circle of radius 5 cm.
No. of
Perimeter in
Batio of perimeter
sides
centimetres
to diameter
3
25-98
2-598
4
28-29
2-829
5
29-39
2-939
6
30-00
3-000
7
30-38
3-038
8
30-61
3-061
9
30-78
3-078
10
30-90
3-090
It will be noticed that the ratio increases with the number of
sides, being always less than ir. If the number of sides is very-
great, the ratio is very nearly equal to tt. E.g. for a polygon of
384 sides the ratio is 3-14156
Ex. 1 1 94. By how much per cent, does the perimeter of a regular decagon
inscribed in a circle differ from the perimeter of the circle?
We have seen that
circumference of circle = diameter x tt
= radius x 27r
— 2-irr, where r is the radius.
Ex. 1195. Calculate the circumference of a circle whose radius ia
(i) 7 in., (ii) 14 cm., (iii) 35 miles. (Take 7r=^.)
Ex. 1196. Calculate the circumference of a circle whose diameter ia
(i) 70 ft., (ii) 21 mm., (iii) 49 miles.
CIRCUMFERENCE OF CIRCLE 233
Ex. 1197. Calculate to three significant figures the cireumference of a
half-penny (diatueter 1 inch).
Ex. 1198. Calculate to three significant figures the circumference of the
earth, measured round the equator^ taking radius = 3963 miles.
Ex. 1199. Calculate to three significant figures the circumference of a
circle whose radius is 5 cm., and compare your result with the perimeters of
regular polygons in the tahle on page 232.
Ex. 1200. How far does a wheel roll in one revolution if its diameter is
28 in.?
Ex. 1201. In fig. 224, AD is divided into three
equal parts and all the arcs are semicircles; show
that the four curved lines which connect A with D
are of equal length.
Ex. 1202. The driving-wheel of an engine is 6 ft.
high, and each of the smaller wheels is 3*5 ft. high ;
how many turns does a small wheel make for one
turn of the large wheel? fig- 224.
Ex. laoa. Calculate the radius of a circle of circumference (i) 22 ft.,
(ii) 40 ft. (to three significant figures).
Ex. ia04. A bicycle wheel makes 7200 turns in an hour while the
cyclist is riding 10 miles an hour : what is the diameter of the wheel (to the
nearest inch) ?
Ex. 1205. Calculate the circumference of a circle whose diameter is
(i) 4-35 in., (ii) 617 mm.
Ex. 1206. Calculate the circumference of a circle whose radius is
(i) 0-346 yards, (ii) 21-7 in.
Ex. 1207. Calculate the radius of a circle whose circumference is
(i) 478 miles, (ii) 27*5 ft
tEx. 1208. Prove that the circumference of a circle is :> three times the
diameter, by inscribing a hexagon in the circle.
tEx. 1209. Prove that the circumference is < four times the diameter by
circumscribing a square round the circle.
Dep. If an arc of a circle subtends, say, 35° at the centre, it
is called an arc of 35°.
Ex. 1210. What fractions of a circumference are arcs of 90°, 60°,
120°, 1°, 35°, 300°?
a. s. II. 16
234 BOOK III
Ex. 1311. Calculate the length of an arc of C0° iu a circle of radius 7 cm.
What is the length of the chord of this arc? Find, to three significant
figures, the ratio -; ?: also the difierenoe of arc and chord.
" chord
Ex. laia. Repeat Ex. 1211 for a circle of radius 2-57 in.
Ex. laiS. Draw a circle of any radius ; mark an arc of 40° ; calculate the
length of the arc, and measure the chord ; then find ratio ^-^ to three
significant figures.
Ex. iai4. Bepeat Ex. 1213, with an arc of 80°.
Ex. laiS. The circimiference of a circle is 7'82 in. and the length of
a certain arc is 1'25 in. What decimal of the circumference is the arc?
What angle does the arc subtend at the centre?
Ex. iai6. The radius of a circle is 10 cm.; a piece of string as long as
the radius is laid along an arc of the circle ; what angle does it subtend at
the centre? Also find the angle subtended at the centre bj a chord of 10cm.
Ex. iai7. In a circle of radius 3 in., what is the chord of an arc of
6 in.? (Calculate the angle at the centre; then draw the figure and measure.)
Ex. iai8. Draw an arc of a circle (any radins and angle). Calculate its
length, and test the accuracy of the following approximate rule: — "To find
the length of an arc, from eight times the chord of half the arc subtract
the chord of the whole arc, and divide the result by three." (It will be
necessary to measure the length of the chords.)
Ex. iai9. Find the length of the minor and major arcs cut off from a
circle of radius 7 cm. by a chord of 7 cm.
Ex. laao. Find the lengths of the two arcs cut from a circle of diameter
4 '37 in. by a chord of 4 in. (Measure the angle at the centre.)
lengths of chords
Theorem 5.
236
V In equal circles (or, in the same circle)
(1) equal chords are equidistant from the centres.
(2) Conversely, chords that are equidistant from the
centres are equal.
fig. 225.
(1) Data ABC, DEF are equal circles; their centres are P and Q.
Chord AB = chord DE.
PG, QH are perpendiculars from the centres P, Q upon
the chords AB, DE.
To prove that
Construction
PG = QH.
Join PA, QD.
Proof
Since PG is j. to AB,
.*. AG = BG,
IIL 1.
.'. AG = |AB.
Sim'y DH = |DE.
But AB=DE,
Data
1
.*. AG = DH.
In the right-angled A* APG,
( L^ Q and H are rt. /. ",
DQH,
Consfr.
•
AP = DQ,
Data
AG = DH,
. Proved
.'. the triangles are congruent,
I. 15.
.-. PG=:QH.
16—2
236
fig. 225.
(2) Converse Theorem
Daia
To prove that
Proof
PG = QH.
chord AB = chord DE.
In the right-angled A' APG, DQH,
z. • G and H are rt, l *,
AP=DQ,
PG = QH,
.'. the triangles are congruent,
.-. AG = DH.
ButAB = 2AG, DE = 2DH,
.•. AB=-DE.
Constr.
Data
Data
I. 15.
Q. E. D.
fEz. 1S21. Prove iii. 6 by means of Pythagoras' theorem.
Ex. 1222. Calculate the distances &0m the centre of a circle (radins
5 om.) of chords whose lengths are (i) 8 cm., (ii) 6 cm., (iii) 5 cm.
Ex. 1223. Calculate the lengths of chords of a circle (radins 2-5 in.)
whose distances from the centre are (i) 2 in., (ii) 1*5 in., (iii) 1 in.
Ex. 1224. Find the locus of the mid-points of chords 6 cm. in length
in a circle of radius 5 cm.
LENGTHS OF CHORDS 237
tEx. laas. Prove that the locus of the middle points of a set of equal
chords of a circle is a concentric circle.
Ex. 1326. A chord CD of a circle, whose centre is O, is bisected at N hy
a diameter AB. 0A = OB = 5 cm., ON =34 cm. Calculate CD, CA, CB.
Ex. 1227. The lengths of two parallel chords of a circle of radius 6 cm.
are 10 cm. and 6 cm. respectively. Calculate the distance between the
chords. (There are two cases.)
Ex. 1228. Calculate the length of (i) the longest, (ii) the shortest chord
of a circle, radius r, through a point distant d from the centre (see Ex. 1238).
Ex. 1229. Calculate the radius of a circle, given that a chord 3 in. long
is 2 in. from the centre.
Ex. 1230. What is the radius of a circle when a chord of length 22 is at
distance d from the centre?
Ex. 1231. Given that a chord 12 cm. long is distant 2-5 cm. from the
centre, calculate (i) the length of a chord distant 5 cm. &om centre, (ii) the
distance from the centre of a chord 6 cm. long.
tEx. 1282. If two chords make equal angles with the diameter through
their point of intersection, they are equal.
[Prove that they are equidistant from the centre.]
tEx. 1288. A straight line is drawn cutting two equal circles and parallel
to the line joining their centres; prove that the chords intercepted by the
two circles are equal.
tEx. 1284. A straight line is drawn cutting two equal circles, and passing
through the point midway between their centres. Prove that the chords
intercepted by the two ckdes are equaL
Ex. 1235. Show how to draw a chord of a circle (i) equal and parallel
to a given chord, (ii) equal and perpendicular to a given chord, (iii) equal to
a given chord and parallel to a given line.
tEx. 1286. If two eborda are at nnetiual distances from the centre,
tbe nearer chord is longer tban tbe more remote.
238 BOOK III
tEx. 1387. Btat* and proT* tli« converse of Sx. 1389.
tEx. 1338. The shortest chord that can be drawn through a point inside
a circle is that which is perpendicular to the diameter through the point.
[Prove that it is furthest from the centre.]
Ex. 1388 a. A wooden ball of 4" radius is planed down till there is a
flat circular face of radius 2". If the block is now made to stand on the
flat face, how high will it stand?
Ex. 1388 b. The distance from the centre of the earth of the plane of
the Arctic circle is 3700 miles (to the nearest 100 miles) ; the radius of the
earth is 4000 miles. Find the radius of the Arctic circle.
Ex. 1338 e. A ball of radius 4 cm. floats in water immersed to the depth
of *J of its diameter. Calculate the circumference of the water-line circle.
Ex. 1388 d. The diameter of an orange is 4", and the thickness of the
rind is |"'. A piece is sliced off just grazing the flesh ; find the radius of
the piece.
Section IV. The Tangent.
The meaning of the term tangent has been explamed on
p. 218. It may be defined as follows: —
Def. a tangent to a circle is a straight line which, how-
ever far it may be produced, has one point, and one only, in.
common with the circla
The tangent is said to touch the circle ; the common point
is called the point of contact.
We shall assume that at a given point on a circle there is one
tangent and one only.
TUB TANGENT
239
Theorem 6.
The tangent at any point of a circle and the radius
through the point are perpendicular to one another.
Data O is the centre of 0 ; A is a point on the circumference;
BC is the tangent at A.
To prove that BC and OA are i. to one another.
Construction If OA be not j. to BC, draw OT ± to BC.
Proof Since L OTA is a rt. z. , Constr.
.-. OT<OA, I. 21.
.*. T is inside the circle,
.♦. the tangent AT, if produced, will cut the circle
in another point.
This is impossible, • Def.
.'. OA is ± to BC,
.'. the tangent at A and the radius through A are j. to
one another.
Q. E. D.
Cor, a straight line drawn through the point of contact
of a tangent at right angles to the tangent will, if produced, pass
through the centre of the circle.
240 BOOK III
To draw the tangent to a circle at a given point on
the circle.
Join the point to the centre, and draw a straight line
through the point perpendicular to the radius.
The proper method of drawing a tangent to a circle from
an external point cannot be explained at the present stage, as
it depends on a proposition that has not yet been proved. In the
meantime it will be sufficient to draw the tangent from an
external point with the ruler (by eye). It is not possible to
distinguish the point of contact accurately without further con-
struction; to find this point, drop a perpendicular upon the
tangent from the centre; the foot of this perpendicular is the
point of contact.
This method is accurate enough for many purposes; the student is
warned, however, that it would not be accepted in most examinations. The
correct construction is given on page 262.
tEx. 1239. Prove tbat the two tangents
drawn to a circle from a point A are (i) equal, _^r~->^
(ii) equaUy inclined to AO. (Fig. 227.) ^.^-^/^^ \
Ex. 1240. P is 4 in. distant from O, the centre ^^^^^v<L„x^
of a circle of radius 3 in. Prom P draw a tangent Q
with your ruler. Determine T, the point of contact, g 227 '
(i) by eye, (ii) by drawing a perpendicular from O.
§
Calculate PT, the length of the tangent (using Pythagoras' theorem).
Verify by measurement.
Ex. 1241. Calculate the lengths of the tangents to a circle of radius r
from a point distant d from the centre when (i) r=6 cm., d=8 cm.;
(ii) r=l in., d = 5 in.
Ex. 1242. At a point A of a circle (radius r, centre O) is drawn a
tangent AP of length I; find OP.
Ex. 1243. At a point P on the circumference of a circle of radius 4 cm.
is drawn a tangent PT 3 cm. in length. Find the locus of T as P moves
round the Q .
THE TANGENT 241
Ex. 1344. Two circles, of radii 3 and 2 in., are concentric. Calculate
the length of a chord of the outer circle which touches the inner.
Ex. 1346. Prove that all chords 8 cm. long of a circle of radius 5 cm.
tonoh a certain concentric circle ; find its radius.
tEx. 1346. All chords of a circle which touch an interior concentric
circle are equal, and are bisected at the point of contact.
tEx. 1347. PQRS is a quadrilateral cireninacrlbed about a eirel*.
Prove that PQ+ RS = QR + SP. (See fig. 219.)
tEx. 1348. Draw a circle and circumscribe a parallelogram about it.
Prove that the parallelogram is necessarily a rhombus (use Ex. 1247).
tEx. 1349. Prove that the point of intersection of the diagonals of
a rhombus is equidistant from the four sides.
lIEx. 1360. Draw a quadrilateral ABCD. What is the locus of the
centres of 0" touching AB, BC ; touching BC, CD? Draw a circle to
touch AB, BC and CD. Does it touch DA? What relation must hold
between the sides of a quadrilateral in order that it may be possible to
inscribe a circle in it?
HEx. 1361. Construct a quadrilateral ABCD, having the sum of one
pair of opposite sides = the sum of the other pair of opposite sides (e.g.
AB = 2 in., BC = 3 in., CD =4 in., DA = 3 in.). Draw a circle to touch three
of the sides. Does it touch the fourth side ? Measure the radius.
HEx. 1353. Repeat Ex. 1251, using the same sides, but altering the shape
of the quadrilateral. Inscribe a circle in it. Is the radius the same as in
Ex.1251?
tEx. 1363. ABCDEF is an irregular hexagon circumscribed about a
oircle ; prove that AB + CD + EF = BC + DE + FA.
tEx. 1364. Two parallel tangents meet a third tangent at U, V ; prove
that UV subtends a right angle at the centre.
tEx. 1366. The angles subtended at the centre of a circle by two
opposite sides of a circumscribed quadrilateral are supplementary.
HEx. 1366. What is the locus of the centres of circles touching two lines
which cross at an angle of 60° ? (Bemember that two lines form four angles
at a point.) Draw a number of such circles.
242 BOOK III
IFEz. ia07. What is the locus of the centres of circles of radius 1 in.
which touch a given line ? Hence draw a circle which has a radius of 1 in.
and touches two given lines incUned at an angle of 60°.
Ex. 1268. Draw four circles of radius 3 cm. to touch two straight lines
which cross at an angle of 140°.
t£z. 1259. A is a point outside a circle, of centre O. With centre O
and radius OA describe a circle. Let OA cut the smaller circle in B. Draw
BC perpendicular to OB, cutting the larger circle in P, Q. Let OP, OQ cut
the smaller circle in S, T. Prove that AS, AT are tangents to the smaller
circle. (This is Euclid's construction for tangents &om an external point.)
tEx. 1260. A chord makes equal angles with the tangents at its
extremities.
Ex. 1261. Each of the tangents, TA, TB, at the ends of a certain
chord AB is equal to the chord ; find the angle between the tangents, and
the angle subtended at the centre by the chord.
t£x. 1262. In fig. 227, the angles PAQ, POQ are supplementary.
Ex. 1263. Show how to draw a tangent to a given circle (i) parallel to
a given line, (ii) perpendicular to a given line, (iii) making a given angle
with a given line.
Ex. 1264. Show how to draw two tangents to a circle (i) at right angles,
(ii) at an angle of 120°, (iii) at a given angle (without protractor).
tEx. 1265. The area of any polygon circumscribing a circle is equal to
half the product of the radius of the circle, and the perimeter of the polygon.
(Divide the polygon into triangles, with the centre for vertex.)
INSCRIBED CIRCLE 243
To inscribe a circle in a given triangle.
Construction It is necessary to find a point equidistant from
the three straight lines AB, BC, CA.
Draw BE, CF to bisect the angles ABC, ACB respectively.
Let these lines intersect at I.
Then I is the centre of the inscribed circle.
Proof Every point on BE is equidistant from AB and BC, and
every point on CF is equidistant from BC, CA. i. 26.
Therefore I is equidistant from AB, BC and CA.
From I draw IX, lY, IZ j.to BC, CA, AB respectively.
Then IX = IY=IZ.
Therefore a circle described with I as centre and IX
as radius will pass through X, Y, Z. Also BC, CA, AB will
be tangents at X, Y, Z. (Why?)
This circle is the inscribed circle of the triangle ABC.
Ex. 1366. Draw the Inscribed circle of a triangle whose sides are
(i) 6, 6, 7 in., (ii) 8, 6, 8 cm. Measure the radii of the circles.
tEx. 1267. Tlie bisectors of tbe three angles of a triangle meet
in a point.
(Join lA, and prove that lA bisects /.A.)
244 BobK in
The escribed circles of a triangle.
Draw Blj, Cli to bisect the angles exterior to ABC and BC/^-
Then Ij is equidistant from AB (produced), BC and AC
(produced).
Drop a perpendicular from Ij to BC. A circle drawn with Ij
as centre and this perpendicular as radius will touch the side BC
and the sides AB, AC produced. This circle is called an escribed
circle of the triangle. There are three such circles (see fig. 229).
Ex. 1368. Draw the inscribed and escribed circles of a triangle whose
sides are 3, 4, 5 in. Measure the radii.
tEIx. 1360. Prove that the internal bisector of ^A and the external
bisectors of ^' B and C meet in a point.
tEx. 1370. Prove that All;^ is a straight line. (I is the centre of the
inscribed circle.)
Ex. 1371. Verify, by drawing, that the circle drawn through the mid-
points of the sides of a triangle touches the inscribed and each of the
escribed circles.
It has been shown that, in general, four circles can be drawn
to touch three unlimited straight lines, namely the inscribed and
escribed circles of the triangle which the three lines enclose.
ESCRIBED CIRCLES
245
^z. 1372. How many circles can be drawn to touoh two parallel
Btraight lines and a third straight line catting them.
^Ex. 1278. How many circles can be drawn to touch three straight lines
which intersect in a point ?
7£x. 1274. How many circles can be drawn to touch three parallel
straight lines?
Section V. Contact of Circles.
Oo
III
The different relative positions which are possible for two
circles are shown in fig. 230.
In Cases II and IV the circles are said to touch, externally
in Case II, internally in Case IV. The formal definition of
contact of circles is as follows : —
Def. If two circles touch the same , line at the same point,
they are said to touch one another.
246
BOOK III
Theorem 7.
If two circles touch, the point of contact lies in the
straight line through. the centres.
fig. 23L
fig. 232.
Data The ©s CMN, CPft touch internally (fig. 231) or externally
(fig. 232) at C.
X, Y are the centres of the 0 s.
AB is the common tangent at C.
To prove timt XY produced (fig. 231), or XY (fig. 232) passes
through C.
Construction Join XC, YC.
Proof Since CA is the tangent at C to 0 CMN, and CX the
radius through C,
.-. L XCA is a. Tt. L, III. 6.
Sim'y L YCA is a rt. ^ ,
••. if the 0s^uch internally, XYC is a straight line,
and if the ©s touch externally, i. XCA + l YCA = 2 rt. iL s.
.•. XCY is a straight line. 1.2.
Q. E. D
CONTACT OF CIRCLES 247
Cor.. If two circles touch externally the distance between
their centres is equal to the sum of their radii; if they touch
internally the distance between their centres is equal to the
difference of their radiL
lIEx. 1276. Draw a figure showing the different relative positions i>os-
sible for two equal circles.
^Ex. 1376. Describe in words each of the relative positions shown in
fig. 230.
lIEx. 1277. Describe the relative position of the two circles in each of
the following cases [d is the distance between the centres, R and r are the
radii). Do this, if you can, without drawing the circles,
(i) d=4'lom., JJ=2*lcm., .r=l"4cni.
(ii) d=0-7cm., iJ=2-2cm., r=l-2cm.
iJ=2*0cm., r=l'4cm.
JJ=2"lcin., r=l'3cin.
i2=l"9cm., r=l-2cm.
i?=2-0cm., r=l-5cm.
JJ=l-7cm., r=l-7cm.
Ex. 1278. What is the distance between the centres of two circles of
radii 15 and 14 in. (i) if they have external contact, (ii) if they have
internal contact ?
Ex. 1279. Show how to draw three circles having for centres the vertices
of an equilateral triangle of side 2 in., so that each circle may touch the two
others externally.
Ex. 1280. Three circles, of radii 1, 2^ 3 in., touch externally, each
circle touching the other two. What are the distances between the centres ?
Draw the circles.
Construction op Circles to satisfy given conditions.
Ii Ex. 1281. 'Wbat Is the locus of the centres of aU circles of radius
1 In., which tonch eztemaUy a fixed circle of radius 2 in.? Draw the
locns, and draw a number of the touching circles.
HEx. 1282. If required to draw a circle to touch a given curcle at a given
point, where would you look for the centre of the touching circle? ixrhat
is the locus of the centres of circles touching a given circle at a given
point? Draw a number of such circles, some enclosing the given circle,
some inside it, some external' to it.
(hi)
d=3-4cm..
(iv)
d=0-8cm.,
(V)
d=Ocm.,
(vi)
d=l*5cm.,
(vii)
d = 2-5cm..
248 BOOK III
HEx. 1388. ^xriiat is Uie loctu of tlie centres of circles wbich toncli
a given line at a given point ?
IFEx. 1384. What is the locus of the centres of ciroles of radias 1 in.,
teaching a given circle of radias 2 in., and lying inside it? Draw a number
of sach circles.
II Ex. 1386. Bepeat Ex. 1284 with 1 in. radius for the touching oircles,
and 3 in. radius for the fixed circle.
ITEx. 1386. Draw a number of circles of radius 3 in. to touch a circle
of radius 2 in. and enclose it.
HEx. 1387. Draw a number of circles of radius 4 in. to touch a given
circle of radias 2 in. and enclose it.
^Ex. 1388. "Vrha-t is tbe locus of centres of circles of given radius
passing ttarougb a given point ?
ITEx. 1389. What is the locus of centres of circles (i) pacing through
two given points, (ii) touching two given lines ?
Each of the following problems is to be solved by finding the
centre of the required circle, (generally by the intersection of
loci). Some of the group have been solved already; they are
recapitulated below for the sake of completeness. In several
cases a numerical instance is given which should be attempted
first, the radius of the resulting circle being measured.
Ex. 1390. Draw a circle (or circles) to satisfy the following conditions : —
(i) To pass through three given points (solved already).
(ii) Of given radius, to pass through two given points (solved
already).
(iii) Of given radius, to pass through a given point and touch a given
line, e.g. take radius 2 in. and a point distant 1 in. from the line. (What is
the locus of centres of 2 in. circles passing through given point? touching
given line ?) When is the general problem impossible ?
(iv) To touch a given line AB at a given point P, and to pass through
a given point Q outside the line. (What is the locus of centres of Q ' touch-
ing line ftt P? passing through P and Q? Let PGl=3 cm., ^QPA=30°.)
CONSTRUCTION OF CIRCLES WITH GIVEN CONDITIONS 249
(v) To touch a given circle at a given point P, and to pass through a
given point Q not on the circle. In what case is this impossible ?
(vi) To touch a given line AB at P, and also to touch a given line
CD, not parallel to AB. (What is the locus of centres of circles touching
AB and CD ?)
(vii) Of given radius, to pass through a given point P and touch a
given circle, e.g. let given radius = 4 cm., radius of given circle = 3 cm.,
distance of P from centre of given circle=5cm. (Compare (iii).)
(viii) Of given radius, to touch a given circle at a given point (how
many solutions ai'e there ?).
(ix) To touch three given lines (solved already).
(x) Of given radius to touch two given lines, e.g. let the lines inter-
sect at an angle of 60°, and radius = 1 in. (How many solutions are there ?)
(xi) Of given radius, to touch a given line and a given circle
(e.g. given radius = 3 cm., radius of given circle = 5 cm., distance of line
from centre of circle — 6 cm.). What is the condition that the general
problem may be possible?
(xii) To touch three equal circles (a) so as to enclose them all,
(6) so as to enclose none of them. (Begin by drawing a circle through the
three centres.)
(xiii) Of given radius, to touch two given circles (e.g. let given
radius = 2 in., radii of given circles = 1 in., 1-5 in., distance between centres
= 3-5 in.).
Ex. 1291. In a semicircle of radius 5 era. inscribe
a circle of radius 2 cm. Measure the parts into which
the diameter of the semicircle is divided by the point of
contact. *See fig. 233.
Ex. laoa. Draw four circles of radius 2 in. , touching
a fixed circle of radius 1 in., and also touching a straight
line 2 in. distant from the centre of the fixed circle.
fig, 233.
Ex. 1393. Show how to inscribe a circle in a sector of 60° of a circle
whose radius is 4 in.
Ex. 1294. Show how to draw three equal circles, each touching the
other two ; and how to circumscribe a fourth circle round the other three.
G. S. II. 17
250 BOOK III
t£x. 1396. Trove that, if circles are described with centres A, B, C
(fig. 228) and radii AY, BZ, CX, the three circles touch.
-fEx. X396. A variable circle (centre O) touches externally each of two
fixed circles (centres A, B). Prove that the difference of AO, BO remains
constant.
fEx. 1297. If two circles touch and a line is drawn through the, point of
contact to meet the circles again at P and Q, the tangents at P and Q are
parallel. (Draw the common tangent at the point of contact.)
fEx. 1298. If two circles touch externally at A and are touched at P, Q.
by a line PQ, then PQ subtends a right angle at A. Also PQ is bisected by
the common tangent at A.
+Ex. 1299. Prove that, in Ex. 1298, the circle on PQ as diameter passes
through A and touches the line of centres.
fEx. 1300. Two circles intersect at A, B; prove tbat tbe line of
centres bisects AB (tbe conunon cbord) at rigbt angles. (See ni. 1 ('or.)
What kind of symmetry has the above figure ?
Ex. 1301. Find the distances between the centres of two circles, their
radii being 5 and 7 cm. and their common chord 8 cm. (There are two
cases.)
Section VI. Angle Properties.
Reflex angles. Take your dividers and
open them slowl)\ The angle between the
legs is first an acute angle, then a right
angle, then an obtuse angle. When the
dividers are opened out flat, tlie angle has
become two right angles (180°). If the di- g„ '2^^
viders are opened still further the angle of
opening is greater than 180° and is called a reflex angle.
Dep. a reflex angle is an angle ,,— .^
greater than two right angles and less
than four right angles. Fig. 235 shows two
straight lines OA, OB forming a reflex
angle (marked), and also an obtuse angle
(unmarked). fig. 235.
ANGLE PROPERTIES
251
HEx. laoa. Account for the necessity of the phrase "less than four right
angles " in the above definition.
HEx. 1303. Open a book to form a reflex angle.
ITEx. 1304. What is the sum of the reflex angle a
and the acute angle b m fig. 236? If Z-6 = 36°, what
is Z.O?
lEx. 1305. What kind of angle is subtended at the
centre of a circle by a major arc?
HEx. laoe. Draw a quadrilateral having one angle fig. 236.
reflex. Prove that the sum of the four angles is 360°.
HEx. 1307. Is it possible for (i) a four-sided figure, (ii) a five-sided figure
to have two of its angles reflex ?
fig. 237.
fig. 238.
fig. 239.
HEx. 1808. Draw a figure like fig. 237, making the radius of the circle
about 2 in. Measure angles x and y.
HEx. 1309. Do the same for figs. 238, 239, 240.
What relation do you notice between the angle x and
the angle y in the four experiments?
HEx. 1310. Draw a circle of radius 5 cm. : place in
it a chord AB of length 9*5 cm. Mark four points
P, Q, R, S in the major arc. Make the necessary joins
and measure the angles APB, AQB, ARB, ASB. What
relation do you notice between these angles ? Can you
connect this with the results of Ex, 1308, 1309?
fig. 240.
llEx. 1311. In the figure of Ex. 1310 mark three points X, Y, Z in the
minor arc; measure the angles AXB, AYB, AZB.
HEx. 1312. Draw a circle and a diameter. Mark four points on the
circle, at random. Measure the angle subtended by the diameter at each of
these points.
fEx. 1313. A side BA of an isosceles triangle ABC is produced, through
the vertex A, to a point D. Prove that Z.DAC=:=2i:.ABC = 2^ACB.
17—2
252
BOOK in
Theoiiem 8.
The angle which an arc of a circle subtends at the
centre is double that which it subtends at any point on
the remaining part of the circumference.
fig. 241.
fig. 242.
Data The arc ACB of 0ACB subtends ^ AOB at the centre O;
and subtends L APB at P, any point on the remaining part
of the circumference.
To prove that L AOB = 2 z. APB.
Construction Join PO, and produce to Q.
Proof. Cask I. When the centre O is inside l APB.
In AAOP, OA = OP (radii)
.'. Z-OPA = z.OAP. 1. 12.
Now L QOA is an exterior /. of A AOP,
.*. z.QOA=Z.OPA+ ^OAP I. 8, Cor. 1.
= 2z.OPA.
Sim'yz.QOB = 2z.OPB,
.'. L QOA + L QOB = 2 ( ^ OPA + L OPB),
.•. z. AOB :^ 2 /. APB.
Case ii. ^Mlen tlie centre O is outside l APB.
As before, l QOB = 2 l OPB,
and z. QOA = 2 z. OPA,
.-. L QOB - Z. QOA = 2 ( Z. OPB - z. OPA),
.*. L AOB = 2 z. APB.
Q. E. D.
ANGLE PROPERTIES
253
lEx. 1314. Prove tlie above theorem for the case in which ACB is a
major arc, and the angle subtended at the centre a reflex angle (see fig.
240). What kind of angle is /. APB in this case?
tEx. 1315. Prove the above theorem for the case in which O lies on
AP (see fig. 238).
tEx. 1316. Prove that in fig 243
La = Lb.
tEx. 1317. If the two circles in figs. 241 and 242 are
equal, and the arcs ACB are equal, prove that the angles
APB are equal.
Ex. 1318. Drai^ a figure for the case of in. 8 in
which arc ACB is a semicurcle. "What does z. ACB be-
come in this case? What does /.APB become?
