SURATION.
..,,-^r::
/*
METRICAL GEOMETRY.
AN
ELEMENTARY TREATISE
ON
MENSURATION
v
BY
GEORGE BRUCE HALSTED
A.B., A.M., AND EX-FELLOW OF PRINCETON COLLEGE; PH.D. AND
EX-FELLOW OF JOHNS HOPKINS UNIVERSITY ; PROFESSOR OF
PURE AND APPLIED MATHEMATICS, UNIVERSITY
OF TEXAS.
FOURTH EDITION.
BOSTON:
PUBLISHED BY GINN & COMPANY.
1890.
CJ>
Entered according to Act of Congress, in the year 1881, by
GEORGE BRUCE HALSTED,
in the office of the Librarian of Congress, at Washington.
TYPOGRAPHY BY J. S. CUSHING & Co., BOSTON, U.S.A.
PRESSWORK BY GINN & Co., BOSTON, U.S.A.
INSCRIBED TO
J. J. SYLVESTER,
A.M., Cam.; F.R.S., L. and E.; Cor. Mem. Institute of France; Mem.
Acad. of Sciences in Berlin, Gottingen, Naples, Milan, St. Peters-
burg, etc. ; LL.D., Univ. of Dublin, U. of E. ; D.C.L.,
Oxford; Son. Fellow of St. John's Col., Cam.;
Savilian Professor of Geometry in the
University of Oxford;
IN TOKEN OF THE INESTIMABLE BENEFITS DERIVED FROM
TWO YEARS' WORK WITH HIM,
BY THE AUTHOR.
PREFACE.
THIS book is primarily the outcome of work on the subject
while teaching it to large classes.
A considerable part, it is believed, is entirely new.
Special mention must be made of the book's indebtedness to Dr.
J. W. DAVIS, a classmate with me at Columbia School of Mines; also
to Prof. G. A. WENTWORTH, who has kindly looked over the proofs.
But if the book be found especially accurate, this is due to the pains-
taking care of my friend H. B. FINE, Fellow of Princeton.
Any corrections or suggestions relating to the work will be thank-
fully received.
GEORGE BRUCE HALSTED.
PRINCETON, NEW JERSEY,
May 12, 1881.
NOTE TO THE THIRD EDITION.
WHATEVER changes have been suggested during continued use
with successive classes in Princeton College and Science School,
are embodied in the present edition, which we have striven to
render free from errors.
October, 1883.
PREFACE TO THE FOURTH EDITION.
IT may be proper to mention that this book has had the honor
of being drawn upon by Prof. WILLIAM THOMSON for his
article " Mensuration " in the Ninth Edition of the Encyclopaedia
Britannica. He has taken from it, among other novelties, the
steregon, the steradian, and the treatment of solid angles associ-
ated with them. But he missed the gem of the book, and the
Encyclopaedia lacks the finest formula, the most surprising rule
of the whole subject, the new or two-term prismoidal formula,
the rule given on page 130.
March, 1889.
2004 WHITIS AVENUE,
AUSTIN, TEXAS.
CONTENTS.
MENSURATION.
INTRODUCTION
THE METRIC SYSTEM
NOTATION AND ABBREVIATIONS
CHAPTER I.
THE MEASUREMENT OF LINES.
§ (A). — STRAIGHT LINES.
ILLUSTRATIVE PROBLEMS 5
(a) To measure a line the ends of which only are accessible . 5
(/3) To find the distance between two objects, one of which
is inaccessible 5
(7) To measure a line when both ends of it are inaccessible . 6
(8) To measure a line wholly inaccessible .... 6
§ (B). — STRAIGHT LINES IN TRIANGLES.
Article.
I. RIGHT-ANGLED TRIANGLES 6
1. To find hypothenuse 6
2. To find side 7
II. OBLIQUE TRIANGLES 7
3. Obtuse 8
4. Acute 8
5. Given perpendicular ........ 9
6. To find medials 9
III. STRAIGHT LINES IN SIMILAR FIGURES . . .10
7. To find corresponding line ....... 10
IV. CHORDS OF A CIRCLE 10
8. To find diameter , 10
Vlll CONTENTS.
Article. Page.
9. To find height of arc 11
10. Given chord and height 12
11. Given chord and radius, to find chord of half the arc . . 12
12. To find circumscribed polygon 13
§ (C).— METHOD OF LIMITS.
V. DEFINITION OF A LIMIT 14
13. Principle of Limits ........ 15
14. The length of the curve 16
§ (D). — THE RECTIFICATION OF THE CIRCLE.
15. Length of semicircumference 19
16. Circumferences are as their radii . . . . . .20
VI. LINES IN ANY CIRCLE 21
17. Value of * 21
18. To find circumference ........ 21
19. Value of diameter 21
CHAPTER II.
THE MEASUREMENT OF ANGLES.
§ (E). — THE NATURAL UNIT OF ANGLE.
VII. ANGLES ARE AS ARCS 22
20. Numerical measure of angle ....... 22
VIII. ANGLE MEASURED BY ARC 23
§(F). CIRCULAR MEASURE OF AN ANGLE.
21. Angle in radians 24
22. To find length of arc 24
23. To find number of degrees in arc 25
IX. ARCS WHICH CORRESPOND TO ANGLES . . . .25
24. Angle at center 25
25. Inscribed angle 26
26. Angle formed by tangent and chord 26
27. Angle formed by two chords 26
28. Secants and tangents 26
CONTENTS. IX
Article. Page.
29. Given degrees to find circular measure 26
30. Given circular measure to find degrees . „ . . .27
31. Given angle and arc to find radius 27
X. ABBREVIATIONS FOR AREAS , 28
CHAPTER III.
THE MEASUREMENT OF PLANE AREAS.
§(G). — PLANE RECTILINEAR FIGURES.
XI. MEASURING AREA OF SURFACE 29
32. Area of rectangle 29
33. Area of square 32
34. Area of parallelogram 33
35. Area of triangle (given altitude) ...... 33
36. Area of triangle (given sides) 34
37. Radius of inscribed circle 35
38. Radius of circumscribed circle ...... 36
39. Radius of escribed circle 37
XII. TRAPEZOID AND TRIANGLE AS TRAPEZOID . . . 38
40. Area of trapezoid ......... 38
XIII. COORDINATES OF A POINT 39
XIV. POLYGON AS SUM OF TRAPEZOIDS . . . .40
41. To find sum of trapezoids 40
42. Area of any polygon ........ 43
43. Area of quadrilateral 45
44. Area of a similar figure ....... 48
XV. CONGRUENT AND EQUIVALENT 48
XVI. PROPERTIES OF REGULAR POLYGON . . . .49
45. Area of regular polygon ....... 49
46. Table of regular polygons 50
§ (H). — AREAS OF PLANE CURVILINEAR FIGURES.
47. Area of circle 51
48. Area of sector ......... 51
49. Area of segment 52
50. Circular zone 54
51. Crescent 54
CONTENTS.
Article. Page.
52. Area of annulus 54
53. Area of sector of annulus 55
XVII. CONICS 56
54. Area of parabola 57
55. Area of ellipse 59
CHAPTER IV.
MEASUREMENT OF SURFACES.
XVIII. DEFINITIONS RELATING TO POLYHEDEONS . . 61
56. Faces plus summits exceed edges by two . . . .61
§(!). — PRISM AND CYLINDER.
XIX. PAEALLELEPIPED AND NORMAL . . . .62
57. Mantel of prism 63
XX. CYLINDEIC AND TEUNCATED 64
58. Mantel of Cylinder 64
§ (J). — PYRAMID AND CONE.
XXI. CONICAL AND FEUSTUM 66
59. Mantel of pyramid 66
60. Mantel of cone 67
61. Mantel of frustum of pyramid ...... 68
62. Mantel of frustum of cone ....... 69
63. Frustum of cone of revolution 70
§ (K). — THE SPHERE.
XXII. SPHEEE A SUEFACE, GLOBE A SOLID . . .71
64. Area of a sphere ......... 71
XXIII. SPHEEICAL SEGMENT AND ZONE . . . .73
65. Area of a zone 73
66. Theorem of Pappus 74
67. Surface of solid ring 74
CONTENTS. XI
§ (L). — SPHERICS AND SOLID ANGLES.
Article. Page.
XXIV. STEEEGON AND STEEADIAN 77
68. Area of a lune 78
XXV. SOLID ANGLE MADE BY Two, THREE, OE MOEE
PLANES 79
XXVI. SPHEEICAL PYEAMID 80
69. Solid angles are as spherical polygons 80
XXVII. SPHEEICAL EXCESS 80
70. Area of spherical triangle 80
71. Area of spherical polygon 81
TABLE or ABBEEVIATIONS , 83
CHAPTER V.
THE MEASUREMENT OF VOLUMES.
§ (M).— PRISM AND CYLINDER.
XXVIII. SYMMETEICAL AND QUADEE . . . .84
XXIX. VOLUME AND UNIT OF VOLUME . . . .84
XXX. LENGTHS, AEEAS, AND VOLUMES AEE RATIOS . 84
72. Volume of quader 85
XXXI. MASS, DENSITY, WEIGHT 86
73. To find density 87
74. Volume of parallelepiped 88
75. Volume of prism . .90
76. Volume of cylinder 91
77. Volume of cylindric shell 92
§ (N). — PYRAMID AND CONE.
XXXII. ALTITUDE OF PYEAMID 93
78. Parallel sections of pyramid 93
79. Equivalent tetrahedra 94
80. Volume of pyramid . . 95
81. Volume of cone 97
Xll CONTENTS.
$ (O). — PRISM ATOID.
Article: Page.
XXXIII.-XLI. DEFINITIONS RESPECTING PRISMATOID, 98-101
82. Volume of prismatoid 101
83. Volume of frustum of pyramid 104
84. Volume of frustum of cone 106
85. Volume of ruled surface 107
86. Volume of wedge 108
87. Volume of tetrahedron 109
§ (P). SPHERE.
88. Volume of sphere 110
89. Volume of spherical segment 112
XLII. GENERATION OF SPHERICAL SECTOR . . .113
90. Volume of spherical sector 113
XLIII. DEFINITION OF SPHERICAL UNGULA . . . 114
91. Volume of spherical ungula ...... 114
92. Volume of spherical pyramid 115
§ (Q). — THEOREM OF PAPPUS.
93. Theorem of Pappus 115
94. Volume of ring 116
§ (R). — SIMILAR SOLIDS.
XLIV. SIMILAR POLYHEDRONS 118
95. Volume of similar solids 118
§ (S). — IRREGULAR SOLIDS.
96. By covering with liquid 119
97. Volume by weighing 120
98. Volume of irregular polyhedron 120
CHAPTER VI.
THE APPLICABILITY OF THE PEISMOIDAL FORMULA.
99. Test for applicability 121
CONTENTS. Xlll
§ (T). — PEISMOIDAL SOLIDS OF REVOLUTION.
Article. Page.
XLV. EXAMINATION OF THE DIFFERENT CASES . . 124
§ (U).— PRISMOIDAL SOLIDS NOT OP REVOLUTION.
XL VI. DISCUSSION OF CASES 127
CHAPTER VII.
THE NEW PRISMOIDAL FORMULA.
§ (V). — ELIMINATION OF ONE BASE.
100. Prismatoid determined from one base . . . .130
CHAPTER VIII.
APPROXIMATION TO ALL SURFACES AND SOLIDS.
§ (W). — WEDDLE'S METHOD.
101. Seven equidistant sections 131
CHAPTER IX.
MASS-CENTER.
$(X). — FOR HOMOGENEOUS BODIES.
102-104. Introductory . . 136
XLVII. BY SYMMETRY 136
105. Definition of symmetric points 136
106-117. Direct PQ deductions 137
118. Triangle .... 138
119. Perimeter of triangle 138
120-122. Tetrahedron . . .138
123. Pyramid and cone 138
XLVIII. THE MASS-CENTER OF A QUADRILATERAL . 139
124. Definition of a sect . . . 139
125. Of an opposite on a sect 139
126. !*C by opposites 139
127. Geometric method of centering quadrilateral . . .139
128. The mass-center of an octahedron 140
CONTENTS.
Article. Page.
XLIX. GENERAL MASS-CENTER FORMULA . . .141
129. Mass-center of any prismatoid 142
130. Test for applicability 142
131. Average haul 144
132. The t*C of a consecutive series 144
EXERCISES AND PROBLEMS IN MENSURATION.
[These are arranged and classed in accordance with the above
132 Articles] .145
LOGARITHMS 225
A TEEATISE ON MENSTJKATION.
TREATISE ON MENSURATION.
INTRODUCTION.
MENSURATION is that branch of mathematics which has
for its object the measurement of geometrical magnitudes.
A Magnitude is anything which can be conceived of as
added to itself so as to double, or of which we can form
the multiples.
Measurement is the process of ascertaining the ratio which
one magnitude bears to some other chosen as the standard;
and the measure of a magnitude is this ratio expressed in
numbers. Hence, we must refer to some concrete standard
to give our measures their absolute meaning.
The concrete standard is arbitrary in point of theory, and
its selection a question of practical convenience.
For the continuous quantity, space, with which we chiefly
have to deal, the fundamental unit, the unit for length, is
so much of a straight line as is contained between two
marked points on a special bar of platinum deposited in
the French archives. This, called the Meter, we choose
because of the advantages of the metric system, which has
correlated units, applies only a decimal arithmetic, and
has a uniform and significant terminology to indicate the
multiples and submultiples of a unit.
MENSURATION.
THE METRIC SYSTEM
convenes to designate multiples by prefixes derived from
the Greek numerals, and submultiples by prefixes from the
a Latin numerals ; as follows :
PREFIX.
DERIVATION.
MKANINII AS USED.
Greek.
inyria-
/j-vpids.
ten thousand.
kilo-
^l\iot.
a thousand.
hecto-
liar**.
a hundred.
deka-
5fKCL.
ten.
Latin.
deci-
decem.
one-tenth.
centi-
centum.
one-hundredth.
niilli-
inille.
one-thousandth.
So a millimeter (mm) is one-thousandth of a
meter.
Thus we are given a number of subsidiary
units. For any particular class of measure-
ments, the most convenient of these may be
chosen.
The kilometer (km) is used as the unit of dis-
tance ; and along roads and railways are placed
kilometric poles or stones.
The centimeter (cm) has been chosen as the
scientific unit for length.
A chief advantage of this decimal system of measures is,
that in it reduction involves merely a shifting of the deci-
mal point.
INTRODUCTION.
NOTATION AND ABBREVIATIONS
TO BE USED IN TIIIS BOOK.
Large letters indicate points ; thus, A, £, and C denote
the three angular points of a triangle, or C may denote the
center of a circle, while A and B are on the circumference ;
then AB will denote the chord joining A to B; and, gen-
erally, AD means a straight line terminated at A and D.
In the formula?, small letters are used to denote the
numerical measures of lines ; so that ab, as in common alge-
bra, denotes the product of two numbers.
The following choice of letters is made for writing a
formula :
[TABLE FOR REFERENCE.]
k = chord.
a, b, and c are the sides of
any triangle, respectively, op-
posite the angular points A,
JB, C.
If the triangle is right-angled,
a = altitude, b = base, c = hy-
pothenuse.
In regard to a circle, c = circum-
ference, d= diameter.
e = spherical excess.
g = number of degrees in an an-
gle or arc.
g° means expressed in degrees
only, g/ minutes, gf/ seconds.
h = height.
i = medial.
j = projection.
I = length.
m = meter.
n = any number.
6 = perigdn.
p = perimeter.
r — radius.
s = £ (a + b + c).
t = tangent.
u = circular measure.
v = a discrete variable.
w = width.
= coordinates of a point.
/. = therefore.
a < b = a is less than b.
CHAPTER I.
THE MEASUREMENT OF LINES.
§ (A). — STRAIGHT LINES.
An accessible piece of a straight line is practically meas-
ured by the direct application of a standard suitably
divided.
If the straight line contain the standard unit n times,
then n is its numerical measure or length.
But, properly speaking, any measurement by actual
application of a standard is imperfect and merely ap-
proximate. Few physical measurements of any kind are
exact to more than six figures, and that degree of accuracy
is very seldom obtainable, even by the most delicate instru-
ments. Thus, in comparing a particular meter with the
standard meter at Paris, a difference of a thousandth of a
millimeter can be detected.
In four measurements of a base line at Cape Comorin, it
is said the greatest error was 0.077 inch in 1.68 miles, or one
part in 1,382,400 ; and this is called an almost incredible
degree of accuracy.
When we only desire rough results, we may readily shift
the place of the line to be measured, so as to avoid natural
obstacles. Still, under the most favorable circumstances,
all actual measurements of continuous quantity are only
approximately true. But such imperfections, with the de-
vised methods of correction, have reference to the physical
measurement of things ; to the data, then, which in book-
questions we suppose accurately given.
THE MEASUREMENT OF LINES.
ILLUSTRATIVE PROBLEMS.
(a) To measure a line the ends of which only are acces-
sible.
Suppose AB the line.
Choose a point C from which
A and B are both visible.
Measure AC, and prolong it
until CD = AC. Measure
BC, and prolong it until CE
= £0. ThenED = AB. E^- ±D
Ww. 95 & 150 ; (Eu. I. 15 & 4 ; Cv. I. 23 & 76).
NOTE. Ww. refers to Wentworth's Geometry, revised edition, 1888.
Eu. refers to Todhunter's Euclid, new edition, 1879. Cv. refers to
Chauvenet's Geometry. These parallel references are inserted in the
text for the convenience of students having either one of these geom-
etries at hand. References to preceding parts of this Mensuration
will give simply the number of the article.
(/?) To find the distance between two objects, one of which
is inaccessible.
Let A and B be the two objects, separated by some
obstacle, as a river.
From A measure any
straight line AC. Fix
any point D in the
direction A B. Pro-
duce AC to F, making
CF == AC; and pro-
duce DC to E, making
CE = CD. Then find
the point G at which the directions of BC and FE inter-
sect ; that is, find the point from which C and B appear in
MENSURATION.
one straight line, and E and F appear in another straight
line. Then the triangles ACD and CEF are congruent,
and therefore ABC and CFG \ whence FG = AB.
Ww. 150 & 147 ; (Eu. I. 4 & 26 ; Cv. I. 76 & 78). '
Hence, we find the length of AB by measuring FG.
(y) To measure a line when both ends of it are inacces-
sible.
At a point C, in the
accessible part of AB,
erect a perpendicular
CD, and take DE=CD.
At E make FG perpen-
dicular to DE. Find in
/'' i \ FG the point F which
falls in the line BD,
and the point G in the line AD. FG = AB.
Ww. 147 ; (Eu. I. 26 ; Cv. I. 78).
(8) To measure a line wholly inaccessible.
If AB is the line, choose a convenient point C from which
A and B are both visible, and measure AC and BC\^j (/?);
then AB may be measured by (a).
§ (B). — STRAIGHT LINES IN TRIANGLES.
I. RIGHT-ANGLED TKIANGLES.
1, Given the base and perpendicular, to find the hypoth-
enuse.
Kule : Square the sides, add together, and extract the
square root.
Formula : a2 + b2 = c2.
Proof: Ww. 379 ; (Eu. I. 47; Cv. IV. 25).
THE MEASUREMENT OF LINES.
EXAM. 1. The altitude of a right-angled triangle is 3,
the base 4. Find the hypothenuse.
a2 - 32 - 9.
6» = 4*-16.
a? + 62 = 32 + 42 = 25 = e2.
.*. c — 5. Answer.
2. Given the hypothenuse and one side, to find the other
side.
Eule : Multiply their sum by their difference, and extract
the square root.
Pormula : # — a2 — (c -f- a) (c — a) = b2.
EXAM. 2. The hypothenuse of a right-angled triangle
is 13, the altitude 12. Find the base.
c + a = 25
c —a = 1
.'. b = 6. Ans.
[For exercises on 1 and 2, see table of right-angled triangles.]
II. OBLIQUE TRIANGLES.
When two lines form an angle, the projection of the first
on the second is the line between
the vertex and the foot of a
perpendicular let fall from the
extremity of the first on to the
second. Thus the projection of .
AC on BC is CD.
Given two sides and the projection of one on the other,
to find the third side :
8
MENSURATION.
3, If the angle contained by the given sides be obtuse.
Eule : To the sum of the squares of the given sides
add twice the product of the projection and the side on
which (when prolonged} it falls ; then extract the square
root.
Formula: a2+ l2+2bj = c2.
Proof: Ww. 343 ; (Eu. II. 12; Cv. III. 53).
EXAM. 3. Given the
sides a = 5, b = 6, con-
taining an obtuse an-
gle, and given j = 4,
the projection of a on
b ; find the third side.
2ft;
.-.""?"
25
36
48
109.
/. c 10-44+. Ans.
4. If the angle contained l>y tin- L^VM sides is acute.
n
Bnle : .From the sum of the squares of the given sides
subtract twice the product of the projection and the side on
which it falls ; the square root of the remainder gives the
third side.
Pormula: a*
b2 — 2bj = c2.
Proof: Ww. 342; (Eu. II. 13; Cv. III. 52).
THE MEASUREMENT OF LINES.
EXAM. 4. Given the sides a = 5,
acute angle, and given j = 4, the
1 projection of a on b ; find the third
side.
= 6, containing an
2bj
= 25 + 36 = 61
= 48
.-. c-2=13
Ans.
3 6
5. If two sides and the perpendicular let fall on one
from the end of the other, are given, the projection can
be found by 2, and then the third side by 3 or 4.
If three sides are given, a projection can be found by 3
or 4, and then the perpendicular by 2.
6, Given three sides of a triangle to find its three me-
dials ; i.e., the distances from the vertices to the midpoints
of the opposite sides.
Eule : From the sum of the squares of any two sides
subtract twice the square of half the base ; the square root
of half the remainder is the corresponding medial.
Formula : a2 + c2 - Itf = 2 i\
Proof: Ww. 344 ; (Eu. Appen. 1 ; Cv. III. 62).
Corollary : Dividing the difference of the squares of two
sides by twice the third side, gives the projection on it of
its medial.
Formula: ?' =
a— c
- —
2b
EXAM. 5. Given two sides, a =7, <? = 9, and the base,
b = 4. Find the medial.
Here
aa + c*
/.*P
49 + 81 = 130
= 122
- 61
.'. i=7'8. ^TIS.
10
MENSURATION.
III. STRAIGHT LINES IN SIMILAR FIGURES.
7, Given two straight lines in one figure, and a line
corresponding to one of them in a similar figure, to find
the line corresponding to the other.
Rule : The like sides of
similar figures are propor-
tional.
Formula : bz = -
al
Ww. 319; (Eu. VI., Def. I;
Cv. III. 24).
EXAM. 6. The height of
an upright stick is '2 meters,
and it casts a shadow 3
meters long ; the shadow of
a flag-staff is 45 meters.
Find the height of the staff.
3:2::45:J2.
90
/. Z>2 = -- = 30 meters. Ans.
o
IV. CHORDS OF A CIRCLE.
Suppose AB any chord in a
circle. Through the center C
a diameter perpendicular to AB
meets it at its middle point D,
and bisects the arc at H. DH
is the height of the arc, and
AH the chord of half the arc.
8, Given the height of an arc and the chord of half the
arc, to find the diameter of the circle.
THE MEASUREMENT OF LINES. 11
Eule : Divide the square of the cJwrd of half the arc by
the height of the arc.
J-2
Kt
Formula : d=-~-
h
Proof : HAF is a right angle.
Ww. 264 ; (Eu. III. 31 ; Cv. II. 59).
/. HF-. HA:: HA: HD.
Ww. 334 ; (Eu. VI. 8, Cor. ; Cv. III. 44).
EXAM. 7. The height of an arc is 2 centimeters, the
chord of half the arc is 6 centimeters. Find the diameter.
7 6x6 -,0 .
d=- =18. Ans.
A
9, Given the chord of an arc and the radius of the
circle, to find the height of the arc.
Eule ,' From the radius subtract the square root of the
difference of the squares of the radius and half the chord.
Formula : h = r — Vr2 — i&2.
Proof: HD = HO- DC.
=r, and
EXAM. 8. The chord of an arc is 240 millimeters, the
radius 125 millimeters. Find the height of the arc.
= r2 - (££)2 = (r + P) (r - $£) = (125 + 120) (125 - 120).
.-. J)C= Vr3 - ±1* = \/245 X 5 = Vl225 = 35.
A = r-.ZK7=125-35 = 90 millimeters
= 9 centimeters. Ans.
12 MENSURATION.
10, Given the chord and height of an arc, to find the
chord of half the arc.
Eule : Take the square root of the sum of the squares of
the height and half the chord.
Formula: £.=
Proof: Alf2 = JID2 + AD2.
Ww. 379 ; (Eu. I. 47 ; Cv. IV. 25).
EXAM. 9. Given the chord — 48, the height — 10. Find
the chord of half the arc.
A2 = 100. (p)2 = (24)8 = 576.
... £,2=676. /. 7-, = 26. Ans.
If, instead of the height, the radius is given, substitute in
10 for h its value in terms of r and k from 9, and we have
= V2/-2 -
From, this folloivs :
Hi Given the chord of an arc and the radius of the
circle, to find the chord of half the arc.
Formula: £4 = ^2i*—r V4r*- It*.
EXAM. 10. Calculate the length of the side of a regular
dodecagon inscribed in a circle whose radius is 1 meter ;
that is, find /-i when r and Jc are each 1 meter long, for lc is
here the side of a regular inscribed hexagon, which always
equals the radius.
Ww. 431 ; (Eu. IV. 15, Cor. ; Cv. V. 14).
THE MEASUREMENT OF LINES.
13
Thus r and Jc being unity, our formula becomes
£, = V2 - vfirj = 0-51763809. Ans.
EXAM. 11. With unit radius, find the length of one side
of a regular inscribed polygon of 24 sides.
£i = V2 - V4 - (.51763809)2 = 0-26105238.
And so on with regular polygons of 48, 96, 192, etc., sides.
12, Given the radius of a circle and the side of a regular
inscribed polygon, to find the
side of the similar circumscribed
polygon.
Formula : t -— — =.
V4r2-/;2
Proof: Suppose AB the given
side k. Draw the tangent at
the middle point If of the arc
AB, and produce it both ways
to the points E and G, where it
meets the radii CA and CB produced ; EG is the side
required, t.
In the similar triangles CEH, CAD,
CN: CD::EH:AD::t:k.
kr
But
CD
CDZ = CA* - AD* = r2 - G it)2.
EXAM. 12. When r=l, find one side of a regular cir-
cumscribed dodecagon.
14 MENSURATION.
With radius taken as unity, t =
From Exam. 10, £ = 0-51763809, and substituting this
value, tu = 0-535898. Ans.
In the same way, by substituting £24 — 0-26105238 from
Exam. 11, we find /24 = 0-263305, and from £43 = 0-13080626
we get ^ = 0-131087, and so on for 96, 192, 384, etc., sides.
§ (C). METHOD OP LIMITS.
V. A variable is a quantity which may have succes-
sively an indefinite number of different values.
DEFINITION OF A LIMIT.
When a quantity can be made to vary in such a manner
that it approaches as near as we please and continually near-
er to a definite constant quantity, but cannot be conceived to
reach the constant by any continuation of the process, then
the constant is called the limit of the variable quantity.
Thus the limit of a variable is the constant quantity
which it indefinitely approaches, but never reaches, though
the difference between the variable and its limit may be-
come and remain less than any assignable magnitude.
EXAM. 13. The limit of the sum of the series,
1 + £ + i + i + irV + sV + *V + etc-. is 2-
EXAM. 14. The variable may be likened to a convenient
ferry-boat, which will bear us just as close as we choose to
the dock, — the constant limit, — but which cannot actu-
ally reach or touch it.
The bridge, the method for passing, in the order of our
knowledge, from variables to their limits, is the
THE MEASUREMENT OF LINES. 15
13. PRINCIPLE OF LIMITS.
If, while tending toward their respective limits, two vari-
able quantities are always in the same ratio to each other,
their limits will be to one another in the same ratio as the
variables*
~ b c B' b> B G' c' C ~~C"
Let the lines AB and AC represent the limits of any
two variable magnitudes which are always in the same ratio
to one another, and let Ab, Ac represent two corresponding
values of the variables themselves ; then Ab :Ac:: AB : AC.
If not, then Ab : Ac :: AB : some line greater or less than
AC. Suppose, in the first place, that Ab : Ac : : AB : AC' ;
AC' being less than AC. By hypothesis, the variable Ac
continually approaches AC, and may be made to differ
from it by less than any given quantity. Let Ab and Ac,
then, continue to increase, always remaining in the same
ratio to one another, till Ac differs from AC by less than
the quantity C'C; or, in other words, till the point c passes
the point C\ and reaches some point, as c', between C' and C,
and b reaches the corresponding point b'. Then, since the
ratio of the two variables is always the same, we have
Ab:Ac: :Ab':Ac'.
By hypothesis,
Ab: Ac: -.AB'.AC';
hence,
Ab':Ac': :AB:AC>
But
Ab'<AB, .-.Ac'<AC';
which is absurd.
* This principle, and the following demonstration of it, are contained
essentially in Eu. XII. 2.
16 MENSURATION.
Hence the supposition that Ab : Ac : : AB : AC,' or to any
quantity less than AC, is absurd.
Suppose, then, in the second place, that Ab : Ac : : AB : A C,"
or to some term greater than A C. Now, there is some line,
as AB', less than AB, which is to AC as AB is to AC'!
If, then, we conceive this ratio to be substituted for that
of AB to A C," we have
Ab:Ac::AB>: AC;
which, by a process of reasoning similar to the above, may
be shown to be absurd. Hence, if the fourth term of the
proportion can be neither greater nor less than AC, it must
be equal to AC; or we must have
Ab: Ac:: AB: AC. Q. E. D.
Cor.: If two variables are always equal, their limits are
equal.
14, 1V//D; /// >• i,i,,l iii'lfntitchj toward zer<\ flir
I a i'ii n <'/>•/• <>f ihc polygon ln*<-nl><-<l increases, tircwn
decreases, toward the same I'nnit, 1lic JrtifjtJt of f//c cw
Proof: Inscribe in a circle any convenient polygon, say,
the regular hexagon.
^ w. 431; (Eu. I\. 15; Cv. V. 14).
Join the extremities of each side, as AB, to the point of
the curve equally distant from them, as II] that is, the
point of intersection of the arc and a perpendicular at the
middle point of the chord. Thus we get sides of a regular
dodecagon. Repeat the process with the sides of the dodec-
agon, and we have a regular polygon of twenty-four sides..
So continuing, the number of sides, always doubling, will
increase indefinitely, while the length of a side will tend
toward zero.
THE MEASUREMENT OF LINES.
17
The length of the inscribed perimeter augments with the
number of sides, since we continually replace a side by two
which form with it a triangle, and so are together greater.
Ww. 33 ; (Eu. I. 20 ; Cv. 5).
Thus AB of the hexagon is replaced by AH and HB in
the dodecagon.
But this increasing perimeter can never become as
long as the circumference, since it is always made up of
chords each of which is shorter than the corresponding
arc, by the axiom, " A straight line is the shortest line
between two points." Therefore, this perimeter increases
toward a limit which cannot be longer than the circum-
ference.
In doubling the number of sides of a circumscribed poly-
gon, by drawing tangents at the middle points of the arcs,
we continually substitute a straight for a broken line ; as,
TU IQT TN-\- NU. So this perimeter decreases.
18
MENSURATION.
But two tangents from an external point cannot be to-
gether shorter than the included arc.
Kg., HT + TB > arc HB.
Therefore this perimeter decreases toward a limit which
cannot be shorter than the circumference.
But the limit toward which the circumscribed perime-
ter decreases is identical with that toward which the corre-
sponding inscribed perimeter increases.
For, in a regular circumscribed polygon of any number
of sides, n, the perimeter is n times one of the sides.
THE MEASUREMENT OF LINES. 19
But. from 12,
2rkn
2 nrkn
But pfn, the perimeter of the corresponding inscribed
polygon, is nkn.
' nf "
pn
the ratio of the perimeters.
Cutting the circumference into n equal parts makes each
part as small as we please by taking n sufficiently great.
But chords are shorter than their arcs ; therefore kn
tends toward the limit zero as n increases.
Thus the limit of — -== is — ===1. The vari-
/v) } sy*
ables ±-S and - - are always equal. Therefore, by
p'n
13, Cor., their limits are equal, arid limit of •£-* = 1.
.'. lim. pn •= lim. p'H. ?*
But we have shown that lim. p'n cannot be longer than c,
and lim. pn cannot be shorter than c. Therefore, the com-
mon limit is c, the length of the curve.
§(D). THE RECTIFICATION OF THE CIRCLE.
15. In a circle whose radius is unity, to find the length
of the semicircumference.
From 14, an approximate value of the semicircumference
in any circle is given by the semiperimeter of every poly-
gon inscribed or circumscribed, the latter being in excess,
and the former in defect of the true value.
In examples 10, 11, and 12, we have already calculated
20
MENSURATION.
for r = 1 , the length of a side in the regular inscribed and
circumscribed polygons of 12, 24, and 48 sides. Continuing
the same process, and in each case multiplying the length
of one side by half the number of sides, we get the follow-
ing table of semiperimeters :
n.
i ni.
2 "An-
\ntn.
6
3-000
3-4641016
12
3-1'
3-2153903
L'l
3-1326286
8-15965
48
31398502
3-1460-
06
3-1410319
3-14271 1U
192
3-111 I.V-'l
3-1418730
384
3-1-H.v
3-1416627
768
3-1415838
3-1416101
1536
3-1415904
3-1415970
3071'
3-1415921
3-1415937
6144
3-1415925
3-l-llf>'.'2!»
1 L'288
34415926
3-1415927
etc.
etc.
etc.
Since the semicircumference, $c, is always longer than
J nkn and shorter than £ ntn, therefore its value, correct to
seven places of decimals, is 3-1415926.
16, The circumferences of any two circles are to each
other in the same ratio as their radii.
Proof : The perimeters of any two regular polygons of
the same number of sides have the same ratio as the radii
of their circumscribed circles.
Ww.413; (Eu. XII. 1, V. 12; Cv. V. 10).
The inscribed regular polygons remaining similar to
each other when the number of sides is doubled, their
THE MEASUREMENT OF LINES. 21
perimeters continue to have the same ratio. Hence, by 13,
the limits, the circumferences, have the same ratio as their
radii.
Cor. 1. Circumferences are to each other as their diame-
ters.
Cor. 2. Since , , 0 0 .
c: c' : : r: r : : 2r: 2r',
. c_ =_ _c^_ = \c
' 2r~ 2r'~ r'
That is, the ratio of any circumference to its diameter is
a constant quantity.
This constant, identical with the ratio of any semicir-
cumference to its radius, is denoted by the Greek letter TT.
But, in circle with radius 1, semicircumference we
have found 3-1415926+. Therefore, the constant ratio
TT = 3-1415926+.
VI. LINES IN ANY CIECLE.
17, ^ = - : = ^- = 3-1415926+.
a r
Multiplying both sides of this equation by d, gives
18. C = C?7T
.'. The diameter of a circle being given, to find the cir-
cumference.
Rule : Multiply the diameter by TT.
NOTE. In practice, for TT the approximation 3f or -272- is generally
found sufficiently close. A much more accurate value is fff ; easily
remembered by observing that the denominator and numerator
written consecutively, thus, 11 3 I 3 55, present the first three odd
numbers each written twice. The value most used is ?r = 3-1416.
19, Dividing in 18 by TT gives
d_2r---<JX-=»cX O3183098+.
TT 7T
CHAPTER II.
THE MEASUREMENT OF ANGLES.
§ (E). — THE NATURAL UNIT OP ANGLE.
To say " all right angles are equal," assumes that the
amount of turning necessary to take a straight line or
ray all around into its first position is the same for all
points.
Thus the natural unit of reference for angular magni-
tude is one whole revolution,
called a perigon, and equal to four
right angles.
VII. A revolving radius de-
scribes equally an angle, a sur-
face, and a curve. Moreover, the
pcrigon, circle, and circumference
are each built up of congruent
parts; and any pair of angles or
sectors have the same ratio as the corresponding arcs.
Ww. 261 ; (Eu. VI. 33; Cv. II. 51).
rrn f any angle its intercepted arc
Therefore, . — ^— — .
pengon circumference
That is, if we adopt the whole circumference as the unit
of arc ;
20, The numerical measure of an angle at the center of a
circle is the same as the numerical measure of its intercepted
arc.
THE MEASUREMENT OF ANGLES. 23
And this remains true, if, to avoid fractions, we adopt,
as practical units of angle and arc, some convenient part
of these natural units. The Egyptian astronomers divided
the whole circle into 360 equal parts, called degrees ; each
of these degrees was divided into 60 parts, called minutes ;
these again into 60 parts, called seconds. These numbers
have very convenient factors, being divisible by 1, 2, 3, 4,
5, 6, etc.
EXAM. 15.
A perigon - 359° 60' of angle.
A circumference = 359° 59' 60" of arc.
VIII. Hence we say, An angle at the center is measured
by its intercepted arc ; meaning, An angle at the center is
such part of a perigon as its intercepted arc is of the whole
circumference.
§(F). CIRCULAR MEASURE OF AN ANGLE.
Half a perigon is a straight angle ; hence, halving the
denominators in VII., and using 2£ to mean angle, gives
any Y. _ its intercepted arc
straight Y semicircumference
But, from 18, in every circle, ^c = rTr.
Therefore, dividing denominators by tr gives
any ^- _ length of its arc
i • straight Y
If, now, we adopt as unit angle that part of a perigon de-
noted by - — 2; that is, the %. subtended at the center
7T
24 MENSURATION*.
of every circle by an arc equal in length to its radius, and
hence named a radian, then, by 20,
21, The number which expresses any angle in radians
also expresses its intercepted arc in terms of the radius.
So, in terms of whatever arbitrary unit of length the arc
and radius may be expressed, if u denote the number of
radians in an angle, then, for every %,
Thus the same angle will be denoted by the same num-
ber, whatever be the unit of length employed.
u, or the fraction arc d by radius, is called the
circular measure of an 2£.
EXAM. 16. Find the circular measure of a st. 2£.
Here, the arc being a semicircumference, its length l = T7r.
nr
.'. u = — — TT. Ans.
r
This is obviously correct, since dividing a straight %. by ir
first gave us our radian.
22, Given the number of degrees in an angle, to find
the length of the arc intercepted by it from a given cir-
cumference.
Eule : Multiply the length of the circumference by the
number of degrees in the angle, and divide the product by
360.
Formula: l=
Proof: From VII. we have 360°: a*: :c:L
j */
CO
NOTE. If the 2£ be given in minutes, the formula becomes ^oT^vy'
If in seconds. I = — c^
1296000"
THE MEASUREMENT OF ANGLES. 25
EXAM. 17. How long is the arc of one degree in a cir-
cumference of 25,000 miles?
Z=£|P = 694-K Ans.
23, Given the length of an arc of a given circumference,
to find the number of degrees it subtends at the center.
Eule : Multiply the length of the arc by 360, and divide
the product by the length of the circumference.
-n o ^360°
Formula : a = - — .
c
NOTE. To find the number of minutes or seconds :
. 121600' „ Z 1296000"
formulae: q'= ; and q"= .
c c
EXAM. 18. Find the number of degrees subtended in
any circle by an arc equal to the radius.
£360° , r360° 180°
Here q — - - becomes =
c 2rir TT
= 57-2957795°+. Ans.
Hence a radian = P = 57° 17'44-8"+ == 206264-8"+.
IX. The arcs used throughout as corresponding to the
angles are those intercepted from circles whose center is
the angular vertex.
These arcs are said to measure the angles at the center
which include them, because these arcs contain their ra-
dius as often as the including angle contains the radian.
Using measured in this sense, we may state the following
Theorems :
24, An angle at the center is measured by the arc inter-
cepted between its sides. Ww ^ . „_
26 MENSURATION.
25, An inscribed angle is measured by half its inter-
cepted arc. Ww 263 _ (Eu< m 2(); Cv> n 5^
26, An angle formed by a tangent and a chord is meas-
ured by half the intercepted arc.
Ww. 269 ; (Eu. III. 32 ; Cv. II. 62).
27, An angle formed by two chords, intersecting within
a circle, is measured by half the sum of the arcs vertically
intercepted. Ww 26g . (Eu Appen 2; Cy n 64)
28, If two secants, two tangents, or a tangent and a se-
cant intersect without the circle, the angle formed is meas-
ured by half the difference of the intercepted arcs.
Ww. 270 ; (Eu. Appen. 3 ; Cv. II. 65).
29. Given the measure of an angle in degrees, to find its
circular measure.
Eule : Multiply the number of degrees by TT, and divide
ly 180.
rnT*TTmlfi * *?/ — • ' — __ "' __ — _ J
~ 180° ~" 10800' ~ 648000"'
Proof : A straight ^ is 180°, and its circular measure is IT.
Hence,
180° TT'
since each fraction expresses the ratio of any given ^ to a
straight X. Therefore,
7Ttf°
11 — **
180°'
and also
75-
THE MEASUREMENT OF ANGLES. 27
This recalls to mind again that the circular measure of
any ~$. is independent of the length of the radius of the circle.
EXAM. 19. Find the circular measure of ^ of one
degree.
Here M = .JL = -01 74532925+. Ans.
EXAM. 20. Find the circular measure of ^ of one
minute.
Dividing the last answer by 60 gives
•000290888208+. Ans.
Of course, this number equally expresses the length of
an arc of one minute in parts of the radius, and in the
same way we obtain
Arc 1" = r x 0-00000484813681+.
30, Given the circular measure of an angle, to find its
measure in degrees.
Eule : Multiply the circular measure by 180, and divide
by TT.
Formula : g° =
7T
EXAM. 21. Find the number of degrees in ^ whose cir-
cular measure is 10.
Here g° = 10 X - = 10 X 57-2957795+
-572-957795+. Ans.
31, Given the angle in degrees and the length of the
arc which subtends it, to find the radius.
Eule : Divide 180 times the length by TT times the number
of degrees.
•n H80°
Formula: r = - — .
28
MENSURATION.