Ex. 1319. Find the magnitude of all the marked
angles in fig. 244. What is the sum of the angles at the
centre? of /.'ACB and ADB? of Z.»CAD and CBD?
fig. 244.
Def. a segment of a circle is the part of the plane
bounded by an arc and its chord.
MAJOR SEGMENT
SEMICIRCLE
MINOR SEGMENT
fig. 245.
Def. An angle in a segment of a circle is an angle sub-
tended by the chord of the segment at a point on the arc
(fig. 245).
Def. a segment is called a major segment or a minor
segment according as its arc is a major or a minor arc. It is
obvious that a major segment of a circle is greater than the
semi-circle (considered as an area) and that a minor segment is
less than the semi-circle.
HEx. 1330. Show by a figure that a minor sector of a circle can be
divided into a segment and a triangle. What is the corresponding
theorem for a major sector ? Is there any fignre which is at the same time
a sector and a segment ?
^54 BOOK III
Theorem 9.
Angles in the same segment of a circle are equal.
P . P .Q
fig. 246.
fig. 247.
fig. 248.
Data L s APB, AQB are two z. s in the same segment APQB of
0APB. (Three figures are drawn, for the three cases in
which the segment >, = or < a semi-circle.)
To -prove that
L APB = L AQB.
Construction Join A, B to the "centre.
Let X be the l subtended at the centre by arc ACB.
Proof In each figure z.a; = 2^APB,
for these angles are subtended by the same arc ACB. iii. 8.
Sim^ ^a;=2z.AClB,
.•. 2. APB = A AQB.
Q. E. D.
Note. Since all the angles in a segment are equal, we may
in future speak of the angle in a segment when we mean the
magnitude of any angle in the segment.
HEx. 1321. Are / ' PAQ, PBQ in fig. 248 equal? Give a reason.
Ex. 1323. Find the angle in a segment of a circle, the chord of the
segment being 6 cm. and the height 2 cm.
Ex. 1828. Eepeat Ex. 1322 with chord=4 ins. and height=2 ins.
Ex. 1824. Repeat Ex. 1322 with chord=6-43 cm., height = 8'61 cm.
ANGLE PROPERTIES
255
Theorem 10.
The angle in a major segment is acute; the angle in
a semi-circle is a right angle ; and the angle in a minor
segment is obtuse.
fig. 249.
fig. 250.
fig. 251.
Case i.
Data
To prove that
Froof
Case ii.
Data
To prove that
Pro(^
Case hi.
Data
To prove that
Proof
APB is a major segment.
/. APB is acute.
Since APB is a major segment,
.'. arc ACB is a minor arc,
.*. LX<2Tt. LS.
But L APB = I z. ax
.*. L APB < 1 rt. z. .
APB is a semi-circle.
L APB is a rt. /. .
Since APB is. a serai-cii'cle, so also is ACB,
.*. ^ a; = 2 rt. l s,
,'. z.APB = lrt. ^.
APB is a minor segment.
z. APB is obtuse.
Since APB is a minor segment,
^.■. arc ACB is a major arc,
.'. Lx>2 rt. L s,
.'. Z- APB> 1 rt. z..
UL 9.
Q. E. D.
2oG BOOK III
Ex. 1326. A regular hexagon iy inscribed in a circle. What is the
angle in each of the segments of the circle which lie outside the hexagon ?
Ex. 1326. Repeat Ex. 1325 for the case of (i) a square, (ii) an equi-
lateral A , (iii) a regular )i-gon.
tEx. 1827. A, B, C, D are points on a oircle ; the diagonals of ABCD
meet at X; prove that a' ABX, DCX are equiangular; as also A' BCX,
ADX.
tEx. 1820. Through X, a point outside a circle, XAB, XCD are drawn
to cut the circle in A, B; C, D. Prove that a' XAD, XCB are equiangular.
Ex. 1320. Copy fig. 952 (on an enlarged scale) ;
join BC. Find all the angles of the quadrilateral ABCD ;
and prove that two of its sides are equal.
tEx. 1830. Prove the following construction for
erecting a perpendicular to a given straight line AB at
its extremity B. With centres A, B describe arcs of
equal circles, cutting at C. With centre C and radius
CA describe a circle. Produce AC to meet this circle
again in D; then BD is X to AB.
tEx. 1331. The circle described on a side of an isosceles triangle as
diameter bisects the base.
tEx. 1332. The circles drawn on two sides of a triangle as diameters
intersect on the base.
tEx. 1888. The four circles drawn with the sides of a rhombus for
diameters have one point in common.
tEx. 1334. Two circles intersect at P, Q. Through P diameters PA, PB
of the two circles are drawn. Show that AQ, QB are in the same straight
line. (Join QP.)
tEx. 1335. AD is X to the base BC of A ABC; AE is a diameter of the
cucumscribing circle. Prove that A'ABD, AEC are equiangular; as also
A'ACD, AEB.
tEx. 1886. The bisector of A, the vertical angle of A ABC, meets the
base in D and the circumscribing circle in E. Prove that A'ABD, AEC are
equiangular. Also prove that A'ACD, AEB are equiangular.
HEx. 1837. Draw four straight lines roughly in the shape of ACBD
(tig. 253), making LC=:LD= 30°. Draw a circle round ACB ; notice whether
it passes through D.
CONCYCLIC POINI'S
257
Theorem 11.
[CONVEKSE OF THEOREM 9.]
If the line joining two points subtends equal angles
at two other points on the sajne side of it, the four
points lie on a circle.
fig. 253. ^8. 251.
Data The line joining AB subtends equal z. s at the points C, D,
which lie on the same side of AB.
To prove tliat the four points A, B, C, D lie on a 0 .
Construction Draw 0 to pass through A, B and C.
It must be shown that this 0 passes through D.
Proof If 0ABC does not pass through D, it must cut AD (or
AD produced) in some other point D'.
Join BD'.
Then z. AD'B = L ACB (in same segment). iii. 9.
But L ADB = L ACB, Data
:. z.ad'b = ^adb.
But this is imjx)ssible, for one of the /. s is an exterior
Z. of ADD'B, and the other is an interior opposite L of
the same A .
Hence 0ABC must pass through D,
i.e. A, B, C, D lie on a © . q. e. d.
Dep. Points which lie on the same circle are said to be
concyclic.
fEx. 1338. BE, CF are altitudes of the triangle ABC ; prove that
B, F, E, C are concyclic. Sketch in the circle.
258
BOOK III
HEx. 1330. Draw a circle (radius about 3 in.) ; take four points
A, B, C, D upon it. By measurement, find the sum of the angles BAD,
BCD; also of the angles ABC, ADC.
Theorem 12.
The opposite angles of any quadrilateral inscribed
in a circle are supplementary.
fig. 256.
Data PQRS is a quadrilateral inscribed in 0 PQR*.
To prove thai (1) l PQR + l PSR = 2 rt. z. s,
(2) Z.SPQ+ ^SRQ = 2rt. Z.S.
Construction Join P and R to the centre of 0.
Proof ^ a = i Lx, iii. 8.
Lh = lLy, ' III. 8.
:. La+ Lh=\{Lx+ Ly).
But Lx-{- ^ y = 4 rt. z. 8,
.'. z. a + z. 6 = 2 rt. ^ s,
i.e. L PQR + L PSR = 2 rt. z. s.
Sim'y it may be shown that L SPQ + l SRQ = 2 rt. z. s.
Q. E. D.
• The two figures represent the two cases in which the centre is (i) inside,
(ii) outside the quadrilateral. The same proof applies to both.
CYCLIC QUADRILATERALS
Ex. 1340. From the given angles, find all the
angles in fig. 257.
Ex. 1341. Repeat Ex. 1340, taking ^ B = 71°,
Z.BCO = 36°, Z.AOD = 108°. Prove that in this case
AD is II to BC.
tEx. 1342. The side PQ of a quadrilateral
PQRS, Inscribed In a circle, Is produced to T.
Prove tbat tbe exterior ;LRQT=tlie Interior opposite/. PS R.
tEx. 1343. If a parallelogram can be inscribed in a circle, it must
be a rectangle.
tEx. 1344. If a trapezimu can be inscribed in a circle, it must be
isosceles.
tEx. 1345. The sides BA, CD of a quadrilateral ABCD, inscribed in a
circle, are produced to meet at O; prove that A' OAD, OCB are equi-
angular.
tEx. 1346. ABCD is a quadrilateral inscribed in a circle, having
Z.A=60° ; O is the centre of the circle. Prove that
Z.OBD + Z.ODB=/.CBD + Z.CDB.
ITEx. 1347. What is the relation between the angles subtended by
a chord at a point in its minor arc, and at a point in its major arc?
TEx. 1348. Draw a quadrilateral ABCD, having zA+zC = 180°.
Draw a circle to pass through ABC ; notice whether it passes through D.
260
HOOK III
Theorem 13.
[Converse op Theorem 12.]
If' a pair of opposite angles of a quadrilateral are
supplementary, its vertices are concyclic.
Xf D
fig. 258.
fig. 259.
Data The L s ABC, ADC of the quadrilateral A BCD are supple-
mentary.
To prove that A, B, C, D are concyclic.
Construction Draw 0 to pass through A, B, C.
It must be shown that this © passes through D.
Proof If 0ABC does not pass through D, it must cut AD (or AD
produced) in some other point D'.
Join CD'.
Then l AD'C + l ABC = 2 rt. ^ s. iii. 12.
But L ADC + L. ABC = 2 rt. z. s, Data
:. L AD'C + L ABC = u ADC + L ABC,
.•. ^AD'C= ^ADC.
]3ut this is impossible, for one of the z. s is an exterior l
of A DD'C, and the other is an interior opposite l of the
same A.
Hence 0 ABC must pass through D,
i.e. A, B, C, D are concyclic.
Q. E. D.
CYCLIC QUADRILATERALS 261
Def. If a quadrilateral is such that a circle can be circum-
scribed round it, the quadrilateral is said to be cyclic.
tEx. 1349. BE, CF, two altitudes of a ABC, intersect at H. Prove
that AEHF is a cyclic quadrilateral. Sketch iu the circle.
Ex. 13SO. ABC, DBC are two congruent triangles on opposite sides of
the base BC. Under what circumstances are A, B, C, D concyclic?
tEx. 1351. ABCD is a parallelogram. A circle drawn through A, B,
cuts AD, BC (produced if necessary) in E, F respectively. Prove that
E, F, C, D are concyclic.
tEx. 1352. ABCD is a quadrilateral inscribed in a circle. DA, CB are
produced to meet at E ; AB, DC to meet at F. Prove that, if a circle can be
drawn through AEFC, then EF is the diameter of this circle; and BD is the
diameter of ©ABCD.
tEx. 1353. The straight lines bisecting the angles of any convex
quadrilateral form a cyclic quadrilateral.
For further exercises on the subject-matter of the ahotie section
see end of section ix.
Section VII Construction op Tangents.
lIEx. 13S4. Stick two pins into the paper 2 in. apart at A and B; place
the set-square on the paper so that the sides
containing the 60° are in contact with the pins ;
mark the point where the vertex of the angle
rests. Now slide the set-square about, keeping
the same two sides against the pins, and plot
the locus of the 60° vertex. What is the locus?
are A, B points in the locus ? Complete the circle,
and measure the angle subtended by A B at a fig. 260.
point in the minor arc.
ITEx. 1365. Repeat the experiment of Ex. 1364 with the 30° vertex.
HEx. 1356. Repeat the experiment of Ex. 1354 with the 90° vertex.
HEx. 1357. What is the locus of points at which a given line subtends
a right angle ?
Ex. 1358. O is the centre of a circle and Q is a point outside the
circle. Construct the locus of points at which OQ subtends a right angle.
Find two points A, B on the first circle, so that /. OAQ=Z. OBQ=90°.
Prove that QA is a tangent to the first circle.
262 BOOK III
To draw tangents to a given circle abc from a given
point T outside the circle.
Construction Join T to O, the centre of 0 ABC.
On OT as diameter describe a 0 cutting the given circle
in A, B.
Join TA, TB.
These lines are tangents.
Proof Join OA, OB.
Since OT is the diameter of 0 OAT,
.*. L OAT is a right angle,
.*. AT, being ± to radius OA, is the tangent at A.
Similarly BT is the tangent at B.
Ex. 1369. Draw tangents to a circle of radios 2 ins. from a point 1 iii«
outside the circle ; calculate and measure the length of the tangents.
Ex. 1360. Draw a circle of radius 3 cm. and mark a point T distant 7 cm.
from the centre. Find where the tangents from T meet the circle (i) by the
method of p. 240, (ii) as above. Calculate the length of the tangents, and
ascertain which method gives greater accuracy.
Ex. 1361. Find the angle between the tangents to a circle from a point
whose distance from the centre is equal to a diameter.
Ex. 1363. Through a point 2 in. outside a circle of radius 2 in. draw a
line to pass at a distance of 1 in. from the centre. Measure and calculate
the part inside the circle.
CONSTRUCTION OF TANGENTS
263
Common Tangents to Two Circlks.
Def. a straight line which touches two circles is called
a common tangent to the two circles.
Pig. 262 shows that when the circles do not
intersect there are four comraon tangents.
If the two circles lie on the same
side of a common tangent, it is called
an exterior common tangent; thus
AB, CD (fig. 262) are exterior common
tangents. If the two circles lie on
opposite sides of a common tangent,
it is c«tUed an interior common tangent;
thus EF, GH are interior common tangents.
fig. 262.
Ex. 1363. Draw sketches to show how many common tangents can be
drawn in cases n., ni., iv., v., of fig. 230 ; in each case state the number of
exterior and of interior common tangents.
HEx. 1364. Draw the tangents to a circle
(centre A ; radius 1 in.) from a point B
(AB=3 in. fig. 263). Draw, parallel to each
tangent, a line ^ in. from the tangent, these
lines not to cut the circle. "With centres A and
B draw circles touching these two lines. Show
that the difference of the radii of these oircles
is equal to the radius of the original circle.
fig. 263.
2G4 BOOK III
To construct an exterior common tangent to two
unequal circles.
fig. 264.
[Analysis Let A, B be the centres of the larger and smaller
circles respectively ; R, r their radii.
Suppose that ST is an exterior common tangent, touching
the 0" at S, T.
Join AS, BT. Then i. ' AST, BTS are right angles,
.'. AS is II to BT.
Tlirough B draw BP || to TS, meeting AS in P.
Tlien BTSP is a rectangle. (Why?)
.•. PS = BT = r.
And AP = AS - PS = R - r.
Also Z.APB is a right angle. (Why?)
.'. DP is a tangent from B to a circle round A, whose
radius is R — r.
Tlie foregoing analysis suggests the following construction.]
Construction With centre A describe a circle having for radius
the difference of the given radii.
From B draw a tangent BP to this circle.
Join AP and produce it to meet the larger 0 in S.
Through B draw BT || to AS to meet the smaller circle in T.
Join ST.
Then this line is a common tangent to the two ©■.
CONSTRUCTION OF COMMON TANGENTS 265
Proof PS is equal and || to BT (why?),
.*. STBP is a parallelogram,
and /. SPB is a right angle (why?),
.'. STBP is a rectangle,
.'. angles at S and T are right angles,
.'. ST is a tangent to each circle.
Ex. 1365. Draw two circles of radii 15 in. and 0*5 in., tho centres
2-5 ins. apart. Draw the two exterior conuuon tangents.
Measure and calculate the length of these tangents (i.e. the distance
between the points of contact). [Use right-angled A APB.]
tEx. 1366. . Where does the above method fail when the two circles are
equal ? Give a construction (with proof) for the exterior common tangents
in this case.
6. s. n. 18
266 BOOK in
To construct an interior common tangent to two
circles.
fig. 265.
[Analysis Suppose that XY is an interior common tangent,
touching the 0' at X, Y.
Join AX, BY. Then j. » AXY, BYX are right angles.
.'. AX is II to BY.
Through B draw BQ || YX, meeting AX produced in Q.
Then BYXQ is a rectangle.
.•. QX = BY = r.
And AQ = AX + XQ = R + r.
Also /. AQB is a right angle,
.'. BQ is a tangent from B to a circle round A, whose
radius is R + r.
Hence the following construction.]
Construction With centre A describe a circle having for radius
the sum of the given radii.
From B draw a tangent BQ to this circle.
Join AQ; let this line cut the (A) circle in X.
Through B draw BY || QA to meet the (B) circle in Y.
Join XY.
Then this line is a common tangent to the two ©■.
CONSTRUCTION OF COMMON TANGENTS 267
Proof (i) Prove that BYXQ is a rectangle.
(ii) Prove that XY is a tangent, to the (A) circle at X,
and to the (B) circle at Y.
Ex. 1367. Draw the two circles of Ex. 1365, and draw the interior
common tangents. Measure and caJcnlate the length of these tangents.
HEx. 1368. Draw two equal circles, not intersecting. Draw the interior
conmion tangents by the above method. Can you suggest an easier method
for this special case ?
Ex. 1369. In the following exercises R, r denote the radii of the circles,
d the distance between their centres. For each pair of circles calculate the
lengths of possible common tangents. {Freehand, )
(i) R = 5cm., r=3cni,, d=8cm.
(ii) R=5cm., r=3cm., d=7cm.
(iii) R=3in., r=lin., d=2in.
(iv) R=3in., r=lin., d=lin.
(v) R = 3-52 cm., r= 1-41 cm., d= 6-29 cm.
HEx. 1370. If the radius of the smaller circle diminishes till the circle
becomes a point, what becomes of the four common tangents?
Ex. 1371. The diameters of the wheels of an old-fashioned bicycle are
4 ft. and 1 ft., and the distance between the points where the wheels touch
the ground is 2^ ft. Calculate the distance between the centres of the
wheels ; check by drawing.
18—2
268 BOOK in
Section VIII. Constructions depending on
Angle Properties.
lIEx. 1372. Draw a line of 2 ins.; on this line as base draw a
triangle with a vertical angle of 40°.
(You will find that it is practically impossible to draw the vertical angle
directly : first draw the angles at the ends of the base. What is their sum ?
Notice that many different triangles may be drawn with the given vertical
angle.)
^Ex. 1373. Draw a line of 2 ins.'; on this line as base, and on the same
side of it, draw a number of triangles (about 10) having a vertical angle of
40°. What is the locus of their vertices ? Complete the curve of which this
locus is a part. Is it possible for the vertex to coincide with an end of the
base, in an extreme case? Does the curve pass through the ends of the base?
HEx. 1374. Repeat Ex. 1373 with base 2 in. and vertical angle 140°.
Compare this with the locus obtained in Ex. 1373.
HEx. 137S. What locus would be obtained if Ex. 1878 were repeated with
an angle of 90° ?
HEx. 1876. (Tracing paper.) Draw, on tracing paper, two straight
lines intersecting at P. On your drawing paper mark two points A, B. Move
your tracing paper about so that the one line may always pass through A,
and the other through B. Plot the locus of P by pricking through.
The foregoing exercises will have prepared the reader for the
following statement : —
The locus of points (on one side of a given straight
line) at which the line subtends a constant angle is
an arc of a circle, the given line being the chord of
the arc.
liEx. 1377. Upon what theorem does the truth of this statement
depend ?
HEx. 1378. What kind of arc is obtained if the angle is (i) acute, (ii) a
right angle, (iii) obtuse?
HEx. 1379. If the constant angle is 45°, what angle is subtended by
the given line at the centre of the circle ? Use this suggestion in order to
draw the locus of points at which a line of 5 cm. subtends 45°, without
actually determining any of the soints.
CONSTRUCTIONS 269
Ex. laso. Show how to construct the locus of points at which a given
line subtends an angle of 30°. Prove that in this case the radius of the
circle is equal to the given line.
Ex. 1381. Show how to construct the locus of points at which a given
line subtends a given angle.
tEx. 1882. On a chord of 3-5 ins. construct a segment of a circle to
contain an angle of 70°. Measure the radius.
Ex. 1383. Repeat Ex. 1382 with chord of 7-24 cm. and angle of 110°.
Ex. 1384. Repeat Ex. 1382 with chord of Sin. and angle of 120°. -
tEx. 1385. Prove that the locus of the mid-points of chords of a circle
which are drawn through a fixed point is a circle.
tEx. 1386. Qf aU. triangles of given base and vertical angle, the
isosceles triangle has greatest area.
tEx. 1387. P is a variable point on an arc AB. AP is produced to Q
so that PQ= PB. Prove that the locus of Q is a circular arc.
To construct a triangle with given base, given alti-
tude, and given vertical angle.
Let the base be 7 cm.; the altitude 6 '5 cm.; the vertical
angle 46°.
Draw the given base.
Draw the locus of points at which the given base sub-
tends 46°.
Draw the locus of points distant 6*5 cm. from the given base
(produced if necessary).
The intersections of these loci will be the required positions of
the vertex.
How many solutions are there to this problem 1
Measure the base angles of the triangle.
Ex. 1388. Construct a triangle having
(i) base =4 in., altitude =1 in., vertical angle = 90°.
(ii) base = 10 cm., altitude = 2 cm., vertical angle = 120°.
(iii) base =8 cm., altitude =5 cm., vertical angle =90°.
(iv) base=3-5in., altitude = 1 in., vertical angle =54°.
In each case measure the base angles.
270 - BOOK III
Ex. 1889. (Without protmclor.) Constract a triangle, given the base,
yertical angle and altitude.
Ex. 1390. Show how to construct a triangle of given base, vertical
angle and median.
Ex. 1391. Show how to oonstmot a triangle, given the base, the vertical
angle and the area.
Ex. 1893. Show how to construct quadrilateral ABCD, given
AB==5-4cm., AC = 9'6cm., AD=5-6om., Z.BAD = 113°, Z.BCD = 70°.
Ex. 1393. Show how to construct a cyclic quadrilateral ABCD, given
AB = l-6in., BC = 30in., CD = 4-9in., Z.B = 125°.
Why are only four measurements given for the construction of this
quadrilateral ?
Ex. 1394. Show how to construct a quadrilateral ABCD, given that
AB = 61 cm.. BC = ll-4 cm., CA=ll-7 cm., AD = ol cm., iLBDC = 76°.
Ex. 1396. Show how to construct a parallelogram with base 2*8 in.
and height 2 in., the angle (subtended by the base) between the diagonals
being 80°. (Try to find the centre of the parallelogram.)
To inscribe in a given circle a triangle with given
angles.
Let the radius of the circle be 2 in. and the angles of the
required triangle 40", 60" and 80°.
[Analysis Draw a sketch of the required figure; join the
vertices of the triangle to the centre of the circle. What are the
angles subtended at the centre by the three sides 1
Knowing these three angles at the centre, it is easy to draw
the required figure.]
Ex. 1396. Draw the figure described above; measure the sides of the
triangle. State the construction formally, and give a proof.
Ex. 1397. Inscribe in a circle of radius 5 cm. a triangle Of angles
30°, 80°, 70°. Measure the sides.
Ex. 1398. Inscribe in a circle of radius 3-5 in. a triangle with angles
60°, 40°. Measure the sides.
CONSTRUCTIONS
271
Ex. 139©. Inscribe in a circle of radius 4 em. an isosceles triangle
having each of the angles at the base double the angle at the vertex.
Measure the basa
Ex. 1400. Inscribe in a circle of radius 2*5 in. a triangle having two
of its angles 35° and 40°. Measure the sides.
Ex. 1401. (Without protractor.) Inscribe in a
circle of radius 6 cm. a triangle equiangular with a
given triangle.
Ex. 1402. Copy fig. 266 on an enlarged scale;
making the radius of the circle 2 in. Check the
angles marked, and measure AC.
fig. 266.
To circumscribe about a given circle a triangle with
given angles.
Let the radius of the given circle be 2-4 in.: the angles of
the required triangle 45°, 70°, 65°.
[Analysis Draw a sketch of the required figure (fig. 267).
Join the centre O to L, M, N the points
of contact of the sides.
If the angles at O can be calculated it
will be easy to draw the figure.
Now Z-sAMO, ANO are right angles,
.*. /. s MAN, MON are supplementary.
Hence calculate z. MON, and similarly
the other angles at C]
Ex. 1403. Draw the figure described above. Measure the longest side
of the triangle. State the construction formally, and give a proof.
Ex. 1404. Circumscribe about a circle of radius 5 cm. a triangle of
angles 30°, 60°, 90°. Measure the longest side.
Ex. 1405. Circumscribe about a circle of radius 3 cm. an isosceles
right-angled triangle. Measure the longest side.
Ex. 1406. Circumscribe about a circle of radius 4 cm. a parallelogram
having an angle of 70°. Measure the sides of the parallelogram, and prove
that it is a rhombus.
Ex. 1407. (Without protractor.) Circumscribe about a circle of radius
2*6 in. a triangle equiangular to a given triangle.
Ex. 140S. (Without protractor.) Circumscribe about a circle of radius
5 cm. a triangle having its sides parcel to three given straight lines.
fig. 2G7.
27-2 BOOK III
Skction IX. "Alternate Segment."
Theorem 14.
If a straight line touch a circle, and from the point
of contact a chord be drawn, the angles which this chord
makes with the tangent are equal to the angles in the
alternate segments.
Data AB touches 0CDE in C; the chord CD is drawn, through
C, meeting © again in D.
To prove that (1) z.BCD=:Z.in alternate segment CED,
(2) L ACD = z. in alternate segment CFD (fig. 269).
(1) Construction Through C draw CE ± to AB, meeting ©in E.
Join CE, DE.
Proof Since CE is drawn ± to tangent AB, at its point of
contact C,
.". CE passes through centre of ©, and is a diameter,
III. 6, Cor.
.'. L CDE is a rt. /. , m. 10.
/. inACDE, ^CED+ z.DCE= Irt. Z-. 1.8.
Now L BCD + L DCE = 1 rt. z. . Constr.
.'. L. BCD + L DCE = L CED + L DCE,
.-. ^BCD = aCED.
•ALTERNATE SEGMENT
273
(2) Construction Take any point F in arc CFD; join CF, DF.
L. BCD + L ACD = 2 rt. z. s.
Also, since CFDE is a quadrilateral inscribed in a circle,
/. CED + L CFD = 2 rt. z- s, ill. 12.
/. I. BCD + L ACD = z. CED + L CFD.
But L BCD = L CED, . Proved
:. Z.ACD= ^CFD.
Q. E. D.
ITEx. 1409. In fig. 269 point out an angle equal to Z. BCF.
ITEx. 1410. Taking CE as the chord (fig, 269), what is the segment
alternate to z. ACE?
liEx. 1411. Find all the angles of fig. 269, supposing that Z. BCD = 60°,
and that Z. FCD==20°. What angles do the chords ED, CD, FC subtend at
the centre ?
Ex. 1412; Find all the angles of fig. 270.
fig. 271.
Ex. 1413. Find all the angles of fig. 271.
ITEx. 1414. Draw the tangent at a given point on a circle without finding
(or using) the centre of the circle.
(For further exercises on "AUemate segment" see Ex. 1434 — 1438.)
274 BOOK m
III. 14 provides alternative methods of dealing with the
constructions of section yiii.
On a given straight line AB to construct a segment of
a circle to contain a given angle x.
Construction At A make L BAG = ^^ X.
Construct a circle to pass through A and B, and to
touch AC at A.
The segment ADB is the segment required.
Proof z. X = z. CAB (between tangent AC and chord AB)
— Lva. alternate segment ADB.
Ex. 1416. Show how to construct on a given straight line a segment
of a circle to contain a given obtase angle. {Freehand.)
Ex. 1416. Show how to construct on a given base an isosceles triangle
with a given vertical angle. (Freehand.)
Ex. 1417. Show how to construct on a given base a triangle of given
vertical angle and given median. Is this always possible ? (Freehand.)
"ALTERNATE SEGMENT" 276
In a given circle to inscribe a triangle equiangular to
a given triangle XYZ.
^Analysis Suppose that the problem has been solved; and that
ABC is the required A.
Let PAQ be the tangent at A.
Then A PAB = /. ACB (in alternate segment)
= AZ,
and L QAC = z. ABC (in alternate segment)
= /.Y.]
Hence : — •
Construction At any point A on the given circle draw the
tangent PAGL
Make i. PAB = LZ; let AB cut 0 in B.
Make z.QAC= /.Y; let AC cut 0 in C.
Join BC.
Then ABC is a triangle equiangular to A XYZ, inscribed
in the given circle.
tEx. 1418. Give the proof of the above construction.
Ex. 1419. (Without protractor.) In a circle of radius 3 in. inscribe
a triangle equiangular to a given obtuse-angled triangle. Test the accuracy
of the angles.
Ex. 1420. In a circle of radius 2 in. inscribe a triangle having its
sides parallel to three given straight lines.
276
BOOK III
Tangent as limit op Chord.
Pigs. 274 (i — iv) show four positions of a chord AB (produced
both ways). Looking at the figures from left to right, the chord
is seen to be turning about the point A; as it turns, the second
point of intersection, B, comes nearer and nearer to A until in fig. v,
B has coincided with A, and the chord has become the tangent
at A.
A tangent therefore may be regarded as the limit of a
chord whose two points of intersection with the circle have come
to coincide.
Fig. 275 suggests another way in which the chord may ap-
proach its limiting position — the tangent.
Looking at the tangent from this point of view, it is
interesting to see that the angle in a segment of a circle de-
TANGENT AS LIMIT OF CHORD
277
velops into the angle between the chord and the tangent at its
extremity. This is shown by fig. 276.
fig. 276.
HEx. 1431. In fig. 275, what becomes of the theorem that "the
perpendicular from the centre on a chord bisects the chord " when B comes
to coincide with A ?
t Ex. 1433. Prove m. 6 by means of fig. 275.
HEx. 1438. In fig. 275, if O is the centre of the circle, what do the
angles OAB, OBA become in the limiting case?
^Ex. 1434. What is the limiting form of Ex. 1300 when the circles
touch?
Miscellaneous Exercises on Sections VI., VIII. and IX,
tEx. 1436. Through P, Q, the points of intersection of two circles, are
drawn chords APB, CQD ; prove that AC is || to BD. [Join PQ.]