Proof:
180C
7T
TIT
EXAM. 22. An arc of 6 meters subtends ^ of 10° find
radius. , ^ „
Here r = - - X — = 0-6 X 57-2957795+
= 34-3775 meters. Ans.
X. One 2£ is called the complement of another, when
their sum equals a rt. ^ ; the supplement, when their sum
= a st. ^ ; the explement, when their sum = d.
REFERENCE TABLE OF ABBREVIATIONS FOR AREAS.
When used alone, as abbreviations, capital letters denote
the area of the figures ; to denote volume, a V is prefixed.
A = annulus.
T = trapezoid.
B = base.
A = triangle.
C = cylinder.
U = volume of quader.
G = circle.
V — volume.
D = volume of prismatoid.
W= volume of wedge.
E = ellipse.
X= volume of tetrahedron.
F = frustum.
Y= pyramid.
# = segment.
Z — zone.
.H"= sphere.
v=n
/ = volume of an irregular poly-
Zft = Ti + T2 + T3 + + Tn.
hedron.
£ = sum of angles.
/= parabola.
A = spherical A.
JT= cone.
N= spherical N.
L = lune.
Y = volume of spherical Y.
J/= midsection.
Q = steregon.
_Z\T= polygon of n sides.
= means equivalent; i.e., equal
0 = solid ring.
in size.
P = prism.
II = parallel.
C3— parallelogram.
_1_ = perpendicular.
Q = quadrilateral.
/•^= similar.
R — rectangle.
/. = therefore.
S = sector.
^ = angle.
CHAPTER III.
THE MEASUREMENT OF PLANE AREAS.
J(Q). PLANE RECTILINEAR FIGURES.
XL The area of a surface is its numerical measure.
Measuring the area of a surface, whether plane or
curved, is determining its ratio to a chosen surface called
the unit of area.
The chosen unit of area is a square whose side is a unit
of length.
EXAM. 23. If the unit of length be a meter, the
unit of area will be called a square meter (m2).
If the unit of length be a centimeter, the unit
of area will be a square centimeter (cm2). centimeter.
32, To find the area of a rectangle.
Eule : Multiply the base by the altitude.
Formula : R = ab.
Proof : SPECIAL CASE. When the base and altitude, or
length and breadth of the rectangle are commensurable.
In this case there is always a line which will divide both
base and altitude exactly.
If this line be assumed as linear unit, a and b are inte-
gral numbers.
30
MENSURATION.
D
B
In the rectangle ABCD divide AD into a, and AB into
b equal parts. Through the points of division draw lines
parallel to the sides of the rec-
tangle. These lines divide the
rectangle into a number of
squares, each of which is a unit
of area. In the bottom row
there are b such squares ; and,
since there are a rows, we have
b squares repeated a times,
which gives, in all, ab squares.
NOTE. The composition of ratios includes numerical multiplica-
tion. Arithmetical multiplication by integers is a growth from
addition. The multiplier indicates the number of additions or repe-
titions. The multiplicand indicates the thing added or repeated.
Therefore, if is not a mutual operation, and the product is always
in terms of the unit of the multiplicand. The multiplicand may be
any aggregate ; the multiplier is an aggregate of repetitions. To
repeat a thing does not change it in kind, so the result is an aggre-
gate of the same sort exactly as the multiplicand. When the rule
says, Multiply the base by the. altitude, it means. Multiply the nu-
merical measure of the base by the number meamiriny the altitude m
terms of the same linear unit. The product is a number which we
have shown to be the area of the rectangle ; that is, its numerical
measure in terms of the superficial unit.
This is the meaning to be assigned whenever we speak of the
product of one line by another.
GENERAL PROOF.
Rectangles, being equiangular parallelograms, have to
one another the ratio which is compounded of the ratios
of their sides. Ww 362 . (Eu. yi. 23 ; Cv. IV. 5).
Let R and JR' represent the surfaces or areas of two rec-
tangles. Let a and a' represent their altitudes ; b and &
their bases.
THE MEASUREMENT OF PLANE AREAS.
31
Thus,
R d o do
R' = ^ X 1? = db1'
' ' = a'6''
For the 'measurement of surfaces, this equation is funda-
mental. To apply it in practice, we have only to select as
a standard some particular unit of area.
R
The equation itself points out as best the unit we have
already indicated. If we suppose a' and b' to be, each of
them, a unit of length, Rl becomes this superficial unit,
and the equation becomes
This shows that the number of units of area in any rec-
tangle is that number which is the product of the numbers
of units of length in two adjacent sides.
This proof includes every case which can occur, whether
the sides of the rectangle be commensurable or incommen-
surable with the unit of length ; that is, whether a and b
are integral, fractional, or irrational.
EXAM. 24. Find the area of a ribbon 1 meter long and
1 centimeter wide.
1 meter is 100 centimeters.
100 square centimeters. Ans.
32 MENSURATION.
33, To find the area of a square.
Kule : Take the second power of the number denoting the
length of its side.
rormtda: D = b\
Proof : A square is a rectangle having its length and
breadth equal.
NOTE. This is the reason why the product of a number into itself
is called the square of that number.
Cor. Given the area of a square, to find the length of a
side.
Kule : Extract the square root of the number denoting the
area.
EXAM. 25. 1 square dekameter, usually called an
Ar (R), contains 100 square meters. Every
unit of surface is equivalent to 100 of the
next lower denomination, because every
unit of length is 10 of the next lower
order. Thus a square hektometer is a
liektar (ha).
EXAM. 26. How many square centimeters in 10 square
millimeters ?
100 square millimeters is 1 square centimeter.
/. 10 square millimeters is -^ of 1 square centimeter. Ans.
Or, 10 square millimeters make a rectangle 1 centimeter
long and 1 millimeter wide.
EXAM. 27. How many square centimeters in 10 milli-
meters square?
One. Ans.
THE MEASUREMENT OF PLANE AREAS. 33
REMARK. Distinguish carefully between square meters
and meters square.
We say 10 square kilometers (kmJ), meaning a surface
which would contain 10 others, each a square kilometer ;
while the expression 5 kilometers square means a square
whose sides are each 5 kilometers long, so that the figure
contains 25 square kilometers.
EXAM. 28. The area of a square is 1000 square meters.
Find its side. /TTV^
VlOOO --= 31-623 meters. Ans.
34, To find the area of any parallelogram.
Kule : Multiply the base by the altitude.
Formula : CU—-ab.
Proof : Any parallelogram is equivalent to a rectangle
of the same base and altitude.
Ww. 365 ; (Eu. I. 35 ; Cv. IV. 10).
Cor. The area of a parallelogram, divided by the base,
gives the altitude ; and the area, divided by the altitude,
gives the base.
EXAM. 29. Find the area of a parallelogram whose base
is 1 kilometer, and altitude 1 centimeter.
b = 1000 meters. a — T^ meter.
.*. ab = 10 square meters. Ans.
35, Given one side and the perpendicular upon it from
the opposite vertex, to find the area of a triangle.
Eule : Take half the product of the base into the altitude.
Formula : A = J ab.
34
MENSURATION.
Proof : A triangle is equivalent to half a parallelogram
having the same base and altitude.
Ww. 368 ; (Eu. I. 41 ; Cv. IV. 13).
Cor. 1. If twice the number expressing the area of a tri-
angle be divided by the number expressing the base, the
quotient is the altitude ; and vice versa.
Cor. 2. Two A's or O's, having an equal 2£, are as the
products of the sides containing it.
EXAM. 30. One side of a triangle is 35-74 meters, and
the perpendicular on it is 6-3 meters. Find the area.
£6 = 17-87 meters.
= 112*581 square meters. Arts.
36. Given the three sides of a triangle, to find the area.
Eule : From half the sum of the three sides subtract each
side separately; multiply the half sum and the three re-
mainders together : the square root of the product will be the
area.
Formula : A =
Proof:
By 4,
whence
(s — a) (s — b)(s — c).
-2#,
26
THE MEASUREMENT OP PLANE AREAS.
35
By 2,
whence,
4 Z>2/t2 = 4 62c2 — (Z>2 4- c2 — a2)2.
= V4TV - (62 + c2 - a2)2.
= V(2 Z>e + Z>2 + c2 - a2) (2 6c - 62 - c2 + a2).
= \/(a + 6 + c) (6 + c — a) (a + 6 — c) (a — b + c).
1 7,7, _
= "
c —
— c)(a — & + c)
~ ~~
But, by 35, i 5A equals the area of the triangle.
Cor. To find the area of an equilateral triangle, multiply
the square of a side by |-"\/^=0-433+.
EXAM. 31. Find the area of an isosceles triangle whose
base is 60 meters and each of the equal sides 50 meters.
Here, from last formula in Proof,
2bh = I V(2a + b)(2a-b) = 60 V160 x 40 - 60 VlBOO x 4.
/. 2bh = 60x40x2.
.*. Area = 1200 square meters. Ana.
B
37, To find the radius of the circle inscribed in a tri-
angle.
Kule : Divide the area of the triangle by half the sum
of its sides.
A
formula : r — — .
s
36 MENSURATION.
Proof.
By 35, area of BOO - — ,
Z
area of CO A = -,
2
area of
•L JJ-4-- A
/. by addition, A =
r — sr.
Cor. The area of any circumscribed polygon is half the
product of its perimeter by the radius of the inscribed
circle.
EXAM. 32. Find radius of circle inscribed in the tri-
angle whose sides are 7, 15, 20.
Here
s = 21, /. A = V21 x 14 x 6 = V3 . 7 . 7 . 2 . 2 . 3.
.-. A = 3 x 7 X 2 = 42.
.'. r = 2. Ans.
38. To find the radius of the circle circumscribing a
triangle.
Eule : Divide the product of the three sides by four times
the area of the triangle.
abc
Formula : $1 =
4A'
Proof: In any triangle the rectangle of two sides is
equivalent to the rectangle of the diameter of the circum-
scribed circle by the perpendicular to the base from the
Vertex< Ww. 350 ; (Eu. VI. C. ; Cv. III. 65).
ac abc
THE MEASUREMENT OF PLANE AREAS.
37
Cor. The side of an equilateral A, b = $ V3 = 2r V3.
EXAM. 33. Find radius of circle circumscribing triangle
7, 15, 20.
Here abc = 2100, A = 42.
..
loo
39, To find the radius of an escribed circle.
Eule : Divide the area of the triangle by the difference
between half the sum of its sides and the tangent side.
Proof: Let rx denote the radius of the escribed circle
which touches the side a. The quadrilateral O^BAC may
be divided into the two triangles, O^AB and
.'. by 35,
its area = -r, +-r,.
2 2
But the same quadrilateral is composed of the triangles
and ABC]
.'. its area = —r1-\ A.
38 MENSURATION.
Thus,
c , b a
2
= A
i
' V
• • ' 1
s — a
In the same way,
A A
7* =
—
7) * S ""•
s-6 s— c
Cbr. 1. Since, by 37, r = — , therefore,
s
36.
s -t - TTTTT - ;
s (s — a) (s — 6) (s — c)
Thus,
/\ — "V •>• T T T
" v ' 'i'2'a-
Ow. 2.
EXAM. 34. Find r1? r2, ?*3, when a — 7, b = 15, c = 20.
42 o 42 ,. 42
<y* .-< n*_ . / /yv U. /. A VI O
/J t->. /o • i « • o:»J. jrl/to.
14 6 1
XII. A trapezoid is a quadrilateral with two sides
parallel.
Cor. A triangle is a trapezoid one of whose parallel sides
has become a point.
40, To find the area of a trapezoid.
Eule : Multiply the sum of the parallel sides by half
their distance apart.
Tormula: T=
THE MEASUREMENT OF PLANE AREAS.
39
Proof: Let E be the midpoint of the side AB. Through
E draw GF parallel to CD. Then A AEG = A BEF.
Ww. 147; (Eu. I. 26; Gv. I. 78).
.-. Trapezoid ASCD = O GFCD. A
That is, by 34, T=GDXx; where a?
is the distance of BC from*4.Z). But Q
BF=AG.
.-. OD =
AD + BO
and calling AD, yv, and BC, y2, we have
m _
_ y\
v
Cbr. The area of a trapezoid equals
the distance apart of the parallel sides multiplied by the
line joining the midpoints of the non-parallel sides.
EXAM. 35. Find the area of a trapezoid whose II sides
or bases are 12-34 meters apart, and 56-78 meters and 9O
meters long.
56-78 + 90- = 146-78
12-34-*- 2 = 6-17
102746
14678
88068
905-6326 sqnare meters. Ans.
XIII. COOEDINATES OF A POINT.
The ordinatc of a point is the perpendicular from it to a
fixed base line or axis,
The corresponding abscissa is the distance from the foot
of this ordinate to a fixed point on the axis called the origin.
40
MENSURATION.
The coordinates of any point are its abscissa x and its
ordinate y.
XIV. If to any convenient axis ordinates be dropped
from the angular points of any polygon, the polygon is
exhibited as an algebraic sum of trapezoids, each having
one side perpendicular to the two parallel sides, and hence
called, right trapezoids.
If triangles occur, as 1Z>2, 6.275, they are considered
trapezoids, y± and y5 being zero.
41, To find the sum of any series of right trapezoids.
Eule : Multiply the distance of EACH intermediate ordi-
nate from the first by the difference between •//* two adjacent
ordinates, always subtracting the one following from the one
preceding in order along the broken line. Also multiply
distance of last ordinate from first by the sum of last two
ordinates. Halve the sum of these products.
Formula : T= J —
Proof: With O as origin, the area of the first trapezoid,
by 40, is (x2 - x,) •/1 + ?/2> and of the second is (ar, - x
THE MEASUREMENT OF PLANE AREAS.
41
Adding the two, we have
Performing the indicated multiplications, x<g/2 is cancelled
by — x<g/2, and
= i [(», - a-J (yt - 2/a)
since here a:^ is balanced by — x-g/z.
Thus we have proved our rule for a pair of trapezoids.
Taking three, we get, by 40,
~ —
a ~ V 4 •' s/
As before, replacing the balancing terms
Xji/4, this becomes
— xg/3 by
This proves the rule for three
trapezoids; and a generalization
of this process proves that if the
rule is true of a series of n trape-
zoids, it is true of ?z-fl.
For, by 40, the area of the
(n + l)th trapezoid
and, adding this to the first n trapezoids, as given by
formula, therefore
- a:,) (yn
42
MENSURATION.
Replacing #n+iyn+i — zn+iyn+i by the balancing terms
— Xil/n+2, this becomes
The same method proves that if the formula applies to n
trapezoids, it must apply to n — 1. Therefore, the rule is
true for any and every series whatsoever of right trape-
zoids.
EXAM. 36. Find the right portion of a railroad cross-
section whose surface line
breaks twice, at the points
(#2, 3/2), (*a, 2/3), to the right
of center line ; (origin being
on grade in midpoint of road-
bed).
This asks us to find the
sum of right trapezoids cor-
responding to the five points (0, c), (x2, y2), (#s, ys)> (^ + ^ r)>
(4, 0).
.'. by formula,
t>=4
£iT* =
In the same way, the portion to the left of center line,
whether without breaks, or with any number of breaks,
is given by our formula, which thus enables us at once to
calculate all railroad cross-sections, whether regular or
irregular.
EXAM. 37. To find the area of a triangle in terms of the
coordinates of its angular points.
THE MEASUREMENT OF PLANE AREAS.
43
Here are three trapezoids, and consequently, four points ;
but A is both 1 and 4, so
xt — xt = 0,
and the formula becomes
A = 2 T. = J [(xt - x,) (yt - y3) + (a;, - art) (y2 - yj].
v=l
Notice the symmetry of this answer.
or
For practical computation this is written
2 A = a?1 (y, - y,) + a-2 (yx - y3) + or, (y, - y,),
2 A = y, (ar, - :r3) + y2 (a?8 - ^ + y3 (xl - x2).
To insure accuracy, reckon the area by each.
42, To find the area of any polygon.
Eule : Take half the sum of the products of the abscissa
of each vertex by the difference between the ordinates of the
two adjacent vertices; always making the subtraction in
the same direction around the polygon.
Formula for a polygon of n sides :
N= J My* - y.) + ^(yj - y») + *3(y2 - y4) + ..... + »»(y»_i - yj]-
Proof: This is only that special case of 41, where the
broken line, being the perimeter of a polygon, ends where
it began.
44
MENSURATION.
Join the vertex 1 of the polygon 1, 2, 3, 4, , n, 1, with
the origin 0. Then the area enclosed by the perimeter is
the same, whether we consider it as starting and stopping
at 1 or at 0. But, under the latter supposition, though we
n-l
have n -f- 3 points, the coordinates (x0, 3/0) of the first and
last are zero, and the second (arlf 3/1) is identical with the
point -next to last; so that formula 41 becomes
NOTE. No mention need be made of minus trapezoids, since the
rule automatically gives to those formed by the broken line while
going forward, the opposite sign to those formed while going back-
ward.
Our expression for the area of any rectilinear figure is
the difference between a set of positive and an equal num-
ber of negative terms. If this expression is negative wrhen
the angular points are taken in the order followed by the
hands of a watch, then it is necessarily positive when they
are taken in the contrary sense, for this changes the order
in every pair of ordinates in the formula.
Observe that each term is of the form xy, and that there
is a pair of these terms, with the minus sign between them,
for each vertex of the figure. Thus, for the vertex
THE MEASUREMENT OF PLANE AREAS. 45
we have the pair xm(ym_l — ym+i) ; °r> pairing those terms
which have the same pair of suffixes, for every vertex m,
we have (xm^ym — xmym+i)- Hence, for twice the area
write down the pair xy — xy for each vertex, and add
symmetrically the suffixes,
1,2 2, 1 ; 2, 3 3, 2 ; 3, 4 4, 3 ; ..... n, I 1, n.
Thus, for every quadrilateral,
2 Q = Xjyt - xfa + x#s - ayy,
But, if any point of perimeter be to the left of origin, or
if, to shorten the ordinates, the axis
be drawn across the figure, then
one or more of the coordinates will
be essentially negative. Thus, if in
a quadrilateral, we take for axis a
diagonal, then
and 2 Q = - xsy2 + .T3?/4.
3
Here, ?/4 being essentially negative,
the two terms have the same sign, and give the ordinary
rule :
43, To find the area of any quadrilateral.
Eule ; Multiply half the diagonal by the sum of the per-
pendiculars upon it from the opposite angles.
EXAM. 38. The two diagonals of a quadrilateral meas-
ure 1-492 and 37-53 meters respectively, and are _L to one
another. Find the area.
Area - 1492 x 37'53 = 55-99476
2 2
= 27-99738 square meters. Ans.
46
MENSURATION.
EXAM. 39. Find the area of the polygon 1234567891,
the coordinates of whose angular points are (0, 90),
(30, 140), (110, 130), (80, 90), (84, 80), (130, 40), (90, 20),
(40, 0), (35, 70).
By 42,
2 N= 30 x - 40 + 110 x 50 + 80 x 50 + 84 x 50 + 130 x 60 + 90 x 40
+ 40 x - 50 + 35 x - 90 = 18750.
.'. area = 9375. Ans.
REMARK. The result of any calculation by coordinates
may be verified by a simple change of origin. If the ori-
'gin is moved to the right through a unit of distance, then
the numerical values of all positive abscissae will be dimin-
ished by one, and all negative abscissae increased by one.
Thus, to verify our last answer, move the origin thirty
units to the right, and the question becomes
EXAM. 40. To calculate the area of polygon whose coor-
dinates are :
By 42, twice the area equals the
x y
I
-30
90
2
0
140
3
80
130
1
50
90
5
54
80
6
100
10
7
60
20
8
10
0
9
5
70
sum of
10 x
-30x-70 = 2100
80 x 50= 4000
50 x 50= 2500
54 x 50= 2700
100 x 60= 6000
60 x 40= 2400
19700
50 and 5 x - 90 = -950
18750
/. area = 9375, as before.
EXAM. 41. From the data of Exam. 40 construct the
figure.
Choose a convenient axis and origin, noticing that the
THE MEASUREMENT OF PLANE AREAS.
47
polygon will lie wholly above the axis, since there are no
minus ordinates. Then, to find first vertex, measure off on
the axis 30 units to the left of origin, and at the point thus
determined, erect a perpendicular 90 units in length. Its
extremity will be the angular point numbered 1. The
extremity of a perpendicular at origin 140 units long
gives vertex 2, and an ordinate 130 long from a point on
axis 80 units to the right of origin gives 3. When all the
angular points have been thus determined, join them by
straight lines in their order of succession.
48
MENSURATION.
44. Given the area and one side of a figure, and the cor-
responding side of a similar figure, to find its area.
Rule : Multiply the given area by the squared ratio of
the sides.
Pormula: A1 =
Proof: The areas of similar figures are to one another as
the squares of- their like sides.
Ww. 376 ; (Eu. VI. 20; Cv. IV. 23).
Cor. The entire surfaces of two similar solids are propor-
tional to the squares of any two homologous lines.
EXAM. 42. The side of a triangle containing 480 square
meters is 8. meters long.
Find area of a similar triangle whose homologous side
is 40.
480 X 1600
A =
64
= 480 X 25 = 12000 square meters. Ana.
XV. Magnitudes which can he made to coincide are
congruent.
Magnitudes which agree in size, but not in shape, are
equivalent.
THE MEASUREMENT OF PLANE AREAS.
49
XVI. A regular polygon is both equilateral and equi-
angular.
The bisectors of any two angles of a regular polygon
intersect in a point equidistant from all the angular points
of the polygon, and hence also
equidistant from all the sides,
and at once the center of an
inscribed and a circumscribed
circle.
Joining this center to every
angle of the polygon cuts it up
into congruent isosceles triangles.
Hence the area of the regular
polygon is the area of any one
of these triangles multiplied by the number of sides of the
polygon.
45. To find the area of a regular polygon.
Kule : Multiply together one
side, the perpendicular from the
center, and half the number of
sides.
Or, in other words :
Take half the product of pe-
rimeter l>y apothem.
Pormula : N=
aln
ap
"2"
EXAM. 43. The side of a regular hexagon is 98 centi-
meters, and its apothem 84*87 centimeters ; find its area.
Area = 3 X 98 X 84-87
= 24951 '78 square centimeters. Ans.
50
MENSURATION.
46, By the aid of a table of polygons, to find the area of
any regular polygon.
Rule : Multiply the square of one of the sides of the poly-
gon by the area of a similar polygon whose side is unity.
Formula: Nt = lnzN^
Proof : This follows from 44, all regular polygons of the
same number of sides being similar.
TABLE OF REGULAR POLYGONS.
Number
of Sides.
Name.
Area when
Side= 1.
of Sides.
Area in Terms of
S<juo,T6 on Si<<> .
3
Triangle
04330127
15
17-642363
4
Square
1-0000000
16
20-109358
5
Pentagon
1-7204771
20
31-568757
6
Hexagon
2-59807i vi
24
46-574*
7
Heptagon
3-633911' 1
49-473844
8
Octagon
4-8284271
30
71-357734
9
Nonagon
6-1818242
81-225360
10
Decagon
7-6842088
10
127-062024
11
Undecagon
9-3656399
48
138-084630
12
Dodecagon
11-1961524
EXAM. 44. The side of a regular hexagon is 98 centi-
meters; find its area.
Area = 98 X 98 X 2-5980762
= 24951-78 square centimeters. Ans.
EXAM. 45. If the side of a regular decagon is 0-6 meters,
its area is
»
0-6 X 0-6 x 7-6942088 = 2-76991524 square meters. Ans.
THE MEASUREMENT OF PLANE AREAS.
51
§(H). AREAS OF PLANE CURVILINEAR FIGURES.
47. To find the area of a circle.
Eule : Multiply its squared radius by TT.
Formula : O = rV.
Proof : If a regular polygon be circumscribed about the
circle, its area, by 45, is
N=
and, by 14, as n increases, pn decreases toward c as limit,
and N toward O. But the variables ^Vand pn are always
in the constant ratio \r ; therefore, by 13, their limits are
in the same ratio, and we have
By 18,
Therefore,
c = 2 rir.
O = rV
EXAM. 46. Find the area of a circle whose diameter is
7-5 meters.
Here r* = 14-0625.
= 44-178+ square meters. Ans.
48. To find the area of a sector.
Eule : Multiply the length of the
arc by half the radius.
Formula : S=%lr=% ur2.
Proof:
Ww. 424;(Cv. V. 44);
52
MENSURATION.
or, as follows : By Eu. VI. 33,
• <3 _
EXAM. 47. Find the area of a sector whose arc is 99-58
meters long, and radius 86-34 meters.
99-58 X 43-17 = 4298-8686 square meters. Am.
EXAM. 48. Find the area of a sector whose radius is 28
centimeters, and which contains an angle of 50° 36!
Here, by 29,
^-28' = 784
3532
7064
8181
2)692-271'
/. $=346-136 square centimeters. Ans.
49, To find the area of a segment less than a semicircle.
Eule : From the sector hav-
ing the same arc as the seg-
ment, subtract the triangle
formed by the chord and lite
//r<> radii from its
Formula :
G = —T-^T
Proof : The segment AJTJ5
*
is the difference between the sector AHBG and the tri-
angle ABC.
THE MEASUREMENT OF PLANE AREAS. 53
By 48, AHBC=\lr.
By 35,
But
HD
Ww. 337 ; (Eu. VI. 13 ; Cv. III. 47).
2A
Substituting this value of r in the expression for 26r, we
obtain
2h 2h
Cor. The area of a segment of a circle is equal to half
the product of its radius and. the excess of its arc over half
the chord of double that arc. For
sector AIISC = llr,
and &A£C = ±rx£L.
/. segment AHB = \r(l- BL). •
Approximate Eule for Segment : Take two-thirds the prod-
uct of its chord and height.
Approximate Formula ; G = $hJc.
EXAM. 49. If the chord of a segment is r X 0-959851 +,
and its height is r X 0- 12241 7+, then an approximation to
its area is
|r« X 0-959851+ X 0-122417+ = £ r2 X 0-117502+ =. r2 X 0-0783+.
54 MENSURATION.
But if, also, we can measure the arc, and here find it
equal to radius, then
r2(0-122417+)2(r+rxO-959851+)+j-r2(0-959851+)2(r-rxO-959851+)
2 ,y,QQfl ,
= t* X 0.0 / 886+. -an*
(2=jr2xO-239919++Ty2xO- 30216+
<3=Jr'(0239919++0-0755i+)
6^x0-315459+.
Proceeding directly by Rule 49, instead of Formula 49,
we here get
S=i»*,
and A = J(r X 0-959851-f) (r - r X 0-122417+)
/. A- r X 0-4799255+ x r.x 0-877583-
.-. A = r*x 0-421171:
.-. O = S- A = 7^x0-07883.
Since, in this example, arc = r, .*. G is, in any O, the
segment whose ^ is p.
50, A circular zone is that part of a circle included be-
tween two parallel chords, and may be found by taking
the segment on the shorter chord out of that on the longer.
•
51, A crescent is the figure included between the cor-
responding arcs of two intersecting circles, and is the dif-
ference between two segments having a common chord,
and on the same side of it.
52, To find the area of $n annulus ; that is, the figure
included between two concentric circumferences.
Eule : Multiply the sum of the two radii by their differ-
ence, and the product l)ij TT.
Formula : A = (rl -f- r2) (i\ — r2) TT.
THE MEASUREMENT OF PLANE AREAS.
55
Proof : By 47, the area of the outer circle is rfir, and of
the inner circle r227r. Therefore, their difference, the an-
nulus, is (rf — r^TT.
Cor. The area of the annular figure will be the same
whether the circles are concentric or not, provided one
circle is entirely within the other.
If the two circles intersect, they form
two crescents, one on each side of the
common chord, and the difference of
the two crescents will always be equal
to the annulus formed by the same
circles.
EXAM. 50. The radii of two con-
centric circles are 39 meters and 11-3 meters. Find the
area of the ring between their circumferences.
Here
A = 50-3 X 27-7 X TT = 4377-2+ square meters. Ans.
53, To find the area of a sector of an annulus.
Eule : Multiply the sum of the bounding arcs by half the
distance between them.
Formula : S. A. = J h (4 -f- 4).
56
MENSURATION.
Proof : The sectorial area ABED is the difference be-
tween the sector ABC and the
sector CDE.
.'. by 48,
8.A.-*(r + fcU-irL
Xi>w, since /i and /2 are arc'q
subtending the same angle at Ct
.'. by IX,
JL.i
r + A. r
Substituting, we have
Cor. By comparison with 40, we see an annular sector
is equivalent to a trapezoid whose parallel sides equal the
arcs, and are at the same distance from one another.
EXAM. 51. The upper arc of a circular arch is 35-25
meters ; the lower, 24-75 meters ; the distance between
the two is 3-5 meters. How many square meters are there
in the face of the arch ?
S. A. = 1-75 X 60 = 105 square meters. Am.
XVII. CONICS.
If a straight line and a point be given in position in a
plane, and if a point move in the plane in such a manner
that its distance from the given point always bears the
same ratio to its distance from the given line, the curve
traced out by the moving point is called a conic.
The fixed point is called the focus, and the fixed line
the directrix of the conic.
THE MEASUREMENT OF PLANE AREAS. 57
When the ratio is one of equality, the curve is called a
parabola.
54, To find the area of a parabolic segment ; that is, the
area between any chord of a parabola and the part of the
curve intercepted.
Kule : Take two-thirds the product of the chord by the
height of the segment.
Formula: J—^hJc.
Proof: A parabolic segment is two-thirds of the triangle
made by the chord and the tangents at its extremities.
If AB, AC, be two tangents to a parabola, to prove that
the area between the curve and the chord EG is two- thirds
of the triangle ABO.
Parallel to EG draw a tangent DPE. Join A to the
point of contact P, and produce AP to cut the chord BC
at N.
By a property of the parabola, deducible from its defini-
tion,
= PN.
.-. BC-2-DE.
Ww. 276 & 279; (En. VI. 2 & 4 ; Cv. III. 15 & 25).
.'. by 35,
This leaves for consideration the two small triangles
PDB, PEC, each made by a chord and two tangents.
With each proceed exactly as with the original triangle :
e.g., draw the tangent FQGf parallel to PB\ join DQ, and
produce it to M] then
58
MENSURATION.
This leaves four little tangential triangles, like PFQ.
In each of these draw a tangent parallel to the chord, etc.,
and let this process be continued indefinitely.
Then the sum of the triangles taken away within the
parabola is double the sum of the triangles cut off without
it. But the sum of the interior triangles approaches, as
its limit, the parabolic segment. For the triangle BPC,
since it is half of AEG, is greater than half the parabolic
area BQPC, and so successively with the smaller interior
triangles. Therefore, the difference between the parabolic
segment and the sum of these triangles can be made less
than any assignable quantity.
Ww. 257 ; (Eu. XII., Lemma; Cv. V. 28).
Therefore, the constant segment is, by definition V., the
limit of this variable sum.
THE MEASUREMENT OF PLANE AREAS. 59
Again, each outer triangle cut off is greater than half
the area between the curve and the two tangents; e.g.,
ADE, being half the quadrilateral ABPC, is more than
half the area ABQPC. Therefore, the limit of the sum of
the outer triangles is the area between the curve and the two
tangents AJ3, AC. But these two variable sums are always
to each other in the constant ratio of 2 to 1. Therefore,
by 13, their limits are to each other in the same ratio,
and the parabolic segment is two-thirds its tangential
triangle.
But the altitude of this triangle is twice the height of
the segment.
.-. A = hk,
and J=$hk.
55, To find the area of an ellipse.
Eule : Multiply the product of the semi-axes by IT.
Formula : E = abir.
Proof: Let ADA'D' be a circle of which AC, CD are
radii at right angles to one another.
In CD let any point B be taken ; then, if this point
move so as to cut off from all ordinates of the circle the
same part that BC is of D C, the curve traced is called an
ellipse.
In one quadrant of the circle take a series of equidistant
ordinates, as Q1 P± Mlt Q2 P2 M>, Q3 P3 Mz, etc. Draw P± R±,
Qifi'i, etc., parallel to the axis A A'. Then, by 32,
area R^ : area R\M^ : : P^M, : Q^ ::BC: AC;
and each corresponding pair being in this constant ratio,
.'.the sum of the rectangles EM is to the sum of R1 M as
BC-.AC. But the sum of RM differs from one-quarter
60
MENSURATION.
of the ellipse by less than the area .Z> J/1; which can be
made less than any assignable quantity by taking (7J/i, the
common distance between the ordinates, sufficiently small.
Hence, A'BCis the limit of the sum of the rectangles RM\
and, in the same way, the quadrant of the circle is the limit
of the sum of R*M. Therefore, by 13,
O AC CL
Cor. The area of any segment of an ellipse, cut off by a
line parallel to the minor axis, will be to the corresponding
segment of the circle upon the major axis in the ratio of
b to a.
EXAM. 52. Find the area of an ellipse whose major axis
is 61-6 meters, and minor axis 44-4 meters.
E = 30-8 X 22-2x3-14159
= 2148-09+ square meters. Ans.
CHAPTER IV.
THE MEASUREMENT OF THE AKEAS OF BROKEN AND
CURVED SURFACES.
XVIII. A polyhedron is a solid bounded by polygons.
A polyhedron bounded by four polygons is called a te-
trahedron; by six, a hexahedron; by eight, an octahedron;
by twelve, a dodecahedron; by twenty, an icosahedron.
The faces of a polyhedron are the bounding polygons.
If the faces are all congruent and regular, the polyhedron
is regular.
The edges of a polyhedron are the lines in which its
faces meet.
The summits of a polyhedron are the points in which its
edges meet.
A section of a polyhedron is a polygon formed by the
intersection of a plane with three or more faces.
A convex figure is such that a straight line cannot meet
its boundary in more than two points.
56, The number of faces and summits in any polyhedron
taken together exceeds l>y two the number of its edges.
Proof : Let c be any edge joining the summits a/3 and
the faces A J5, and let e vanish by the approach of ft to a.
If A and B are neither of them triangles, they both re-
62
MENSURATION.
main, though reduced in rank and no longer collateral,
and the figure has lost one edge c and one summit ft.
If £ is a triangle and A no triangle, B vanishes with e
into an edge through a, but A remains. The figure has
lost two edges of B, one face B, and one summit /?. If B
and A are both triangles, B and A both vanish with e,
five edges forming those triangles are reduced to two
through a ; and the figure has lost three edges, two faces,
and the summit ft.
In any one of these cases, whether one edge and one
summit vanish, or two edges disappear with a face and a
summit, or three edges with a summit and two faces, the
truth or falsehood of the equation
ft+£=e+2
remains' unaltered.
By causing all the edges which do not meet any face to
vanish, we reduce the figure to a pyramid upon that face.
Now, the relation is true of the pyramid ; therefore it is
true of the undiminished polyhedron.
PRISM AND CYLINDER.
XIX. A prism is a polyhedron two of whose faces are
congruent, parallel polygons, and the other faces are paral-
lelograms.
AREAS OF BROKEN AND CURVED SURFACES.
63
The bases of a prism are the congruent, parallel polygons.
A parallelepiped is a prism whose bases are parallelo-
grams.
A normal is a straight line perpendicular to two or more
non-parallel lines.
The altitude of a prism is the normal distance between
the planes of its bases.
A right prism is one whose lateral edges are normal to
its bases.
57. To find the lateral surface or mantel of a prism.
Eule ; Multiply a lateral edge by the perimeter of a right
section.
Formula : P = Ip.
Proof: The lateral edges of a prism are all equal.
The sides of a right section, being perpendicular to the
lateral edges, are the altitudes of the parallelograms which
form the lateral area of the prism.
Cor. The lateral area of a right prism is equal to its alti-
tude multiplied by the perimeter of the base.
EXAM. 53. The base of an oblique prism is a regular
pentagon, each side being 3 meters, the perimeter of a
right section is 12 meters, and the length of the prism 14
meters. Find the area of the whole surface.
64 MENSURATION.
By 46, the area of the pentagonal base is
9 x 1-7204774 = 15-4842966.
Doubling this for the base and top together, and adding
the lateral area of the prism, which, by 56, is
12x14 = 168,
the total surface
= 168 -f 30-9685932 = 198-9686- square meters. Ans.
XX. A cylindric surface is generated by a straight line
BO moving that every two of its positions are parallel.
The generatrix in any position is called an element, of
the surface.
A cylinder is a solid bounded by a cylindric surface and
two parallel planes.
The axis of a cylinder is the straight line joining the
centers of its bases.
A truncated cylinder is the portion between the base
and a non-parallel section.
58, To find the curved surface or mantel of a right cir-
cular cylinder.
Eule : Multiply its length by the cir-
cumference of its base.
Pormula : C = cl = 2 irrl.
First Proof : Imagine the curved
surface slit along an element and then
spread out flat. It thus becom*-- ;>
rectangle having for one side the circumference and for
the adjacent side the length of the cylinder.
Second Proof : Inscribe in the right cylinder a right
prism having a regular polygon as its base. Bisect the
AREAS OF BROKEN AND CURVED SURFACES.
65
arcs subtended by the sides of this polygon, and thus in-
scribe a regular polygon of double the
number of sides, and construct on it, as
base, an inscribed prism.
Proceeding in this way continually to
double the number of its sides, the base of
the inscribed prism, by 14, approaches the
base of the cylinder as its limit, and the
prism itself approaches the cylinder as its
limit. But, by 57,
P = lp,
and always the variable P bears to the variable p the
constant ratio I. Therefore, by 13, their limits are in the
same ratio, and
C = cl.
Cor. 1. The curved surface of a truncated circular cyl-
inder is the product of the circumference of the
cylinder by the intercepted axis. For, by sym-
metry, substituting an oblique for a right section
through the same point of the axis alters neither
the curved surface nor the volume, since the
solid between the two sections will be the same above and
below the right section.
Cor. 2. The curved surface of any cylinder on any curve
equals the length of the cylinder multiplied by the perime-
ter of a right section.
EXAM. 54. Find the mantel of a right cylinder whose
diameter is 18 meters and length 30 meters.
C= 30 X 18 X 3-14159 = 1696-4586 square meters. Ans.
66
MENSURATION.
§(J). PYRAMID AND CONE.
XXI. A regular pyramid is contained by congruent
isosceles triangles whose bases form a regular polygon.
A conical surface is generated by a straight line moving
so as always to pass through a fixed point called th>
A cone is a solid bounded by a conical surface and a
plane.
The frustum of a pyramid or cone is the portion in-
cluded between its base and a cutting plane parallel to
the base.
r\
59, To find the area of the lat
surface or mantel of a regular pvra-
mid.
Bule : ^TnJf/jJi/ tlr. perimeter of
///,- base l>ij Jin ft' ///>• S/K,!/. lt< i<j]tf.
Formula : Y - ~ \ 1ij>.
Proof : The altitude of each of the
equal isosceles triangles is the slant
height of the pyramid, and the sum
of their bases is the perimeter of its base.
EXAM. 55. Find the lateral area of a regular heptagonal
pyramid whose slant height is 13-56224 meters, and basal
edges each IT meters.
One quarter of 13-56224 is 3-39056. Adding these and
dividing their sum by 2 gives 8-4764 for the area of one
triangular face. The lateral area is 7 times this, or
59-3348 square meters. Ans.
AREAS OF BROKEN AND CURVED SURFACES.
67
60, To find the area of the curved surface or mantel of
a right circular cone.
Kule : Multiply the circum-
ference of its base by half the
slant height.
formula : K~%ch = trrh.
First Proof : The distance A
from the vertex of a cone . of
revolution to each point on the circum-
ference of its base is the slant height of
the cone. Therefore, if the surface of the
cone be slit along a slant height and
spread out flat, it becomes the sector of a
circle, with the slant height as radius and
the circumference of cone's base as arc.
.*. by 48, its area is \ch.
Second Proof : About the base of the cone circumscribe
a regular polygon, and join its vertices and points of con-
tact to the vertex of the cone.
Thus is circumscribed about
the cone a regular pyramid
whose slant height equals the
slant height of the cone.
By drawing tangents, cir-
cumscribe a regular polygon of
double the number of sides,
and construct on it, as before,
a circumscribed regular pyra-
mid. Thus proceeding continually to double the number
of sides, the base of the circumscribed pyramid, by 14, ap-
proaches the base of the cone as its limit, and the pyramid
itself approaches the cone as its limit.
68
MENSURATION.
But, by 59,
and always the variable Y has to the variable p the con-
stant ratio %h. .*. by 13, their limits are in the same
ratio, and
K = j en.
Cor. 1. In the Proof of 47 we find
.'. the slant height of a right circular cone has the same
ratio to the radius of the base that the curved surface has
to the base, or
K : B : : h : r.
Cor. 2. Calling X the sector angle of the cone, we have
A : 360 : : r : h.
EXAM. 56. Given the two sides of a right-angled tri-
angle. Find the area of the surface described when the
triangle revolves about its hypoth-
enuse.
Calling a and b the given alti-
tude and base, and x the length
of the perpendicular from the
right angle to the hypothenuse, by
59, the area described by a is irxa, and described by b is
•n-xb. Thus the whole surface of revolution is IT (a -f- b)x.
But
a:z:: Va*T6*:6. Eu. I. 47 & VI. 8.
61, To find the lateral surface or mantel of the frustum
of a regular pyramid.
AREAS OF BROKEN AND CURVED SURFACES.
69
Eule ! Multiply the slant height of the frustum by half
the sum of the perimeters of its bases.
Formula : F=%h (pi -f p%).
Proof : The base and top being similar regular polygons,
the inclined faces are congruent trapezoids, the height of
each being the slant height of the frustum. If n be the
number of faces, by 40,
( P P \
area of each face = $h ( — * + — ).
.'. area of lateral surface = F = % h(pl
/7k
EXAM. 57. Find the lateral area of a regular pentagonal
frustum whose slant height is 11-0382 meters, each side of
its base being 2-f meters, and of its top 1£ meters.
The sum of a pair of parallel sides is •1^-.
11-0382 X 13 - 143-4966.
143-4966 -H 6= 23-9161,
the area of one trapezoidal face. The lateral area is five
times this, or
119-5805 square meters. Ans.
62, To find the curved surface or mantel of the frustum
of a right circular cone.