What does this theorem become if A, C are made to coincide ?
tEx. 1436. Through P, Q, the points of intersection of two circles, are
drawn parallel chords APB, CQD ; prove that AB = CD.
tEx. 1437. If two opposite sides of a cyclic quadrilateral are equal, the
other two are parallel.
tEx. 1438. Each of two equal circles passes through the centre of the
other: AB is their common chord. Through A is drawn a line cutting the
two circles again in C, D ; prove that A BCD is equilateral.
tEx. 1430. ABC is an equilateral triangle inscribed
in a circle ; P is any point on the minor arc BC. Prove
that PA=PB+PC. [Make PX = PB. Then prove XA=
PC]
tEx. 1430. In fig. 229, B, C, Ij, and the centre of the
inscribed circle are concyclic.
278 BOOK III
fEz. 1481. From a point on the diagonal of a square, lines PR, QS ar«
drawn parallel to the sides, P, Q, R, S being on the sides. Prove that these
four points are concyclic.
fEx. 1432. O is the centre of a circle, CD a diameter, and AB a chord
perpendicular to CD. If B is joined to any point E in CD, and BE pro-
duced to meet the circle again in F, then A, O, E, F are concyclic.
Ex. 1433. Show how to construct a right-angled triangle, given the
radius of the inscribed circle, and an acute angle of the triangle.
tEx. 1434. Two circles touch at A. Through A are drawn straight
lines PAQ, RAS ; cutting the circles in P, Q and R, S. Prove that PR is
parallel to QS. (Draw tangent at A. Compare Ex. 1425.)
tEx. 1436. Two circles cut at P, Q. A, a point on the one circle, is joined
to P, Q; and these lines are produced to meet the other circle in B, C.
Prove that BC is parallel to the tangent at A. (Compare Ex. 1425.)
tEx. 1436. A chord AB of a circle bisects the angle between the
diameter through A, and the perpendicular from A upon the tangent at B.
tEx. 1437. ABCD is a cyclic quadrilateral, whose diagonals intersect at
E : a circle is drawn through A, B and E. Prove that the tangent to this
circle at E is parallel to CD.
tEx. 1438. AB, AC are two chords of a circle} BD is drawn parallel to
tb'j tangent at A, to meet AC in D; prove that Z.ABD is equal or supple-
mentary to L BCD. Hence show that the circle through B, C and D
touches AB at B.
Section X. Arcs and Angles at the Circumference.
HEx. 1439. Draw a circle of radius 2-5 in.; draw a diameter OPg and
a tangent AOB as in fig. 278. Divide L AOPg into five
equal parts; also ^ BOPj. Measure the chords OPj,
P^Pj, ... etc. What angle does PjPg subtend at the centre
of the circle? Prove that OPiPj...etc. are the vertices
of a regular decagon.
ITEx. 1440. In the fig. of Ex. 1439 draw any straight
line cutting across the set of lines OPj, OPj, OP3, etc.
Is this line divided into equal parts ? °8* ^^*
^%i.. 1441. Take a point O, lin. from the centre of a circle of radius
2*5 in.; draw through O a diameter and a set of chords making angles of
18° with one another. Find by measurement whether these chords divide
the circumference into equal arcs.
Angles at the circumference 279
HEx. 1442. Woiild the circamferenoe be divided into equal arcs if the
point O in Ex. 1441 were taken at the centre? How many arcs would there
be?
tEx. 1443. Prove that equal ares or chords of a circle subtend
equal (or supplementary) angles at a point on the circumference.
Draw a figure to illustrate the case of supplementary angles.
Prove the converse.
Note. In thefoUxnmng exercises {Ex. 1445 — 1462) the stucl«rU
is advised to make use of '^the angle subtended at the circumference.'^
tEx. 1444. Draw a regular pentagon ABODE in a circle. Prove that
the angle A is trisected by AC, AD.
tEx. 1445. ABODE is a regular pentagon.
(i) Prove that AB is parallel to EO. (Join AO.)
(ii) At what angle do BD, OE, intersect?
(iii) Prove that A AOD is isosceles, and that each of its base angles
is double its vertical angle.
(iv) If BD, OE meet at X, prove that A'OXD, ODE are equiangular.
(v) Prove that the tangent to the circle at A is parallel to BE.
[Use m. 14.]
tEx. 1446. AB, OD are parallel chords of a circle. Prove that
aro AO^arc BD.
tEx. 1447. On a circle are marked off equal arcs AO, BD. Prove that
AD is parallel or equal to OB.
tEx. 1448. AOB, OOD are two chords of a circle, intersecting at right
angles. Show that arc AO + arc B D = arc O B + arc DA.
tEx. 1449. Through a given point draw a chord of
a given circle so that the minor segment cut off may be
the least possible.
/V
tEx. 1450. Prove that in fig. 278
arcOPi=arcPiP2.
Ex. 1451. In fig. 279 what fractions of the circum-
ference are the arcs AB, BO, OD, DA, BOD?
280
BOOK 111
Ex. 1463. In %. 280 what fractions of the oirouiu-
ference are the arcs AB, BC, CD, DA?
Ex, 1453. ABC and ADEFG are respectively an
equilateral triangle and a regolar pentagon inscribed in a
circle. What fraction of the circumference is the arc BD ?
Ex. 1464. PQRS is a quadrilateral inscribed in a
circle ; the two diagonals intersect at A. PQ is an arc
of 30° (see p. 233), QR 100°, RS 70°. Find all the angles
in the figure.
Ex. 1455. In the figure of Ex. 1454 find two pairs of equiangular
triangles.
Ex. 1466. If in fig. 268 arc CD = 2 arc DE, what is /.BCD?
tEx. 1457. In fig. 268, the bisector of ^BCD bisects arc CD.
Ex. 1468. If in fig. 268 arc ED were | arc DC, what would be the
magnitude of z. BCD ?
Ex. 1469. The two tangents OA, OB from a point O are inclined at
an angle of 48°. How many degrees are there in the minor and major arc
AB respectively ? What is the ratio of the major to the minor arc ?
tEx. 1460. P is a variable point on an arc AB. Prove that the
bisector of l APB always passes through a fixed point.
[Begin by finding the probable position of the fixed point by experi-
ment.]
tEx. 1461. A, B, C are three points on a circle. The bisector of
/ ABC meets the circle again at D. DE is drawn || to AB and meets the
circle again at E. Prove that DE= BC.
tEx. 1462. A tangent is drawn at one end of an arc ; and from the mid-
point of the arc perpendiculars are drawn to the tangent, and the chord of
the arc. Prove that they are equal.
Regular Polygons*.
Def. a polygon which is both equilateral and equiangiilar is
said to be regular.
ITEx. 1463. What is the name for a quadrilateral that is (i) equi-
lateral and not equiangular, (ii) equiangular and not equilateral, (iii) regular?
HEx. 1464. Draw a hexagon that is equiangular but not equilateral.
ITEx. 1465. Is there any polygon which is necessarily regular if it is
either (i) equilateral, or (ii) equiangular ?
* The section on regular polygons should be omitted at a first reading.
REGULAR POLYGONS 281
Theorem 15. t
If the circumference of a circle be divided into n equal
arcs, (1) the points of division are the vertices of a regular
jvgon inscribed in the circle; (2) if tangents be drawn
to the circle at these points, these tangents are the sides
of a regular w-gon circumscribed about the circle.
fig. 281.
(1) Ikiia The circumference is divided into n equal arcs at
the points A, B, C, D, E, F, G.
The chords AB, BC, etc. are drawn forming the inscribed
n^gon ABCDEFG.
To prove that ABCDEFG is regular.
Proof Since arcs AB, BC, eta are equal,
.*. chords AB, BC, etc. are equal, lii. 4.
.*. ABCDEFG is equilateral.
Again, arc GA = arc BC, Data
.'. adding arc AB to both,
arc GAB = arc ABC,
.". L GAB = L ABC, these angles being contained in equal arcs.
Sim^ it may be shown that aU the z. s of the polygon are
equal ; i.e. that ABCDEFG is equiangular.
.'. ABCDEFG, being equilateral and equiangular, is i-egular.
c. s. II, 19
282
BOOK 111
(2) Data The tangents at A, B, C, etc. ace drawn forming
the circumscribed 7*-gon PQRSTUV.
To prove thai PQRSTUV is regular.
Construction 'Join P, Q, R, etc. to the centre O.
4 Proof Show that
(i) adjacent As, similarly numbered, are congnient.
(ii) adjacent As, differently numbered, are congruent
(by (i) L POB - J ^ Abe, l QOB ^^l COB,
.•. L POB = L QOB).
(iii) all the numbered As are congruent,
(iv) PQRSTUV is equilateral
(v) PQRSTUV is equiangular.
.". PQRSTUV is regular.
Q. Bl. D
tEx. 1466 (a). (Alt«rnatiT« proof of Tfa. 16 (a).)
Join ED, DC.
Prove that
(1) AS ESD, DRC are isoBoeles,
(2) zEDS=zCDR (by means of angles in alternate segments),
(3) AS ESD, DRC are congruent.
.•. etc.
REGULAR POLYGONS 283
Ex. 1466. Construct a regular pentagon of side 2 in. (see Ex. 398);
draw the circumscribed and inscribed circles and measure their radii.
Ex. 1467. Repeat Ex. 1466 with a regular octagon of side 2 in. (Without
jwotractor.)
Ex. 1468. Find the perimeter and area of a regular 6-gon circumscribed
about a circle of radius 5 cm.
fEx. 1469. Prove that an equilateral polygon inscribed in a circle must
'also be equiangular.
Ex. 1470. Is the converse of Ex. 1469 true?
fEx. 1471. Prove that an equiangular polygon ciicumscribed about a
circle must also be equilateral.
Ex. 1472. Is the converse of Ex. 1471 true?
Ex. 1473. The area of the square circumscribed about a circle is twice
the area of the square inscribed in the same circle.
Ex. 1474. Prove that the area of the regular hexagon inscribed in
a circle is twice the area of the inscribed equilateral triangle. Verify this
fact by cutting a regular hexagon out of paper, and folding it.
Ex. 1475. The side of an equilateral triangle circumscribed about a
circle is twice the side of an inscribed equilateral triangle.
tEx. 1476. The exterior angle of a regular n-gon is eqiial to the angle
which a side subtends at the centre.
fEx. 1477. The Lines joining a vei-tex of a regular H-gon to the other
vertices divide the angle into {n - 2) equal parts.
19—2
284
BOOK III
Section XI. Akea of Clrclk.
P
Let PQRST be a polygon (not necessarily regular) circum-
scribing a circle.
Join the vertices of the polygon to the centre of the circle.
The circle is thus divided into a number of triangles, having for
bases the sides of the polygon, and for vertex the centre of the
circle.
Draw perpendiculars from the centre to the sides of the
polygon. These meet the sides at their points of contact and are
radii of the circle. Thus the triangles OPQ, OQR, etc. are all of
height equal to the radius of the circle.
Let r be the radius of the circle, a, b, c, d, e the sides of the
polygon (PQ = a, QR = 6, etc.).
The area of A OPQ is ^ ar; A OQR = ^br, etc.
.'. area of polygon — ^a/r + ^br + ^cr + \dr + ^er
— ^r(a + b + c + d + e)
= ^ radius x perimeter of polygon.
This is true for any polygon circumscribing the circle.
AREA OF CIRCLE
285
If we draw a polygon of a very great number of sides, it
is difficult to distinguish it from the circle itself. The area of
the polygon approaches closer and closer to the area of the circle;
and the perimeter of the polygon to the circumference of the
circle. Hence we conclude that
area of a circle = ^ radius x circumference of circle
= ^ r X 2irr
[In the following exercises it vnll generally be sufficient if answers are
given correct to three significant figures."]
Ex. 1478. Calcnlate the area of a ourde whose radius is 1 inch. Also
draw the circle on inch paper and find the area by .counting the squares.
Ex. 1479. Eepeat Ex. 1478 for a circle of radius 2 in. Check your
result by squared paper.
Ex. 1480. The radius of one circle is twice the radius
of another; how many times does the area of the greater
contain the area of the smaller? Fig. 284 shows that the
area of the greater is more than double the area of the
smaller. Find the area of the shaded part of fig. 284, taking
the diameter of the small circles to be 1 cm.
Ex. 1481. Find the ratio of the area of a circle to the area of the
circumscribing square.
Ex. 1482. Squares are inscribed and circumscribed to . ^^y^^ i
a circle (fig. 285); how many times does the circumscribed // \\
square contain the inscribed square?
Ex. 1483. What is the ratio of the area of the circle to
the area of the iascribed sauare?
fig. 286.
Ex. 1484. Find the area of a circle, given (i) radius = 5*72 cm.,
(ii) diameter =1 in. (the size of a halfpenny), (iii) r= 0*59 in.
Ex. 1485. Find, in square inches, the area of one side of a penny.
Ex. 1486. Draw an equilateral triangle of side 10 cm. and its circum-
scribing circle ; make the necessary measurements and calculate the area of
the circle. Find the ratio of the area of the circle to that of the triangle.
286 BOOK III
Ex. 1487. Find the ratio of the area of a circle to the area of the in-
scribed regular hexagon. (Compare result with those of Ex. 1483 and 1486.)
Ex. 1488. In the centre of a circular pond of radius 100 yards is a
circular island of radius 20 yards. Find the area of the surface of the water.
Ex. 1489. Find whether the area in Ex. 1488 is greater or less than the
area of a circular sheet of water of 80 yards radius.
Ex. 14 OO. The radius of the inside edge of a circular running track is
a feet ; and the width of the track is 5 feet ; find the area of the track.
Ex. 1491. From a point P, on the larger of two concentric circles,
a tangent PT is drawn to the smaller. Show that area of the circular ring
between the circles is v . PT".
Ex. 1493. Show how to draw a circle equal to (i) the sum, (ii) the
difference of two given circles.
Ex. 1498. Calculate the radius of a circle whose area is 1 sq. in.
Ex. 1494. Calculate the diameter of a circular field whose area is
1 acre (=4840 sq. yards).
Ex. 1495. Let A = area of circle, c= circumference, r=radias, d=dia-
meter.
(i) Express c in terms of r,
(ii) c d,
(iii) A : r,
(iv) A d,
W r c,
(vi) d c,
(vii) r A,
(viii) A A,
(ix) A c,
(x) c A.
Ex. 1496. Find the radius and circumference of a circle whose area is
(i) 6 sq. in., (ii) 765 sq. cm.
Ex. 1497. Calculate the area of a circle whose circumference is
25,000 miles. (Fmd r first.)
AREA OF CIRCLE '287
Ex. 1408. Prove that in fig. 224 the three portions into which the (drole
is divided by the curved lines are of equal area.
tEx. 1499. Prove that if circles are described with the hypotenuse and
the two sides of a right-angled triangle for diameters, the area of the greatest
is the sum of the areas of the other two.
tEx. 1500. In fig. 286 L BAG is a right angle, and
the curves are semicircles. Prove that the two shaded
areas are together equal to the triangle.
Area of sector of circle.
If through the centre of a circle were drawn 360 radii making
equal angles with one another, 360 angles of 1 degree would be
formed at the centre of the circle. The area of the circle would
be divided into 360 equal sectors. A sector of angle 1° has
therefore -^^ of the area of the circle; and a sector of angle,
say, 53° contains ^^ of the area of the circle.
Ex. 1501. Find the area of a sector of 40° in a circle of radius 5 in.
Ex. 150a. Find the area of a sector of 87° in a circle of radios 12*4 cm.
Ex. 1603. Find the areas of the two sectors into which a circle of
diameter 12-5 inches is divided by two radii inclined at an angle of 60°.
Ex. 1604. Calculate the area of a sector whose chord is 3 in. in a
circle of radius 4 in. (find the angle by measurement).
Ex. 1&05. Prove that tbe area of a sector of a circle la half tba
Vroduct of the ra^Utis and the arc of the sector.
Area of segment of circle.
In fig. 287,
segment AG B = sector PAG B — triangle PAB.
Ex. 1506. Find the areas of the two segments
into which a circle radius 10 cm. is divided by a chord
of 10 cm.
Ex. 1507. Repeat Ex. 1506 with the same circle and a chord of
20 cm.
288
BOOK III
Ex. 1608. Bepeat Ex. 1506 with a ohord that subtends 90° at the
centre.
Ex. 1609. Find the area of a segment AvhoBe chord is 12 cm. and
height 3 cm. Also find the ratio of the segment to the rectangle of the
same base and height.
Ex. 1510. Find the area of a segment of base 10 cm. and height 5 cm.
Ex. 1611. Find the area of a segment of base 4 cm. and height 8 cm.
Ex. 16ia. A square is inscribed in a circle of radius 2 in. Find the
area of a segment cut off by a side of the square.
Ex. 1613. From a point outside a circle of radius 10 cm., a pair of
tangents are drawn to the circle ; the angle between the tangents is 120°.
Find the area included between the two tangents and the circumference.
Section XII. Further Examples of Loci.
Ex. 1614. Plot the locus of points the sum of whose distances from
two fixed points remains constant
(Mark two points S, H, say, 4 in. apart. Suppose that the point P
moves so that SP+ HP = 5 in. Then the following are among the possible
pairs of values :
SP
4-5
4-0
3-5
3-0
2-5
2-0
1-5
1-0
0-5
HP
0-5
1-0
1-5
2-0
2-5
3-0
3-5
4-0
4-5
Plot all the points corresponding to all these distances, by means of inter-
secting arcs. Why were not values such as SP=4-7, HP=0-3 included in
the above table? Draw a neat curve, free-hand, through all these points.
The locus is an oval curve called an ellipse.)
HEx. 1616. What kinds of symmetry are possessed
by an ellipse?
Ex. 1616. Describe an ellipse mechanically as
follows. Stick two pins into the paper about 4 in. apart ;
make a loop of fine string, gut or cotton and place it round
the pins (see fig. 288).. Keep the loop extended by means ^' ^^'
of the point of a pencil, and move the point round the pins. It will, of
course, describe an ellipse.
LOCI
289
Ex. 1617. Plot the locus of points the difference of whose distances
from two fixed points remains constant.
(For example, let the two fixed points S, H be 4 in. apart, and let the
constant difference be 2 in. Make a table as in Ex. 1514. Eemember to
make SP > H P for some points, H P > SP for other points.)
This curve is called a hyperbola.
Ex. 1518. Plot the locus of points the product of whose distances from
two fixed points remains constant.
(For example, mark two potnte S, H exactly 4 in. apart. First, to plot
the locus SP. HP = 5.
Fill np the blanks in the following table:
SP
5
4-8
4
3
s/5
2
HP
3
4
4-8
5
Secondly, plot the locus SP . HP=4; thirdly, plot the locus SP . HP=8.
All three loci should be drawn in the same figure.
The first locus will be found to resemble a dumb-bell, the second a
figure of 8 ; the third consists of two separate ovals.)
Ex. 1519. Plot the locus of a point which moves so that the ratio of
its distances from two fixed points remains constant.
(For example, let the two fixed points S, H be taken 3 in. apart ; and let
- = 2.)
HP '
Ex. 1520. OP is a variable chord passing through a fixed point O on
a circle; OP is produced to Q so that PQ=OP; find the locus of CL
Ex. 1521. A point moves so that its distance from a fixed point S
is always equal to its distance from a fixed line M N : find its locus.
(This is best done on inch paper. Take the point S 2 in. distant
from the line MN. Then plot points as follows. What is the locus of
points distant 8 in. from M N ? distant 3 in. from S ? The intersection of
these two loci gives two positions of the required point. Similarly find
other points.)
The curve obtained is called a parabola. It is the same curve as
would be obtained by plotting the graph y = — +1, taking for axis of x
the line MN, and for aris of y the perpendicular from S to MN. It is
remarkable as being the curve described by a projectile, e.g. a stone or a
cricket-ball. Certain comets move in parabolic orbits, the sun being situated
at the point S.
290
BOOK III
Ex. X623. A point moves in a plane subject to the condition that its
distance from a fixed point S is always in a fixed ratio to its distance from
a fixed straight line MN. Plot the curve described.
(i) Let the distance from S be always half the distance from MN.
Take S Sin. from MN.
(ii) Let the distance from S be sClways twice the distance from MN.
Take S Sin. from MN.
These curves will be recognized as having been obtained already.
Ex. IS 23. Plot the loous of a point on the oonnecting-rod of a steam-
engine..
fig. 289.
(The upper diagram in fig. 289 represents the cylinder, piston-rod (AB),
connecting-rod (BC), and crank (CD) of a locomotive. In the lower
diagram the different parts are reduced to lines. B moves to and fro
along a straight line, C moves round a circle. Take BC = 10 cm., CD =3 cm.
Plot the locus of a point P on BC, where BP is (i) 1 cm., (ii) 5 cm.,
(iii) 9 cm. This may be done, either by drawing a large number of different
positions of BC; or, much more easily, by means of tracing paper.
Draw BD and the circle on your drawing paper, BC on tracing paper. Keep
the two ends of BC on the straight line and circle respectively, and prick
through the different positions of P.)
Ex. 1524. A rod moves so that it always passes through a fixed point
while one end always lies on a fixed circle. Plot the locus of the other end.
(Tracing paper should be used. A great variety of curves may be ob-
tained by varying the position of point and circle, and the length of the
rod. It will be seen that this exercise applies to the locus of a point on
the piston-rod of an osciUatiug cylinder; also to the locus of a point on
the stay-bar of a casement window.)
LOCI
291
X
/^
Ex. 1635. The ends of a rod slide on two mres which cross at right
angles. Find the locus of a point on the rod.
(Represent the rod by a line of 10 cm. ; take the
point 3 cm. from one end of the rod; also plot the
locus of the mid-point. Use tracing paper.)
Ex. 1526. Two points A, B of a straight line
move along two lines intersecting at right angles.
Plot the locus of a point P, in AB produced. [Tracing
paper.] fig. 290.
Ex. 1527. Draw two intersecting lines. On tracing paper mark three
points A, B, C. Make A slide along one line and B along the other ; plot
the locus of C.
Ex. 1528. Draw two equal circles of radius 4 cm., their centres being
10 cm. apart. The two ends of a line PQ, 10 cm. in length, slide one
along each circle. Plot the locus of the mid-point of PQ ; also of a point
1cm. from P.
(Most quickly done with tracing paper. It is easy to construct a model
machine to describe the curve.)
Ex. 1529. Draw two circles. On tracing paper mark three points
A, B, C. Make A slide along one circle, B along another, and plot the locus
of C. (Experiment with different circles and arrangements of points. You
will find that in at least one case the locus-curve shrinks to a single point.)
Ex. 1530. OA, AP are two rods jointed at A. OA revolves about a
hinge at O, and AP revolves twice as fast as OA, in the same direction. Find
the path of a point on AP. (Make 0A=2 in., AP=2 in. Plot the locus of
fig. 291.
P; also of Q. and R, taking AQ=1 in., AR = ^ in. To draw the different
positions of the rod, notice that when OA has turned through, say, 30°, AP has
turned through 60° and therefore makes an angle of 30° with OA produced.)
The loci 'are different forms of the limafon ; the locus of Ql is heart-
shaped, and is called a cardlold. The locus of P has a small loop in it.
292 BOOK III
Ex. 1631. Bepeat Ex. 1530, mth the difference that, as OA revolves,
AP remains parallel to its original position.
Ex. 1532. Two equal rods OA, AQ, jointed as in Ex. 1530, revolve in
opposite directions at the same rate. Find the locus of Q and of the mid-
point of AQ.
Ex. 1533. O is a Axed point on a circle of radius 1 in. OP, a variable
chord, is produced to Q, PQ being a fixed length ; also PQ' ( = PQ) is marked
off along PO. Plot the locus of Q and Q! when PQ is (i) 2iin., (ii) 2 in.,
(iii) 1^ in.
(Draw a long line on tracing paper, and on it mark P, Q and Q'.)
Ex. 1684. Through a fixed point S is drawn a variable line SP to meet
a fixed line MN in P. From P a fixed length PQ is measured off along SP
(or SP produced). Find the locus of Q.
(Use tracing paper. Take S 1 in. from MN. Plot the locus of Q
(i) when PQ=1 in., measured from P away from S,
(ii) when PQ=1 in., measured from P towards S,
(iii) when PQ=2 in., measured from P towards S.)
The curves obtained are different forms of the eoncboid.
Ex. 1535. A company of soldiers are extended in a straight line. At a
given signal, they all begin to move towards a certain definite point, at the
regulation pace. Are they in a straight line after 3 minutes ? If not, what
curve do they form ?
Ex. 1536. XOX', YOY' are two fixed straight
lines, C is a fixed point (see fig. 292). A variable
line PQ is drawn through 0 to meet XOX', YOY'
in P, Q respectively. Plot the locus of the mid-
point of PQ.
(Let XOX', YOY' intersect at 60°, and take C
on the bisector of Z.XOY, 5 cm. from O.)
Ex. 1537. (Inch paper.) Draw a circle of
radius 2 in. and a straight line distant 6 in. from ^„ 292.
the centre of the circle. P is a variable point on
the circle ; Q is the mid-point of PN, the perpendicular from P upon the
line. Plot the locus of GL
LOCI — ENVELOPES
293
Envelopes.
We have seen that a set of points, plotted in any regular
way, marks out a curve which is called the locus of the points.
In a rather similar manner, a set of lines (straight or curved)
drawn in any regular way, marks out a curve which is called the
envelope of the lines. Each of the lines touches the envelope.
Let a piece of paper be cut
out in the shape of a circle, and
a point S marked on it. Then
fold the paper so that the circum-
ference of the circle may pass
through S. If this is done many
times, the creases left on the paper
will envelope an ellipse (fig. 293).
fig. 293.
Ex. 1538. Take a piece of cardboard with one edge straight; drive
a pin through the cardboard into the paper underneath ; then turn the
cardboard round the pin, and in each position use the straight edge of the
cardboard to role a line. What is the envelope of these lines?
Ex. 1539. One edge of a flat ruler is made to pass through a fixed
point, and lines are drawn with the other edge. Find their envelope.
Ex. 1540. Prove that the envelope of straight lines which lie at a
constant distance from a fixed point is a circle.
Ex. 1541. Find the envelope of equal circles whose centres lie on a
fixed straight Une.
Ex. 1543. Find the envelope of a set of equal circles whose centres are
on a fixed circle when the radius of the equal circles is (i) less than, (ii) equal
to, (iii) greater than, the radius of the fixed circle.
294 . BOOK III
Ex. 1548. Draw a straight line MN and drive a pin into your paper at
a point S ^ in. from M N (see fig. 294).
Keep the short edge (AB) of your set-squsire
pressed against the pin, and keep the right
angle (B) on the line MN. Rule along BCj
and thus plot tlie envelope of BC, as the
set-square slides on the paper, (Lines must
of course be draWn with the set-square
placed on the left of S, as well as on the right.)
Ex. 1644. Bepeat Ex. 1545 using the 30° angle instead of tlio right
angle, and putting the pin 1 in. from MN.
Ex. 1545. Draw a circle of radius 5 cm. and mark a point S 4 cm. from
the centra Let a variable line SP meet the circle in P and let PQ be
drawn perpendicular to SP. Find the envelope of PQ. (The part of PQ
inside the circle is the important part.)
Ex. 1546. Bepeat Ex. 1545 with the point S on the circle.
Ex. 1547. Find the envelope of circles passing through a fixed point
O, and having their centres on a fixed circle.
(i) Take radius of fixed circle =4 cm., distance of O from centre of
fixed circle =3-2 cm.
(ii) Take radius of fixed circle=:4 cm., distance of O froiu centre of
fixed circle = 4 cm.
(iii) Take radius of fixed circle = 3 cm., distance of O from centre of
fixed circle =5 cm.
Ex. 1548. Find the envelope of circles passing through a fixed point,
and having their centres on a ^xed straight line.
Ex. 1540. Plot the envelope of a straight line of constant length
whose ends slide upon two fixed lines at right angles. *
EXEKCISES ON BOOK m 295
MISCELLANEOUS EXERCISES.
Ex. 1550. (Without protractor.) Trisect an arc of 90°.
Ex.1551. (Without protractor.) Trisect a given semicircular arc.
tEx. 1552. There are two fixed concentric circles; AB is a variable
diameter of the one, and P a variable point on the other. Prove that
AP^+ BP^ remains constant.
[Use Apollonios' theorem, Ex. 1133.]
Ex. 1553. In a circle of radius 2-5 in. inscribe an isosceles triangle of
vertical angle 40°. Measure its base.
tEx. 1554. Points A, P, B, Q, C, R are taken in order on a circle
so that arc AP=arc BQ=:arc CR. Prove that the triangles ABC, PQR are
congruent.
Ex. 1555. The railway from P to Gl consists of a circular arc AB and
two tangents PA, BQ. AB is an arc of 28° of a circle whose radius is J mile ;
PA =1 mile, BQ=^mile. Draw the railway, on a scale of 2 inches to the
mile, and measiure the distance from P to Q as the crow flies. Also calculate
the distance as the train goes.
tEx. 1556. From a point P on a circle, a line PGl of constant length is
drawn parallel to a fixed line. Plot the locus of Q, as P moves round the
circle. Having discovered experimentally the shape of the locus, prove it
theoretically.
- tEx. 1557. YZ is the projection of a diameter of a circle upon a chord
AB ; prove that AY = BZ.
tEx. 1558. Through two given points P, Q on a circle draw a pair of
equal and parallel chords. Give a proof.
tEx. 1559. AOB, COD are two variable chords of a circle, which are
always at right angles and pass through a fixed point O. Prove that
AB^ + CD^ remains constant. ,
tEx. 1560. Through A, a point inside a circle (centre O), is drawn a
diameter BAOC ; P is any point on the circle. Prove that AC> AP>-AB.
Ex. 1561. What is the length of (i) the shortest, (ii) the longest
chord of a circle of radius r, drawn through a point distant d from the
centre?
296 BOOK III
Ex. 1602. Two chords of a circle are at distances from its centre
equal to | and ^ of its radius. Find how many times the shorter chord
is contained in three times the longer chord.