70
MENSURATION.
Rule : Multiply the slant height of the frustum by half
the sum of the circumferences of its bases.
Formula : F = = \ h (^ + c2) — vh (ri + '> 2) •
Proof: Completing the cone and
slitting it along a slant height, the
curved surface of the frustum develops
into the difference of two similar sec-
tors having a common angle, the arcs
of the sectors being the circumferences
of the bases of the frustum. By 53,
the area of this annular sector = F = $ h(ct + c2).
EXAM. 58. Find the mantel area of the frustum of a
right cone whose basal diameter is 18 meters ; top diame-
ter, 9 meters; and slant height, 171-0592 meters.
3-1416 X 9 = 28-2744 = circumference of top.
Twice top circumference = 56-5488 = circumference of base.
Half their sum is 42-4116, and this multiplied by the
slant height 171-0592, gives, for the curved surface,
7254-89 square meters. Ans.
63, To find the curved surface of a frustum of a cone of
revolution.
Eule : Multiply the projection of the frustums slant
height on the axis by twice TT times a perpendicular erected
at the midpoint of this slant height and terminated by the
axis.
Pormula : F = 2 iraj.
Proof: By 62, the curved surface of the frustum whose
•slant height is PR and axis MG is
AREAS OF BROKEN AND CURVED SURFACES.
71
But, by 40, Cor.,
But the triangle RPL is equiangular to CQO, since the
three sides of one are perpen- p M
dicular to the sides of the other.
Ww. 279 & 259 ; (Eu. VI. 4 & 16 ;
Cv. III. 25 & 5).
.-. F=
= 2irja.
O
c
Cor. This remains true, if either PM or EN vanish, or
if they become equal ; that is, true for a cone or cylinder
of revolution.
§ (K). THE SPHERE.
XXII. A sphere is a closed surface all points of which are
equally distant from a fixed point within called its center.
A globe is the solid bounded by a sphere.
64, To find the area of a sphere.
Eule : Multiply four times its squared radius by IT.
Pormula : H =4 r*ir.
Proof : In a circle inscribe a regular polygon of an even
number of sides. Then a diameter through one vertex
passes through the opposite vertex, halving the polygon
symmetrically. Let PR, be one of its sides. Draw PM,
UN perpendicular to the diameter ED. From the center
C the perpendicular CQ bisects PR.
Ww. 232 ; ( Eu. III. 3 ; Cv. II. 15).
Drop the perpendiculars PL, QO.
72
MENSURATION.
Now, if the whole figure revolve around BD as axis, the
semicircumference will generate a sphere, while each side of
B the inscribed polygon, as PR, will
generate the curved surface of the
frustum of a cone. By 63, this
and the sum of all the frustums, that
is, the surface of the solid generated
c by the revolving semi-polygon, equals
2-n-CQ into the sum of the projections.
As we double n the number of
sides of the inscribed polygon, by 14,
its semiperimeter approaches the
semicircumference as limit, and its surface of revolution ap-
proaches the sphere as limit, while CQ or a, its apothem,
approaches r the radius of the sphere as limit. But the
variable sum bears to the variable a the constant ratio 4 TTT.
Therefore, by 13, their limits have the same ratio, and
Cor. 1. A sphere equals four times a circle with same
radius.
Cor. 2. A sphere equals the curved surface of its circum-
scribing cylinder.
EXAM. 59. Considering the earth as a sphere whose
radius is 6-3709 X 108 centimeters, find its area.
H= 4 x 40-58836681 x 3-14159265 x 1016.
H = 5,100,484,593,831,997,860 square centimeters. Ans.
Or, about 510 million square kilometers.
AREAS OF BROKEN AND CURVED SURFACES. 73
XXIII. A spherical segment is the portion of a globe
cut off by a plane, or included between two parallel
planes.
A zone is the curved surface of a spherical segment.
The Proof of 64 gives also the following rule for the area
of a zone :
65. To find the area of a zone.
Eule : Multiply the altitude of the segment by twice ir
times the radius of the sphere.
Pormula : Z = 2 -n-ra.
Cor. 1. Any zone is to the sphere as the altitude of its
segment is to the diameter of the sphere.
Cor. 2. Let the arc BP generate a calot
or zone of a single base. By 65, its area /',
» / '
Ww. 334 ; (Eu. VI. 8, Cor. ; Cv. III. 44). j \
\
Hence, a calot or zone of one base is \ \
equivalent to a circle whose radius is the \ \
chord of the generating arc. \^ \
D
EXAM. 60. Find the area of a zone of one base, the di-
ameter of this base being 60 meters, and the height of the
segment 18 meters.
Using Cor. 2, the square of the chord of the generating
arc is
(30)2 + (18)2=1224,
which, multiplied by TT, gives, for the area of the calot,
3845-31 square meters. Ans.
74 MENSURATION.
66, THEOREM or PAPPUS.
If a plane curve lies wholly on one side of a line in its
oiun plane, and revolving about that line as axis generates
thereby a surface of revolution, the area of the surface is
equal to the product of the length of the revolving line into
the path described by its center of mass.
Scholium. The demonstration given under the next rule,
though fixing the attention on a single representative case,
applies equally to all cases where the generatrix is a closed
figure, has an axis of symmetry parallel to the axis of revo-
lution, and so turns as to be always in a plane with the
axis of revolution, while its points describe circles perpen-
dicular to both axes.
EXAM. 61. Use the Theorem of Pappus to find the dis-
tance of the center of mass of a semicircumference from
the center of the circle, by reference to our formula for the
surface of a sphere.
By 64, H = 4i*7r.
By 66, J2"=r7rx2z7r.
Equating, we get 2r=z7r. o
7T
67, To find the area of the surface of a solid ring.
Kule : Multiply the generating circumference by the path
of its center.
Formula : 0 = 4 -n2 r± r2.
Proof: Conceive any plane to revolve about any straight
line in it. Any circle within the plane, but without the
axis, will generate a solid ring.
AREAS OF BROKEN AND CURVED SURFACES.
75
Draw the diameter BCD parallel to the axis AO. Di-
vide the semicircumference BPD into n equal arcs, and
call their equal chords each k. From the points of divis-
ion drop perpendiculars to the axis, thus dividing the
other semicircumference BQD into n corresponding parts.
Let BP, BQ, be a pair of arcs. If we draw their chords,
we have a pair of right-angled trapezoids, ABPN and
ABQN, which, during the revolution, describe frustums
whose curved surfaces, by 62, are
and
If MLGF is the medial line, then, by Proof to 40,
and
.-. F: + F2 = 2irk(FM + LM] = 2nk(FG + GM+GM- QL).
But the diameter BCD is an axis of symmetry, and
76 MENSURATION.
the radius of the path of O.
This is the expression for each pair ; and, as we have n
pairs, therefore, the whole surface generated by a symmet-
rical polygon of 2n sides equals
since 2>nJc is p the whole perimeter.
But, as we increase 2n the number of sides of the in-
scribed polygon, by 14, p approaches c as its limit, and the
sum of frustral surfaces approaches the surface of the ring
as limit. But the variable sum bears to the variable pe-
rimeter the constant ratio 27r?*2. Therefore, by 13, their
limits have the same ratio, and
0 = c 2 TTT, = 2 TiTj 2 7rr2,
where rt is the radius of the generating circle, and ra the
radius of the path of its center.
EXAM. 62. Find the surface of a solid ring, of which the
thickness is 3 meters, and the inner diameter 8 meters.
Here TI is 1£ meters, and r2 is 5£ meters.
/. 4 Tr2^ = 3-1416 x 3 x 3-1416 X 11
= 9-4248 x 34-5576
= 325-698 square meters. Ans.
EXAM. 63. Find the area of the surface of a square ring
described by a square meter revolving round an axis
parallel to one of its sides, and 3 meters distant.
Here the length of the generating perimeter is 4 meters.
The path of its center is 7?r, since r2 is 3? meters.
.'. 0 = 28 TT = 87-9648 square meters. Ans.
AREAS OF BROKEN AND CURVED SURFACES. 77
EXAM. 64. A circle of 1-35 meters radius, with an in-
scribed hexagon, revolves about an axis 6-25 meters from
its center and parallel to a side of the hexagon. Find the
difference in area of the generated surfaces.
Here
^ = 1-35 and r2 = 6-25.
Therefore, area of circular ring is
47r2r1r2 = 7T2 x 2-7 X 12-5 = 9-8696 X 33-75 = 333-099.
For the hexagonal ring
Ww. 431 ; (Eu. IV. 15, Cor. ; Cv. V. 14)
the length of the generating perimeter is
6x1-35 = 8-1.
The path of its center is
77 x 12-5 = 39-27-.
Therefore, its area is
39-27 X 8-1 = 318-087.
Thus the difference in area is
15-012 square meters. Ans.
§(L). SPHERICS AND SOLID ANGLES.
XXIV. A great circle is a section of a globe made by
a plane passing through the center.
A lune is that portion of a sphere
comprised between two great semi-
circles.
The angle of two curves passing
through the same point is the angle
formed by the tangents to the two
curves at that point.
A spherical angle is the angle included between two
arcs of great circles.
78 MENSURATION.
A plane angle is the divergence between two straight
lines which meet in a point.
A solid angle is the spread between two or more planes
which meet at a point.
Two polyhedral angles, having all
their parts congruent, but arranged
in reverse order, are symmetrical.
A steregon, the natural unit of solid
angle, is the whole amount of solid
angle round about a point in space.
As a perigon corresponds to a
circle and its circumference, so a
steregon corresponds to a globe and
its sphere.
The steregon is divided into 360 equal parts, called
spherical degrees of angle, and these divide the whole
sphere into 360 equal parts, each called a degree of spheri-
cal surface.
A stcradian is the an^le subtended at the center by that
part of every sphere equal to the square of its radius, and
so by unit surface on a sphere of unit radius.
68, To find the area of a lune.
Eule : Multiply its angle in radians by twice its squared
radius.
Formula : L = 2 r\i.
Proof: Let PAQBP, PBQCP be two lunes having
equal angles at P] then one of these lunes may be placed
on the other so as to coincide exactly with it : thus lunes
having equal angles are congruent. Then, by the process
of Eu. VI. 33 ; (Ww. 769 ; Cv. VIII. 95), it follows that a
lune is to the sphere as its angle is to a perigon;
L _u_ . j
A ,.2— O _ '
4 T^TT ZTT
AREAS OF BROKEN AND CURVED SURFACES. 79
Cor. 1. A lune contains as many degrees of spherical
surface as its angle contains degrees.
Cor. 2. A lune measures twice as
many steradians as its angle contains
radians. A
EXAM. 65. Find the area comprised
between two meridians one degree
apart on the earth's surface.
Assuming as the earth's surface 196,625,000 square
miles, dividing by 360, gives for the lune,
546,180-5+ square miles. Ans.
XXV. Suppose the angular point of a solid angle is
made the center of a sphere ; then the planes which form
the solid angle will cut the sphere in arcs of great circles.
Thus a figure will be formed on the sphere, which is
called a spherical triangle if it is bounded by three arcs of
great circles, each less than a semicircumference.
If the solid angle be formed by the meeting of more than
three planes, the corresponding figure on the sphere is
bounded by more than three arcs of great circles, and is
called a spherical polygon.
The solid angle made by only two planes corresponds
to the lune intercepted on any sphere whose center is in
the common section of the two planes.
The plane angle of two planes is the amount of rotation
which one plane must make about their intersection in
order to coincide with the other.
The angles of a spherical polygon equal these plane
angles of its solid angle.
The sides of the polygon measure the face angles of this
polyhedral angle.
80 MENSURATION.
From any property of polyhedral angles we may infer
an analogous property of spherical polygons. Reciprocally,
from any property of spherical polygons we may infer a
corresponding property of polyhedral angles.
XXVI. A spherical pyramid is a portion of a globe
bounded by a spherical polygon and the planes of the sides
of the polygon.
The center of the sphere is the vertex of the pyramid ;
the spherical polygon is its base.
69, Just as plane angles at the center of a circle are
proportional to their intercepted arcs, and also sectors ;
so solid angles at the center of a sphere are proportional to
their intercepted spherical polygons, and also spherical
pyramids. '
XXVII. The spherical excess of a spherical triangle is
the excess of the sum of its angles over a straight angle.
The spherical excess of a spherical polygon is the excess of
the sum of its angles above as many straight angles as it
has sides less two.
70. To find the area of a spherical triangle.
Eule : Multiply its spherical excess in radians by its
squared radius.
Formula : A = er2.
Proof: Let ABC be a spherical triangle. Produce the
arcs which form its sides until they meet again, two and
two. The A ABC now forms a part of three lunes, name-
ly, ABDCA,BCEAB, and CAFBC.
Since the A's CDE and FAB subtend vertical solid
AREAS OF BROKEN AND CURVED SURFACES. 81
angles at 0, they are equivalent, by 69. Therefore, the
lune CAFBC equals the sum of the two triangles ABC
and CDE. Thus the lunes
whose angles are A, J3, and C,
are together equal to a hemi-
sphere plus twice A ABC.
Subtracting the hemisphere,
which equals a lune whose
angle is a straight angle, we
.
lune whose $- is (A + B + C— st. Y).
.'. A = lune whose 2(1 is \ e.
/. by 68, A = er2.
GOT. 1. A A contains half as many degrees of spherical
surface as its e contains degrees.
*-\
Cor. 2. A A measures as many steradians as its e contains
radians.
Cor. 3. Every £ of a A is > \e.
EXAM. 66. Find the area of a tri-rectangular A.
Here
e = rt 2 = TT.
or, a tri-rectangular triangle is one-eighth of its sphere.
By Cor. 1, a tri-rectangular A contains forty-five de-
grees of spherical surface.
71, To find the area of a spherical polygon.
Kule : Multiply its spherical excess in radians by its
squared radius.
Formula : 7V"= [« — (n — 2) TT] r*.
82 MENSURATION.
Proof : From any angular point divide the polygon into
(TO — 2) A's. /. by 70,
This expression is true even when the polygon has reen-
trant angles, provided it can be divided into A's with each
%. less than a st. ^.
Cor. 1. On the same or equal spheres, w-gons of equal
angle-sum are equivalent ; or,
N = N if £ = c
•"i •"!» 5-1 hr
Cor. 2. To construct a dihedral solid ^ equal to any
polyhedral ^ ; that is, to transform into a lune any sphe-
rical polygon ; add its angles, subtract (w— 2) st. 2£, and halve
the remainder.
EXAM. 67. Find the ratio of the vertical solid angles
of two right cones of altitude a\ and «2, but having the
same slant height h.
These solid angles are as the corresponding calots on the
sphere of radius h.
Therefore, from 65, the required ratio is
the ratio of the calot-altitudes. For the equilateral and
right-angled cones this becomes
2-V3
2-V2
AREAS OF BROKEN AND CURVED SURFACES.
83
THIRD REFERENCE TABLE OF ABBREVIATIONS.
= angles.
= density.
= edge.
= V. of paraboloid.
= V. of ellipsoid.
= V. of prolate spheroid.
= 2£ of cone.
= mass.
= approximation.
= radian.
= V. of oblate spheroid.
= distance.
= V. of spherical ungula.
= function.
= V. of hyperboloid.
= V. of mid F. of spindle.
= weight.
= varies as.
= = congruent.
= = approaches.
PC = mass-center.
CHAPTER V.
THE MEASUREMENT OF VOLUMES.
g(M). 'PRISM AND CYLINDER.
XXVIII. Two polyhedrons are ,<*>///> //><•//•/'>// whoso faces
are respectively congruent, and whose polyhedral angles
are respectively symmetrical; e.g., a polyhe-
dron is symmetrical to its image in a mirror.
A quader is a parallelepiped whose six faces
are rectangles.
A cube is a quader whose
six faces are squares.
XXIX. The volume of a
solid is its ratio to an assumed
unit.
The unit for measurement of volume is a cube whose
edge is the unit of length.
Thus, if the linear unit be a meter, the unit of volume,
contained by three square meters at right angles to each
other, is called a cubic meter (m3).
XXX. Using length of a line to mean its numerical
measure, lengths, areas, and volumes, are all three quantities
of the same kind, namely ratios. All ratios, whether ex-
pressible as numbers or not, combine according to the same
simple laws as ordinary numbers and fractions.
Ww. Bk. III. ; (Eu. Bk. V. ; Cv. Bk. II.).
THE MEASUREMENT OF VOLUMES. 85
Therefore, we may multiply lengths, areas, and volumes
together promiscuously, or divide -one by the other in any
order.
If we ever speak of multiplying a line, a surface, or a
solid, we mean always the length of the line, the area of
the surface, or the volume of the solid.
72, To find the volume of a quader.
Kule : Multiply together the length, breadth, and height
of the quader.
Or, in other words,
Multiply together the lengths of three adjacent edges.
Formula : If= abl.
Proof : By 32, the number of square units in the base of
a quader is the product of two adjacent edges, LI.
If on each of these square units we place a unit cube,
for every unit of altitude we have a layer of bl cubic
units ; so that, if the altitude is a, the quader contains abl
cubic units.
Cor. 1. The volume of any cube is the third power of
the length of an edge ; and this is why the third power
of a number is called its cube.
Cor. 2. Every unit of volume is equivalent to a thousand
of the next lower order.
Cor. 3. The arithmetical or algebraic extraction of cube
root makes familiar the use of the equation
(a + b}3 = a3 + 3 a*b + 3 ab* + b3.
86
MENSURATION.
Its geometric meaning and proof follow from inspection
of the figure of a cube*on the edge a -f- 5, cut by three
planes into eight quaders.
The cube a8 of the longer rod a, taken out, had faces a2
in common with three quaders of altitude b ; had edges a
in common with three quaders of base b2, and one corner
the corner point of the smaller cube bs.
XXXI. MASS, DENSITY, WEIGHT.
The unit of capacity is a cubic decimeter, called the
liter 0).
The quantity of matter in a body is termed its mass.
The unit of mass is called a gram (g). Pure water at tem-
perature of maximum density is 1-000013 gram per cubic
centimeter (cm8). So, in physics, the centimeter is chosen
as the unit for length, because of the advantage of making
the unit of mass practically identical with the mass of
unit-volume of water ; in other words, of making the value
of the density of water practically equal to unity ; density
being defined as mass per unit-volume.
THE MEASUREMENT OF VOLUMES. 87
The second is the fundamental unit of time adopted with
the centimeter and the gram.
Though the weight of a body, that is, the force of its at-
traction toward the earth, varies according to locality, yet
weight being proportional to mass, the number expressing
the mass of a body expresses also its weight in terms of the
weight of the mass-unit at the same place. Thus, in terms
Gram
Liter = Cubic Decimeter. Centimeter. Weight. Liter (common form).
of the gram and centimeter, or of the kilogram (kg) and liter,
the mass, weight, and volume of water are expressed by the
same number.
So the density of any substance is the number of times
the weight of the substance contains the weight of an equal
bulk of water. Therefore, the density of a substance is the
weight in grams of a cubic centimeter of that substance, or
the weight in kilograms of a liter. Hence,
73, To find the density of a body.
Kule : Divide the weight in grams by the bulk in cubic
centimeters.
-n i u> tog wkg
Formula: 8 = — ==_.
EXAM. 68. If 65 cubic centimeters of gold weigh 1251-77
grams ; find its density. _^ ^
88 MENSURATION.
EXAM. 69. How many cubic centimeters (cm3) in one
hektoliter (h1)?
Since 1 liter = 1000 cubic centimeters,
.'. 1 hektoliter = 100,000 cubic centimeters. Ans.
EXAM. 70. If the density of iron is 7-788, find the in.
of a rectangular iron beam 7 meters long, 25 centimeters
broad, and 55 millimeters high.
The volume of the beam in cubic centimeters is
700 x 25 x 5-5 = 96,250 cubic centimeters.
Therefore, its mass is
96,250 X 7-788 = 749,595 grams. Ans.
74, To find the volume of any parallelepiped.
Eule : Multiply its altitude by the area of its base.
Pornmla : V. P = abl
Proof : Any parallelepiped is equivalent to a quader of
equal base and altitude. For, supposing AB an oblique
parallelepiped on an oblique base, prolong the four ed.m-rf
parallel to AB, and cut them normally by two parallel
planes whose distance apart, CD, is equal to AB. This
gives us the parallelepiped CDE, which is still oblique,
but on a rectangular base. Prolong the four edges parallel
to DE, and cut them normally by two planes whose dis-
tance apart FG is equal to DE. This gives us the qua-
der FG.
Now, the solids AC and BDE are congruent, having all
their angles and edges respectively equal. Subtracting
each in turn from the whole solid ADE leaves CDE
equivalent to AB.
THE MEASUREMENT OF VOLUMES.
89
Again, the solids CDF and EG are congruent. Taking
from each, the common part EF leaves CDE equivalent
G
to FG. Therefore, the parallelepiped AB is equivalent
to the quader FG of equal base and altitude.
EXAM. 71. The square of the altitude of a parallelepiped
is to the area of its base as 121 to 63, and it contains
1,901,592 cubic centimeters. Find its altitude.
Here
aB = 1,901,592 and 63 a2 = 121 B.
.'. 63 a3 = 121 x 1,901,592 = 230,092,632.
•. a3 = 230,092,632 -*- 63 = 3,652,264.
.*. a = 154 cubic centimeters. Ans.
75. To find the volume of any prism.
Eule : Multiply the altitude of the prism by the area of
its base.
Formula \ V. P =
90
MENSURATION.
Proof : For a three-sided prism this rule follows from 74,
since any three-sided prism is half a parallelepiped of the
same altitude, the base of the prism being half the base of
this parallelepiped. To show
this, let ABCafty be any
three-sided prism. Extend-
ing the planes of its bases,
and through the edges Aa,
Oy, drawing planes parallel to
the sides, By, A/3, we have the
parallelepiped ABCDaftyS,
whose base A BCD is double
the base ABC of the prism.
Wr-
Ww.l78;(Eu.L34; Cv. 1. 105).
Also, this parallelepiped it-
self is twice the prism. For,
its two halves, the prisms, are congruent if its sides are all
rectangles. If not, the prisms #re symmetrical and equi-
valent. For, draw planes perpendicular to Aa at the
points A and a. Then the prism ABCafly is equivalent
to the right prism AEMa-^p., because the pyramid
AEBCM is congruent to the pyramid ar//3y/x. In the
same wray, ADCa&y equals ALMaXp..
But AEMarjfji and ALMaXp. are congruent, Therefore,
ABCafiy and ADCaSy are equivalent, and the parallele-
piped ABCDa(3y& is double the prism ABCafiy.
Thus, the rule is proved true for triangular
prisms, and consequently for all prisms ; since,
by passing planes through any one lateral edge,
and all the other lateral edges, excepting the
two adjacent, we can divide any prism into a
number of triangular prisms of the same alti-
tude, whose triangular bases together make the given
polygonal base.
7
THE MEASUREMENT OF VOLUMES.
91
Cor. 1. The volume of any prism equals the product of
a lateral edge by the cross-section normal to it.
Cor. 2. Every parallelepiped is halved by each diagonal
plane.
Cor. 3. Every plane pass-
ing through two opposite
corners, halves the paral-
lelepiped.
Cor. 4. The volume of
a truncated parallelepiped
equals half the sum of two
opposite lateral edges multiplied by the cross-section nor-
mal to them.
EXAM. 72. The altitude of a prism is 5 meters, and its
base a regular triangle. If, with density 4, it weighs 1836
kilogrammes, find a side of its base.
Its volume is 1836 -f- 4 = 449 cubic decimeters.
The area of its base = 459 -s- 50 = 9-18 square decimeters.
By 36, Cor., the square of a side of this regular triangle is
9-18 -H 0-433 = 21-2 square decimeters.
Therefore, a side equals . ar\A^. j • A
4-504+ decimeters. Am.
76, To find the volume of any cylinder.
Eule : Multiply the altitude of the cylinder by the area
of its base.
Formula when Base is a Circle : V. 0 —
92
MENSURATION.
Proof : In 58, Second Proof, we saw the cylinder to be
the limit of an inscribed prism when the number of sides
of the prism is increased indefinitely, ami
the breadth of each side indefinitely dimin-
ished, the base of the cylinder being coi
quently the limit of the base of the prism.
But, by 75, always V. P is to B in the con-
stant ratio a ; hence, by 13, their limits will
be to one another in the same ratio ; ami
V. 0 = aB.
. This applies to all solids whose
cross-section docs not vary, whatever be the shape of the
cross-section.
Cor. 1. Between any two parallel planes, the volume of
any cylinder equals the product of its axis by the
cross-section normal to it.
Cor. 2. By 58, Cor. 1, the volume of any trun-
cated circular cylinder equals the product of its
axis by the circle normal to it.
EXAM. 73. A gram of mercury, density 13-0, fills a cylin-
der 12 centimeters long; find the diameter of the cylinder.
Volume of cylinder = — 1 — = -OT.Vc".!-!- culiic centini'
»
Area of base of cylinder = r2* = •oy:;."!".) + li>
= -OOGll'T 1 1 fi sijuuri- centime:
Therefore r* = •OOiilL'Tlljj -:- 3-1416 - -OOlorxi.
r — -044 centimeters = -44 millimeters,
and d = 2r = -88 millimeters. An*.
77, To find the volume of a cylindric shell.
Rule: Multiply the sum of the inner and outer r<i<ln l>/
their difference, and this product by TT times tltc altitude of
the shell.
THE MEASUREMENT OF VOLUMES. 93
Formula : V. GI — V. C2 = a-n- (rx + r2) (TI — r2).
Proof: Since a cylindric shell is the difference between
two circular cylinders of the same altitude, its volume
equals
a?i TT — arg TT = CJTT (r*i — rs").
EXAM. 74. The thickness of the lead in a pipe weighing
94-09 kilograms is 6 millimeters, the diameter of the open-
ing is 4-8 centimeters ; taking TT = ^-, and density 11, find
the length of the pipe.
Here
r, = 2-4 centimeters and r\ = 3 centimeters.
z i
94,090 = !H7r(r1 + r2) (r± - r2)
= ^784-08;
.-. 658,630 = 784-08 1.
.'. ^=658,630 -T- 784-08 = 840 centimeters
= 8-4 meters. Ans.
g(N). PYRAMID AND CONE.
XXXII. The altitude of a pyramid is the normal dis-
tance from its vertex to the plane of its base.
78, Parallel plane sections of a pyramid are similar fig-
ures, and are to each other as the squares of their distances
from the vertex.
Proof : The figures are similar, since their angles are
respectively equal, Ww> m . (Eu XI 1Q . ^ yi ^
and their sides proportional.
Ww. 321 ; (Eu. VI. 4; Cv. III. 25).
94 MENSURATION.
By 44, they are to each, other as the squares of homolo-
gous sides, and hence as the squares
of the normals from the vertex.
Ww. 469; (Eu. XI. 17; Cv. VI. 37).
Scholium. This is the reason why
the intensity of gravity, light, heat,
magnetism, electricity, and sound, de-
creases as the square of the distance
from the source.
Image part of the rays from a
luminous point as a pyramid of light. If a cutting plane
is moved away parallel to itself, the number of units of
area illuminated increases as the square of the distance.
But the number of rays remains unchanged. Therefore,
the number of rays striking a unit of area must decrease
as the square of the distance.
79, Tetrahedra (triangular pyramids) having equiva-
lent bases and equal altitudes are equivalent.
Proof: Divide the equal altitudes a into n equal parts,
and through each point of division pass a plane parallel to
the base. By 78, all the sections in the first tetrahedron
are triangles equivalent to the corresponding sections in
the second.
Beginning with the base of the first tetrahedron, con-
struct on each section as lower base a prism - high with
n
lateral edges parallel to one of the edges of the tetrahe-
dron.
In the second, similarly construct prisms on each section
as upper base.
Since the first prism-sum is greater than the first tetra-
hedron, and the second prism-sum less than the second
THE MEASUREMENT OF VOLUMES. 95
tetrahedron, therefore the difference of the tetrahedra is
less than the difference of the prism-sums.
But, by 75, each prism in the second tetrahedron is
equivalent to the prism next above it on the first' tetrahe-
dron.
So the difference of the prism-sums is simply the lowest
prism of the first series, whose volume, by 75, is - —
n
As n increases this decreases, and can be made less than
any assignable quantity by taking n sufficiently great.
Hence the tetrahedra can have no assignable difference ;
and, being constants, they cannot have a variable differ-
ence.
Therefore the tetrahedra are equivalent.
Scholium. This demonstration indicates a method of
proving that any two solids having equivalent bases and
equal altitudes are equivalent, if every two plane sections
at the same distance from the base are equivalent.
80. To find the volume of any pyramid.
Eule : Multiply one-third of its altitude by the area of its
base.
Formula: V. Y
96
MENSURATION.
Proof : Any triangular prism, as ABC-FDE, can be
divided into three tetraliedra, two (B-DEF and
D—ABCT) having the same altitude as the prism, and its
top and bottom respectively as bases, while the third
(BCDF) is seen to have an altitude and bnso equal to
each of the others in turn by resting the prism first on its
side CE and next on its side AF. Hence, by 79, these
three tetrahedra are equivalent, and therefore, by 75, the
volume of each is \aB.
The rule thus proved for triangular pyramids is true for
all pyramids, since, by passing planes through
any one lateral <•'!;_:<>. and all the other lateral
edges excepting the two adjacent to this one,
we can exhibit any pyramid as a sum of te-
trahedra having the same altitude whose bases
together make the given polygonal base.
EXAM. 75. If the altitude of the highest Egyptian pyra-
mid is 138 meters, and a side of its square base 228 meters,
find its volume.
Here V. Y == £138 (228)2
=-46x51,984
= 2,391,264 cubic meters. Ans.
81. To find the volume of any cone.
THE MEASUEEMENT OF VOLUMES.
97
Kule s Multiply one-third its altitude by the area of its
base.
Formula when Base is a Circle : V. K = i ar*ir.
Proof : In 60, Second Proof, we saw that the base of a
cone was the limit of the base of the circum-
scribed or inscribed pyramid, and therefore
the cone itself the limit of the pyramid.
But, by 80, always the variable pyramid is
to its variable base in the constant ratio
ia.
Therefore, by 13, their limits are to one
another in the same ratio and
Scholium. This applies to all solids determined by an
elastic line stretching from a fixed point to a point de-
scribing any closed plane figure.
Cor. The volume of the solid generated by the revolu-
tion of any triangle about one of its sides as axis is one-
third the product of the triangle's area into the circumfer-
ence described by its vertex.
F=§7rrA.
EXAM. 76. Find the volume of a conical solid whose al-
titude is 15 meters and base a parabolic segment 3 meters
high from a chord 11 meters long.
By 54, here
V.K==il5x 13x11
= 5x2x11
= 110 cubic meters. Ans.
98
MENSURATION.
EXAM. 77. Required the volume of an elliptic cone,
the major axis of its base being 15-2 meters ; the minor
axis, 10 meters ; and the altitude, 22 meters.
By 55, here
V. K=\2-7-67r5
= 387r7£
-7T278I
= 875-45+ cubic meters. Ans.
EXAM. 78. The section of a
right circular cone by a plane
through its vertex, perpendicu-
lar to the base is an equilateral
triangle, each side of which is 12 meters ; find the volume
of the cone.
Here
a = Vl^-G2 - VT08.
= 391-78 cubic meters. Ans.
$ (O). PRISMATOID.
XXXIII. If, in each of two parallel planes is construct-
ed a polygon, in the one an w-gon, e.g., ABCD; in the
other, an n-gon, e.g., A'B'C1 ; then, through each side of
one and each vertex of the other polygon a plane may be
passed.
Thus, starting from the side AB, we get the n planes
ABA', AEB\ ABC'\ again, with the side EC, the n
planes EC A', BOB', EGC\ etc. Using thus all m sides
of the polygon A BCD, we get mn planes. Also, combin-
ing each of the n sides A*B\ B'C', C'D' with each of the
m points A, B, C, D, gives nm planes A'B'A, A'B'B,
THE MEASUREMENT OF VOLUMES.
99
A'B'C, A'B'D; B'C'A, B'C'B, etc. ; so that, altogether,
2 mn connecting planes are determined by the two poly-
gons. Among these are m -f- n outer planes, which to-
B
D
gether enclose the rest. These outer planes form the sides,
and the given polygons the bases of a solid called a pris-
matoid. Our figure is a case of this body when
m = 4 and n •= 3.
The midcross-section IJ is given to show the seven
sides.
XXXIV. A prismatoid is a polyhedron whose bases are
any two polygons in parallel planes, and whose lateral
faces are determined by so joining the vertices of these
bases that each line in order forms a triangle with the pre-
ceding line and one side of either base.
EEMARK. This definition is more general than XXXIII. ,
and allows dihedral angles to be concave or convex, though
neither base contain a reentrant angle. Thus, BB' might
have been joined instead of A'G.
100
MENSURATION.
From the prismatoids thus pertaining to the same two
bases, XXXIII. chooses the greatest.
XXXV. The altitude of a prismatoid is the normal dis-
tance between the planes of its bases. Passing through
the middle point of the altitude a plane parallel to the
bases gives the midcross-section. Its vertices halve the
lateral edges of the prismatoid. Hence, its perimeter is
half the sum of the basal perimeters. But, if one base
reduces to a straight line, this line must be considered a
digon, i. e., counted twice.
XXXVI. In stereometry the prism, pyramid, and pris-
matoid correspond respectively to the parallelogram, tri-
angle, and trapezoid in planimetry.
XXXVII. Though, in general, the lateral faces of a
prismatoid are triangles, yet if two basal edges which form,
THE MEASUREMENT OF VOLUMES. 101
with the same lateral edge, two sides of two adjoining
faces are parallel, then these two
triangular faces fall in the same
plane, and together form a trape-
zoid.
XXXVIII. A prismoid is a pris-
matoid whose bases
have the same num-
ber of sides, and
every corresponding pair parallel.
XXXIX. A frustum of a pyramid is a
prismoid whose two bases are similar.
Cor, Every three-sided prismoid is the frustum of a
pyramid.
XL. If both bases of a prismatoid become lines, it is a
tetrahedron.
XLI. A wedge is a prismatoid whose lower base is a
rectangle, and upper base a line parallel to a basal edge.
82. To find the volume of any prismatoid.
Enle : Add the areas of the two bases and four times the
midcross-section ; multiply this sum by one-sixth the alti-
tude.
Old Prismoidal Formula : D = J a (B^ -f 4 M+ -B2).
Proof: In the midcross-section of the prismatoid take
a point JVj which join to the corners of the prismatoid.
These lines determine for each edge of the prismatoid a
102
MENSURATION.
plane triangle, and these triangles divide the prismatoid
into the following parts :
1. A pyramid whose vertex is JV and whose base is _Z?2»
the top of the prismatoid. Since the altitude of this pyra-
mid is half that of the prismatoid, therefore, by 80, its vol-
ume s
2. A pyramid whose vertex is JV and whose base is £lt
the bottom of the prismatoid. Since the altitude of thia
pyramid also is ? a, therefore its volume is
THE MEASUREMENT OF VOLUMES.
103
3. Tetrahedra, like ANFG, each, of which can have its
volume expressed in terms of its own part of the midcross-
section. For, let NS and NK be the lines in which the
two sides ANF, ANG of the tetrahedron cut the mid-
cross-section ; and consider the part ANHK of the tetra-
hedron ANFG. This part ANHK is a pyramid whose
base is the triangle NHK, and whose altitude is \ a, half
the altitude of the prismatoid. Hence, by 80, the volume
of ANHK is \a(NHK}, But, drawing KF, by 79,
and
Therefore,
ANHK=%ANFKt
ANFK = %ANFG.
ANFG = ±a(NHK).
In like manner, the volume of every such tetrahedron
is fa times the area of its owrn piece of the midcross-sec-
tion, and their sum is %aM. Now, combining 1, 2, and 3,
which together make up the whole volume of the prisma-
toid, we find
D =
aif =
H
EXAM. 79. Given the plan of an embankment cut per-
pendicularly by the plane AEID, its top the pentagon
EFGHI, its bottom the D
trapezoid ABCD, with $
the following measure- J
ments : For the lower K
base, AB = 90 meters,
jii
CD = 110 meters, AD L<
= 65 meters ; for the T'
upper base EF = 70
meters, EI= 30 meters, MF '= MH — MG = 15 meters ;
the breadth of the scarp AE = 20 meters, DI= 15 meters ;
the altitude of the embankment a = 15 meters. Find its
volume.
104 MENSURATION.
Here, for the midcross-section, we get
TS=80 meters,
LR = 87'5 meters,
IQ = 97'5 meters,
NP= 90 meters,
TL = 7'5 meters,
Z7 — 32'5 meters,
IN= 7-5 meters.
Thus, the areas are
Bv — 6500 square meters,
Bt = 232") square meters,
M= 4337-5 square meters,
and for the whole volume we get
D = 65,437-5 cubic meters. Ans.
NOTE. In a prismoid the midcross-section has always the same
angles and the same number of sides as each base, every side being
half the sum of the two corresponding basal edges. The rectangular
prismoid has its top and bottom rectangles ; hence, by '32, its volume
/x^ Cor, If a prism has trapezoids for
bases, its volume equals half the sum
of its two parallel side-faces multiplied by their normal
distance apart.
83, To find the volume of a frustum of a pyramid.
Kule : To the areas of the two ends of the frustum add the
square root of their product ; multiply this sum by one-third
the altitude.
Pormula : V. F =
THE MEASUREMENT OF VOLUMES. 105
Proof : If Wi and w2 are two corresponding sides of the
bases £{ and -Z?2> then a side of the midsection is 1 — -•
Since in a frustum £lt Bz, and M are similar, by 44, we
have
whence ^ + w2 :
Hence
and
Substituting this in 82 gives
VTj^ 1 -. /o 7? i O
• JP = -f d I £ x*1 ~7" *-i
Cor. By 44,
Substituting this for _Z?2 gives
106 MENSURATION.
EXAM. 80. The area of the top of a frustum is 160
square meters ; of the bottom, 250 square meters ; and its
altitude is 24 meters. Find its volume.
Here
V. F = 8(250 + 200 + 160)
= 4880 cubic meters. Ans.
If, instead of the top, we are given Wi : w2 : : 5 : 4, then,
by our Corollary,
V. F = 2000(1+ | + H)
— 4880 cubic meters. Ans.
84, To find the volume of a frustum of any cone.
Eule : To the areas of the two ends add
the square root of their product;
multiply this sum by one-third the
altitude.
Formula for Circular Cone :
V F = *a
T • A tJ tt
Proof: As in 81, so the frustum of a cone is the limit of
the frustum of a pyramid.
EXAM. 81. The. radius of one end is 5 meters; of the
other, 3 meters ; the altitude, 8 meters. Find the volume.
V. F = £8(25 + 15 + 9)3-1416
= 410-5024 cubic meters. Ans.
85, To find the volume of any solid bounded terminally
by two parallel planes, and laterally by a surface generated
by the motion of a straight line always intersecting the
planes, and returning finally to its initial position.
THE MEASUREMENT OF VOLUMES. 107
Rule : Add the areas of the two ends to four times the
midsection; multiply this sum by one- sixth the altitude.
Prismoidal Pormula : D = i a (Bl + 4 M + JB2).
Proof: Join neighboring points in the top perimeter of
such a solid to form a polygon, likewise in the perimeter
of the bottom. Take the two polygons so formed as bases
of a prismatoid. Then when the number of basal edges
is indefinitely increased, each edge decreasing indefinitely
in length, as thus its bases approach to coincidence with
the bases of the solid, the sides of the prismatoid approach
the ruled surface, and its volume and midsection approach
the volume and midsection of the solid as limit. But
always the variable volume is to the variable sum
(J3i-|-4Jf -j-j5») in the constant ratio la. Therefore, by
108 MENSURATION.
13, their limits will be to one another in the same ratio ;
and D = $a(£l + ±M + £2)
for the prismoidal solid.
EXAM. 82. The radius of the minimum circle in a hyper-
boloid is 1 meter. Find the volume contained between
this circle of the gorge and a circle 3 meters below it whose
radius is 2 meters.
/Solution : The hyperboloid of revolution of one nappe
is a ruled surface generated by the rotation of a straight
line about an axis not in the same
plane with it. All points of the
generatrix describe parallel circles
'' whose centers are in the axis. The
shortest radius, a perpendicular both to axis and genera-
trix, describes the circle of the gorge, which is a plane of
symmetry. Hence, taking this circle as midsection, and
for altitude twice the distance to the base below it, the
Prismoidal Formula gives twice the volume sought in
Exam. 82.
27= D = i6[2*7r + 4(l)27r + 2V] = \2ir.
Therefore' V= 67T cubic meters. Ans.
86, To find the volume of any wedge.
Rule : To twice the length of the base add the opposite
edge; multiply the sum by the width of the base, and this
product by one-sixth the altitude of the wedge.
rormula : W -- = i aw (2 ^ + &2).
Proof: Since the upper base of a wedge is a line, so, by
the Prismoidal Formula,
THE MEASUREMENT OF VOLUMES. 109
Therefore, by 32,
W =
Cor. 1. If the length of edge equals the length of base ;
9 s>
'bl = 62, then W= $ awb,
the simplest form of wedge.
Cor. 2. The volume of any truncated triangular prism is
equal to the product of its right section by one-third the
sum of its lateral edges.
EXAM. 83. Find the volume of a wedge, of which the
length of the base is 70 meters ; the width, 30 meters ; the
length of the edge, 110 meters ; and the altitude, 24-8
meters.
Here W= (140 +110) *^8
= (140 +110) 10 x 12-4
= 2500 x 12-4
= 31,000 cubic meters, ^ws.
87. To find the volume of any tetrahedron.
Eule : Multiply double the area of a parallelogram whose
vertices bisect any four edges by one-third
the perpendicular to both the other edges.
Formula : X=
Proof: When, the bases being lines,
= = 0 then D = X
110
MENSURATION.
Since M bisects the line perpendicular to the two basal
edges, it bisects the four lateral edges ; and is a parallelo-
gram.
EXAM. 84. The line perpendicular to both basal edges
of a tetrahedron is 2 r long; the length of the top edge is
2r, and of the bottom edge, 2?rr. The midsection is a rec-
tangle. Find the volume of the tetrahedron.