Ex. 1563. The star-hexagon in fig. 295 is formed by
producing the sides of the regular hexagon. Prove that the
area of the star-hexagon is twice that of the hexagon.
tEx. 1664. Chords AP, BQ are drawn X to a chord AB
at its extremities. Prove that AP= BQ. ^8- 295.
fEx. 1666. The line joining the centre of a circle to the point of inter-
section of two tangents is the perpendicular bisector of the line joining the
points of contact of the tangents.
tEx. 1666. Find the locus of the point of intersection of tangents to a
circle which meet at an angle of 60°.
Ex. 1667. Show how to construct a right-angled triangle, given that
the radius of the inscribed circle is 2 cm. and that one of the sides about the
right angle is 5 cm.
Ex. 1668. Construct an isosceles triangle, given the radius of the in-
scribed circle, and the base.
tEx. 1660. A is a point outside a given circle (centre O, radius r).
With ceutre O and radius 2r describe a circle ; with centre A and radius AG
describe a circle; let these two circles intersect at B, C. Let OB, OC cut
the given circle at D, E. Prove that AD, AE are tangents to the given
circle.
tEx. 1670. A circle is drawn having its centre on a side AC (produced)
of an isosceles triangle, and touching the equal side AB at B. BO is
produced to meet the circle at D. Prove that the radius of the circle
through D is perpendicular to AC.
Ex. 1671. Find the angles subtended at the centre of a circle by the
three segments into which any tangent is divided by the sides (produced if
necessary) of a circumscribed square.
tEx. 1672. An interior common tangent of two circles cuts the two
exterior common tangents in A, B. Prove that AB is equal to the length
intercepted on an exterior tangent between the points of contact.
tEx. 1573. The radius of the oiroumcircle of an equilateral triangle is
twice the radius of the in-circle.
EXERCISES ON BOOK III
297
fig. 296.
Ex. 1674. Show how to inscribe three equal circles
to touch cue another in an equilateral triangle, of side
6 in. (fig. 296).
Ex. 1575. Show how to inscribe in a square, of side
6 in., four equal circles, each circle to touch two others.
tEx. 1576. Two circles touch externally at E ;
AB, CD are parallel diameters drawn in the same sense
(see page 78, footnote) ; prove tnat AE, ED are in the
same straight line ; as also are BE, EC.
tEx. 1577. Two circles touch at A; T is any point on the tangent at
A; from T are drawn tangents TP, TQ to the two circles. Prove that
TP=TQ. What is the locus of points from which equal tangents can be
drawn to two circles in contact ?
tEx. 1578. S is the circumcentre of a triangle ABC, and' AD is an
altitude. Prove that ^ B A D = Z. CAS.
tEx. 1579. Through a given point on the circumference of a circle
draw a chord which shall be bisected by a given chord. Give a proof.
Ex. 1580. From the given angles, find
all the angles of fig. 297.
Draw the figure, making the radius of the
circle 2 in. Check the marked angles, and
measure CD.
tEx. 1581. Two circles intersect at B, C ;
P is a variable point on one of them. PB,
PC (produced if necessary) meet the other
circle at Q, R. Prove that QR is of constant
length.
[Show that it subtends a constant angle at B.]
Ex. 1582. Show how to find a point O inside A ABC so that
Z.AOB = 150°, /lAOC = 130°.
Ex. 1583. Show how to find a point O inside a ABC, such that the
three sides subtend equal angles at O.
Ex. 1584. Show how to construct a triangle, having given the vertical
angle, the altitude and the bisector of the vertical angle (terminated by the
base).
tEx. 1585. A is one of the points of intersection of two circles whose
centres are C, D. Through A is drawn a line PAQ, cutting tlie circles again
in P, Q. PC, QD are produced to meet at R. Prove that the locus of R is
a circle through C and D.
G. s. 11. 20
fig. 297.
298
BOOK III
tEx. 1A08. A, C are two fixed points, one upon each of two circlefi
which intersect at B, D. Through B is drawn a variable chord PBQ,
cutting the two circles in P, CL PA, QC (produced if necessary) meet at R.
Prove that the locus of R is a circle.
tEx. 1687. Two equal circles cut at A, B ; a straight line PAQ meets
the circles again in P, Q. Prove that BP = BQ. [Consider the angles sub-
tended by the two chords.]
i Ex. 1S88. C is a variable point on a semicircle whose diameter is AB,
centre O; CD is drawn X to AB ; OX is the radius X to AB. On OC a
point M is taken so that OM = CD. Prove that the locus of M is part of a
circle whose diameter is OX.
t£x. 1589. ABC, DOB are two congruent triangles on the same side of
the base BC. Prove that A, B, C, D are concyclic.
tEx. 1A90. D, E, F are the mid-pomts of the sides of BC, CA, AB of
A ABC; AL is an altitude. Prove that D, E, F, L are concyclic (see
Ex, 1589).
fEx. 1691. Prove that tbe drele throngb the mid-points of the
sides of a triangle also passes through the feet of the altitudes (sec
Ex. 1590).
tEx. 159a. The altitudes BE, OF of a ABC intersect at H ; prove
(i) that AEHF is a cyclic quadrilateral,
(u) that Z.FAH = ^FEH,
(iii) that ^FEH = ^FCB,
(iv) that, if AH is produced to meet BC in D,
AFDC is cyclic,
(v) that AD is ± to CB.
Hence : The three altitudes of a triangle meet in a point ; which is
called the orthocentre.
tEx. 1593. In fig. 298 AD is X to-BC and
BE is X to CA ; S is the centre of the circle.
Show that •
BF = AH,
and that AB, FH bisect one another.
[Prove AHBF a parallelogram.]
Ug* ^Jo.
tEx. 1594. BE, CF, two altitudes of A ABC, intersect at H. BE
produced meets the circumcircle in K. Prove that E is the mid-point
of HK.
[Show that BFEC is a cyclic quadrilateral, .-. Z.FCE=Z.FBE. But
^KCE=^FBE (why?), .-.etc.]
EXERCISES ON BOOK III 299
tEx. 1595. I is the centre of the inscribed circle of a ABC; 1^ is the
centre of the circle escribed outside BC. Prove that BICI^ is cyclic.
tEx. 1596. An escribed circle of A ABC touches BC externally at D, and
touches AB, AC produced at F, E respectively; O is the centre of the circle.
Prove that
(i) Z.BOC = JZ-FOE=:90°-^,
(ii) 2AE=2AF = BC + CA + AB.
tEx. 1597. Prove that
Z-BICr=90° + ^,
where I is the inscribed centre of A ABC.
Hence find the locus of the inscribed centre of a triangle, whose base and
vertical angle are given.
tEx. 1598. I is the centre of the inscribed circle of a ABC; Al pro-
duced meets the cireumcircle in P; prove that PB = PC=PI.
tEx. 1599. P is any point on cireumcircle of . |^
A ABC. PL, PM, PN are 1 to BC, CA, AB re- y\j^
spectively. Prove that
(i) Z.PNL = 180°-aPBC,
(ii) ^PNM = Z.PAM,
(iii) aPNL+/LPNM = 180°,
(iv) LNM is a straight Une.
Verify this result by drawing. „ „gq
LNM is called Simson's Une.
tEx. 1600. ABCDEF is a regular hexagon ; prove that BF is trisected
by AC, AE.
tEx. 1601. In fig. 300, BC is ± to PA.
Prove that PA bisects i^QPR.
tEx. 1602. Through A, a point of intersec-
tion of two circles, lines BAC, DAE are drawn,
B, D being points on the one circle, C, E on the
other. Prove that the angle between DB and
CE (produced if necessary) is the same as the
angle between the tangents at A.
tEx. 1603. Two circles touch internally at A; BC, a chord of the
larger circle, touches the smaller at D ; prove that AD bisects Z.BAC.
[Let BC meet the tangent at A in T.]
tEx. 1604. A radius of one circle is the diameter of another; prove
that any straight line drawn from the point of contact to the outer circle is
bisected by the inner circle.
20—2
soo
BOOK m
fig. 301.
tEx. 10O6. In fig. 801 AB is a tangent;
00 = DA = AB.
BO outs the oiroumferenoe at E. Prove that aro AE is ^
and aro EF ^ of the oiroumferenoe.
tEz. 1606. Join O, the oironmoentre of a triangle,
to the vertioes A, B, C. Through A draw lines || to OB,
OC ; through B lines || to 00, OA ; through 0 lines || to OA, OB. Prove
that these lines form an equilateral hexagon; that each angle of the
hexagon is equal to one other angle, and double an angle of the triangle.
Ex. 1607. Power is being transmitted from one shaft to another parallel
shaft by means of a belt passing over two wheels. The radii of the wheels
are 2 ft. and 1 ft. and the distance between the shafts is 6 ft. Assuming the
belt to be taut, find its length (i) when it does not cross between the shafts,
(ii) when it does cross.
tEx. 1608. PQ is a chord bisected by a diameter AB of a circle
(centre O). PG bisects the Z.OPGL Prove ^at it bisects the semi-circle on
which Q lies.
tEx. 1609. If through C, the mid-point of an arc AB, two chords arc
drawn, the first cutting the chord AB in D and the circle ra E, the second
cutting the ohord in F and the circle in G, then the quadrilateral DFGE is
cyclic.
tEx. leiO. P is a point on an aro AB. Prove that the bisector of
Z. APB and the perpendicular bisector of the ohord AB meet on the circle.
Ex. 1611. P, Q are two points on a circle; AB is a diameter. AP,
AQ are produced to meet the tangents at B in X, Y. Prove that A' APGl,
AYX are equiangular ; and that P, G, Y, X are concyclic.
tEx. 1612. In fig. 302 the angles at O are
allequal; andOA = AB = BC = CD = DE. Prove
that O, A, B, C, D, E are concyclic.
tEx. 1613. From a point A on a circle, two
chords are drawn on opposite sides of the dia-
meter through A. Prove that the line joining
the mid-points of the minor aro of these chords
cuts the chords at points equidistant from A.
tEx. 1614. Two equal chords of a circle intersect;
prove that the segments of the one chord are respectively
equal to the segments of the other.
Ex. 1615. In fig. 303 O is the centre of the arc
AB ; and Q is the centre of the arc EC ; L ADC is a
right angle. DA = 3, DC = 5, DQ = 3. Find OA and
QC; and draw the figure. (Let OD = a;.)
fig. 303.
BOOK IV.
Similarity.
Ratio and Propobtion.
To measure a length is to find how many times it contains
another length called the unit of length.
The unit of length may be an inch, a centimetre, a millimetre, a mile,
a light-year ^, or any length you choose. Hence the importance of always
stating your unit.
If you have two lines, one 4 in. long, the other 5 in., you say
that the first is ^ of the second.
The ratio of a length XY to a length PQ is the quotient
measure of XY
measure of PQ *
the two measurements being made with respect to the same unit
of length.
The practical way then, to find the ratio of two lengths, is to
measure them in inches or centimetres or any other convenient
unit^ and divide.
The ratio of a to & is written j- , or a/6, or a : &, or a-^b.
^ Astronomers sometimes express the distances of the fixed stars in terms
of the distance traversed by light in a year. This distance is called a light-
year, and is 63,368 times the distance of the earth from the sun. The
nearest star is o Centauri, whose distance is 4-26 hght-years.
302 ]j(joK IV
HEx. 1616. Find the ratio ^^ (fig. 304); measure (i) in inches, (ii) in
centimetres. Work out the ratios to three significant figures. Why might
you expect your results to differ?
fig. 304. •
DC
Ex. 1617. Find the ratio ^^ as in Ex. 1616.
If the lengths are determined approximately, the ratios can
he calculated only approximately.
If you measure two lines and find their lengths to he 5*82 in., and
8'65 in., the last figure in each case is doubtful; you are not sure, for
example, that the second length is not nearer to 3'64 in. or 3*66 in.
Now 111=1.602-, and|||=1.6874-.
You see that the results differ in this instance by '015 (i.e. about 1 °/o)'
As a general rule, work out ratios to three significant
figures.
Ex. 1618. Express the following ratios as decimals: —
<^>8l|' <"^IS' <^)W' (i^) 9310:3-35, (v) -0128 : -00637.
Hitherto we have only considered the ratio of two lengths.
In the case of other magnitudes, ratio may be defined as
follows : —
Def. The ratio of one magnitude to another of the same
kind is the quotient obtained by dividing the numerical measure
of the first by that of the second, the unit being the same in each
case.
RATIO AND PROPORTION 303
The ratio of two magnitudes is independent of the
unit chosen.
For example, the ratio of a length of 5 yds. to a length of 2 yds. is 5 : 2;
if these lengths are measured in feet the measures are 15 and 6, and the
ratio is 15 : 6. Now we know that 5 : 2=15 : 6.
Def. li a :b =c : d, the four magnitudes a, h, c, d are said
to be in proportion.
1[Ex. 1619. Are the following statements correct?
(i) 3 yds. : 1 yd. = 3 shillings : 1 shilling,
(ii) 3 yds. : 3 8hillings=l yd. : 1 shilling.
ITEx. 1620. Fill in the missing terms in the following: —
(i) 1=1,, (iv) 5:2=7: ,
(ii) ^=h (V) 2^ = |.
(iii) 7: =3:11, (vi) ^ = g.
The following algebraical processes will be used in the course
of Book lY.
I Tf "l ~'L ^^' Conversely if
then ^ X 6d = -; X 6d, then — r = — .
b d ' hd hiV
.*. ad = 6c a c
[e-g-l=i^. .■.SxW = 5x9l 'b~d-
m. If ^ = 5, IV. If ^ = ^,
0 d b d
.: ad=bc, .; ad=bc,
ad _bc - ad _bc
" cd cd' " ac ac'
, a _b d b
" c d' " c~ a'
V. If J=^; VI. If ^=^=^=...=/;,
b d'
y
a. , c _ iu a + b + c+... ,
. -=tl = -±l, then = fc.
0 d x+y+z + ...
aJzb _csfcd
304 BOOK IV
Ex. l«ai. Draw two straight lines SVT and XZY.
sv xz
Prove fully that, if ST ~ XY ' *^®"
... ST XY .... VT ZY ..... VT ZY
^'^ 5V-XZ' ^"^ SV-XZ' ^""^ ST = )(Y-
What rectangle properties oan be obtained from the above results by
clearing of fractions ?
Ex. 1622. State and prove the converses of the properties proved
in Ex. 1621.
^Ex. 1623. From each of the following rectangle properties deduce a
ratio property :
(i) AB.CD=PQ.QR,
(ii) XY2=XZ.XW.
Ex. 1624. In fig. 4, find what fraction AC is of AB.
Internal and External Division*.
If in a straight line AB a point P is taken, AB is said to be
PA
divided internally in the ratio — (i.e. the ratio of the distances
of P from the ends of the line). In the same way, if in A B
produced a point P is taken, AB is said to be divided externally
PA
in the ratio — (i.e. the ratio of the distances of P from the ends
PB ^
of the line).
In the latter case, it must be carefully noted that the ratio is not ■^=. .
BP
Suppose the points A, B connected by an elastic string; take hold of the
string at a point P and, always keeping the three points in a straight line,
vary the position of P; whether P is in AB or AB produced, the ratio in
which AB is divided is always the ratio of the lengths of the two parts
of the string.
* The discussion of oases of external division may be postponed.
INTERNAL AND KXTERNAL DIVISION 305
^Ex. 1625. In fig. 305, name the ratios in which (i) H divides AB,
(iij A divides BH, (iii) C divides KA.
1[Ex. 1626. In fig. 317, what lines are divided (i) by D in the ratio
BD ZY BC
g^, (ii) by Z in the ratio ^tw, (iii) hy B in the ratio ^g?
Proportional Pivisiow of Straight Links.
Bevise pp. 142, 143.
ITEx. 1627. Draw a triangle ABC and draw HK parallel to BC (see
fig. 305). What fraction is AH of AB ? What fraction is AK of AC ?
[Express these fractions as decimals.]
Ex. 1628. In the figure of Ex. 1627, calculate
AH AK HB KC
^^' HB' KC ^"' AB' AC
TEx, 1629. (On inch paper.) Mark A (1, 2), B (1, 0), O (2, 0); draw
the triangle ABC. In AB mark the point H (1, 0'7), through H draw HK
parallel to BC cutting AC at K. The horizontal lines of the paper divide
AC into 20 equal parts (why are they equal?); how many of these parts
AK AH
does AK contain? What are the values of -jt-a* im^
AC AB
Ex. 1630. (On inch paper.) Bepeat Ex. 1629 with A (1, 1), B (1, 0),
C (3, 0), H (1, 0-3).
300
BOOK IV
Theorem ].
If a straight line HK drawn parallel to the base BC
of a triangle ABC cuts AB, AC in H, K respectively,
AH AK
then — = -r:: .
AB AC
fig. 305.
AH p
Proof Suppose that — = - , where ^ and q are integers. Then
if A B is divided into q equal parts, AH contains p of these
parts.
Through the points of division draw parallels to BC.
Now AB is divided into equal parts.
.'. these parallels divide AC into equal parts ; I. 24.
AC contains q of these parts, and AK contains p of these parts.
AK
Ac"
P
AH
ab'
AK
^ AC*
Q. KD.
PROPORTIONAL DIVISION 307
Cor. 1. If a strai^t line is drawn parallel to one
side of a triangle, the other two sides are divided
proportionally.
rp ., , AH AK
io prove that — = — .
^ HB KC
Firs6 proof * In the figure, AB is divided into q equal parts;
AH contains p of these equal parts ; •
.*. HB „ q^p „ „
• — _ P
' ' HB ~ q—p'
Sim'y
AK p
KC q-p
AH _ AK
HB ~ KC*
AH AK
HB KC
i.e. - — = — ■ ,
AH AK'
. AH AK
"HB ~ KC'
Q. E. D.
AH AK
Second proof * Since - = — , Proved
. AB _ AC
"AH ~ AK'
"AH~^~AK ^'
AB - AH AC - AK
Q. E.D.
These proofs apply to the first figure : see Ex. 1631.
308 BOOK IV
Cor. 2. If two straight lines are cut by a series of
parallel straight lines, the intercepts on the one have to
one another the same ratios as the corresponding inter-
cepts on the other.
tEx. 1631. Write out the two proofs of Cor. 1 for the second and third
figures of page 30G.
tEx. 1683. ' Triangles of the same height are to one another as their
bases.
[Suppose one base is f of the other.]
Ex. 1633. Divide a given straight line so that one part is | of the
whole line-
Ex. 1634. Divide a given straight line in the ratio | (i.e. so that the
ratio of the two parts =|).
Ex. 1635. Show bow to divide a given straigbt line AB in tbe
ratio of two given straight lines p, q.
[Through A draw AC, from AC cut ofiE AD=p, DE = q; join BE j draw a
AD
line through D to divide AB in the ratio g^ ; in what direction must this
line be drawn ?]
4"2 3'7
Ex. 1636. Find the value of x, when 07^ = — , (i) graphically,
(ii) by calculation.
[Make an Z.POQ; from OP cut ofi 00=42 in., DE=2'5 in.; from OQ
out off OF = 3-7; draw EG i| to OF. Which is the required length?]
Ex. 1637. Find, both graphically and by calculation, the value of x
in the following cases :
... 2-25 3-05 ,.., -935 1-225
(iii) a;:2-63=5'05:2-84, (iv) 8 36 : -026= a; : -037.
Def. If X is such a magnitude that ^ = - (or a :b = c:x),
X is called the fourth proportional to the three magnitudes
Of b, c. .
PROPORTIONAL DIVISION 309
To find the fourth proportional to three given straight
lines.
Let o, b, c be the three given, straight lines.
Construction Make an angle POGL
From OP cut off OD = a, and DE = 6.
From OQ cut off OF = c.
Join DF.
Through E draw EG || to DF, cutting OQ in Q.
Then FG (x) is the fourth proportional to a, h, c.
Proof Since FD is || to EG,
a c
IV. 1.
Def. If X is such a magnitude that j-= - (or a : ft = 6 : x),
X is called the third proportional to the two magnitudes a, h.
Note. If x is the third proportional to a, b, it is also the
fourth proportional to a, b, b.
Ex. 1638. Show bow to find tbe tbird proportional to two given
straigbt lines.
[See note above. ]
Ex. 1639. Find graphically the fonrth proportional to 3, 4, 5. Check
by calculation.
Ex. 1640. Find graphically the third proportional to 6-32, 8*95. Check
by caloalation.
310 BOOK IV
tEx. 1641. Jnatliy tbe fbUowing eonstmetioii for flndins tbe
fonrtli iiroportional to p, q, r : — Make an ^BAC; from AB cut ofl
AX=p, AY = g ; from AC cut off AZ=r ; join XZ, aud draw YW || to XZ.
Then AW is tbe fourth proportionaL
Ex. 1642. Using the oonstractioa of Ex. 1641, find the fourth propor-
tional to 1, 1*41, 4-23. Check your result.
Ex. 1643. Explain and justify a construction, analogous to that of
Ex. 1641, for finding the third proportional to p, q.
Ex. 1644. Using the construction of Ex. 1643, find the third propor-
tional to 1, 1-73. Check your result
Ex. 1640. Given that the circumference of a circle of 1 in. radius is
6*28 in., find graphically the circumferences of circles whose radii are
(i) 3-28 cm., (ii) 16-7 in., (iii) 8-37 miles, (iv) 4-28 km.
Also find the radii of circles whose circumferences are (i) 3-36 in.,
(ii) 12-35 in., (iii) 8-66 cm., (iv) 11 yards.
Ex. 1646. (On inch paper.) Mark four points A (1, 1), B (1, 4),
0 (4, 1), D (3, 3) ; join AB, and mark P (1, 2). Produce AC, BD to meet
AP CQ
at V ; join VP ; ^et it cut CD at Q. Find =q and 7.^7^; arc they equal?
Pd GLO
Ex. 1647. Make a copy of the points A, B, C, D, P in Ex. 1646, by
AP CR
pricking through. Divide CD at R so that 55 = sn •
PB RD
[Begin by dividing CB in the required ratio.]
Ex. 1648. Draw a straight line AB, on it take two point? P, Q ; draw
another straight line CD ; divide CD similarly to AB. (Freehand)
PROPORTIONAL DIVISION
311
' Theorem 2.
[Converse of TnEORb^M L]
If H, K are points in the sides ab, ac of a triangle abc,
AH_ AK
AB~ AC'
AH AK
such that ^^ = -—, then HK is parallel to ec.
Construction
To "prove that
Proof
fig. 308.
Draw HK' parallel to BC.
HK and HK' coincida
Since HK' is || to EC.
AH _ AK'
■ * a'b ~ AC '
,. AH AK
J>ut — = — .
AB AC
AK' AK
" AC ~ AC '
.•. AK' = AK,
.'. K and K' coincide,
.'. HK and HK' coincide,
.". HK is parallel to BC.
IV. 1.
Dafn
Q. E. D.
Cor. 1. If ^-
AH
AC
AK
, then HK and BC are parallel.
Cor, 2. If a straight line divides the sides of a triangle
proportionally, it is parallel to the base of the triangle
312 BOOK lY
tEx. 1640. Prove Cor. 1 without assuming iv. 2. '
tEx. 1660. Prove Ck>r. 2 wiUiout assuming rv. 2.
tEx. 1651. O is a point inside a qnadrilateral ABOD ; OA, OB, OO,
CD are divided at A', B', C, D'
,, ^ DA' OB' oc ozy ,
so that 55rA- = ;cs- = sv^ =:*nk- = 8.
OA OB OC OD ^
Prove that A'B' is parallel to AB.
Also prove that L O'A'B' = L DAB.
tEx. 1652. Draw a large quadrilateral ABCD ; in it take a point O,
and join OA, OB, OC, OD ; in OA take a point A', through A' draw A'B'
parallel to AB to cut OB at B', through B' draw B'C parallel to BC to out
OC at C, through C draw CD' parallel to CD to cut OD at D'. Prove that
D'A', DA are parallel. Are they parallel in your figure? Keep your figure
for Ex. 1663.
Ex. 1653. In the figure of Ex. 1652, calculate
A'B' B'C CD' DW
AB ' BC ' CD ' DA •
Ex. 1664. Bepeat Ex. 1652 for (i) a triangle, (ii) a pentagon.
tEx. 1655. A variable line, drawn through a fixed point O, outs two
fixed parallel straight lines at P, Q ; prove that OP : OQ is constant.
tEx. 1656. O is a fixed point and P moves along a fixed line. OP is
divided at Q (internally or externally) in a fixed ratio. Find the locus of GL
tEx. 1667. D is a point in the side AB of A ABC; DE is drawn
parallel to BC and cuts AC at E ; EF is drawn parallel to AB and cuts BC
at F. Provethat AD :DB=BF:FC.
tEx. 1658. D is a point in the side AB of A ABC; DE is drawn
parallel to BC and cuts AC at E ; CF is drawn parallel to EB and cuts AB
produced at F. Prove that AD : AB= AB : AF.
tEx. 1650. AD, BC are the parallel sides of a trapezium ; prove that a
line drawn pejrallel to these sides outs tne other sides proportionally.
tEx. 1660. From a point E in the common base AB of two triangles
ACB, ADB, straight lines are drawn parallel to AC, AD, meeting BC, BD at
P, Q ; show that FG is parallel to CD.
PROPORTIONAL DIVISION 313
tEx. 1661. In three straight lines OAP, OBQ, OCR the points are
chosen so that AB is parallel to PQ, and BC parallel to QR. Prove that AC
is parallel to PR.
tEx. 1663. AB, DC are the parallel sides of a trapezium. P, Gl are
points on AD, BC, so that AP/PD = BQ/ClC. Prove that PQ is || to AB and
DC. (Use reductio ad absurdnm.)
Similar Triangles.
Def. Polygons which are equiangular to one another and
have their corresponding sides proportional are called similar
polygons.
1[Ex. 1663. Draw a quadrilateral ABCD; draw a straight line parallel
to CD to cut BC at P and AD at Q. Prove that ABCD, ABPQ are equi-
angular. Are they similar ?
ITEx. 1664. Draw a quadrilateral ABCD having AB = 3in., BC = 2in.,
CD = 3in., DA = 2in., Z.B=30°; draw a quadrilateral PQRS having PQ
= 6cra., QR = lcm., RS = 6cm., SP = 4cm., /LQ = 90°. Are ABCD, PQRS
similar ?
HEx. 1665. Draw a quadrilateral XYZW having XY=^3in., YZ = 2in.,
ZW=lin., WX = 4in., Z.Y = 120°. Outside XYZW describe a quadri-
lateral X'Y'Z'W' having its sides parallel to the sides of XYZW and 1 in.
away from them. Are the two quadrilaterals similar ? Find
xy rz/ z/w wx'
XY ' YZ ' zw ' wx •
HEx. 1666. Draw two equiangular triangles; find the ratios of their
corresponding sides.
Revise Ex. 146—151.
G. s. II. 21
314 BOOK IV
Theorem 3.
If two triangles are equiangular, their corresponding
sides are proportional.
Data ABC, DEF are two triangles which have
Lk- lD, LBi^L^y and lC-lF. (See i. 8, Cor. 5.)
To prove that
Construction
Proof
BC CA AB
EF~ FD~ DE'
From AB cut off AH = DE,
From AC cut off AK = DF.
Join HK.
In the As AHK, DEF,
AH = DE,
Constr.
AK = DF,
Gonstr.
Lk= LD,
Data
:. AAHK = ADEF,
I. 10.
/./.AHK^aE,
-Z.B.
Data
SIMILAR TRIANGLES 315
/. HK is II to BC, I. 4.
. AH _ AK
' AB~ AC'
DE _DF
■ AB ~ AC"
IV. 1.
Sim'y by cutting off lengths from BA, BC,
ED_ EF
BA~ BC'
— , — , — are all equal.
BC' CA' AB ^
BC _ CA _ AB
EF" FD~ DE'
Q. K. D.
ED EF
tKx. 1667. Write out the complete proof that oi^=H7^'
Ex. 1668. ABC Is a triangle having BC = 3 in., CA = 4 in., AB = 5 in.;
DEF is an equiangular triangle having EF=:2-2 in. Calculate DE, DF and
check by measurement.
Ex. 1669. Repeat Ex. 1668 vrithBC = 5-8 cm., CA=7-7cm., AB=8-3cm.,
EF = l-8in.
IFEx. 1670. If P is any point on either arm of an angle XOY, and PN
FN
is drawn perpendicular to the other arm, ^^ has the same value for all
positions of P.
PN P N
[Take several different positions of P and prove that ^ = — J= ...]
PN
^^ is the sine of Z.XOY; this exercise might have been stated as
follows : — the sine of an angle depends only on the magnitude of the angle.
/0N\ /PN \
ITEx. 1671. Prove that the cosine ( pcp I and tangent ( Tsrjrj ) of an
angle depend only on the magnitude of the angle.
21—2
316 BOOK IV
Ex. 1673. On a base 4 in. long draw a quadrilateral ; on a base 3 in.
long construct a similar quadrilateral. Calculate the ratio of each pair of
corresponding sides.
[Draw a diagonal of the first quadrilateral.]
fEx. 1673. PQRS is a quadrilateral inscribed in a circle whose diagonals
intersect at X ; prove that the a' XPS, XQR are equiangular. Write down
the three equal ratios of corresponding sides.
PQ XP
tEx. 1674. In the figure of Ex. 1673, prove that or = ve*
[If you colour PG, SR red, and XP, XS blue, you will see which two
triangles you require.]
•
tEx. 1675. XYZW is a cyclic quadrilateral; XY, WZ produced inter-
PY PZ
sect at a point P outside the circle ; prove that ^^ = ^^ .
tEx. 1676. ABC is a triangle right-angled at A ; prove that the altitude
AD divides the triangle into two triangles which are similar to A ABC.
Write down the ratio properties you obtain from the similarity of a* BDA,
BAC.