Here
Therefore,
and
Ans.
$(P). SPHERE.
88. To find the volume of a sphere.
Kule : Multiply the cube of its radius by 4-1888— .
Formula: V. H-
Proof: Any sphere is equal in volume to a tetrahedron
whose midsection is equivalent to a great circle of the
sphere, and whose altitude equals a diameter.
Let the diameter DC be normal to the great circle AB
at C. Let Q be the point in which the midsection LN
bisects the altitude JK at right angles. In both solids
THE MEASUREMENT OF VOLUMES. Ill
take any height CI = QR, and through the points I and
R the sections PS parallel to AH, and MO parallel to LN.
Then, in the sphere, by 47,
OAB:OPS::AC*:PI2,
or, by Ww. 337 ; (Eu. VI. 8, Cor. ; Cv. III. 44),
OAB:QPS::AC2:TIxID ..... (1).
In the tetrahedron, by Ww. 321 & 499 ; (Eu. VI. 4, &
XL 17; Cv. IV. 25, & VI. 37),
LV-.MZ ::GL-.OM:'.KQ:KR-
and since, by Ww. 362 ; (Eu. VI. 23 ; Cv. IV. 5),
::LUxLV: MW X MZ,
therefore, aLN : a MO : : JQ x QK : JR x RK . . (2).
But now, by hypothesis and construction, in proportions
(1) and (2), the first, third, and fourth terms are respec-
tively equal, therefore
and since these are corresponding sections at any height,
therefore, by Scholium to 79, the sphere and tetrahedron
are equal in volume.
Thus, by 87,
V. H = X= f aM = f SrrV = f Trr3.
EXAM. 85. If, in making a model of the tetrahedron
EFGH, we wish the midsection LN to be a square, and
the four lateral edges equal, find in terms of radius their
length and that of the two basal edges.
By hypothesis,
square LN = r2?r ; .-. L U = r \Ar.
112
MENSURATION.
But
and
EF =2
For any one of the equal lateral edges,
EG2 = GK* + KE* = G~K* + KJ* + JE* = LU* + DT* + LV*.
.'. EG = VrV + 4^ + r*7r =-y4r*( 1 + ~ - 2ryl + -.
So it is not a regular tetrahedron.
89i To find the volume of any spherical segment.
Eule ; To three times the sum of the squared radii of the
two ends add the squared altitude; multiply this sum by
the altitude, and the product by -5236—.
Formula : V. G = * a* [3 (r? + r22) + a2].
Proof: In 88, we proved any spherical segment equal in
volume to a ' prismoid of
equivalent bases and alti-
tude. Therefore, by 82,
V. G-ia(r1!7r4-4r32rr4-r2V).
To eliminate rB call x the
distance from center of
sphere to bottom of seg-
ment, and r the radius of
sphere ; then, by Eu. II. 10,
Doubling and subtracting both members from ^r2, gives
2 r2 - 2 ( a + x)2 + 2 r2 - 2 xs = 4 r2 - 4 (^ + x V - a2,
or, by 2, 2 r22 + 2 rt2 + a2 - 4 r32.
Substituting, V. G = £ arr (3 r,2 + 3 r32 + a2).
THE MEASUREMENT OF VOLUMES. 113
Cor. In a segment of one base, since r2 = 0, we have
V. G = kair(3r* + a*).
But now, by Ww. 337 ; (Eu. VI. 8, Cor. ; Cv. III. 44),
r-j2 = a(2r — a).
Substituting,
V. G = £a7r[3a(2r-a) + a2] = |a7r(6ar - 3a8 + a2) = a27r(r - Ja).
EXAM. 86. If the axis of a cylinder passes through the
center of a sphere, the sphere-ring so formed is equal in
volume to a sphere of the same altitude.
For, since the bases of a middle segment are equidistant
from the center, V. G = * aTr (6 tf + «2)
-j- i Tra3.
But, by 76, the volume of the cylinder cut out of the
segment is a^V, and the remaining ring i?ra3 is, by 88,
the volume of a sphere of diameter a.
XLII. When a semicircle revolves about its diameter,
the solid generated by any sector of the semicircle is called
a spherical sector.
90, To find the volume of any spherical sector.
Kule : Multiply its zone by one-third
the radius.
Formula : V. S = t -jrar2.
Proof : If one radius of the gene-
rating sector coincides with the axis
of revolution, the spherical sector is
the sum of a spherical segment of one
base and a cone on same base, with vertex at center of
sphere.
114 MENSURATION.
By 89, GOT., V. G = «V(r -i«).
By 81, V. K= l(r-a)r?Tt-
or, substituting for rf, its value used in 89, Cbr.,
V. K = £(r-a)a(2r-a)7r = £
Adding, we have
V. S = V. G + V. K^
Any other spherical sector is the difference of two such
sectors.
V. S = V. S, - V. S2 = § r**av - I r2™, = § rV (a, - a,).
But at — a, = a,
the altitude of $'s zone, whose area, by 65, is 2-n-ra. Thus,
for every spherical sector the volume is zone by $r.
Cor. If i\ and r2 are the radii of the bases of the zone, its
altitude,
a = x/r2 — rs2 — Vr2 — r*.
EXAM. 87. Find the diameter of a sphere of which a
sector contains 7-854 cubic meters when its zone is 0-6
meters high.
V. S - | 0-6 Trr2 = 0-4 irr2 = 7'854.
Dividing by TT = 31416,
0-4^-2-5. /. 4ra=25.
/. 2r — 5 meters. Ans.
XLIII. A spherical ungula is a part of a globe bounded
by a lune and two great semicircles.
91. To find the volume of a spherical ungula.
THE MEASUREMENT OF VOLUMES.
115
Kule : Multiply the area of its lune by one-third the ra-
dius.
Pormula: £ =
Proof :
By 69, v:V. TL::L:H.
.-. v : | Trr3 : : 2 r'u :
Cor. On equal spheres, ungulae are as their angles.
92. To find the volume of a spherical pyramid.
Eule : Multiply the area of its base by one-third the ra-
dius.
Formula : Y = i i*e.
Proof:
By 69,
Y: V. H : : N: H.
g(Q). THEOREM OF PAPPUS.
93, If a plane figure, lying wholly on the same side of a
line in its own plane, revolves about that line, the volume
of the solid thus generated is equal to the product of the
revolving area by the length of the path described by its
center of mass.
Scholium. As for 66, so we give under 94, by a single
representative case, the general demonstration for all fig-
ures having an axis of symmetry parallel to the axis of
revolution.
116 MENSURATION.
EXAM. 88. Find the distance of the center of mass of a
semicircle from the center of the circle.
.
By 93, V. H =
Equating, we get $ ^ =
94, To find the volume of a ring.
r
= — . Ans.
07T
Eule : Multiply the generating area by the path of its
center.
Formula for Ellipse : V. 0 = 2?^^?*.
Proof : Conceive any ellipse to revolve about an exterior
axis parallel to one of its axes. Divide the axis of symme-
try AE into n equal parts, as
and from these points of division drop perpendiculars on
the axis of revolution PO. Join the points where these
perpendiculars cut the ellipse by chords FD, DA, AG,
OH, etc.
The volume generated by one of the trapezoids thus
formed, as DGHF, is the difference between the frustums
generated by the right-angled trapezoids QDFW and
QGHW. Therefore, by 84,
V. by DGHF
= $zTr(FW*+FW, DQ+D^-^zTr^W^+HW, GQ + GQ?)
= $z7r[(CO + FT)* + (CO + FT) (CO + DV) + (CO + DV)2
-(CO-FT)*-(CO-FT) (CO-D V] - (CO-D 7)2]
- £z7r(6<70, FT + 6(70, D V)
THE MEASUREMENT OF VOLUMES.
117
Thus, by 40, the volume generated by the polygon
HGADF, etc., equals its area multiplied by the path of
the center. But, as we increase n, and thus decrease z
indefinitely, as shown in 55, the area of the polygon ap-
proaches the area of the ellipse as its limit. But always
the variable volume is to the variable area in the constant
ratio 2-Trr ; therefore, by 13, their limits will be to one an-
other in the same ratio ; and
V. 0 = 27rra&7r.
EXAM. 89. Find the volume of the ring swept out by an
ellipse whose axes are 8 and 16 meters, revolving round an
axis in its own plane, and 10 meters from its center.
V. 0-4x8x10 X27T2
= 6316-5 cubic meters. Ans.
118
MENSURATION.
§(R). SIMILAR SOLIDS.
XLIV. Similar polyhedrons are those bounded by the
same number of faces respectively similar and similarly
placed, and which have their solid angles congruent.
95. Given the volume and one line in a solid, and the
homologous line in a similar solid, to find its volume.
Eule : Multiply the given volume by the cubed ratio of
homologous lines.
Formula : V\ =
n J
#2
Proof : The volumes of two similar solids are as the
cubes of any two corresponding dimensions.
W\v. 022; (Eu.XI. 33; Cv. VII. 73).
Thus, for the sphere, by 88,
NOTE. If a tetrahedron is cut by a plane parallel to one of its
faces, the tetrahedron cut off is similar to the first. If a cone be cut
by a plane parallel to its base, the whole cone and the cone cut off
are similar.
THE MEASUREMENT OF VOLUMES.
119
EXAM. 90. The edge of a cube is 1 meter ; find the edge
of a cube of double the volume.
The cube of the required number is to the cube of 1
as 2 is to 1; or,
or5: 1:: 2:1.
... x = </2 = 1-25992+. Ans.
Thus a cube, with its edge 1*26 meters, is more than
double a cube with edge 1 meter.
EXAM. 91. The three edges of a quader are as 3, 4, 7,
and the volume is 777,924 ; find the edges.
By 72, the volume of the quader, whose edges are 3, 4, 7,
is 84; then 84 : 777,924 :: 3* : 250,047,
and
3 : 7 : : 63 : 127.
Therefore, the edges are 63, 84, 127. Ans.
V250,047 = 63 ;
3: 4:: 63: 84,
3 : 7 : : 63 : 127.
§(S). IRREGULAR SOLIDS.
96, Any small solid may be
estimated by placing it in a vessel
of convenient shape, such as a
quader or a cylinder, and pouring
in a liquid until the solid is quite
covered ; then noting the level,
removing the solid, and again
noting the level at which the li-
quid stands. The volume of the
solid is equal to the volume of the vessel between the two
levels.
120 MENSURATION.
97, If the solid is homogeneous, weigh it. Also weigh a
cubic centimeter of the same substance. Divide the weight
of the solid by the weight of the cubic centimeter. The
i quotient will be the number of cubic centimeters in the
solid.
From 73, we have the
Pormula: Vccm = ~.
8
EXAM. 92. A ball 5 centimeters in diameter weighs
431-97 grams. An irregular solid of the same substance
weighs 13-2 grams ; find its volume.
The volume of the ball is
53 x 0-5236 = 65-45.
.-. 431-97 -f- 65-45 = 6-6 grams,
the weight of a cubic centimeter.
.'. 13-2 -f- 6-6 = 2 cubic centimeters. Ans.
98, To find the volume of any irregular polyhedron.
Eule : Cut the polyhedron into prismatoids by passing
parallel planes through all its summits.
Formula for n consecutive prismatoids:
J= i x*Bi ~ .5 x*B* ~ -#4 + etc.
(xs — x2)M2 + (.r4 — xs)M3 -f etc.
NOTE. <r2 is the distance of B2 from B^ and xa is the distance of
B. from B,, etc.
•> 1
Proof: This formula is obtained directly by the method
of 41.
CHAPTER VI.
THE APPLICABILITY OF THE PBJSMOIDAL FORMULA.
99, To find whether the volume of any solid is deter-
mined by the Prismoidal Formula.
Kule : The Prismoidal Formula applies exactly to ALL
SOLIDS contained between two parallel planes, OF WHICH
the area of any section parallel to these planes can be ex-
pressed by a rational integral algebraic function, of a degree
not higher than the third, of its distance from either of these
bounding planes or bases.
Test: Ax = q-\- mx -\- nx* -\- fy? .
NOTE. Ax is the area of any section of the solid at the distance x
from one of its ends. The coefficients q, ra, n,f, are constant for the
same solid, but may be either positive or negative ; or any one, two,
or three of them may be zero.
Proof: Measuring x on a line normal to which the sec-
tions are made, let <j> (x} be the area of the section at the
distance x from the origin.
The problem then is, What function <£ will fulfil the con-
ditions of the Prismoidal Formula ?
For any linear unit, the segment between <£ (0) and <£ (4)
is the sum of the segments between <j> (0) and <£ (2) and be-
tween <£(2) and ^(4). Therefore, if <£ is such a function
122 MENSURATION.
as to fulfil the requirements of the Prismoidal Formula, we
have identically
/, 0(0) -40 (1) + 60(2) -40(3) + 0(4) = 0.
But for 00*0 — q + mx + nz? +/J^ + <7#*>
0(0) - 40(1) + 60(2) - 40(3) + 0(4)
becomes + q
— 4<2 — 4m— 4n— 4/- 4g
+ 6^ + 12™ + 24n + 48/ + 96$r
-4q-l2m-36n- 108/- 324 g
+ q+ 4ra + 16n + 64/+256$r
0 0 0 0+240
So the conditions are satisfied only by functions which
have no fourth and higher powers. Hence <j>(z) must be
an algebraic expression of positive integral powers not ex-
ceeding the third degree.
Thus, in general, the cubic equation
Ax = q + mx + nx* +/T3
expresses the law of variation in magnitude of the plane
generatrix of prismoidal spaces; i.e., solids to which the
Prismoidal Formula universally applies,
Cor. 1. Since for prismoidal solids
therefore, 0 (0) + 4 0 Q a) + 0 (a) =
no
+ 4 n0 -f 2 a7^1 + a2n2 + ^ a.3n3
= 6 n0 + 3 anj + 2 a2/i2 +
THE APPLICABILITY OF THE PRISMOIDAL FORMULA. 123
Thus, D = l
= i
= an0 + $ aanx + £ a3n2 + J a4>?3.
Cor. 2. Of any solid whose
Ax = 0(x) = n0 + T^B + w,3* + ftgZ3 + ntof 4-
the volume is
arc0 + \ a*nt + % a?n2 + J- a4/i8 + ^ a5n4 + ..... + - am+1 nm.
m + 1
For the volume of the prism whose base is the cross-sec-
tion <f> (x), and whose altitude is the nth part of the altitude
of the whole solid, is -</>(V).
n
The limit of the sum of all the prisms of like height
when n becomes indefinitely great, is the volume of the
whole solid.
But
lr + 2*- + ..... + (n - IY I ,
— , when n = oo.
nr + 1 r + 1
[For full proof of this Cor. see page 233.]
§ (T). — PEISMOIDAL SOLIDS OF REVOLUTION.
The general expression
Ax = q + mx + nx? +/X3,
has as many possible varieties as there are combinations
of four things taken one, two, three, and four together ;
that is, 24 — 1, or 15 varieties.
Corresponding to each of these there will be at least one
solid of revolution generated by the curve whose equation
is, in the general case,
Try2 = q + mx + nx* +fx3.
For, if y be the revolving ordinate of any point in the
curve, then Try2 is the area of the section at distance x from
one end of the solid.
124
MENSURATION.
XLV. EXAMINATION OF THE DIFFERENT CASES.
(1) Let Try2= q\ .'.y is constant, and the solid is a circular
cylinder.
(2) Let 7r^=mx; .'. y2 oc x, and
we have a paraboloid of rev-
olution; for, in a parabola,
the square of the ordinate
varies as the abscissa.
(3) Let Try2 = no;2; .*. y oc x, and
the solid is a right
circular cone.
(4) Let Try2=/#3; .*. y2 oc Xs, and we have a semi- /
cubic paraboloid of revolution. ^"
(5) Let Try2 = q + mx ; .'. y2 oc (A -f- x) where h
is constant, and we have a frustum of a paraboloid of
revolution, h being the height of the segment cut
off.
(6) Let Try2 = q -f- nx* ; supposing q and n posi-
tive, this is the equation to a hyperbola,
the conjugate axis being the axis of x, and
the center the origin. Hence, we have a
hyperboloid of one nappe.
(7) Let Try2 = q -f-/#3. In this case, the solid is generated
by the revolution of a curve,
somewhat similar in form to the
semicubic parabola, round a line
parallel to the axis of x, and at
a constant distance from it.
(8) Let Tryi = mx-\-nx'i. In this case,
we may have a sphere, a prolate
spheroid, an oblate spheroid, a hyperboloid of revolu-
tion, or its conjugate hyperboloid.
THE APPLICABILITY OP THE PRISMOIDAL FORMULA. 125
(9) Let Try2 = q -f fmx -f- nx1. In this case, we will have
a frustum of a circular
cone, or of the sphere,
spheroids, or hyperbo-
loids of revolution, made
by planes normal to the
axis. In the frustum of
the cone q, m, and n are
all positive. The other
solids in (8) and (9) are
distinguished by the val-
ues and signs of the constants m and n.
(10) Let Try2 = q + mx -f nx2 -f-/#3. In this case, we have
a frustum of a semicubic paraboloid of revolu-
tion. For, if x be the distance of the section from
the smaller end of the frustum, and A the height
of the segment cut off, .*. y2 oc (h-{-xf. .'. Try2 is
of the form q -f- w# + no? -{-fx3.
(11) Let Try2 = mx +/T5.
(12) Let Try2 = nx2 +/X3.
(13) Let Try2 — q -f mx -fjfc3.
(14) Let Try2 = q + nx2 +fx*.
(15) Let Try2 = mx -f- nx2 -{-fx3.
EXAM. 93. Since, for an oblate spheroid,
^ = 0, £2 = 0, 41f=47ra2, and h = 25,
therefore its volume
Similarly, the volume of a
prolate spheroid
Thus, in volume, each is, like the sphere, two-thirds of
the circumscribed cylinder.
126 MENSURATION.
EXAM. 94. The volume of the solid generated by the
revolution round the conjugate axis of an arc of a hyper-
bola, cut off by a chord = and II to the conjugate axis, is
twice the spheroid generated by the revolution of the
ellipse which has the same axes.
Making the conjugate the axis of x,
jBt is the value of Try2 when x = b ; /. Bl = 2 Tret2.
Ez is the value of Try8 when x = — b ; /. j5a = 27ra2.
M is the value of Try2 when x = 0 ; .'. 4 M = 47ra2.
Since the conjugate axis is the height of the solid,
.-. h = 26.
Hence its volume X = |-7ra26.
EXAM. 95. To find the volume of a paraboloid of revo-
lution. Let h be its height, that is, the length of the axis,
r the radius of its base, and p the parameter of the gene-
rating parabola, y2 = px.
Then ,
j = 0 £ = iri* = Trh 4 M
.'. £= %vph*. Ans.
Cor. Since
= ph, .'. i, = J TrpM = | Trr2^,
or a paraboloid is half its circumscribing cylinder.
EXAM. 96. To find the volume of any frustum of a
paraboloid contained between two planes normal to the
axis.
Trp V = J *p ( V - Ai2) =
THE APPLICABILITY OF THE PEISMOIDAL FORMULA. 127
•But r'2 = phiy
and n2 = ph«
2~hl = a,
and
the altitude of the frustum.
P
= $ Tra (TI -f~
• Ans.
§ (U). — PBISMOIDAL SOLIDS NOT OF REVOLUTION.
We may now consider the same fifteen possible varieties
when Ax is not of the form Try2.
XLVI. DISCUSSION OF CASES.
(1) Let Ax •= q. In this case all the transverse sections
are constant. This is the property of all prisms and
cylinders; also, of all solids uniformly twisted, e.g.,
the square-threaded screw.
(2) Let Ax = mx. This is a property of the elliptic para-
boloid, or the solid generated by the motion of a
variable ellipse whose axes are the double ordinates
of two parabolas which have a common axis and a
128 MENSURATION.
common vertex, the plane of the ellipse being always
normal to this axis ; for, in this solid, the area of the
section at distance x from the vertex will be tryy\
where y and y1 are the ordinates of the two parabo-
las, and since both y2 and y'2 vary as x, :. Try?/
varies as x.
(3) Let Ax— nxz. This is a property of
all pyramids and cones, whatever
may be their bases.
(4) Let Ax=fx\ The solid will be
bounded by an elliptic semicubic
paraboloid. Substitute semicubic
for common parabolas in (2).
(5) Let Ax = q -\- mx. This is a prop-
erty of a frustum of an elliptic
paraboloid.
(6) Let Ax = q + nx*. This is a property of a groin, a
simple case of which is the square groin seen in the
vaults of large buildings.
This solid may be generated by a variable square,
which moves parallel to itself, with the midpoints of
two opposite sides always in a semicircumference,
the plane of which is perpendicular to that of the
square.
If y is a side of the square when at a distance x
from the centre ; /. y2 = 4 (r2 — x2).
(7) Let Ax = rnx -j- nx2. This is a property of the ellip-
soid, and of the elliptic hyperboloid.
(8) Let Ax = q -j- mx -f- nx2. This is a property of a pris-
moid, and of any frustum of a pyramid or cone,
whatever may be the base ; also, of any frustum of
an ellipsoid, or elliptic hyperboloid made by planes
perpendicular to the axis of the hyperboloid, or to
any one of the three axes of the ellipsoid.