[See Ex. 132—134.]
tEx. 1677. The altitude QN of a triangle PQR right-angled at Q. cuts
QN PN
RP in N ; prove that r^ = qj^«
[Find two equiangular triangles ; colour the given lines ; see Ex. 1674.]
tEx. 1678. XYZ is a triangle inscribed in a circle, XN is an altitnde
of the triangle, and X D a diameter of the circle ; prove that
XY:XD = XN:XZ.
tEx. 1673. XYZ is a triangle inscribed in a circle ; tiie bisector of Z.X
meets YZ in P, an 1 the circle in Q, ; prove that XY : XQ = XP : XZ.
tEx. 1680. PQRS is a quadrilateral inscribed in K
circle; PT is drawn so that z SPT= / QPR. (See fig.
810.) Prove that (i) SP:PR = ST:QR,
(ii) SP:PT=SR:TQ.
tEx. 1681. Three straight lines are drawn from a
point O ; they are cut by a pair of parallel lines at X , Y , Z
and X', Y', Z'. Prove that X Y : YZ = X'Y' : Y'Z'. fig. 310.
SIMILAR TRIANGLES
317
On a given straight line to construct a figure similar
to a given rectilinear figure. (First Method.) t
fig. 311.
Let ABODE be the given figure and A'B' the given straight line
Constmction Join AC, AD.
On A'B' make AA'B'C' equiangular to A ABC.
On A'C' make AA'C'D' equiangular to AACD.
On A'D' make AA'D'E' equiangular to AADE,
Then A'b'C'd'E' is similar to ABCDE.
Proof This may be divided into two parts :
(i) the proof that the figures are equiangular; this is
left to the student.
(ii) the proof that
AB
DE
JEA
E'A'
BC _ CD
A'B' Fc' ~ C'D^ D'E'
Since A* ABC, A'b'C' are equiangular,
. _AB^ _ Bb AC
A^ ~ B^' ~ aW •
Since A* ACD, A'C'D' are equiangular,
. AC _ CD _ AD
** A*C' ~C^' ~A^'
Again since A* ADE, A'D'E' are equiangular, Constr,
. AD _ DE _ EA
' * A^ ~ dT' ~ E'A' '
Constr.
IV. 3.
Constr.
iv. 3.
IV. 3.
AB
a¥'
BC
Fc^
CD
C^'
DE
Fe'
EA
Fa'*
818 BOOK IV
Ex. 1683. On inch-paper, mark the points O (0, 0), P (3, 0), Q (5, 2),
R (4, 6), S (1, 4); join OP, PQ, QR, RS, SO. On plain paper, draw
0'P'= 1-5 in. ; on O'P' describe a similar polygon. Check by measuring the
angles and finding the ratios of corresponding sides. (Keep your figures for
the next exercise.)
Ex. 1683. On inch-paper, describe a polygon similar to OPQRS
of Ex. 1682, having its base O'P' =1*5 in. Do this by halving the co-
ordinates of the points O, P, Q, R, S. Make a copy on tracing paper of the
smaller polygon obtainec} in Ex. 1682, and compare with the polygon
obtained in the present exercise.
Ex. 1684. On inch-paper, mark the points A (1, 0), B (4, 0), 0 (1, 3),
D (8, 4) ; join AB, BC, CD, DA. On plain paper draw A'B' = 2-5 in.; on
A'B' describe a figure similar to ABCD. Check by calculating the ratios of
corresponding sides.
Ex. 168S. Draw a pentagon ABODE; draw A'B' || to AB; on A'B'
construct a pentagon similar to ABODE. (This should be done with set-
square and straight edge only.)
Revise Ex. 146.
Ex. 1686. Draw four parallel lines AP, BQ, OR, DS ; draw two straight
lines ABCD, PQRS to cut them. "With AB, BC, CD as sides, describe
a triangle ; with PQ, QR, RS describe a triangle. Measure and compara
the angles of the two triangles.
similar triangles 319
Theorem 4.
[Converse of Theorem 3.]
BC CA AB
If; in two triangles abc, def, ^ = e^= gf > t^®^ *^®
triangles are equiangular.
A
B c
X
fig. 312.
Const/ruction Make z. FEX = /. B and L EFX = z. C, X and D being
on opposite sides of EF.
Proof In the As ABC, XEF,
f ^ B = ^ FEX,
t^C = ^EFX,
.'. the third angles are equal,
and the triangles are equiangular.
BC _ CA _ AB
" EF~ FX~ XE'
IV. 3.
But — = — = — , D(Ua
EF FD DE'
CA _ CA 1 AB _ AB
■ • FX ~ FD ^'^ XE ~ DE '
.'. FX = FD and XE = DE.
In the As XEF, DEF,
j'XE=DE,
I FX = FD,
(. and EF is common,
.-. AXEFsADEF. • I. 14.
But the As ABC, XEF are equiangular,
.'. the As ABC, DEF are equiangular.
a E. D.
320
BOOK IV
tEx. 1687. Draw a quadri-
lateral ABCD ; join AC. Make
an angle XOY; from OX out off
OP = AB, OQ=BC, OR = CD,
OS = DA, OT = CA; through P,
Q, ... draw a set of parallel lines
cutting OY in P*, Q', .... Con-
struct a quadrilateral A'B'C'D'
having A'B' = OP', B'C'=OQ'
Prove and verify that ABCD and
A'C'C'iy are equiangular.
The diagonal scale (fig.
313), depends in principle on
the properties of similar tri-
angles.
HEx. 1688. Are the triangles
•whose comers are marked 0, d, 10
and 0, e, 6 equiangular?
TEx. 1689. What fraction is
the distance between the points 6,
c of the distance between 10, d ?
The distance between 10, d is
*1 in. ; what is the distance betwe^
6,c?
liEx. 1690. What are the dis-
tances between the points (i) a, 6,
(ii) 6, c, (iii) c, 6?
What is the whole distance be-
tween a, 6 ?
ITEx. 1691. Draw a triangle
ABC; make an ^XOY=^A; from
OX, OY cut off OP = §AB, 0Q =
§ AC ; join PQ; measure Z.' P, Q,
and compare them with A* B, C.
—
1 —
=
—
O^
}
-^
.6,
K>
*■
a
OB
e
o
fig. 313.
similar triangles 321
Theorem 5.
If two triangles have one angle of the one equal to
one angle of the other and the sides about these equal
angles proportional, the triangles are similar.
fig. 314.
Data ABC, DEF are two triangles which have Lt<= z. D, and
AB AC
DE~DF'
To prove that the As ABC, DEF are similar.
Construction From AB cut off AH = DE.
From AC cut off AK = DF.
Join HK.
Proof In As AHK, DEF,
j- AH = DE, Cmiatr.
J AK = DF, Cmifst/r.
[ z. A = /. D, Data
.*. AAHKsADEF. L 10.
,T. AB AC
Now — = — .
DE DF
. AB _ AC
*' AH ~ AK'
.-. HK is II to BC. JV. 2, Cor. 1.
z. H = ^ B and /. K = z. C,
.'. As AHK, BCA are equiangular.
Hence As DEF, ABC are equiangular,
and therefore have their corresponding sides
proportional, iv. 3.
.*. As DEF, ABC are similar. Q. E. D.
322
BOOK IV
Note. In iv. 3 and 5, if DE>AB and DF>AC, H, K lie in
AB, AC produced; the proofs hold equally well for these cases.
tEx. lesfl. S is a point in the side PQ of A PQR ; ST is drawn
parallel to QR and of saoh a length that ST : QR = PS : PGL Prove that
T lies in PR.
[Prove Z.SPT = ^QPR.]
Ex. 1698. (Inch paper.) Prove that the pomts (0, 0), (2, 1), (6, 2-5)
axe in a straight line. In what ratio is the line divided?
tEx. 1604. In a triangle ABC, AD is drawn perpendicular to the hase ;
if BD : DA = DA : DC, prove that a ABC is right-angled.
tEx. 1605. AX, DY are medians of the two similar triangles ABC, DEF ;
prove that they make equal angles with BC, EF, and that AX : DY= AB : DE.
(Compare Ex. 411.)
tEx. 1606. The bases, BC, EF, of two similar triangles, ABC, DEF,
are divided in the same ratio at X, Y. Prove that AX : DY = BC : EF.
Fig. 315 represents a pair of propor-
tional compasses. AB = AC and AH = AK,
AH AK
.•. — = — , and I. BAG = z. HAK,
AB AC
.*. A' ABC, AHK are similar.
HK AH
Hence — — — , which is constant for any
BC AB' ^
fixed position of the hinge. In fig. 315 the
AH
hinge is adjusted so that — = ^; thus, what-
ever the angle to which the compasses are
opened, HK = ^ BC.
fig. 316.
AKEAS OF SIMILAR TRIANGLES
323
ITEx. 1697. On bases of 6 in. and 3 in. describe two similar triangles;
calculate their areas, and find the ratio of their areas. Is it 5 : 3 ?
What is the ratio of their altitudes ?
A|f — ^ — A — I — A-i
fN J M
A'
^~~
'B'
E'
F'
fig. 316.
In fig. 316,
AABD = ^ ||°'P^ ABCD, and AA'B'D' = | ||08ram a'B'C'D'.
The parallelograms ABCD, A'B'C'd' are divided up into con-
gruent parallelograms; the squares are divided up into congruent
squares.
. AABD _ |ABCD _ ABCD _ 25 small \\°«^^' _ 25
■ * AA'B'D' ~ |A'B'C'D' ~ A'B'O'D' "" 9 small \\'>«'^^ ~ "9 '
But
sq. AE _ 25 small squares 25
sq. A'E' 9 small squares 9 '
AABD square on AB
A A'B'd' ~ square on A'B' '
tEz. 1698. The ratio of corresponding altitudes of similar triangles
is equal to the ratio of corresponding sides.
324
BOOK IV
Theorem G.
The ratio of the areas of similar triangles is equal
to the ratio of the squares on corresponding sides.
Data
To prove that
Construction
Proof
D ^ Y W Z
fig. 317.
ABC, XYZ are two similar triangles.
AABC
YZ^'
to BC,
AXYZ
Draw AD
and XW ± to YZ.
AABC = ^BC. AD,
and AXYZ = |YZ.XW,
AABC BC.AD
II. 2.
YZ . XW *
It remains to prove that
V i
{
AXYZ
AD BC
XW "^ YZ
/ Now in the As ABD, XYW,
il. B = Z. Y,
A. D = /- W (rt. /. s),
". the third angles are equal,
and the A s are equiangular,
AD _ AB
*' XW ~XY'
AB BC
XY ~YZ'
AD BC
XW ~YZ'
]•
But
Data
IV. 3.
Data
AREAS OF SIMILAR TRIANGLES 325
T^ ^ AABC BC AD
isut = — . ■ Iroved
AXYZ YZ XW
BC BC
YZ ■ YZ
BC*
YZ-'
Q. R. D.
^Ex. 1690. What is the ratio of the areas of two similar triangles on
bases of 3 in. and 4 in.?
HEx. 1700. The area of a triangle with a base of 12 cm. is 60 sq. cm. ;
find the area of a similar triangle with a base of 9 em.
What is the area of a similar triangle on a base of 9 in.?
HEx. 1701. The areas of two similar triangles are 100 sq. cm. and
64 sq. cm. ; the base of the greater is 7 cm. ; find the base of the smaller.
Ex. 1702. The areas of two similar triangles are 97'5 sq. cm. and
75*3 sq. cm. ; the base of the first is 17*2 cm. ; find the base of the second.
lIEx. 1703. The sides of a triangle ABC are 7-2 in., 3-5 in., 5'7 in. ; the
sides of a triangle DEF are 7*2 cm., 3*5 cm., 5*7 cm. ; find the ratio of the
area of the first triangle to that of the second.
HEx. I704. Find the ratio of the bases of two similar triangles one of
which has double the area of the other.
Show how to draw two snch triangles, without using a graduated ruler.
Ex. 1705. Describe equilateral triangles on the side and diagonal of a
square; find the ratio of their areas. {Freehand.)
Ex. 1706. Show how to draw a straight line parallel to the base of
a triangle to bisect the triangle.
326 BOOK IV
Ex. X707. Describe equilateral triangles on the sides of a right-angled
triangle whose sides are 1*5 in., 2 in., 2-5 in. What connection is there
between the areas of the three equilateral triangles ? (Freehand)
tEx. 1708. Prove that, if similar triangles are described on the three
Bides of a right-angled triangle, the area of the triangle described on the
hypotenuse is equal to the sum of the other two triangles.
tEx. 17O0. ABC, DEF are two triangles in which ^ B = ^ E ; prove that
A ABC : A DEF=AB . BC : DE . EF.
[Draw AX 1 to BC, and DY X to EF.]
Ex. 1710. What is the ratio of tlie areas of two circles whose radii are
R, r? 3in., 2in.?
^Ex. 1711. Draw two similar quadrilaterals ABCD, PQRS; calculate
their areas (join AC, PR) ; find the ratio of their areas, and compare this
with the ratio of corresponding sides.
R-BCTANOLE PROPERTIES.
fEx. 1713. XYZ is a triangle inscribed in a circle, XN is an altitude of
XY YD
the triangle and X D a diameter of the circle ; prove that rrw. = rr^ • Express
XN NZ
this as a result clear of fractions. What two rectangles are thus proved equal ?
fEx. 1718. With the same construction as in Ex. 1712, prove that
XZ.NY=XN. ZD.
[Ton will have to pick out two equal ratios from two equiangular
triangles. If you colour XZ, NY red and XN, ZD blue you will see which
are the triangles.]
fEx. 1714. ABCD is a quadrilateral inscribed in a circle ; its diagonals
intersect at X. Prove that (i) AX . BGs AD . BX, (ii) AX . XC» BX . XD.
RECTANGLE PROPERTIES 327
tEz. 1715. ABCD is a qaadrilateral inscribed in a circle; AB, DC pro-
duced intersect at Y. Prove that
(i) YA.BD = YD.CA, (ii) YA . YB = YC . YD.
tEx. 1716. The rectangle contained by two sides of a triangle is equal to
the rectangle contained by the diameter of the circumcircle and the altitude
drawn to the base.
[Draw the diameter through the vertex at which the two sides intersect.]
tEx. 1717. The bisector of the angle A of a ABC meets the base in P
and the tsircumcircle in Q. Prove that the rectangle contained by the sides
AB, AC=rect. AP . AQ.
tEx. 1718. In Ex. 1680, prove that PQ . SR = PR . TQ.
tEx. 1719. The sum of tbe rectangles contained by opposite sides
of a cyclic quadrilateral Is equal to tlie rectangle contained by Its
diagonals. (Ptolemy's theorem.)
[Use the construction of Ex. 1680.]
ITEx. 1720. Draw a circle of radius 7 cm.; mark a point P 3 cm. from
the centre O; through P draw five or six chords A PB, CPD, .... Measure
their segments and calculate the products PA.PB; PC.PD;.... Take
the mean of your results and estimate by how much per cent, each result
differs from the mean. (Make a table.)
ITEx. 1731. Draw a circle of radius 7 cm. and mark a point P 10 cm.
from the centre O ; through P draw a number of chords of the circle, and
proceed as in Ex. 1720.
[Remember that if P is in the chord AB produced, PA, PB are still
regarded as the segments into which P divides AB ; you must calculate
PA . PB, not PA . AB.]
HEx. 1732. What will be the position of the chord in Ex. 1721 when the
two segments are equal ?
328
BOOK IV
Theorem 7 (i).
If AB, CD, two chords of a circle, intersect at a point P
inside the circle, then pa . pb = PC . PD.
fig. 318.
Construction
Proof
Join BC, AD,
In the As PAD, PCB,
L APD = L CPB (vert, opp.)
/. B = z. D (in the same segment),
.*. the third angles are equal,
and the As are equiangular.
PA
PC
PD
PB'
.•. PA . PB :^ PC . PD.
To calculate the area of the rectangle
PA . PB in IV. 7 (i).
Suppose EPF is the chord bisected at P.
Then PA . PB = PE . PF = PE^ = OE^ - OP^.
IV. 3.
fig. 319.
RECTANGLE PROPERTIES 329.<
Theorem 7 (ii).
If AB, CD, two chords of a circle, intersect at a point P
outside the circle, then pa . pb = PC . pd.
fig. 320.
Construction
Join BC, AD.
Proof
In the As PAD, PCB,
r z. P is common,
\lB = lD (ia the same segment)
.'. the third angles are equal,
and the As are equiangular,
PA PD
*• PC"" PB'
IV. 3.
PA . PB = PC . PD.
^. E. D.
Note. Theorems 7 (i) and 7^ (ii) are really two different
cases of the same theorem; notice that the proofs are nearly
identical. For alternative proofs, not depending on similarity,
see Appendix I, pages 354, 355.
G. s. IL 22
^30
hooft 1^^
tEx. 17aa. If CT is a tangent to a elreU
and AB a chord of tba drela passing tlirongb
P, then Pr»= PA . PB. (See fig. 321.)
To calculate the area of the rect-
angle PA. PB in IV. 7 (ii).
TJse the fact that
PA . PB = Pr» = OP» - OT".
ITEx. 1724. What becomes of iv. 7 when P in
a point on the circle ? When P is the centre ?
fig. 321.
Ex. 1726. Calculate (and check graphically) the areas of the rectangles
contained by the segments of chords passing through P when (i) r=5in.,
OP=3 in., (ii) r=5 cm., OP = 13 cm., (iii) r=3-7 in., OP = 2-3 in.,
(iv) r=2-9 in., OP =33 in.
Ex. 1726. Find an expression for the areas in Ex. 1725, r being the
radius, and d the distance OP (i) when d<r, (ii) when d>-r. Explain fully.
Ex. 1727. Draw two straight lines APB, CPD intersecting at P;
make PA = 4cm., PB = 6cm., PC = 3cm. Describe a circle through ABC,
cutting OP produced inT). Calculate PD, and check by measurement.
What would be the result if the exercise were repeated with the same
lengths, but a different angle between APB, CPD?
Ex. 1728. From a point P draw two straight lines PAB, PC; make
PA =4 cm., PB = 9cm., PC = 6 cm. Describe a circle through ABC; let it
cat PC again at D. Calculate PD, and check by measurement.
+Ex. 1729. APB, CPD intersect at P; and the lengths PA, PB, PC,
PD are so chosen that PA . PB= PC . PD. Prove that A, B, C, D are con-
cyclic. (Draw G through ABC; let it cut CP produced in D'.) Make up
a numerical instance, and draw a figure. What relation does this exercise
bear to iv. 7 (i) ?
tEx. 1730. State and prove the converse of rv. 7 (ii).
tEx. 1731. P is a point outside a circle ABC and straight lines PAB, PC
are drawn (A, B, C being on the circle) ; prove that, if PA . PB= PC, PC is
the tangent at C.
[Use reductio ad absurdum.']
RECTANGLE PROPERTIES S3l
tEx. I9da. ABC id a triangle right-angled at A ; AD is drawn perpen-
dicular to BC ; prove that AD^s BD . DC.
[Produce AD to cut the circumcircle of A ABC]
tEx. 1733. If the common chord of two intersecting circles be produced
to any point T, the tangents to the circles from T are equal to one another.
tEx. 1734. The common chord of two intersecting circles bisects their
common tangents.
tEx. 1735. The altitudes BE, CF of a triangle ABC intersect at H,
prove that
(i) BH . HE=CH . HF, (ii) AF . AB=AE . AG, (iii) BH . BE=BF . BA.
tEx. 1786. Two circles intersect at A, B; T is any point in AB, or AB
produced; TCD, TEF are drawn cutting the one circle in C, D, the other
in E, F. Prove that C, D, E, F are concyclic.
Ex. 1737. ABC ia a triangle right-angled at A; AD is an altitude
of the triangle. Prove that a' ABD, CDA are equiangular. Write down the
three equal ratios; and, by taking them in pairs, deduce the corresponding
rectangle properties.
Def. If X is such a quantity that a : x = x :b, then x is
called the mean proportional between a and 6.
ITEx. 1738. Prove that, if a; is the mean proportional between a and 6,-
II Ex. 1739. Find the mean proportional between
(i) 4 and 9, (ii) 1 and 100, (iii) J and 2,
(iv) f and |, (v) 1 and 2, (vi) 2 and i.
22—2
332
BOOK IV
To find the mean, proportional between two given
straight lines.
Proof
Let a, 6 be the two given straight liiitw.
Construction Draw a straight line PGL
From PQ cut off PR = a, and RS = h.
On PS as diameter describe a semicircle.
Through R draw RT x to PS to cut the semicircle at T.
Then RT (x) is the mean proportional between a, h.
Join PT, TS.
A* PRT, TRS are equiangular. (Why?)
.•. RP : RT= RT : RS,
.". a : flj = a; : 6,
.*. X Ls the mean proportional between a and 6.
tEx. 1740. Prove the above construction by completing the circle, and
producing TR to meet the circle in T'.
Ex. 1741. (On inch paper.) Find graphically the mean proportionals
between (i) 1 and 4, (ii) 1 and 3, (iii) 1-5 and 2-5, (iv) 1-3 and 1-7.
Check by calculation.
Note. If -=tj (x^ — ab, and therefore x= Jah; thus the
X o
mean proportional between two numbers is the square root of the
product.
Ez. 1742. (On inch paper.) Find the square roots of (i) 2, (ii) 3, (iii) 6,
(iv) 7.
[Find the mean proportionals between (i) 1 and 2, (iii) 2 and 3.]
Ex. 174S. Draw a triangle; and constrxtan equivalent rectangle.
[What is the formula for the area of a triangle ?]
MEAN PROPORTIONAL 333
To describe a square equivalent to a given rectilinear
figure.
Construction (i) Reduce the figure to a triangle (see p. 178).
(ii) Convert the triangle into a rectangle,
(iii) Find the mean proportional between the sides
of the rectangle.
This will be the side of the required square.
Proof If a, h are the sides of the rectangle, x the side of the
equivalent square, then
area of rectangle = ah =a?.
Ex. 1744. (On inch paper.) Find the side of the square equivalent to
the triangles whose angular points are
(i) (1, 0), (5, 0), (4, 3),
(ii) (0, 0), (0, 2), (5, 0-5),
(iii) (0, 0), (3, 1), (2, 3).
Ex. 1745. Constroct a square equivalent to a regular hexagon of
side 2 in, ; measure the side of the square.
Ex. 1746. Bepeat Ex. 1745 for a regular octagon of side 2 in.
Ex. 1747. Find the side of a square equivalent to the quadrilateral
A BCD, when
(i) DA=1 m., Z.A=100°, AB=2-3 in., ^8 = 64°, BC = l-5 in.
(ii) AB=5-7 cm., BC=5-2 cm., CD = l-7 cm., DA=3-9 cm., Z.A=76°.
tEx. 1748. In fig. 322, prove that (i) PTa= PR . PS, (ii) S-n=SR . SP.
tEx. 1749. Prove Pythagoras' theorem by drawing the altitude to the
hypotenuse and using similar triangles (see Ex. 1748).
liEx. 1750. Draw a large scalene triangle ABC; draw the bisector of
j.tK and let it cut BC at D. Calculate AB : AC and DB : DC.
HEx. 1761. Bepeat Ex. 1750 with a triangle of different shape.
ITEx. 1752. Draw a large scalene triangle ABC; draw the bisector of
the external angle at A; let it cut the base produced at D. Calculate
AB : AC and the ratio in which D divides the base BC (see p. 804).
UEx. 1753. Bepeat Ex. 1^5^ with a triangle of different shape.
334
BOOK IV
Theohem 8 (i).
The internal bisector of an angle of a triangle divides
the opposite side internally in the ratio of the sides
containing the angle.
Data
fig. 323.
ABC is a triangle,
AD bisects l BAG internally and cuts BC at D.
To prove that — = — .
^ DC AC
Construction Through C draw CE || to DA to cut BA produced
at E.
Proof Since DA is || to CE,
. DB _ AB
" DC ~ AE '
[It remains to prove that AE = AC].
•/ DA is 11 to CEj
.■. L BAD =: corresp. z. AEC,
and L DAC= alt. l ACE.
But L BAD= L DAC,
.'. Z.AEC=z.ACE,
.•.AE = AC,
DB _ AB
" DC ~AC*
IV. 1.
I. 5.
I. 5.
Data
I. 13.
tEx. 1764. State and prove the converse of this theorem.
[Use rednctio ad dbsurdum.^
Q. K. D.
BISECTOR OF ANGLE OF TRIANGLE
335
Theorem 8 (ii).
The external bisector of an angle of a triangle divides
the opposite side externally in the ratio of the sides
containing the angle.
F
fig. 824.
Daia ABC is a triangle,
AD bisects z. BAG externally (i.e. AD bisects l CAF) and
cuts BC produced at D.
m ,1 , DB AB
To prove t/iat — = — .
^ DC AC
Construction Through Cdraw CE || to DA to cut BA at E.
Proof Since DA is || to CE,
. DB _ AB
" DC~ AE'
[It remains to prove that AE = AC].
/ •.• DA is II to CE,
■ .'. L FAD = corresp. l. AEC,
and L. DAC = alt. i. ACE.
But z. FAD = z. DAC,
.-. z.AEC = z.ACE,
.'. AE = AC,
DB AB
■* DC" AC"
IV. 1.
1.5.
1.5.
Data
L 13.
Q. E. D.
Note. There is a very close analogy between theorems 8 (i)
and 8 (ii) ; notice that the proofs are nearly identical.
tEx. 1755. State and prove the converse of this theorem.
386 BOOK IV
Ex.1756. In a aABC, BC = 3-5in., CA = 3in., AB = 4in. and the
internal bisector of /.A cuts BC at D; calculate BD, DC; check by
drawing.
Ex. 1767. The internal bisector of /. B of A ABC cuts the opposite
side in E; find EA, EC when BC = 8-9cm., CA = ll-5cm,, AB = 4-7cm.
Ex.1768. In a aABC, BC = 3-5in., CA = 3in., AB = 4in. and the
external bisector of ^A cuts the base produced at D; calculate BD, DC.
Ex. 1769. Bepeat Ex. 1758 with
(i) BC = 5-2in., CA=4-lin., AB=4-5in.,
(ii) BC = ll-5cm., CA = 4-7cm., AB=8-9cm.
Ex. 1760. Calculate the distance between the points in which AB,
a side of a A ABC, is cut by the bisectors of'/.C, having given that
BC=6-9cm., CA = ll-4cm., AB = o-8cm.
tEx. 1761. The base BC of a triangle ABC is bisected at D. DE, DF
bisect /L" ADC, ADB, meeting AC, AB in E, F. Prove that EF is || to BC.
tEx. 1762. Prove that the bisectors of an angle of a triangle divide
the base internally and externally in the same ratio.
Ex. 1763. The internal and external bisectors of the ^P of a aPQR
cut the base at X, Y respectively; what is Z.XPY?
tEx. 1764. A itoint P moves so that the ratio of its distances £roin
two fixed points Q, R is constant; prove that the locns of P is a
circle. (The Circle of Apollonlns.)
[Draw the internal and external bisectors of Z.P, and use Ex. 1763.]
tEx. 1766. O is a point inside a triangle ABC. The bisectors of
Z.8 BOC, CCA, AOB meet BC, CA, AB in P, Q, R respectively. Prove that
BP CQ AR
PC'^QA^RB"^-
Bevise Ex. 1651.
SIMILAR POLYGONS
337
Theorem 9. t
If the straight lines joining a point to the vertices of a
given polygon are divided (all internally or all externally)
in the same ratio the points of division are the vertices
of a similar polygon.
:~-5^o
fig. 325. fig. 326,
DcUa ABODE is a polygon ; the st. lines joining a pt. O to
its vertices are all divided in the same ratio at A', B', C',
D', E'.
To prove that Z. A = Z. A', Z. B = A B', ... ,
A'B' B'C'
and
AB
BC
in the same
Proof Since OA, OB, ... are divided at A', B',
ratio, it follows that
In the As OA'b', OAB
OA' _ OB^
■ OA ""ob'
and z. AOB is common,
.*. As OA'B', OAB are similar,
.'. L OA'B' = L OAB.
Sim'J^ L. OA'E' = L. OAE,
. .'. ^B'A'E'=^BAE.
Sim'y the other z. s of A'B'C'D'E' are equal to the corre-
, sponding /. s of ABODE.
IV. 5.
338
BOOK IV
fig. 325. fig. 326.
^ Again, since As OA'b', OAB are similar,
■ ' AB ~ OA ~ ■
Sim'J' , , . , , , each = k,
BC * CD ' ' '
a;b'
AB
E2-
BC
CD'
'.ABODE, A'B'C'D'E' are similar.
Proved
Q. E. D.
NoTK This theorem is the principle of the magic lantern;
every part of the figure is magnified outwards from a point.
HEx. 1766. Draw a figure to show that equiangolar pentagons are not
necessarily similar.
^Ex. 1767. Draw a figure to show that a pentagon whose sides taken
in order are halves of the sides of another pentagon is not necessarily similar
to the other pentagon.
IfEz. 1768. A rectangular picture frame is made of wood 1 in. wide ; are
the inside and outside of the frame similar rectangles ?
TEx. 1769. Draw a figure for iv. 9 for the case in which O coincides
with B.
ITEz. 1770. Draw a figure for iv. 9 for the case in which O is on AB.
tEx. 1771. Assuming that the polygons ABODE in figs 325, 326 are
congruent, and that the ratio of division is the same for the two figures, prove
that the two polygons A'B'O'D'E' are congruent.
SIMILAR POLYGONS 339
On a given straight line to construct a figure similar
to a given rectilinear figure. (Second Method.) t [See
p. 317.]
D
fig. 327.
Construction Let ABODE be the given figure, A'B' the given,
straight line (see figs. 325, 326, 327.)
Place A'B' parallel to AB, and produce AA', BB' to meet
at O; join OC, OD, OE.
Divide OC, OD, OE at C', D', E* in the same ratio as OA
and OB are divided. [This is most easily done by drawing
parallels.]
Join B'C', C'D', D'E', E'A'.
Then A'B'c'd'E' is similar to ABODE. iv. 9.
Note. The method (used in Ex. 1683) of dividing coordinates
in a given ratio is substantially the same as the above.
Ex. 1772. (On inch paper.) Mark the pomts (0, 0), (3, 0), (8, 3), (1, 4),
(0, 3) ; join them in order. On the line (1, 1), (2, 1) describe a similar
pentagon by the method just explained. From your figure, read off the
coordinates of its vertices.
Ex, 1773. Eepeat Ex. 1772, with (0, 0), (-5, OJ, (-7, -3), (-1, -6) as the
comers of the given figure, and (1, 1), (3-2, 1) as the ends of the given line.
Ex. 1774. Bepeat Ex. 1772, with (-2, -2), (2, -2), (3, 3), (-1, 2) as
the comers of the given figure, and (-1, -1), (1, -1) as the ends of the
given line.
Ex. 177S. (On inch paper.) Draw the triangle ABO, A (2, 0), B (2, 3),
O (0, 1) ; on PQ, P (3, 3), Q (3, 0-2), as base describe a triangle similar to
A ABC. Find the coordinates of the vertex.
[Take O as the point of intersection of AP, BQ.]
840
BOOK IV
Theorem 10. +
If a polygon is divided into triangles by lines joining
a point to its vertices, any similar polygon can be
divided into corresponding similar triangles.
Data ABODE, PQRST are two equiangular polygons which
have
PQ _ OR _ RS _ ST _ TP_ ,
AB " BO ~ CD ~ DE ~ EA " ^^^''
ABODE is divided into As by lines joining its vertices to
a pt. O.
To prove that there is a point X such that the As formed by
joining X to the vertices of PQRST are similar to the corre-
sponding As into which ABODE is divided.
Construction Divide OA, OB, ... at A', B', ... so that
OA'_OB^_
0A~ 0B~-~'^
Join A'B', B'O'
SIMILAR POLYGONS 341
'' Proof [First to jMX)ve A'B'C'D'E , PQRST congruent].
OA' OB'
Since — = — = .... (Joftistr.
OA OB
As OA'B', OB'C', ... are similar to As OAB, OBC, ...
respectively. Prwed in iv, 9.
For the same reason A'B'C'D'E' is equiangular to
ABODE, IV. 9.
but ABODE is equiangular to PQRST, Dctta
PQRST.
IV. 9.
A'B' _ PQ
■ ■ AB ~ AB*
.-. A'B' = Pa
Sim'y B'O' = QR, O'D' = RS, . . . ,
.*. .A'B'O'D'E', PQRST have all their corresponding angles
and sides equal and are therefore congruent.
Apply A'B'C'D'E'O to PQRST; since A'B'O'D'E', PQRST
are congruent, they must coincide; let X be the point on
which O falls.
Join XP, XQ, ....
Then AXPQ= AOA'B'.
But As OA'B', OAB are similar,
.". As XPQ, OAB are similar.
Likewise the other pairs of corresponding A s in the two
.polygons are similar.
<j. K. D.
A'B'O'D'E'
»
»
Again,
since
OA'
OB'
OA
~ OB
"•■
A'B'
OA'
= k.
AB
OA
PQ
-K
342 BOOK IV
KoTE. The praotioal way to And the point X is to make
L. QPX = L BAO and L. PQX = L ABO.
O and X are called corresponding points.
CoK. If in two similar figures whose sides are in the
ratio 1 : h, Oj, o, correspond to x^, Xj, then OiOj : XjX, =- 1 : ^.
Ex. 1776. (Inch paper.) O is a point inside a triangle ABC. A is
(-.S, 3), B is (-2, -1), C is (2, -2), O is (-1, 0). PQR is a similar
triangle; P is (-1-5, 1-5), Gl is (-1, -0-5), R is (1, -1). Find the co-
ordinates of the point X which corresponds to O.
tEx. 1777. O is the circumcentre of a ABC ; X is the "corresponding"
point in a similar triangle PQR. Prove that X is the circumcentre of
A PQR,
Ex. 1778. Construct aABC, given iLA=:70°, Z.B=46°, /.C=65°, and
altitude AD= 8 cm. Measure BC.
[First construct aA'B'C having its angles equal to the given angles;
draw the altitude A'D'. Magnify a ABC in the ratio AD : A'D'.]
Ex. 1770. Construct aABC, given ^A=45°, Z.B=25°, ^C = 110°, and
median BM =7*5 cm. Measure BC. [See note to Ex. 1778.]
Ex. 1780. Show how to describe a triangle, having given its angles
and its perimeter. ^
Ex. 1781. Show how to describe a triangle, having given its angles
and the difference of two of its sides.
Ex. 1782. Show how to inscribe in a given triangle a triangle which
has its sides parallel to the sides of a given triangle.
Ex. 1783. Show how to inscribe a square in a given triangle.
Ex. 1784. Show how to inscribe a square in a given sector of a circle.
Ex. 1786. Show how to inscribe an equilateral triangle in a given
triangle.
Ex. 1786. Show how to describe a circle to touch two given straight
lines and pass ' through a given point.
Ex. 1787. Show how to inscribe a regular octagon in a given square.
similar polygons 343
Theorem 11.+
The ratio of the areas of similar polygons is equal to
the ratio of the squares on corresponding sides.
R
Data ABODE, PQRST are two similar polygous; let ^ = ^•
area of ABODE AB'
^ area of PQRST PQ'
Construction In ABODE take any point O.
Let X be the corresponding point in PQRST. iv. 10.
Join OA, OB, ... ; XP, XQ, ....
Proof Since O, X are corresponding points,
.'. As OAB, XPQ are similar,
. AOAB _ AB2
'*A^PQ~P^~
AXQR QR'*
.'. AOAB=^.AXPQ,
and AOBO = P.AXQR,
AOAB + AOBO + ...=Ar'{AXPQ + AXQR+ ...},
.-. ABODE = A* . PQRST,
ABODE , AB*
PQRST ~ " PQ'' '
Q. K. D.
344 BOOK IV
HEx. 1788. "What is the ratio of the area of a room to the area by which
it is represented on a plan whose scale is 1 in. to 1 ft.?
HEx. 1780. On a map whose scale is 1 mile to 1 in., a piece of land is
represented by an area of 20 sq. in. ; what is the area of the land ?
IfEx. 170O. On a map whose scale is 2 miles to I in., a piece of land is
represented by an area of 24 sq. in. ; what is the area of the land ?
Ex. 1791. What is the acreage of a field which is represented by an
area of 3 sq. in. on a map whose scale is 25 in. to the mile ? (640 acreu
= 1 sq. mile.)
Ex. 1793. What areas represent a field of 1 acre on maps in which
1 mile is represented by (i) 1 in., (ii) ^ in., (iii) 6 in., (iv) 2-5 in. ?
If the field were square, what would be the length of a line representing
a side of the field ?
Ex. 1798. Two similar windows are glazed with small lozenge-shaped
panes of glass, these panes being all identical in size and shape. The
heights of the windows are 10ft. and 15 ft. The number of panes in the
smaller window is 1200; what is the number in the larger?
Ex. 1794. A figure described on the hypotenuse of a right-angled-
triangle is equal to the sum of the similar figures described on the sides
of the triangle. (This is a generalisation of Pythagoras' theorem.)
Ex. 1795. Similar figures are described on the side and diagonal of
a square ; prove that the ratio of their areas is 1 : 2.
Ex. 1796. Similar figures are described on the side and altitude of
an equilateral triangle; prove that the ratio of their areas is 4: 3.
To construct a figure equivalent to a given figure A
and similar to another figure B. t
Construction Reduce both figures to squares (see p. 333).
Let a and b be the sides of these squares.
Let ^ be a side of the figure B.
SIMILAR POLYGONS 345
Construct a length x so that b : a = l : x.
On X describe a figure C similar to B; the side a; of C
corresponding to the side I of B.
Proof The area of C : area of B = a:^ : P
= area of A : area of B,
.'. area of C = area of A.
Ex. 1797. Show how to construct an equilateral triangle equivalent
to a given square.
Ex. 1798. Show how to construct an equilateral triangle equivalent
to a given triangle.
Ex. 1799. Show how to construct a rectangle having its sides in a
given ratio and equivalent to a given square.
MISCELLANEOUS EXERCISES.
+Ex. 1800. One of the parallel sides of a trapezium is double the other;
show that the diagonals trisect one another.
fEx. ISOl. A straight line drawn parallel to the parallel sides of a
trapezium divides the other two sides (or those sides produced) pro-
portionally.
fEx. 1802. Find the locus of a point which moves so that the ratio of
its distances from two intersecting straight lines is constant.
Ex. 1803. Show how to draw through a given point within a given
angle a straight line to be terminated by the arms of the angle, and divided
in a given ratio (say ^) at the given point.
+Ex. 1804. Prove tbat two medians of a triangle trisect one
another. Hence prove that the three medians pass throngh one point.
fEx. 1805. The bisectors of the equal angles of two similar triangles are
to one another as the bases of the triangles.
G. s. II. 23
34^ BOOK IV
tEx. ia06. In two similar triangles, the parts lying within the
triangle of the perpendicular oisectors of corresponding sides have the
same ratio as the corresponding sides of the triangle.
tEx. 1807. ABC, DEF are two similar triangles ; P, Q. are any two
points in AB, AC; X, Y are the corresponding points in DE, DF. Prove
that PGl:XY=AB:DE.
tEx. 1808. The sides AC, BD of two triangles ABC, DBC on the same
base BC and between the same parallels meet at E ; prove that a parallel to
BC through E, meeting AB, CD, is bisected at E.
Ex. 1809. Show how to divide a parallelogram into five equivalent
parts by lines drawn through an angular point.
tEx. 1810. Show how to divide a given line into two parts such that
their mean proportional is equal to a given line. Is thi? always possible ?
Ex. 1811. Show how to construct a rectangle equivalent to a given
square, and having its perimeter equal to a given line. [See Ex. 1810.]
tEx. 1813. A common tangent to two circles cuts the line of centres
externally or internally in the ratio of the radii.
Ex. 1813. Show how to construct on a given base a triangle having
given the vertical angle and the ratio of the two sides.
Ex. 1814. Show how to construct a triangle havmg given the vertical
angle, the ratio of the sides containuag the angle, and the altitude drawn to
the base.
tEx. 1815. TP, TQ are tangents to a circle whose centre is C, CT cuts
PQ in N ; prove that CN . CT=CP2.
tEx. 1816. In fig. 318, prove that A PBC : A PAD = BC« : AD«. Is the
same property true for fig. 319 ?
tEx. 1817. In fig. 318, prove that PB .PC : PA . PD = BC^: AD^.
tEx. 1818. ABCDE is a regular pentagon; BE, AD intersect at F;
prove that EF is a thurd proportional to AD, AE.
Ex. 1819. In fig. 295, the area of the regular hexagon obtained by
joining the vertices of the star is three times that of the small hexagon.
tEx. 1820. In fig. 320, PQ is drawn paraUei to AD to meet BC pro-
duced in Q.; prove that PQ is a mean proportional between QB, QC.
MISCELLANEOUS EXERCISES 347
tEx. 1821. The angle BAG of a a ABC is bisected by AD, which cuts
EC in D ; DE, DF are drawn parallel to AB, AC and cut AC, AB at E, F
respectively. Prove that BF : CE= AB^ : AC^.
+Ex. 1822. ABC is a triangle right-angled at A; AD is drawn perpen-
dicolar to BC and produced to E so that DE is a third proportional to
AD, DB ; prove that a ABD= aCDE, and aABD is a mean proportional
between a'ADC, BDE.
tEx. 1823. Two circles touch externally at P ; Q, R are the points of
contact of one of their common tangents. Prove that QR is a mean propor-
tional between their diameters.
[Draw the common tangent at P, let it .cut QR at S ; join S to the
centres of the two circles.]
tEx. 1824. Two church spires stand on a level plain ; a man walks on
the plain so that he always sees the tops of the spires at equal angles of
elevation. Prove that his locus is a circle.
tEx. 1825. The rectangle con^^ed by two sides of a triangle is equal to
the square on the bisector of the angle between those sides together with the
rectangle contained by the segments of the base. [See Ex. 1717.]
tEx. 1826. The tangent to a circle at P cuts two parallel tangents at
Q, R ; prove that the rectangle QP . PR is equal to the square on a radius of
the circle.
tEx. 1827. ABCD is a quadrilateral. If the bisectors of L* A, C meet
on BD, then the bisectors of /.' B, D meet on AC.
Ex. 1828, Prove the validity of the following method of solving a
qttadratic equation graphically: —
Suppose that ax^ -^ 6a; + c =0 is the equation ; on squared paper, mark
, the origin, from OX cut off OP = a, from P draw a perpendicular PQ
upwards of length h, from Q draw to the left QR = c (regard must be paid
to the signs of a, 6, c ; e.g. if h is negative PQ will be drawn downwards) ; on
OR describe a semicircle cutting PQ at S, T ; the roots of the equation are
PS , PT
"OP^-OP-
[C5onsider a» OPS, SQR.]
23—2
348 BOOK IV
Ex. 1839. Solve the following eqnations graphically as in Ex. 1828,
and check by calculation : —
(i) 2«« + 6x + 1 = 0,
(ii) a:" + 3x - 2 = 0,
(iii) 2x» - « + 1 = 0.
tEx. 1837. Find a point P in the aro AB of a circle such that chord AP
is three times the chord PB.
tEx. 1838. Show how to draw through a given point D in the side AB
of a triangle ABC a straight line DPQ cutting AC in P and BC produced
in Q so that PQ is twice DP.
tEx. 1830. A straight line AB is divided internally at C; equilateral
triangles ACD, CBE are described on the same side of AB; DE and AB
produced meet at F. Prove that FB : BC = FC : CA.
fEx. 1831. ABC is an equilateral triangle and from any point O in AB
straight lines DK and DL are drawn parallel to BC and AC respectively.
Find the ratio of the perimeter of the parallelogram DLCK to the perimeter
of the triangle ABC.
tEx. 1833. If from each of the angular points of a quadrilateral perpen-
diculars are let fall upon the diagonals, the feet of these perpendiculars are
the angular points of a similar quadrilateral.
fEx. 1833. ABCD is a parallelogram, P is a point in AC produced;
BC, BA are produced to cut the straight line through P and D in Q, R
respectively. Prove that PD is a mean proportional between PQ and PR,
tEx. 1834. ABCD is a quadrilateral inscribed in a circle of which AC
is a diameter; froni any point P in AC, PQ and PR are drawn perpendicular
to CD and AB respectively. Prove that DQ : PR = DC : BC.
tEx. 1836. Two circles ABC, ADE touch internally at A; through
A straight lines ABD, ACE are drawn to cut the circles. Prove that
AB . DE=AD . BC.
tEx. 1836. In the sides AD, CB of a quadrilateral ABCD points P, Q
are taken so that AP: PD = CQ;QB. Prove that a ADQ+ a BPC = ABCD.
tEx. 1830. Show how to draw through a given point O a straight line
to cut two given straight lines in P and Q respectively so that OP : PQ is
equal to a given ratio.
tEx. 1840. O is a fixed point inside a circle, P is a variable point on
the circle ; what is the locus of the mid -point of OP?
MISCELLANEOUS EXERCISES 349
tEx. 184 1. ABCD is a quadrilateral ; through A, B draw parallel straight
lines to cut CD in X, Y so that CX = DY. [X and Y are both to be between
C and D, or one in CD produced and the other in DC produced.]
fEx. 1842. Show how to construct a triangle having given the lengths
of two of its sides and the length of the bisector (terminated by the base) of
the angle between them.
tEx. 1848. From any point X in a chord PR of a circle, XY is drawn
perpendicular to the diameter PQ, prove that PX : PY = PQ : PR.
tEx. 1844. Through the vertex A, of a triangle ABC, DAE is drawn
parallel to BC and AD is made equal to AE; CD cuts AB at X and BE cuts
AC at Y ; prove XY parallel to BC.
tEx. 1845. ABCD is a parallelogram; a straight line through A cuts
BD in O, BC in P, DC in Q. Prove that AO is a mean proportional
between OP and OQ.
tEx. 1846. A triangle PQR is inscribed in a circle and the tangent to
the circle at the other end of the diameter through P cuts the sides PQ, PR
produced at H, K respectively; prove that the as PKH, PQR are similar.
tEx. 1847. Two circles ACB, ADB intersect at A, B; AC, AD touch
the circles ADB, ACB respectively at A; prove that AB is a mean pr6-
portional between BC and BD.
tEx. 1848. A variable circle moves so as always to touch two fixed
circles; prove that the straight line joining the points of contact cuts the
line of centres of the fixed circle in one of two fixed points.
tEx. 1849. ABC is an equilateral triEingle and D is any point in BC.
On BC produced points E and F are taken such that AB bisects the angle
EAD and AC bisects the angle DAF. Show that the triangles ABE and ACF
are similar and that BE . CF = BC^.
tEx. 1850. (i) In a A ABC, AB=^AC, CX is drawn perpendicular to
the internal bisector of the l BAC ; prove that AX is bisected by BC.
(ii) State and prove an analogous theorem for the external bisector of
the / BAC.
tEx. 1861. Two circles touch one another externally at A, BA and AC
are diameters of the circles ; BD is a chord of the first circle which touches
the second at X, and CE is a chord of the second which touches the first
atY. Prove that BD.CE=4DX.EY.
350 BOOK IV
tEx. lasa. Two straight lines AOB, COD intersect at O; prove that,
if OA : OB = OC : CD, then the as AOD, BOO are equivalent.
tEx. 1863. The sides AB, AD of the rhomhus ABOD are bisected in
E, F respectively. Prove that the area of the triangle CEF is three-eighths
of the area of the rhombus.
tEx. 1864. ABC is a triangle right-angled at A, the altitude AD is
produced to E so that DE is a third proportional to AD, DC; prove that
AS BDE, ADC are equal in area.
tEx. 1866. Two circles ABC, AB'C, whose centres are O and O', touch
externally at A; BAB' is a straight line; prove that the triangles OAB',
O'AB are equal in area.
tEx. 1866. ABC is a triangle, and BC is divided at D so that
BD2=BC . DC. A line DE parallel to AC meets AB in E. Show that the
triangles DBE, ACD are equal in area.
tEx. 1867. PA, PB are the two tangents from P to a drde whose centre
is O ; prove that a PAB : a 0AB = PA^ : OA^.
tEx. 1858. Two triangles ABC, DEF have / A and L D supplementary
and the sides about these angles proportional, prove that the ratio of the
areas of these triangles is equal to AB* : DE*.
1Ex. 1869. Through the vertices of a triangle ABC, parallel straight
lines are drawn to meet the opposite sides of the triangle in points a, /3, 7 ;
prove that A aj87= 2 a ABC.
Ex. I860. Through the vertices A, B, C of an equilateral triangle
straight lines are drawn perpendicular to the sides AB, BC, CA respectively,
so as to form another equilateral triangle. Compare the areas of the two
triangles.
tEx. 1861. A square BCDE is described on the base BC of a triangle
ABC, and on the side opposite to A. If AD, A E cut BC in F, G respectively,
prove that FG is the base of a square inscribed in the triangle ABC.
tEx. 1863. Prove that the rectangle contained by the hypotenuses of
two similar right-angled triangles is equal to the sum of the rectangles
contained by the other pairs of corresponding sides.
tEx. 1863. The sides AB, AC of a triangle are bisected at D and E
respectively; prove that, if the circle ADE intersect the line BC, and P be
a point of intersection, then AP is a mean proportional between BP and CP.
MISCELLANEOUS EXERCISES 361
tEx. 1804. Circles are described on the sides of a right-angled triangle
ABC as diameters, and through the right angle A a straight line APQR is
drawn catting the three circles in P, Q, R respectively. Show that AP= QR.
tEx. 1865. The bisector of the angle BAG of a triangle ABC meets the
side BC at D. The circle described about the triangle BAD meets CA again
at E, and the circle described about the triangle CAD meets BA again at F.
Show that BF is equal to CE.
tEx. 1866. D, E, F are points in the sides BC, CA, AB of a a ABC
such that AD = BE=CF. From any point O within the a ABC, OP, OQ,
OR are drawn parallel to AD, BE, GF to meet BC, CA, AB in P, Q, R
respectively. Show that OP + OQ+OR=AD.
tEx. 1867. ABCD is a quadrilateral with the angles at A and C right
angles. If BK and DN are drawn perpendicular to AC, prove that AN = CK.
tEx. 1868. The angle BAC of a triangle is bisected by a straight line
which meets the base BC in D; a straight line drawn through D at right
angles to AD meets AB in E and AC in F, Prove that EB : CF = BD : DC.
tEx. 1869. If the tangents at the ends of one diagonal of a cyclic
quadrilateral intersect on the other diagonal produced, the rectangle con-
tained by one pair of opposite sides is equal to that contained by the other
pair.
tEx. 1870. Two circles ABC, APQ (of which APQ is the smaller) touch
internally at A ; BC a chord of the larger touches the smaller at R ; AB, AC
cut the circle APQ at P and Q respectively. Prove that AP : AQ= BR : RC.
tEx, 1871. AB is a fixed chord of a circle; CD is the diameter per-
pendicular to AB; P is a variable point, on the circle; AP, BP cut CD
(produced if necessary) in X, Y ; if O is the centre of the circle, prove that
OX . OY is constant.
tEx. 1872. . Any point P is taken within a parallelogram ABDC; PM
and PN are drawn respectively parallel to the sides AC and AB and ter-
minated by AB and AC; NP produced meets BD in E; AE is joined meeting
PM iu P'; P'Q is drawn parallel to AB meeting the diagonal AD in Q.
Prove that AQ : AD = parallelogram AMPN : parallelogram ABDC.
tEx. 1873. A straight line HK is drawn parallel to the base BC of a
triangle ABC to cut AB, AC in H, K respectively; BK, HC intersect at X,
AX cuts HK, BC at Y, Z respectively. Prove that YX : XZ = AY : AZ.
tEx. 1874. ABCDEFG is a regular heptagon ; BG outs AC, AD in X, Y
respectively ; prove that AX . AC = A Y . AD.
862 BOOK IV
fEx. 1876. P, Q, R, S are four consecutive corners of a regular polygon;
PR, QS intersect at X ; prove that QR is a mean proportional between PR
and RX.
t£x. X876. Two straight lines BGE, CGF intersect at G bo that
GE=j^BE and GF = JCF; BF and CE are produced to meet at A; prove
that BF=FA and CE=EA.
tEx. 1877. In two circles ABC, DEF, / BAC= / EDF, prove that the
ratio of the chords BC, EF is equal to Ihe ratio of the diameters of the
circles.
tEx. 1878. ABCDEF is a hexagon with its opposite sides parallel, CF
is parallel to AB (and DE), and AD is parallel to BC (and EF) ; prove that
BE must be parallel to CD (and AF).
APPENDIX I.
Euclid ii. 14*.
To describe a square equal to a given rectangle Abcd.
G
Jig. 330.
Construction Produce AB to E, so that BE = BC.
Bisect AE at F.
With centre F and radius FA describe a semicircle AGE.
Produce CB to meet the semicircle at G.
Then, if a square is described on BG, this square is
equal to rect. AC.
Proof Since F is the centre of © AGE,
.'. FA = FG = FE = a:.
Let FB=y, BG=».
Then AB = a; + y,
BC= BE = a;-y,
.*. area of rect. AC = AB . BC = (a; + y) (a; — y) = ar* — ^.
Again, since A FBG is rt. /.'* at B,
.*. a:* = y^ + 2i^, Pythcoyoras-
:. rect. AC = square on BG.
For Exercises see p. 333.
* The two propositions given below have been treated, in the present
work, as applications of the theory of similar figures. For examinations
in which only the first three books are required, an independent proof of
these propositions is desirable : the proofs in the Appendix are substantially
those of Euclid.
354 APPENDIX I
Euclid hi. 36, 36.
If two chords of a circle intersect, the rectangle
contained by the segments of the one is equal to the
rectangle contained by the segments of the other.
Case i. Let the chorda AB, CD intersect at P, a point inside
the circle.
To prove that rect. PA . PB = rect. PC . PD.
Construction From O, the centre of the O, draw OM x to AB.
Join OA, OP.
Proof Since OM is i. to chord AB,
.'. AM = BM=a;, say. ni. 1.
Let PM=y, OA«=r, OM=s, OP=-t.
Then Pk=x + y, PB = a;-y,
/.rect. PA. PB = (a! + y)(a;— y)
Now A OMA is rt. iL ** at M,
.'. af + ^ = r^. Pythagoras
8]m^y' + sr' = t%
.*. subtracting, af — y^ = r^ — t^,
:. rect PA. PB=r^-t^
= radius' -OP*.
Sim'y by drawing a perpendicular to chord CD it may be shown
that
rect. PC . PD = radius^ — OP*,
,*. rect. PA . PB = rectw PC . PD.
Q. E. D.
APPENDIX I
365
Case il Let the chords AB, CD intersect at P, a point outside
the circle.
To prove that
Construction
Proof
fig. 332.
rect. PA . PB = rect. PC . PD.
Draw CM X to AB.
As in Case i., AM = BM =ai
PA = y + ic, PB = y — a;,
.*. rect. PA . PB = (y + a;) (y — a;)
= 2/2 -ar*.
Again, as in Case i.,
a:» + /»» = r',
.'. rect. PA. PB = «2-r*
= OP' — radius''.
Sim^ it may be shown that
rect. PC . PD = OP* — radius*,
.'. rect. PA. PB = rect. PC . PD.
Q. E. D.
For the discussion of the case in which C, D infi^. 882 coincide, and PCD
becomes a tangent, see Ex. 1723. Exercises on the above theorem will be found
on page 330.
APPENDIX II.t
Thb Pentagon.
To divide a given straight line into two parts such
that the square on the greater part may be equal to the
rectangle contained by the whole line and the smaller
part.
[Analysis. Iiet the whole line contain a units of length.
Let the ratio of the greater part to the whole line be a; : 1.
Then the greater part contains ax units; and the smaller
a — ax units.
The square on the greater part contains a^a^ units of area
and the rectangle contained by the whole line and the smaller
part contains a (a — ax) units of area,
.-. a^x' = a^-ah^
.'. a^=l —Xy
.-. or' + « - 1 = 0.
Solving this equation, we find
^ = ±2-2-
For the present* we reject the lower sign, which would give
a negative value for x ; and we are left with
-=^-^ = 0-618....]
* It will be seen below (p. 358) that a meaning can b« found for the
negative valae of x.
THE PENTAGON
357
In order to construct this length with ungraduated ruler and
compass only, we proceed as follows : —
Let AB be the given straight line.
Construction At A erect AC ± to AB, and equal to ^AB.
Join CB.
From CB cut off CD = CA.
From BA cut off BE= BD.
Then AB is divided as required.
Ptoof BC^ = AB^ + ACl
But AB -- a and AC - ^a,
.-. BC^ = a?+ {a"
IT Vs
,-. BC= vf a--jj-a.
To verify that this length satisfies the given conditions.
— (^-l)-(H-i>.
.-. AE.AB=(l^-^jaxa = BE''.
358 APPENDIX II
Extreme and mean ratio. The relation AE . AB = BE* may
be written AE : BE = BE : AB. Thus the straight line AB has been
divided so that the larger part is the mean proportional
between the smaller part and the whole line. In other
words, the larger part is the mean, while the smaller part and
the whole line are the extremes of a proportion. For this reason,
a line divided as above is said to be divided in extreme and
mean, ratio. This method of dividing a line is also known as
medial section.
/5 I
Note. The solution x = — ^ ^ was rejected. Strictly
speaking, however, it is a second solution of the problem. The
fact that this value of a; is negative indicates that BE must be
measured from B in the other direction — away from A rather than
towards A — as BE' in fig. 334.
A ^B E*
fig. 334.
Ex. 1890, With ruler and compass, divide a straight line one decimetre
long in extreme and mean ratio. Calculate the correct lengths for the two
parts, and estimate the percentage error in your drawing.
Ex. 1801. Devise a geometrical construction for dividing a line ex-
ternally as in the above note (fig. 334),
tEx, 1893. ~ Prove that, if E' is constructed as in the note (fig. 334), then
AB . AE'= BE'2; and hence that the line AB is divided externally in extreme
and mean ratio.
tEx, 1803. Prove that if AB is divided externally in extreme and mean
ratio at E', then AE' is divided internally in extreme and mean ratio at B.
Ex. 1804. Show how to divide a straight line AB at C so that
(i) AB,AC = 2CB2, -
(ii) 2AB.AC = CB2,
(iSi) AC''=2CB2.
THE PENTAGON 359
To construct an isosceles triangle such that each of
the base angles is twice the vertical angle.
A
fig. 335.
Construction Draw a straight line AB of any length.
Divide AB at C so that AB . BC = ACl
With centre A and radius AB describe a circle.
In this circle place a chord BD = AC.
Join AD.
Then ABD is an isosceles A having Z.B — z.D = 2z.A.
Proof Join CD.
Since BC . BA = AC^ = BD^
.•. BC : BD = BD : BA.
Thus, in the A' BCD, BDA, the z. B is common and the
sides about the common angle are proportional.
.". A* are similar. IV. 5.
But A BDA is isosceles (•.• AB = AD),
.'. A BCD is isosceles,
.•. CD = BD = CA.
.". L CDA =: z. A.
Now Z. BCD (ext. z.of ACAD)
= Z. A + Z. CDA
= 2 z. A,
.'. Z. B = 2 z. A.
360 APPENDIX n
Ex. 1805. Perform the above oonstmotion. Calculate what should be the
magnitudes of the angles of the triangle, and verify that your figure agrees
with your calculation. (To save time, it will be best to divide AB in the
required manner arithmetically, i.e. by measuring off the right length.)
tEx. 1806. Show that, in fig. 335, BD is the side of a regular decagon
inscribed in the circle.
tEx. 1807. Show that, if Q ACD is drawn, BD will be a tangent to that
circle.
fEx. 1808. Prove that AC and CO are sides of a regular pentagon in-
scribed in oACD.
tEx. 1800. Let DC be produced to meet the circle of fig. 335 in E
Prove that BE is the side of a regular 5-gon inscribed in Q A.
tEx. lOOO. Prove that AE = EC. (See Ex. 1899.)
tEx. lOOl. Prove that AE is || to BD. (See Ex. 1899.)
tEx. lOOa. Prove that a s AED, CAD are similar. (See Ex. 1899.)
tEx. lOOa. Prove that DE is divided in extreme and mean ratio at C.
(See Ex. 1899.)
tEx. 10O4. Prove that, if oABD is drawn, BD is the side of a regular
pentagon inscribed in the Q.
tEx. 10O6. Let the bisectors of /.s B, D meet O ABD in F, G. Prove
that AGBDF is a regular pentagon.
THE PENTAGON
361
To describe a regular pentagon.
A
fig. 336.
Construction Construct an isosceles A ABC with each of its
base angles twice the vertical angle.
Draw the circumscribing © of A ABC.
Then BC is a side of a regular 5-gon inscribed in ©ABC.
Proof Since z.ABC=z.ACB = 2iL BAG,
.-. L BAC = 1 of 2 rt. ^ s= 36°.
.'. BC subtends 36° at the circumference and 72° at the
centre.
.'. BC is a side of a regular 5-gon inscribed in the ©
The pentagon may now be completed. (How ?)
Practical method of describing a regular pentagon.
The above method is interesting theoretically, but inconvenient,
in practice. The practical method is as follows.
J3
Draw AOB, COD, two perpendicular diameters of a circle.
Bisect OA at E.
With centre E and radius EC describe a © cutting OB in F.
Then OF is equal to a chord of a regular pentagon inscribed
in the © O.
(The proof of this needs some knowledge of Trigonometry.)
G. s. II. 24
362
APPENDIX II
tEx. 1906. Prove that in fig. 836 AB, CE divide each other in extreme
and mean ratio.
tEx. 10O7. In fig. 336, show that A DCX is similar to A ABC.
tEx. 1908. Show that aCXY is similar to A ABC.
tEx. 1909. Prove that BY is divided in medial section at X.
tEx. 1910. Prove that BY is the mean proportional between BX and BD.
To prove that sin 18° = 7 *.
Let ABD be an isosceles A having z.B = iiLD — 2^A (see
page 359) ; let AC = BD as in fig. 335, and let AE l>e drawn to
bisect BD at rt. L s.
Then AB is divided in extreme and mean ratio at C.
/5 — 1
Thus, if AB = a, AC = '^ — a (see p. 356).
Now ^ BAD =36° (p. 361),
.'. aBAE = 18°,
BE BD v'S-1
.-. sin 18° =
AB 2AB
Ex. 1911. Calculate sin 18° as a decimal; and verify the value by
measurement.
• See Ex. 1670, p. 315.
363
REVISION PAPERS
PAPER I (ON Book I).
tl. In the given figure AB is parallel to CD ;
and AF and CG are the bisectors of z BAE and
L DCA. Prove that AF and CG are parallel.
2 ABC and DEF are two triangles. If the
following facts hold, are the triangles congruent
(give reasons for your answers) :
(i)
AB=DE,
AC=DF,
1A=|D;
(")
AB = DF,
AC=DE,
|A = [D;
(iii)
AB = EF,
AC = DF,
|A = |F;
(iv)
DE = BC,
!A = |F,
11=|D;
(V)
BC=DE,
|A = [E,
IB = |F?
fS. ABCD and ABPQ are a parallelogram and a rectangle on opposite
sides of a straight line AB ; join DQ, CP : prove that CDQP is a parallelo-
gram.
t4. The triangles ABC and A'BC are on the same side of their common
base BC, and the angle A'BC equals the angle ACB, and the angle A'CB
equals the angle ABC ; also AB and A'C intersect in O. Prove that the
triangles AOC and A'OB are congruent.
5. In a triangle AB = 12cm., BC = 9cm. and the perpendicular from
B to AC is 5-7 cm. Show that there are two triangles that fulfil these
conditions and draw them both. State how the two triangles are related.
6. A tower, whose base is a circle of diameter 40 feet, is surmounted by
a spire. The distance bf the shadow of the point of the spire from the
nearest point of the base is measured to be 33 feet, while the line joining
the top of the spire with its shadow makes an angle of 60° with the ground.
Draw a sketch to scale, and measure to the nearest foot the height of the
top of the spire above the ground.
24—2
364 REVISION PAPERS
PAPER II (ON Book I).
fl. AB and CD are two parallel straight lines and a straight line is
drawn to cut them in E and F. If B and D are both on the same side of
EF, prove that the biseotors of / BEF and z DFE are at right angles to one
another.
2. How many sides has a regular figure the angle of whioh contains
162°?
Is it possible for a regular figure to have angles of 180° ?
-|-3. ABCD is a parallelogram (not rectangular), and AL and CM are
the perpendiculars from A and C on to the diagonal BD. Prove (otherwise
than by a mere appeal to symmetry) that ALCM is a parallelogram.
+4. In the given figure YM is perpendicular to
XZ, ZN is perpendicular to XY, and YM and ZN
intersect at O, also O Y = OZ. Prove that X Y = X Z .
5. Construct a parallelogram ABCD whose sides
AB, AD are 6 -4" and 3", and the distance between
AB and CD is 2'6". Measure the angle ADC.
Find a point P in AB which is equidistant from
D and B.
Measure PB. State your constmction.
6. Find, by drawing, the length of the shadow of a man 6 feet high,
when the altitude of the sun is 57°.
PAPER III (on Book I).
1. Give careful definitions of the following, and draw simple figures
to make your definitions more clear: — Supplementary angles; angle of
depression.
Explain, with sketches, the meaning of prism; triangular pyramid.
+2. ABC is an isosceles triangle (AB = AC) ; through C is drawn CD at
right angles to BCj CD outs BA produced at D. . Prove that ACD is an
isosceles triangle.
+3. ABCD is a parallelogram ; E is the mid-point of AB ; CE and DA
are produced to meet at F. What angles in the figure are equal to angles
ECDandECB? Give reasons. Also prove AF = AD.
REVISION PAPERS 365
4. Is each of the following statements true for any parallelogram ? If
not, state in each case a kind of quadrilateral for which it is true. No proofs
are required, (a) The diagonals bisect one another, (b) The diagonals
bisect the angles, (c) The opposite angles together make two right angles,
(d) The diagonals are equal.
t5 PQR is an isosceles triangle having PQ=PR. A straight line is
drawn perpendicular to QR and cuts PQ, PR (one of them produced) in X, Y.
Prove that the triangle PXY is isosceles.
6. Two points of land, A, B, on the shore are 2-8 miles apart, A being
S. 71° W. of B. A ship at sea observes A to bear N. 17° E., and B to bear
N. 42° E. Find the distance of the ship from A and from B.
If the ship's course is N. 50° E., at what distance will she pass B ?
PAPER IV (ON Book I).
1. Draw a figure of a cuboid showing three of its faces. If you placed
a cuboid with one of its faces vertical, how many of its faces (i) must be
vertical, (ii) might be vertical? If you placed it with one of its edges
horizontal, how many of its edges (i) must be horizontal, (ii) might be
horizontal, (iii) must be vertical? How many diagonals has a cuboid?
t2. Prove that, if all the sides of a quadrilateral are equal, the figure is
a parallelogram and its diagonals cut at right angles.
fS. D is the middle point of the side BC of a triangle ABC. If DA is
equal to half BC, prove that the angle BAC is equal to the sum of the angles
B and C.
4. A and B are two points on paper. What is the locus of the point C
under the following conditions '.—firstly, when C is restricted to being in the
plane of the paper ; secondly, when C may be anywhere in space? — (i) Angle
ACB is 90°. (ii) C is equidistant from A and B. (iii) Angle CAB is 20°.
(iv) C is always 1 inch from AB (which may be produced indefinitely in
both directions).
5. Construct a triangle whose base is 7*3 cms. long, with vertex 3 cms.
away from the base line and 4 cms. away from the middle point of the base ;
measure the sides of the triangle.
6. A ship steaming N. 55° W. at 18 knots sights a lighthouse bearing
N. 42° W., distant 8-5 miles at noon. Find, by drawing, how near the ship
will pass to the lighthouse, if she keeps on her course. Find also at what
time (to the nearest minute) she will pass the lighthouse.
366
REVISION PAPERS
PAPER V (ON Book I).
+1. If the bisector of an exterior angle of a triangle is parallel to one of
the sides, prove that the triangle is isosceles.
2. Draw freehand diagram of
(i) a quadrilateral with only two sides parallel which has equal
diagonals,
(ii) any other quadrilateral which has equal diagonals.
What is the name of the quadrilateral (i) ?
3. What is the name of the geometrical solid whose surface is traced out
(i) by one arm of a pair of dividers being rotated about the other when the
latter is kept vertical ? (ii) by one edge of the piece of paper on which you
are writing being rotated about the opposite edge ?
+4. ABCD is a parallelogram; DA and DC are produced to X and Y
respectively so that AX = DA and CY= DC ; XB and BY are drawn. Prove
that X BY is a straight line.
5. P, Q are two points 6 cms. apart. PS is a straight line making an
angle of 40° with PQ. Find a point (or points) equidistant from P and Q.
and 4 cms. from PS. Measure the distance of your point (or points) from P.
Is this problem always possible whatever the angle SPQ?
6. Three ships, A, B and C, start together from a port. A proceeds due
North ; B, N.E., and C, East. If the ships always keep in a line, A going
20 knots and B 12 knots, what is C's speed ?
PAPER VI (ON Book I).
1. In the figure, which is not drawn to scale,
find y if AB and CD are parallel and a; is 18; also
prove that AB and CD will meet if produced
towards B and D (whatever x may be), provided
that x+y is greater than 32.
2. ABC is a triangle in which ABC is 50°
and ACB is 70°. CB is produced beyond B to D,
so that BD = BA, and BC is produced beyond C to
E, so that CE=:CA. Determine from theoretical
considerations the angles of the triangle ADE.
Construct the triangle ABC when its perimeter
is 3".
REVISION PAPERS
367
+3, ABC is an isosceles triangle, the equal sides AB, AC are produced to
D, E respectively; the bisectors of / DBC, / ECB intersect at F. Prove
that A FCB ia isosceles.
4. A right cylinder of diameter 6 '8 cm. and height 7'6cm. is divided
into two parts by a plane through its centre at 28° to its base : measure the
length of the section.
t5. PS, the bisector of the angle P of a triangle PQR, cuts QR at S;
through S, ST and SU are drawn parallel to PR and PQ, thus forming
a quadrilateral TPUS: prove that the sides of TPUS are all equal to one
another
6. A captive balloon is observed from two positions A and B on a
horizontal plane, A being due north of the balloon and B due south of it.
A and B are 2 miles apart. From A the angle of elevation of the balloon is
27° and from B it is 18°. Find by drawing the height of the balloon.
PAPER VII (on Books I, II).
tl. ex, the bisector of an exterior angle of a ABC, which is not
isosceles, meets AB in X. Prove that i AXC is equal to half the difference
between the angles A and B.
2. Draw XOX', YOY', two straight lines intersecting at O, so that
/. XOY = 50°. Make OX = 4 in., OY = 3 m. Find a point P equidistant from
X, Y, and at the same time equidistant from XOX', YOY'. Find another
such point, Q. Explain (in two or three lines) how these points are found.
Measure OP, OQ in inches.
+3. Rays ©flight proceeding from a point
O fall on a mirror AB and are reflected,
making an equal angle with the mirror.
Prove that the reflected rays, if produced
backwards, would all be found to pass
through a point P, such that ON = PN and
OP is perpendicular to AB.
4. ABC is a triangle having a fixed base BC, 5 cm, long, and a moveable
vertex A. What is the locus of A
(i) when ABC is isosceles (AB = AC)?
(ii) when ABC has a fixed area ( = 10sq.cm.)?
(iii) when the median AM has a fixed length (=6 cm.)?
368 REVISION PAPERS
5. Draw a parallelogram having Bides of 4 cm. and 6*5 om. and an angle
of 75°. Find its area.
t6. Oall the corners of a rectangular sheet of paper A, B, C, D (AB being
a long side of the rectangle); if it were folded along the diagonal AC, then
AB and CD would cut at a point we will call O. Make a freehand sketch of
the figure you would obtain and prove triangles ADO, BCO equal in area.
7. The range of a gun is 2| miles. If it is stationed 1^ miles from a
straight road, what length of the road can it command?
PAPER VIII (ON Books I, II).
1. A destroyer, steaming N. 10° E. 30 knots, sights a cruiser, 11 miles
off, bearing N. 62° E. Half an hour later the cruiser is 6 miles off, bearing
N. 70° E. Find the course and speed of the cruiser. If the destroyer then
alter course to N. 70° E., how far astern of the cruiser will she pass?
t2. With centre A and radius AB a
circle is drawn. With centre B and equal
radius an arc is drawn intersecting' the first
circle at C. Similarly from centre C the
point D is determined, and from centre D
the point E. Prove that BAE is a straight
line.
3. OA is the vertical line which is the junction between two walls of a
room, OB and OC the horizontal lines running along the junction between
the walls and the floor. What is the locus of the point P which moves about
in the room under the following conditions : — (i) so as to be always 6 feet
above the floor? (ii) so that the angle AOP is always 50°? (iii) so as to be
always equidistant from OB and 00? (iv) so as to be 4 feet from O and
2 feet from the plane AOC?
4. How is the area of a parallelogram measured?
Construct a rhombus whose area is 8 sq. ins. and whose sides are each
3-2" long.
5. The area of a triangle is 24 square inches, the altitude is 8" ; find the
length of the base, and on it describe a parallelogram, the area of which
shall be 48 square inches.
REVISION PAPERS 369
t6. A four- sided field is to be divided into two parts of eqnal area ; prove
the accuracy of the following constniction. Draw a quadrilateral ABCD to
represent the field ; draw the diagonal AC ; find E, the mid-point of AC ;
join BE, DE; then the areas ABED and CBED are equal.
t7. HVQ is a triangle right-angled at V, HVT is a triangle on the
opposite side of HV having L THV a right angle ; prove that the squares on
HT, HQ are together equal to the squares on TV, VQ.
PAPER IX (ON Books I, II).
1. In the shear-legs shown in the figure AD is
30 feet, BD is 50 feet, and the angle BAD is 130°.
The load is supported by a chain passing over a
pulley at D and controlled by a winch at A. If the
end B of the tie-rod BD is moved away from A until
D is brought vertically over A, find (1) the distance
through which B is moved, and (2) the length of
chain which must be let out so that the load remains at the same height
above the ground.
2. A, B are points 3" apart on an unlimited straight line ; state carefully
and fully the locus of the following points : —
(i) points equidistant from A and B, (ii) points 3" from AB, (iii) the
middle points of chords of a circle (centre C), which are parallel to AB,
(iv) the middle points of chords of a circle (centre C), which are equal to AB,
(v) the centres of circles of radius 3" which pass through A, (vi) points at
which AB subtends a right angle, (vii) the centres of circles passing through
A and B, (viii) the centres of spheres passing through A and B.
3. (i) How many faces has a prism on a six-sided base? (ii) How
many vertices has a cone ? (iii) How many edges has a pyramid on a base
of 5 sides?
4. ABC is a triangle having AC = 7-2", BC = 9-6". AX, BY are drawn
perpendicular to BC, AC respectively. If AX = 2 -4", find length of BY.
5. Draw a parallelogram of base 8 cm., angle 70°, and area 56 sq. cm.
Transform this parallelogram into an equivalent rhombus on the same base.
Measure the acute angle of the rhombus.
f6. Draw a quadrilateral ABCD having the angles at A and D both
acute; from B and C draw BE and CF perpendicular to AD. Prove that
the area of the quadrilateral ABCD is equal to the sum of the areas of the
triangles ABF and ECD.
370 REVISION PAPERS
7. A Tolcanio moantaia is in the shape of a cone 4000 ft. high : the base
is a circle of 8000 ft. radius. Calculate (to the nearest tenth of a mile) the
length of a rack-and-pinion railway which takes the shortest way to the top.
PAPER X (ON Books I, II).
1. CD and EF are two given parallel straight lines, and A is a given
point in CD. B is a given point on the side of EF remote from A. It is
required to determine the position of a point P in CD, such that, when the
straight line PB is drawn crossing EF at Q, then PQ may be equal to AP.
Prove that the perpendicular distance of the line PB from the point A is equal
to the distance between the parallel lines. Hence solve the problem, and
show that there are in general two possible positions for the point P.
Draw the figure, making the perpendicular distance between the parallel
lines equal to 3 cm., AB = 8 cm. and angle DAB = 60°. Determine from your
drawing the two possible lengths of PQ.
t2. The triangles OAB, OPQ are con- . Q
gment. QP produced meets AB in X. /^K
Prove that OX bisects the angle AX P. / / \ \
3. How many edges has (i) a cube; X / \ \
(ii) a cuboid; (iii) a square pyramid? /^/ / \ \
Draw a freehand sketch of each. v\ — /■ ^ -^ O
X / P
4. Construct a triangle, given BC=9"2 g
cms.; CA=8-2cms. ; AB = 10cms.
On AB construct an equivalent isosceles triangle. Measure the equal
sides and find the area.
5. A triangular field ABC has to be divided into four parts which are to
be equal to one another in area. Draw any triangle to represent the field
and show how to divide it so that the given conditions may be satisfied.
Give a proof.
6. Prove that the area of a trapezium is obtained by multiplying half
the sam of the parallel sides by the altitude.
Draw a trapezium having its parallel sides 8 cms. and 6 cms., and altitude
5 cms., and one of its acute angles 68°. Transform the trapezium into an
equivalent triangle. Describe your construction briefly and show how the
above rule for finding the area of a trapezium follows directly from your new
figure.
REVISION PAPERS
371
t7. In the right-angled triangle ABC, BC is the hypotenuse and the side
AB is double the side AC. A square is described on BC and is divided into
two rectangles by a line through A perpendicular to BC. Prove that one
rectangle is four times the other.
PAPER XI (ON Books I, II).
1. A vessel steaming at uniform speed finds that the bearings of a
lighthouse at 3, 4 and 5 p.m. are N. 20° E., N. 25° W., and N. 50° W.
respectively. Its distance from the lighthouse at 4 p.m. is 10 miles. Find,
by drawing, the ship's course.
2. Show that it is usually possible to draw two circles each of which
touches two given sides of a given triangle (produced if necessary) and has
its centre on the third side, but that under certain circumstances only one
such circle can be drawn.
3. The figure shows three jointed rods, BC having a length of 6 cm.,
CD of 10 om., and DA of 3-6 cm. B moves in a slot XY and BC is aUcays
perpendicular to XY. The rod AD revolves in
the plane of the paper about the point A, which
is fixed.
Draw the figure full size, with C at its greatest
possible distance from A. Now imagine AD to
revolve clockwise at the rate of 60° per second,
and show the position of B at the end of each second. Give the greatest and
the least distances of C from A and the range of movement possible for B.
Show in a table the distance of B from its original position at the end
of each second, and illustrate by a graph.
4. The figure represents a field to scale, 1 centimetre denoting a chain.
Estimate the area of the field in acres.
d
X
a
372 REVISION PAPERS
t5. A straight line is drawn parallel to the base BC of a triangle ABC
cutting AB at X and AC at Y; prove (i) that triangles XBC, YBC are equal
in area, and (ii) that triangles ABY, ACX are equal in area.
6. By means of a sketch-figure, with a very brief explanation, illustrate
the identity
{a+x)(a + y)sa^+ax + ay + xy.
7. A shelter trench of rectangular section is 3 ft. wide and 4 ft. deep ;
the earth excavated is piled up in front as a rampart ; if the vertical section
is a right-angled isosceles triangle, how high is the rampart?
PAPER XII (ON Books I, II).
1. ■ If the line joining two points P, Q. is bisected perpendicularly by a
given straight line, then Q is said to be the image of P in the given straight
line. Given a point and a straight line, show how to find the image of the
point using compasses only (no proof is required).
2. A corner shelf ABC is to be made from a board and to consist of two
pieces ABED and CDE glued together along DE. The depth 8" of each
piece is to be the same as the breadth of the board. Determine the greatest
breadth AB of the shelf. What is the shortest length of board which will
sufi&ce for the job?
C
X
ct>.
"A ^^ /
1 /
/4A.E
;^
^' A
\v
V /"l^ \
.
:4r\
A B
3. (i) A shot is fired from an airship high overhead. Assuming that
sound travels through the atmosphere at a uniform rate of 1100 feet per
second, what ia the solid-locus of points at which the report will be heard
in one second ?
(ii) What is the space-locus of points equally distant from two given
points?
REVISION PAPERS
373
t4. E and F are the middle points of AD and BC, the sides of a parallelo-
gram ABCD. Prove that the lines BE, DF divide the diagonal AC into
three eqnal parts.
5. In a field in the form of a quadrilateral ABCD, B is due North of
A and D is due East of A. Also AB = 7-5 chains, BC = 8-4 chains, CD = l-3
chains, DA =4 chains. Find the area of the field in acres.
t6. Assuming that the medians of a triangle ABC pass through one
point, prove that the six triangles into which they divide the triangle ABC
are equal in area to one another.
7. Find, in centimetres, the base-radius of a cone of slant side 3 deci-
metres and height 12 cm.
PAPER XIII (ON Books I— III).
tl. If two pairs of straight railway lines cross one another, prove that
the figure they enclose is a rhombus. [You are to assume that the perpen-
dicular distance between one pair of lines is the same as the perpendicular
distance between the other pair of lines.]
2. The figure represents a coal-box.
figure shown.
Find the volume of the solid
31"
3. Treasure is known to be buried in a field 20 yards from a straight
hedge, and 80 yards from a cairn, this being inside the field and 40 yards
from the hedge. Show that it may be in either of two positions. Find
the distance apart of these positions (i) by measurement ; (ii) by calculation.
374 REVISION PAPiJRS
4. A model boat sails in a straight line acrosB a circular pond, towards
a point 50 yards away. The greatest distance across the pond is 70 yards.
How near to the centre of the pond will the boat go? What will be the
boat's least distance from the point on the pond's edge exactly opposite the
starting-point? [Both answers by calculation.]
5. Describe a triangle with sides 4-5, 6, 7*5 in. ; find the centres of the
inscribed and circumscribed circles, and measure the distance between them.
t6. ABC is a triangle having the sides AB, AC equal; perpendiculars
drawn to AC at A and to BC at B meet at D. Prove that AD bisects the
angle between CD and BD produced.
7. A circle whose centre is O is touched internally at A by a circle of
half its radius. A radius OQ of the former circle cuts the smaller circle at P.
Prove that arc AQ=aro A P.
PAPER XIV (ON Books I— III).
tl. Prove that the line joining the middle points H, K of the sides AB,
AC of a triangle ABC is parallel to BC.
2. A ship ia situated 4*5 miles from a straight shore. Two piers are
respectively 6 miles and 8*9 miles from the ship. Calculate the distance
between the piers.
3. The figure shows three equal bars AB, BC, CD, jointed at B and C.
The three are placed on a table, and the bar AB
is kept fixed while the point D is gradually
moved along AB from A to B, the joint C moving
in consequence across the table. Prove that, if
the straight line AC is drawn, in all positions of
D, the triangle ADC has one of its angles double 1^/' -jo. \a.
of another. /'
If each bar is of length a, obtain an expression /'
for the length k of AC when D has been moved a -^
distance h from A towards B. Also calculate k ^ ^ B
when h=^a, taking a =10 cm.
4. In a circle a chord 24 in. long is 5 in. distant from the centre.
Calculate (i) the radius of the circle ; (ii) the length of a chord which is 10 in.
distant from the centre.
a-h
REVISION PAPERS 375
5, A triangle ABC is inscribed in a circle, centre O, and radius 4".
If the angles of the triangle are A = 72°, B = 55°, C = 53°, -what are the
angles BOC, COA, AOB? Hence find the sides of the triangle by dravdng.
+6. AB is a chord of a circle whose centre is O, and AB is parallel to
the tangent at P. If the tangents at A and P intersect at T, prove that the
angles POA, TAB are equal to one another.
7. RS is a fixed chord of the circle RLNS ; a chord LN of given length
is placed in the arc RNS, and RN and SL meet in O. Show that the
magnitude of the angle ROS is independent of the position of the chord LN,
in the arc.
PAPER XV (ON Books I— III).
fl. Draw a triangle ABC, bisect AB at D; draw DE parallel to BC and
let it cut AC at E ; prove that E is the mid-point of AC.
2. ABC is a right-angled triangle. The angle A is 90°. A circle is
circumscribed round the triangle. Its radius is found to be 6". AN is drawn
perpendicular to the base BC. O is the mid-point of BC and ON = 2'4".
Calculate the lengths of AN, AB, and AC.
fS. If the diagonals of a quadrilateral intersect at right angles, prove
that the sum of the squares on one pair of opposite sides is equal to the sum
of the squares on the other pair of opposite sides.
4. Construct a triangle ADE such that AD, DE, EA measure 5, 6*1,
9-7 cm. respectively. Construct the circumscribing circle and the circle
escribed to DE. Measure the radii of the circles and the distance between
their centres.
5. ABC is a tangent to a circle at B, BD is a diameter and BE, BF are
chords such that Z ABE=20°, z CBF = 60°; DE, DF, EF are joined. Find
all the angles of the figure.
+6. A triangle ABC is right-angled at A, O is the mid-point of BC, and
AP is drawn perpendicular to BC : prove that the angle CAP is equal to the
difference between the angles at B and C.
376
REVISION PAPERS
7. In order to avoid the shoals shown in the fignre, the navigator is
instructed to take bearings of the fixed objects A and B and to take care
that the angle subtended by AB never exceeds 130°. Explain the reason for
this instraction.
PAPER XVI (ON Books I— III).
+1. ABODE is a five-sided figure in •which BO, CD are respectively
equal to AE, DE and ^BCD = z.DEA. Prove that AC = BE.
2. ABCD is a rectangle in which AB = 4 in., BC = 6 in. A circle with A
as centre passes through the middle point of BC and cuts AD at F. Calcu-
late the length of CF.
3. ABC is a triangle in which AB is 7 in., BC is 5 in., CA is 3 in.
The circle whose centre is A and radius is AC cuts BC again in D. Prove
that ACD is an equilateral triangle.
4. In playing with coins of the same size a boy observed that he oould
arrange six coins round a centre one, each touching the centre one and two
others. Show the possibility of this by drawing a careful figure in which
each circle has a radius of 2 cm. State clearly how you determine the
centres of the circles and what help you get in this construction from
considerations of symmetry; then justify by general reasoning the method
you have adopted.
5. P and Q are two points on the circumference of a circle, and the
tangents to the circle at P and Q intersect at an angle of 56°. What fraction
of the whole circumference is the minor arc PGt ? and what is the ratio of the
major arc PQ to the minor arc PQ?
REVISION PAPERS 877
+6. AOB is & diameter of a circle; through A and B parallel chords of
the circle are drawn. Prove that these chords are eqaaL
t7. Two circles cut one another in the points A and B. Through A any
line is drawn which cuts the circles again in the points P, Q. and the
tangents at P, Q, cut in T. Prove that the four points B, P, T, Gl are on a
circle.
PAPER XVII (ON Books I— III).
tl. PQR is a triangle, and S is the mid-point of QR. From S, ST is
drawn parallel to QP, meeting PR in T, and SU parallel to RP, meeting PQ
in U. Prove SU = RT and also SU =TP.
2. Construct (without any calculation) a square which shall be equal in
area to the difference between the areas of two squares whose sides are
7 and 4 cm.
t3. PQR is a triangle right-angled at Q, S is the iMd-point of PQ;
prove that PR2=RS2 + 3GIS2.
4. A, B, C, D are four points on a circle of which O is the centre. AC
is a diameter and / BAG =35°, / DBC=40°. Find /ODC, /ODB, giving
your reasons briefly.
5. The line CD measures 14 cm.; with centres C and D describe
circles of radii 3 and 7 cm. respectively. Construct one of the interior
common tangents, and measure the perpendiculars upon this from the nearer
points at which the line joining the centres cuts the circimiferences.
t6. A triangle ACE is inscribed in a
circle ABCDEF; prove that the sum of
the angles ABC, CDE, EFA is equal to
four right angles.
t7. In the figure, BC touches the circle
ABD. Show that CE touches the circle
ADE at E. (You may assume the converse
of the "alternate segment" theorem.)
G. S. II. 25
378 REVISION PAPERS
PAPER XVIII (ON Books I— III).
1. Take a line AB, 9 cm. long (Fig.), and through B draw CD at right
angles, making BC = 6 cm., and BD = 3 cm.
Join AC and complete the rectangle AFDEC.
Denote by x the number of degrees in the angle
BDF, and write in each angle of the figure its
value in degrees.
You are told that if the parts of the figure
marked Y and "Z were out out they could be
placed against the part marked X bo that the
three parts would form a square.
Give the area of the square.
2. AB and XY are unlimited parallel straight lines 2 cm. apart; AB is
8 cm. long and C is its middle point. P is a point on XY, PN is perpendicular
to AB, and CN is a; cm. long. Find expressions for AN, BN, AP, BP, and
simplify the last two as far as possible. Hence find an equation for x when
AP is three times BP; solve it, and test the accuracy of your result by
drawing a figure to scale.
•f-S. A straight line AB is produced to C, so that AC = SAB; on BC an
equilateral triangle BCD is described. Prove that the square on AD is seven
times the square on AB.
4. What would be the radius of a circle in which an arc 11" in length
subtended an angle of 31j° at the circumference?
5. Construct a quadrilateral OPQR, given 0P = 6 cm., OR = 6 cm.,
angle O = 74°, angle P = 83°, and angle R = 97°. Draw a circle to pass through
O, P, and R.
+6. ABC is a triangle. Points X, Y are taken in AC, BC respectively
such that the angle XYC is equal to angle BAC. Prove that the angle XYA
= the angle ABX.
t7. O is the centre of the inscribed circle of a triangle ABC, and AC is
produced to meet at D the circle circumscribed to the triangle. Show that
DBsDCsDO.
REVISION PAPERS 379
PAPER XIX (ON .Books I— IV).
1. A rectangular sheet of paper ABCD 12 in. by 10 in. is folded along
XY, a line 4 inches from the shorter side BC. Find by calculation to three
significant figures, and illustrate by rough sketches, the shortest distance of
A from C : (a) before folding. (6) when the two parts of the sheet are at
right angles.
2. On a fixed line AB, 8 cm. long as base, construct a triangle ABC,
whose area is 24 sq. cm., such that the vertical angle is 63°. Measure
(1) the smallest angle, (2) the radius of the circumcircle of the triangle.
3. A position X lies 4000 yards N. 68° E. of Y, whilst Z lies 3000 yards
due S. of Y. Find the distance and bearing from X of a position which is
equidistant from the three positions X, Y and Z.
t4. ABC is a triangle inscribed in a circle; the bisector of the angle
BAG meets the circumference in D. A circle described with centre D and
radius DC cuts AD in E. Prove that BE bisects angle ABC.
t5. AB is a chord of a circle and AD the tangent at A. A chord QP is
drawn parallel to AB, meeting the tangent AD at D. Prove that the triangles
DPA and AQB are equiangular.
6. A pendulum swings through an angle of 10° on either side of the
vertical : calculate the length of a scratch made on the clock-case by the back
of the pendulum-weight, given that the pendulum is 4' 4" long. If the
pendulum were to swing through twice as large an angle, would- the scratch
be twice as long? Would the distance between the ends of the scratch be
doubled?
t7. The sides BA, CD of a cyclic quadrilateral ABCD are produced to
meet in O ; the internal bisector of the angle BOC meets AD in L and BC
in M. Prove that
AL:LD = MC:BM.
PAPER XX (ON Books I— IV>
1. A room is 20 ft. long, 16 ft. wide, and 12 ft. high. A string is
stretched diagonally from one corner of the floor to the opposite corner
of the ceiling. By drawing and measurement determine approximately in
degrees the inclinations of the string (i) to the floor of the room, (ii) to one
of the longer sides of the floor.
25—2
880 REVISION PAPERS
2. Draw two straight lines making an angle of 60** with one another
and intersecting at O. On one of the lines take the two points X, Y on
opposite sides of O such that XO = 2in., OY = 4-5in. Draw the circle
throngh the points X and Y which will cut the other line in two points
equidistant from O. Measure the distance of each of these points
from O.
3. (Jonstruct a triangle ABC in which a=S", c = i", and B = 29°.
Draw its inscribed circle, and also the escribed circle which touches AC
between A and C. Measure the radius of each circle, and show theoretically
that the line joining their centres must pass through B.
4. What is the length of the edge of the largest equilateral triangular
piece of paper which, when lying perfectly flat, will just float on the surface
of a hemispherical bowl, filled with water, of 6*2 cm. radius ?
t5. Show that the four points A, Q, X, R lie on a circle.
A
t6. QR is a chord ol a circle, TR is the tangent at R ; a straight line
through Q. perpendicular to this tangent meets it in T and the circumference
of the circle again in P; PM is the perpendicular from P on QR. Prove
that the angles QPM, TPR, TMR, TRM are all equal.
■\7. In a triangle PQR, PQ= PR =2 inches, and QR = 1 inch. In the
side PQ a point S is taken such that QS = | inch. Prove that the triangle
QRS is isosceles.
PAPER XXI (ON Books I— IV).
1. A sphere of 6" diameter rests on the top of an open hollow cylinder
whose inner diameter is 4". To what distance will the sphere project above
the top of the cylinder?
2. State, without actually performing any construction, how you would
solve the problem of drawing two tangents to a circle, which should include
a given angle and intersect upon a given straight line. How many solutions
of the problem would you expect to get?
EEVISION PAPERS 381
3. AB is a fixed line. Through A a line AC is drawn, of length 2-4 in.,
making an angle of 40° with AB. Draw the figure to full scale, and construct
a circle to touch AB at A and to pass through C. Explain your construction.
Measure the radius of this circle. Verify by calculation.
4. Two circles of radii 4 cm. and 7 cm. have their centres 9 cm. apart.
Calculate the length of the common tangent to the two circles.
15. AB is arc of a circle and C its middle point. Prove that the
angle ABC is one-quarter of the angle which the arc AB subtends at the
centre of the circle.
t6o ABC is a right-angled triangle in which C=90°. A square is
described on AB so as to be on the opposite side of AB from C. The
diagonals of the square intersect in D. Prove that CD bisects the angle C.
t7. ABCD is a parallelogram. From any point E in the diagonal AC,
EH is drawn parallel to AD to meet DC at H, and EF parallel to DC to meet
BC at F. Prove that the triangles ABD, EFH are similar.
PAPER XXII (ON Books I— IV).
1. A penny falls into a cup whose shape is an exact hemisphere of radius
5 cm. If the penny lies symmetrically at the bottom and its diameter is
3 cm., calculate how far below the penny the lowest pomt of the cup is.
2. A, B, C are 3 landmarks. B is 200 yards due East of A, and C is
200 yards N. 26° E. of B. An observer in a ship which is due North of B,
observes that AC subtends an angle of 90° at his eye. Find, by drawing,
the distance of the ship from A, B, and C.
3. Draw a circle of radius 7 cm. and a chord PQ distant 4 cm. &om the
centra
Now draw a circle of radius 5 cm. to touch your first circle internally, and
also to touch PGl. State your construction.
t4. To two circles, centres O and C, an internal and an external
common tangent are drawn, meeting in P. Prove that P lies on the circle
on 00' as diameter.
5. ABCD is a circle: AC, BD meet in X. Given that /ABD=33°,
/ ADB = 27°, z BAC = 45°, calculate the angles BXC, ACD, ABC, showing
your reasoning clearly but shortly.
382 REVISION PAPERS
6. AB is a breakwater, 2000 yards long, B being dae East of A. The
breakwater subtends an angle of 50° at each of two ships, x and y. If x
bears N. 10° E. from A, and y is 800 yards to the Eastward of x, find the
distance of each ship from the breakwater. [Scale 400 yards to 1 inch.]
7. The height of the Great Pyramid is 149 metres; an exact model of
the pyramid is made of height 1*49 metres, its side faces being triangles
similar to the side faces of the pyramid. What is the ratio of the total slant
Borfaoe of the pyramid to that of the model?
PAPER XXIII (ON Books I— IV).
1. A dirty football was found to leave a circle of mud, 11" in circum-
ference, when bouncing ; if the radius of the ball was 6J", find the depth to
which it was squashed in by the impact.
2. Draw a circle of 4 cm. radius to touch two circles of radii 3 cm.
and 2 cm. respectively, whose centres are 6 cm. apart. The 3-cm. circle
is to lie entirely inside the 4-cm. circle, and the 2-cm. circle is to lie entirely
outside.
3. P is a point on the circumference of a circle of centre O and radius
1^ in. Q is taken so that l POQ=40° and OQ = 3 in. Construct a circle
to touch the given circle at P and to pass through Q. Measure the radius
of this circle.
t4. If AB is a tangent to a circle of radius 5", where A is any point on
the circumference and B is 12" from A, find the locus of B as A moves
round the circle.
5. O and P are points 1000 yards apart, P being due East of O. At
Q the line OP subtends an angle of 63°. If Q is 450 yards from the line
OP, draw a figure to scale, and find the distance and bearing of Q. from O.
6. BAG is an equilateral arch, B being the centre of the arc AC and C
the centre of arc BA ; BED, CFD are similar A
arches, B being the centre of DE and C the
centre of DF and D of BE and OF. What is
the locus of the centres of circles touching (i) arcs
AB and AC, (ii) arcs DE and DF, (iii) arcs BA
andDF?
Hence explain how to construct with your
instruments a circle (shown dotted in the figure)
which will touch the arcs BA, AC, DE, DF. B
Draw the figure carefully, taking BC 5 inches long.
REVISION PAPERS 383
+7. In any triangle ABC, P is a point in BC such that BP is one-third
of BC. Join AP and take on it a point Q such that AQ is one-third of AP.
Then prove that the area of the triangle ABQ is one-ninth of that of ABC.
What is the ratio of the areas of the triangles ABQ and ACQ? Give
your reason.
PAPER XXIV (ON Books I— IV).
1. A paper pyramid on a square base is made as follows. On each side
of a square of side 3 inches is constructed an isosceles triangle of height
5 inches, the triangles lying outside the square A 4-pointed star is thus
formed, which is cut out of paper. By folding the triangles upwards a
pyramid is formed. Find its height, and the length of each of its sloping
edges.
t2. ABC is a triangle inscribed in a circle and the tangents at B and C
meet in T. Prove that, if through T a straight line is drawn parallel
to the tangent at A meeting AB, AC produced in F and G, then T is the
mid-point of FG.
3. OY, OX are two straight lines at right angles. On OX two points
A, B are marked so that OA=r', OB = 3". By construction find a point
(or points) on OY at which AB subtends an angle of 25°. Explain your
construction and measure the distance of the point (or points) from O. Find,
by drawing o/* otherwise, the position of the point on OY at which AB
subtends the greatest possible angle.
t4. Two chords of a circle AB, CD intersect at a point X.
If XB=XD, show that AB = CD, and that ACBD is a trapezium.
+5. A given point D lies between two given straight lines AB and AC.
Find a construction for a line through D terminated by AB and AC, such
that D is one of its points of trisection. Prove also that there are two such
lines,
6. Draw two straight lines AB, AC enclosing an angle of 48°. Take a
point D in AB such that AD = 2-6in. Construct a circle DEF to touch AB
in D and also to touch AC.
Construct another circle to touch AB, AC and also to toach the circle
DEF. State the steps of this construction. .
384 REVISION PAPERS
7. Tlie mouth of a stable bucket (Fig. ) is 13 inches in diameter, the
base 8^ inches in diameter, and the slant side measures
9 inches. Draw a vertical section through the axis of the .•- r>
bucket, and find by calculation the height of the bucket and \ /
the height of the cone formed by producing the slant sides V/
beyond the base. - \/
V
Assuming the volume of a cone to be a third of the
product of the base and the height, find how many gallons the bucket will
hold. (1 cubic foot = 6^ gallons.)
PAPER XXV (ON Books I— IV).
l.~ A cubical block of edge 4 ft. rests on a table ; the base is ABCD, and
P, Q, R, S are the corners above A, B, C, D respectively. If the edge CD is
raised 2 ft., AS remaining on the table, find by drawing to scale the height
of R above the table, and the inclination of A R to the table.
2. A paper cone (like an electric light reflector) is slit down straight
from the vertex to the base, and opened out flat ; sketch the figure produced,
and name it.
t3. ABC is a triangle inscribed in a circle. BD, CE are drawn perpen-
dicular to AC, AB, and are produced to cut the circle in F and G. Prove
that FG is parallel to DE.
t4. The side BC of a triangle ABC is divided at D so that BD = 2DC;
AD is bisected at E; and CE meets AB in F. Prove that CE=2EF.
5. Draw a straight line AB of length 5 cm. Find a point P at which
AB subtends an angle of 54° and such that AP is 4 cm. Measure the
distance PB.
6. From a point P outside a circle of radius a, are drawn the tangent
PQ (of length a;), and the straight line PAB through the centre cutting the
circle in points A and B, A being nearer to P and PA being of length y.
Write down the relation connecting the lengths of the lines PQ, PA, PB, and
express it in terms of x, y, a.
If the circle is taken as representing a section of the earth through its
centre, PQ will be the range of vision of a person situated at a height y above
the surface. Take a = 4000 miles, use the relation in the approximate form
2ay=x% and find in miles and in feet to what height it is necessary to
ascend in order to have a range of vision of 50 miles.
REVISION PAPERS 385
PAPER XXVI (ON Books I— IV).
1. "What is the locus of centres of
(a) circles which touch a fixed line PQ at a fixed point P;
(b) circles of radius 3" which touch a fixed line PQ ?
Also of the following points: —
(c) the points of contact of tangents drawn from a fixed point to
a fixed sphere;
(d) points on the earth which are 3000 nautical miles N. of the
equator?
2. The figure represents a bridge, whose span A B is 80 ft., supported
on an arch in the form of an arc of a circle.
AC = 16 ft. = BD. Let r ft. denote the rad.us of
the circular arc and 6 ft. the heij^ht of the
roadway AB above the highest point of the
arch. Find an algebraic equation connectinfj r
with 6. Then use it to calculate (i) the value of b when r=65, (ii) the value
ofrwheni = l.
3. A stick, 4' long, is leant up against a cylindrical wooden roller of
18* diameter. The axis of the cyUnder is perpendicular to the vertical
plane in which the stick lies. The point where the stick touches the ground
is 3' away from the point of contact of the cylinder with the ground.
Without drawing to scale, find (i) the distance between the two points of
contact which the stick makes respectively with the ground and the roller,
(ii) the distance of the axis of the roller from the point of contact of the
stick with the ground.
t4. ABCD is any parallelogram. From A a straight line is drawn
cutting BC in E and BD in F. Prove that AF : FE = BO : CE.
t5. The side BC of an equilateral triangle ABC is produced to D so
that CD = BC. Prove that the perpendiculars to AC drawn through B and
C respectively trisect AD.
6. Constrnct a triangle ABC, in which BC is 2", the angle BAC is 60",
and the sided AB and AC are in the ratio 3 : 4.
25—5
386 REVISION PAPERS
PAPER XXVII (ON Books I— IV).
tl. ACC, BOD are chords of a circle; the tangents at A and B meet at
P ; the tangents at C and D meet at Q. Prove that the sum of the angles P
and Q is twice the angle BOC.
2. AB is a diameter of a circle of radius 5 cm. Draw a chord CD of the
circle perpendicular to AB and 6 cm. in length. Also, through O, the point
of intersection of AB and CD, draw a chord of the circle 8 cm. long. State
the steps of your construction.
t3. ABC is a triangle. O is the middle point of BC, and AO is
produced to T. The lines bisecting internally the angles BOT, COT cut
externally the sides AB, AC in D, E. Prove that DE is parallel to BC.
4. Construct a square equal in area to an equilateral triangle of side
3 inches. Measure the side of the square.
f5. D is the middle point of the base BC of a triangle ABC, E is a
point in AC such that the angle ADE is equal to the angle ABC. EF is drawn
parallel to BC and meeting AD in F. Prove that the rectangle AF . FD is
equal to the square on EF.
6. Draw a circle of radius 5 cm. and take a point O at a distance of
10 cm. from its centre. From O draw a line cutting the circle in P and Q
each that P is the middle point of OGl
PAPER XXVIII (ON Books I— IV).
1. A and B are two forts 5 miles apart. The effective range of A's guns
is 3^ miles, and of B's, 3 miles. Draw the circles bounding the area covered
by the two forts, and let C be one of the points of intersection of these
circles. An enemy's ship comes to C so as to be able to bombard a town
lying between A and B without being within the range of the guns from
either fort. By measurement, find bow far C is from the coast-line.
Measure the angles CAB and CBA and hence calculate approximately the
number of square miles covered by the zone of effective fire from the two
forts.
t2. O is the centre of a circle of 2 in. radius, A is a point 8 in. from O,
AP is a tangent from A. If OP is produced to Q so that PQ=2AP, prove
that the circle whose centre is A and radius AQ will touch the given circle.
REVISION PAPERS 387
t3. In a triangle ABC, AB = AC and / A is a right angle; if the bisector
of / C cuts AB in D, prove that BD2=2DA«.
4. Show how to construct a triangle similar to and double the area of a
given triangle.
5. A point R is taken on the side AB of a triangle ABC of area z, so
that AR = a;. AB, where x>l/2. RQ is drawn parallel to BC to meet AC at
Q, RH parallel to AC to meet BC at H, and QK parallel to AB to meet BC
at K. Prove that the areas ofARQandBRH are x-z and ( 1 - r) "^z respectively
(notice that they are similar to ABC), and find in similar form the area of
CQK; use these results to find the area of QRHK. Verify yoiu: result for
QRHK by giving x the values 1 and 1/2.
t6. A B is an arc of a circle, of radius 4 inches, subtending an angle of
45° at its centre C. Let the tangents at A and B meet at T, aud produce
CA and BT to meet at S. Prove that AS = AT = TB, and, denoting each of
these equal lengths by x inches, calculate the value of x.
Now suppose that AT and STB represent two railway lines crossing each
other at T. The points A and B are connected by a loop-line represented by
the arc AB of radius 400 yards. Determine in yards the distances of the
points A and B from the crossing T.
PAPER XXIX (ON Books I— IV).
1. The two equal circles, centres P and Q, are so drawn that each
passes through the centre of the other : they intersect at A aud B.
The radius of each circle is r. Prove that (i) arc AQB subtends an
angle of 120° at P. (ii) the sector of arc AQB and centre P is of area — - .
r2\/3
(iii) AB=rv3. (iv) area of aPAB = — - — . (v) the area common to the
two circles = j-^ i— — j . (vi) the ratio of this common area to the area
of each circle is 0"39.
t2. ABCD is a quadrilateral in a circle. One side BC is produced to E.
Prove that the bisectors of the angles BAD, DCE meet on the circumference.
3. Draw a straight line OBC, making 0B=^2-5cm., OC = 6-4cm.
Through O draw a line OA making the angle AOB = 42°. Then draw a
circle passing through B and C and touching OA. (Describe the steps
of your construction.)
388 REVISION PAPERS
+4. The bisector of the angle A of a triangle ABC meets BC at D; and
DE, DF are drawn respectively perpendicular to the external bisectors of the
angles B, C, to meet AB, AC produced at E, F respectively. Prove that EF
is parallel to BC.
5. B, C, D are three points in order on a straight line, such that BC = 2",
and CD = 4", Construct a triangle ABC, such that AB + AC = 5" and the
bisector of the external angle at A passes through D.
PAPER XXX (ON Books I— IV).
1. Draw a circle whose diameter is 7 cm. long, and a line 2-5 cm. distant
from the centre. Mark off on the line a point which is 6 cm. distant from tlie
centre, and then describe a circle touching the line at this point and also
touching the circle.
2. C is the middle point of a straight line A B, 12 cm. long. On AC, CB
and AB semicircles are described. What is the radius of the circle which
can be described in the space enclosed by the three semicircles touching all
three of them ?
f 3. Draw any triangle ABC. It is required to inscribe in this triangle
an equilateral triangle one side of which is parallel to AB, and the opposite
vertex lies on AB.
, Show how this can be done by employing the properties of similar
triangles.
+4. At two points A, B of a straight line perpendiculars AC, BD are
erected and AD, BC meet in a point E; from E a perpendicular EF is drawn
to AB. Prove that
EF~AC"*'BD'
+5. ABCD is a rhombus; a straight line through C meets AB and AD,
both produced, at P and Q respectively. Prove that PB : DQ= AP2 : AQ2.
+6. AD is the bisector of the angle BAC of the triangle ABC, and F is
the middle point of AB; also AD and CF intersect in P, and PH is parallel
to AB cutting BC in H, Prove that
PHAC
BH~BC*
REVISION PAPERS
389
PAPER XXXI (ON Books I— IV).
■fh O, A, B, C, D are points on the circumference of a circle such that
the angles AOB, BOC, COD are equal. Prove that the angle between the
chords AC, OB is equal to that between AD, OC.
•f2. Through the vertices B, C of a triangle ABC two parallel lines BL
and CM are drawn, meeting any straight line through A in L and M
respectively. If LO is drawn parallel to AC and meets BC in O, prove that
CM is parallel to AB.
t3. Prove that, if circles are described passing through two given points
A and B and cutting a given circle in P and Q., the chord PQ cuts AB in a
fixed point.
4. PX, PY are two straight lines intersecting at an angle 45° : A, B are
points on PX such that PA= AB=5 cm. Constmct in one figure the points
on PY at which the segment AB subtends angles 20°, 30°, 40°, explaining
your method. How would you find the point K on PY at which AB subtends
the greatest angle? Construct this point in any way you please, and measure
this greatest angle.
5. CDEF is a rectangle (Fig.) in which CD = a and CF = 6. A circle,
whose centre is at O, the middle point
of EF, is described to cut CF at H and
DE at K. If the radius r of the circle
is such that CH = HO, express r in terms
of a and b.
Suppose that the two right-angled tri-
angles A and B are cut away from the
rectangle and placed in the positions
Aj and Bj, thus converting the rectangle
into an equilateral hexagon. Show that if
a/2= &/^3, the resulting hexagon is regular,
i.e. has also its angles all equal.
PAPER XXXII (ON Books I— IV).
1. ABC is any triangle. Show how to inscribe a square PQRS in the
triangle so that P lies on AB, Q, on AC, and the side RS on BC.
390
REVISION PAPERS
2. Draw a circle 3 inches in diameter and place in it a chord AB
2*5 inches in length; draw the diameter BC and produce BC to D so that D
is 1 inch distant from the circle; and through D draw DE perpendicular to
BD. Then draw a circle touching DE and also touching the former circle at
A. State the steps of jour construction.
t3. If ABCD is a cyclic quadrilateral and AB, DC be produced to meet
at Q, and BC, AD to meet at R, prove that QP, RP, the bisectors of the
angles BQC, CRD, are at right angles to one another.
t4. A straight line is drawn parallel to the side BC of a triangle ABC
cutting AC in P and AB in Q; BP cuts CQ in T. Prove that AT produced
bisects BC.
5. In Fig. ABCD is a cross-section showing a railway cutting made in
ground, the surface of which slopes in a dirfection at right angles to the
cutting as shown by the line AED. BC is the trace of the horizontal plane
on which the track will be laid, and EF is a vertical centre line bisecting BC.
The side AB of the cutting is to have the same slope to the horizontal as the
side CD.
Calculate the dimensions x and y. Find the volume, in cubic yards, of
the earth which must be excavated per chain length of track.
1i-X~^t<- 20
LIST OF DEFINITIONS 391
List of Definitions.
Acnte ansle, obtuse angle, reflex angle. An angle less than a right
angle is said to be acute ; an angle greater than a right angle and less than
two right angles is said to be obtuse (p. 64) ; an angle greater than two and
less than four right angles is said to be reflex, (p. 250.)
Acute-angled triangle. A triangle which has all its angles acnte is
called an acute-angled triangle, (p. 82.)
Adjacent angles. When three straight lines are drawn from a point,
if one of them is regarded as lying between the other two, the angles which
this line makes with the other two are called adjacent angles, (p. 64.)
Alternate angles, corresponding angles. (See p. 70.)
Altitude. See triangle, parallelogram.
Angle. Wh^n two straight lines are drawn from a point, they are said
to form, or contain, an angle. The point is called the vertex of the angle,
and the straight lines are called the anus of the angle, (p. 64.)
Angle in a segment. An angle in a segment of a circle is the angle
subtended by the chord of the s^ment at a point on the arc. (p. 253.)
Angle of elevation, of depression. (See p. 48.)
Arc of a circle. (See p. 218.)
Base. See triangle, parallelogram.
Cbord of a circle. (See p. 218. )
Circle. A circle is a line, lying in a plane, such that all points in the
line are equidistant from a certain fixed point, called the centre of the
circle. The fixed distance is called the radius of the circle, (p. 217.)
Clrcumcentre. The centre of a circle circumscribed about a triangle is
called the clrcumcentre of the triangle, (p. 224.)
Circumference of a circle. (See p. 215.)
Circumscribed polygon. If a circle touches all the sides of a polygon,
it is said to be inscribed in the polygon ; and the polygon is said to be
circumscribed about the circle, (p. 224.)
Common tangents, exterior and interior. (See p. 263.)
ConcycUc. Points which lie on the same circle are said to be concyclic.
(p. 257.)
392 LIST OF DEFINITIONS
Oon«. (See p. 215.)
Congruent. Figures which are equal in all respects are said to be
congruent, (p. 85.)
Contact of circles. If two circles touch the same line at the same point,
they are said to touch one another, (p. 245.)
Converse. (See p. 76.)
Coordinates. (See p. 152.)
Cube. (Seep. 42.)
Cuboid. (See p. 43.)
Cyclic quadrilateral. If a quadrilateral is such that a oircle can be
circumscribed about it, the quadrilateral is said to be oyclio. (p. 261.)
Cylinder. (See p. 217.)
BiagonaL See qnadrilateral.
Diameter of circle. (See p. 218.)
Envelope. If a line moves so as to satisfy certain conditions, the curve
which its different positions mark out is called its envelope. (See p. 293.)
Equilateral triangle. A triangle which has all its sides equal is called
an equilateral triangle, (p. 82.)
Equivalent. Figures which are equal in area are said to be equivalent,
(p. 168.)
Escribed circles of a triangle. (See p. 244.)
Fourth proportional. If x is such a magnitude that a ih=c : x, then
X is called the fourth proportional to the three magnitudes a, 6, c. (p. 309.)
Heigbt. See triangle, parallelogram.
Heptagon. See pentagon.
Hexagon. See pentagon.
Hypotenuse. See right-angled triangle.
Inscribed polygon. If a circle passes through all the vertices of a
polygon, the circle is said to be circumscribed about the polygon ; and the
polygon is said to be inscribed in the circle, (p. 224.)
Isosceles triangle. A triangle which has two of its sides equal is called
fin isosceles triangle, (p. 82.)
LIST OF DEFINITIONS 393
Xalne. The boundary between any two parts of a surface is called a line.
A line has length bat no breadth or thickness.
Iiocns. If a point moves so as to satisfy certain conditions, the path
traced out by the point is called its locus, (p. 144.)
Major arc, minor are. (See p. 218.)
Major segment, minor segment. (See p. 253.)
Mean proportionaL If x is such a magnitude that a:x=x :b, then x
is called the mean proportional between a and b. (p. 331.)
Median. See triangle.
Wet. (See p. 27.)
Obtnse angle. See acute angle.
Obtuse-angled triangle. A triangle which has one of its angles an
obtuse angle is called an obtuse-angled triangle, (p. 81.)
Octagon. See pentagon.
Parallel straight lines are straight lines in the same plane, which do
not meet however far they are produced in either direction, (p. 70.)
Parallelogram. A quadrilateral with its opposite sides parallel is called
a parallelogram, (p. 73.)
Any side of a' parallelogram may be taken as the base. The perpen-
dicular distance between the base and the opposite (parallel) side is called
the height, or altitude, (p. 167.)
Pentagon, hexagon, heptagon, octagon, etc. — a polygon of 5, 6,
7, 8, ... sides; 5-gon, 6-gon, 7-gon, 8-gon..,. (p. 18.)
Perimeter. The perimeter of a figure is the sum of its sides, (p. 18.)
Perpendicular. See right angle.
Plane. A surface which is such that the straight line joining every pair
of points in it lies wholly in the surface is called a plane surface, or, briefly,
a plane.
Point. ^ The boundary between any two parts of a line is called a point.
A point has no length, breadth, or thickness, but it has position.
Polygon. A plane figure bounded by straight lines is called a polygon,
or, a rectilinear figure, (p. 83.)
Prism. (See p. 44.)
Projection. (See p. 210.)
Proportion. (See pp. 302, 303.)
394 LIST OF DEFINITIONS
Pyramid. (See p. 27.)
QnadrllateraL A plane figure bounded by four straight lines is called
a quadrilateral, (p. 73.)
The straight lines which join opposite comers of a quadrilateral are
called its diagonals, (p. 73.)
K«din«. See circle.
Batlo. (See pp. 302, 303.)
Bectangle. A parallelogram which has one of its angles a right angle
is called a rectangle, (p. 135.)
Rectilinear figure. A figure contained by straight lines.
BeducUo ad absnrdtun. (See p. 122.)
Beflex ang^e. See acnte angle.
Begular polygon. A polygon which has all its sides equal and all its
angles equal is called a regular polygon, (p. 84.)
KbombTU. A parallelogram which has two adjacent sides equal is
called a rhombus, (p. 135.)
Bigbt angle, perpendicular. When one straight line stands on another
straight line and makes the adjacent angles equal, each of the angles is called
a right angle ; and the two straight lines are said to be at right angles, or
perpendicular to one another, (p. 64.)
Bight-angled triangle. A triangle which has one of its angles a right
angle is called a right-angled triangle.
The side opposite the right angle is called the bypotennse. (p. 81.)
Scalene triangle. A triangle which has no two of its sides equal is
called a scalene triangle, (p. 82.)
Sector of a circle. (See p. 219.)
Segment of a circle. (See p. 219.)
«
Semicircle. (See p. 219.)
Similar. Figures which axe equiangular to one another and have their
corresponding sides proportional are said to be similar, (p. 313.)
Solid. Any limited portion of space is called a solid. A solid has
length, breadth and thickness, (pp. 55 — 59.)
Spbere. (See p. 217.)
LIST OF DEFINITIONS 395
Square. A rectangle which has two adjacent sides equal is called
a square, (p. 135.)
Straight line. If a line is such that any part, however placed, lies
wholly on any other part if its extremities are made to fall on that other
part, the line is called a straight line.
Supplementary angles. When the sum of two angles is equal to two
right angles, each is called the supplement of the other, or is said to be
supplementary to the other, (p. 66.)
Sur£ace. The boundary between two parts of space is called a surface.
A surface has length and breadth but no thickness.
Symmetry. (See p. 51.)
Tangent. A tangent to a circle is a straight line which, however far
it may be produced, has one point, and one only, in common with the
circle.
The tangent is said to toucb the circle ; the common point is called
the point of contact, (p. 238.)
Tetrahedron. (See pp. 26 — 27.)
Third proportionaL If x is such a magnitude that a:b = b:x, then x
is called the third proportional to the two magnitudes a, b. (p. 309.)
Trapezium. A quadrilateral which has only one pair of sides parallel
is called a trapezium. A trapezium in which the sides that are not parallel
are equal is called an isosceles trapezium, (p. 135.)
Triangle. A plane figure bounded by three straight lines is called a
triangle, (p. 73.)
Any side of a triangle may be taken as base. The line drawn perpen-
dicular to the base from the opposite vertex is called the height, or
altitude, (p. 172.)
The straight line joining a vertex of a triangle to the mid-point of the
opposite side is called a median, (p. 110.)
Vertically opposite angles. The opposite angles made by two inter-
secting straight lines are called vertically opposite angles {vertically opposite
because they have the same vertex), (p. 68.)
Vertices. The corners of a triangle or polygon are called its vertices,
(p. 16.)
Wedge. A 3-sided prism. (See p. 44.)
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