THE APPLICABILITY OF THE PRISMOIDAL FORMULA. 129
EXAM. 97. To find the volume of an ellipsoid. Let a, 6,
c be the three semi-axes, a the greatest ;
Ans.
CHAPTER VII
THE NEW PRISMOIDAL FORMULA.
§ (V). ELIMINATION OF ONE BASE.
For all solids whose section is a function of degree not
higher than the second, or
q, m, n, and consequently Ax, for all values of x, are deter-
mined if the value of Ax for three values of x is known.
Measuring x from B^ we have
Supposing we know the section at - the height of the
z
solid above £i} we have for determining m and n the two
equations,
ma + no
a a
.
• 2
Hence,
(z-l)a2
130 MENSURATION.
For the volume of the solid we have, by Cor. 2, page 123,
V= B^ + $ ma? + na3,
or F=[(22
For z = 3, this gives
V=*(Bt
Again, for z = 1£,
These give the following theorem :
100, To find the volume of a prismatoid, or of any solid
whose section gives a quadratic :
Bnle : Multiply one fourth its altitude by the sum of one
base and three times a section distant from that base two-
thirds the altitude.
NOTE. For an easy synthetic proof of this two-term prismoidal
formula, the first and only one ever printed, see Halsted's Geometry,
page 344, where the formula is written
J) = -(B + 3T).
Cor. If J32 reduces to an edge or a point,
CHAPTER VIII.
APPROXIMATION TO ALL SURFACES AND SOLIDS.
§ (W). WBDDLE'S METHOD.
101 • To find the content between the first and seventh
of equidistant sections :
"Weddle's Kule : To five times the sum of the even sections
add the middle section, and all the odd sections; multiply
this sum by three-tenths of the common distance between
the sections.
Formula: ^ = -^A [5 (y
Applicability : We know from Cor. 2, page 123 (proved
in full in Note on page 233), that of any figure whose
every section parallel to a base is expressible in terms of
x, its distance from that base, as
<(> (x) = 7iQ + UjX + n2x* + nzs? + n4x4 + ngp,
the exact content is
132 MENSURATION.
and this will be found identical with the result of apply-
ing Weddle's Kule to such a figure, for this gives
106
5 [ n0 + n^a+ n.2 ($ a)2+ n3 (f a)3 f n< (£ a)4+r?5 (jj a)5]
n0 + rij | a + n2 (| a)2 + n3 (| a)3+ w4 (f a)4+ r?5 (f a)5
n + ria + r?a2 + rja3 + wa4 + na5
na + na + na
23
EXAM. 98. Between yl and y19
+2/9 +yn + yis + yi5 + yi7
+2/8
EXAM. 99. Find the volume of the middle frustum of a
parabolic spindle.
A spindle is a solid generated by the revolution of an
arc of a curve round its chord, if the curve is symmetric
about the perpendicular bisector of the chord. Hence a
parabolic spindle is generated by the revolution of an arc of
a parabola round a chord perpendicular to the axis.
Let the altitude of the middle frustum be divided into
six parts each equal to h, and let A^ A2, A3, A4, A5, A6, A7
be the areas of the sections.
APPROXIMATION TO ALL SURFACES AND SOLIDS. 133
Taking origin at center and r the longest radius, or
radius of mid-section, by the equation to a parabola, we
have for any point on the curve
a a
will be the area of the section at the distance x from O\
and this being a rational integral function of x of the
fourth degree, Weddle's Rule will determine the volume
exactly.
Now, if TI is the shortest radius, or radius of the two
bases, we have
^ = 7 =
A -A -
A-A-,
a o
A, - Trr2.
Therefore, by 101,
i2 + 32-i-
a a?) \ a a
r
But
a 9
20 Q / \ . n / \on
_ _*!_p, 7* ( 7* jn— 7* I I. s, \ 7* ._ 1* V* I
134
MENSURATION.
Making TI = 0, gives for the volume of the entire spindle,
But 6 hiri2 is the volume of the circum-
scribing cylinder ; hence volume of spin-
dle is -j^- circumscribing cylinder. The
middle frustum of a parabolic spindle is
a very close approximation to the general
form of a cask, and hence is used in cask-gauging.
EXAM. 100. The interior length of a cask is 30 decime-
ters ; the bung diameter, 24 decimeters ; and the head di-
ameters, 18 decimeters. Find the capacity of the cask.
Here r-12, rt=9, *-5;
/. lc = TT (2304 -f 864 + 486) = TT x 3654. Ans.
EXAM. 101. The two radii which form a diameter of a
circle are bisected, and ordinates are raised at the points
of bisection. Find approximately the area of that portion
of the circle between them.
Here
Hence, by 101,
60
2 + 3V3 + V35].
x 0-956608. Ans.
APPROXIMATION TO ALL SURFACES AND SOLIDS. 135
But the exact area is the difference between a semicircle
and the segment whose height is half the radius. Taking
TT = 3-1415927 and V3 = 1-7320508
gives for this area r2 X 0-956612, so that our approxima-
tion is true to five places of decimals. In all approximate
applications, it is desirable to avoid great differences be-
tween consecutive ordinates ; applied to a quadrant of a
circle, Weddle's Rule leads to a result correct to only two
places of decimals.
CHAPTER IX.
MASS-CENTER.
§(X). — FOB HOMOGENEOUS BODIES.
102. The point whose distances from three planes at
right angles to one another are respectively equal to the
mean distances of any group of points from these planes,
is at a distance from any plane whatever equal to the
mean distance of the group from the same plane.
103. The mass-center of a system of equal material
points is the point whose distance is equal to their average
distance from any plane whatever.
104. A solid is homogeneous when any two parts of
equal volume are exactly of the same mass. The determi-
nation of the mass-center of a homogeneous body is, there-
fore, a purely geometrical question. Again, in a very thin
sheet of uniform thickness, the masses of any two portions
are proportional to the areas. In a very thin wire of uni-
form thickness, the masses of different portions will be
proportional to their lengths. Hence we may find the
mass-center of a surface or of a line.
XLVII. BY SYMMETRY.
105. Two points are symmetric when at equal distances
on opposite sides of a fixed point, line, or plane.
106. If a body have a plane of symmetry, the mass-cen-
ter PC lies in that plane. Every particle on one side cor-
MASS-CENTER. 137
responds to an equal particle on the other. Hence the
of every pair is in the plane, and therefore also the PC of
the whole.
107, If a body have two planes of symmetry, the PC lies
in their line of intersection ; and if it have three planes of
symmetry intersecting in two lines, the PC is at the point
where the lines cut one another.
108, If a body have an axis of symmetry, the PC is in
that line.
109, If a body have a center of symmetry, it is the PC.
110, The PC of a straight line is its midpoint.
111, The PC of the circumference or area of a circle is
the center.
112, The PC of the perimeter or area of a parallelo-
gram is the intersection of the diagonals.
113, The PC of the volume or surface of a sphere is the
center.
114, The PC of a right circular cylinder is the midpoint
of its axis.
115, The PC of a parallelepiped is the intersection of
two diagonals.
116, The PC of a regular figure coincides with the PC of
its perimeter, and the PC of its angular points.
117, The PC of a trapezoid lies on the line joining the
midpoints of the parallel sides.
138 MENSURATION.
118, Since in any triangle each medial bisects every line
drawn parallel to its own base, therefore the PC of any tri-
angle is the intersection of its medials. By similar triangles,
this point lies on each medial two-thirds its length from its
vertex, and so coincides with the PC of the three vertices.
119, The PC of the perimeter of any triangle is the cen-
ter of the circle inscribed in the triangle whose vertices are
the midpoints of the sides.
Proof: The mass of each side is proportional to its length,
and its PC is its midpoint. So the PC of M2 and M3 is at
a point D, such that
DMa b $ AC
Hence, the PC of the whole perimeter is in the line
and since DMt divides the base M2MS into parts propor-
tional to the sides, it bisects ^ M± Similarly the PC is in
the line bisecting
120, The PC of the surface of any tetrahedron is the
center of the sphere inscribed in the tetrahedron whose
summits are the PC's of the faces.
121, If the vertex of a triangular pyramid be joined
with the PC of the base, the PC of the pyramid is in this
line at three-fourths of its length from the vertex. (Proof
by similar triangles.)
122, The PC of a, tetrahedron is also the PC of its four
summits.
123, The PC of any pyramid or cone is in the line join-
ing the PC of the base with the apex at three-fourths of its
length from the apex.
MASS-CENTER. 139
XLVIII. THE MASS-CENTER OF A QUADRILATERAL.
124, A sect is a limited line or rod.
125, The opposite to a point P on a sect AB is a point
jF", such that P and P' are at equal distances from the cen-
ter of AB, but on opposite sides of it.
126, The PC of any quadrilateral is the W of the tri-
angle whose apices are the intersection of its two diagonals
and the opposites of that intersection on those two diagonals
respectively.
Proof: Construct the PC's E and F of the triangles
AB C and AGD made by the diagonal A C of the quadri-
lateral ABCD; then the point on the line EF, which
divides it inversely as the areas of these triangles, will be
the ^0 of the quadrilateral. But if the diagonal BD is
cut by AC in the point G, then ABC :ACD = BG: GD.
So the sought point is the PC of BG X E and DG X F\
that is, of BG xA, BGx B, BG X C, and GD X A,
GD X D, GDx C. But we may substitute BD X A for
BGx A and GDxA; also BD X C for BGxC and
GD X C. For BGx B and GD X D we may substitute
BD X K, where K is the opposite of G on the sect BD.
Therefore the sought point is the PC of A, C, and K, that
is, of G, J, and JT, where J is the opposite of G on ,4 (7.
Cbr. Calling the PC of the quadrilateral I/, we have
127, The PC of the four angular points of a quadrilat-
eral is the intersection of the lines joining the midpoints of
pairs of opposite sides. Let this point 0 be called the mid-
center, and G the intersection of the two diagonals be
140
MENSURATION.
called the cross-center; then the '"C'of the quadrilateral is
in the line joining these two centers produced past the mid-
center, and at a distance from it equal to one-third of the
distance between the two centers.
That is, LOG will be a straight line, and
128,
THE MASS-CENTER OF AN OCTAHEDRON.
Let AF, BG, CH be three sects (finite lines) not meet-
ing. By an octahedron understand the solid whose eight
faces are ABC, ACG, AGH, AHB, A
FBC, FCG, FGII, FEB. The solid
is girdled by the perimeters of three
skew quadrilaterals, BCGH, CAIIF,
ABFG. Now the mid-points of the B(
sides of any skew quadrilaterals are
in one plane. Draw, then, three
planes bisecting the sides of these
quadrilaterals, and let them meet in a point A' which call
the cross-ccnf<'/'.
Let, also, M (mid-center) be the mean point of the six
vertices A, B, C, F, G, H; it is the PC of the triangle
formed by the mid-points of AF, BG, CH. To find /Sy, the
PC of the solid, join KM, and produce it to S so that
Proof : The solid is the sum of the four tetrahedra
AFBC, AFCG, AFGH, AFHB.
Now the PC of a tetrahedron is the PC or mean point of
its vertices ; consequently the line joining the P'CafAfBC
to the mid-point of GHis divided by the point M in'ihe
ratio 1 : 2. The same is true of the other three tetrahedra
and the mid-points of HB, BG, CG. Hence, the mass-
MASS-CENTER. 141
centers of the four tetrahedra are in one plane passing
through the point $ found by the above construction, and
therefore the ^Cof the whole solid is in this plane. So,
also, it is in the other two planes determined by dividing
the solid into tetrahedra having the common edge BG and
the common edge CH respectively. Therefore it coincides
with the point /&
Cor. By making the pairs of faces ABH, AHG; ACGf,
CFG; CBF, BHFio be respectively coplanar, we pass to
a truncated triangular pyramid. If we join its cross-center
JTwith its mid-center M, and produce KM to S, making
= i KM, then S will be the I1 Oof the trunc.
NOTE. This corollary was Sylvester's extension of the geometric
method of centering the plane quadrilateral, and suggested to Clifford
the above.
XLIX. GENERAL MASS-CENTER FORMULA.
In any body between parallel planes, we can reckon
the distance of its PC above its base, if every cross-section
is a given function <f>(x) of its distance x from the
base.
The prism whose base is the section <£(#), and whose
height is the nth part of the altitude a of the body, has for
volume -<l>(x\ Its PC is #+^- from the base of the body,
n 2n
and has the coefficient -<£(#). Now suppose the f*O of the
n
body distant T from its base, and form the product of r with
the sum D of the values which - </> (x) takes when x equals
1 9 —1 n
0, -a, -a, ..... , - - a; also form the sum jS of the values
n n n
which [x -f — ) - <f> OP) takes for the same worths of x. The
\, 2njn
142 MENSURATION.
product rD equals the sum /S when the arbitrary number n
is taken indefinitely great.
rD = S, for n = oo.
But for n = oo, the sum D expresses the volume of the
body. a
The sum S consists of the sum C of terms from -x<f>(x\
n
and of the sum E of terms from o~~ • -<f>(x). But the sum
Zn n v '
E has the value -— D, and vanishes for n = oo.
2n
Therefore, for the determination of T, we have the equa-
tion
129, MASS-CENTER OF ANY PRISMATOID.
If <£(#) is of degree not higher than the second,
X(f>(x), which we will call /(a:), is not higher than the third.
Therefore, by 99,
But /(0)=0, /(Ja)=
Therefore,
a
, a2[0(a) -0(0)1
Therefore, r = £ a + 1on '
j. _ / '
, aflB.-B.
T = "+
130, For the applicability of the Mass-Center Formula
the fes^ is ^lz = + mx + nx2.
MASS-CENTER. 143
For an examination of the possible varieties, see 99, § (T),
(1), (2), (3), (5), (6), (8), (9) ; and § (U), (1), (2), (3), (5),
(6), (7), (8).
Of course it applies also to the corresponding plane
figures.
EXAM. 102. For trapezoid
b^la 3(6,
EXAM. 103. For pyramid or cone
from J$i, or \ a from Bz.
EXAM. 104. The V-Q of a pyramidal frustum is in the line
joining the PC's of the parallel faces, and
T —
So, if any two homologous basal edges are as I to A,
the distance of the frustral PC from one base will be
a /A2
, and from the other -( -
EXAM. 105. From Exam. 95, for paraboloid
For frustum of this we obtain an expression similar to
that for trapezoid. It is
r = 1 a 4- o
144 MENSURATION.
EXAM. 106. For PC of half-globe, from center,
1 ""7** Q
131. The average haul of a piece of excavation is the dis-
tance between the ^C of the material as found and its '"C'as
deposited.
132, The ^O of a series of consecutive equally-long quad-
ratic shapes may be found by assuming the PC of each
shape to be in its mid-section, then compounding, and to the
distance of the point thus found from Bl adding - - 2 l'.
NOTE. It is a singular advantage of the PC formula that its second
or correction term remains as simple: for any number of shapes in the
series as for one.
In consequence, the error of the assumption that the PC of each
shape is in its mid-section, is less as the scries is longer. No error
whatever results from this assumption \vhen the end areas £l and J5t
are equal. For instance, in finding PC of a spherical sector whose
component cone and segment have equal altitude, we may* assume
that the PC of each is midway between its bases.
EXERCISES AND PROBLEMS IN MENSURATION.
PROBLEMS AND EXERCISES ON CHAPTER I.
EXERCISE 1. Find the diagonal of a square whose side
is unity. V2 = 141421+. Answer.
2. Find the diagonal of a cube whose edge is unity.
V3 = 1-732050+. Ans.
3. To draw a perpendicular to a line at the point C.
Measure CA = 3 meters, and fix at A and C the ends of
a line 9 meters long ; which stretch by the point B taken
4 meters from C and 5 meters from A.
BO is JL to AC because 32 + 42 = 52.
4. The whole numbers which express the lengths of the
sides of a right-angled triangle, when reduced to the lowest
numbers possible by dividing them by their common divi-
sors, cannot be all even numbers. Nor can they be all odd.
For, if a and b are odd, a2 and b2 are also odd, each
being an odd number taken an odd number of times.
Then a2 + 62 is even, and /. c is even.
2. <*-c? = (c + a)(c-a) = P.
5. To obtain three whole numbers which shall represent
the sides of a right-angled triangle.
146 MENSURATION.
w2— 1
Rule of Pythagoras. Take n any odd number, then — p —
= the second number, and - — = the third number.
2
Plato s Rule. Take any even number m, then also - —1,
and — • + 1.
4
Euclid's Rule. Take x and y, either both odd or both
even, such that xy = &2, a perfect square ; then a = — ^-^
and c = _*
()/" Maseres. Of any two numbers take twice their
product, the difference of their squares, and the sum of
their squares.
Proof: If m and n are whole numbers, so also are
m2 -f- r*2 and m2 — w2 and 2 mn.
But always
(m2 + nj = (m* - nj -f (2mn)\
EXERCISES AND PROBLEMS.
147
TABLE I. — DISSIMILAR RIGHT-ANGLED TRIANGLES.
Sides.
Area.
Sides.
Area.
345
6
57 176 185
5,016
5 12 13
30
85 132 157
5,610
8 15 17
60
36 323 325
5,814
7 24 25
84
29 420 421
6,090
9 40 41
180
60 221 229
6,630
12 35 37
210
119 120 169
7,140
20 21 29
210
31 480 481
7,440
11 60 61
330
84 187 205
7,854
16 63 . 65
504
104 153 185
7,956
13 84 85
546
95 168 193
7,980
28 45 53
630
40 399 401
7,980
15 112 113
840
69 260 269
8,970
33 56 65
924
33 544 545
8,976
20 99 101
990
68 285 293
9,690
17 144 145
1,224
133 156 205
10,374
48 55 73
1,320
44 483 485
10,626
36 77 85
1,386
35 612 613
10,710
39 80 89
1,560
105 208 233
10,920
19 180 181
1,710
75 308 317
11,550
24 143 145
1,716
96 247 265
11,856
21 220 221
2,310
140 171 221
11,970
65 72 97
2,340
120 209 241
12,540
44 117 125
2,574
37 684 685
12,654
60 91 109
2,730
76 357 365
13,566
28 195 197
2,730
48 575 577
13,800
23 264 265
3,036
115 252 277
14,490
51 140 149
3,570
39 760 761
14,820
25 312 313
3,900
52 675 677
17,550
32 255 257
4,080
87 416 425
18,096
52 165 173
4,290
160 231 281
18,480
88 105 137
4,620
136 273 305
18,564
27 364 365
4,914
161 240 289
19,320
148
MENSURATION.
TABLE I. — Continued.
Sides.
Area.
Sides.
Area.
56 783 785
21,924
319 360 481
120
93 476 485
±U34
121 957 in ;5
59,3:5 1
207 224 305
23,184
231 520
60,060
120 391 409
23,460
200 609 611
60,900
135 352 377
760
279 440 r>:M
01,380
92 525 533
24,150
185 672 li'.i 7
82,160
1 7-") 288 337
25,200
336 377 505
63,336
204 253 325
25,806
80 1,599 1,601
63,960
152 345 .",77
26,220
308 435 533
66,990
180 299 349
26,910
195 748 773
72,930
tin 899 910
26,970
396 403 565
7' 'I
145 408 433
29,580
259 660 709
85,470
225 272 353
30,600
368 465 593
85,560
100 621 629
31,050
336 527 625
88,536
105 608 617
31,920
315 572
90,090
189 340 389
32,130
273 7: '.6 785
100,464
64 1,023 1,025
32,736
400 ;V>1
112,200
252 275 373
34,05t >
1 627 725
11 1,114
168 425 457
35,700
155 528 697
120,120
155 468 493
36,270
407 '''-'I 715
126,984
228 325 397
37,050
301 900 949
135,450
111 680 689
37,740
468 595 757
139,230
108 725 733
39,150
432 665 793
143,640
68. 1,155 1,157
39,270
369 800 881
147,600
203 396 445
40,194
429 700 821
150,150
165 532 557
43,890
315 988 1,037
155,610
297 304 425
45,144
555 572 797
158,730
72 1,295 1,297
46,620
540 629 829
169,830
184 513 545
47,196
451 780 901
175,890
116 837 845
48,546
504 703 865
177,156
280 351 449
49,140
329 1,080 1,129
177,660
217 456 505
49,476
420 851 949
178,710
261 380 461
49,590
464 777 905
180,264
76 1,443 1,445
54,834
533 756 925
201,474
EXERCISES AND PROBLEMS.
149
TABLE I. — Concluded.
Sides.
Area.
Sides.
Area.
616 663 905
204,204
748 1,035 1,277
387,090
473 864 985
204,336
893 924 1,285
412,566
580 741 941
214,890
560 1,551 1,649
434,280
496 897 1,025
222,456
884 987 1,325
436,254
615 728 953
223,860
792 1,175 1,417
465,300
330 644 725
107,226
684 1,363 1,525
466,830
557 840 1,009
234,780
740 1,269 1,469
469,530
696 697 985
242,556
931 1,021 1,381
474,810
660 779 1,021
257,070
833 1,056 1,345
481,474 '
645 812 1,037
261,870
969 1,120 1,481
542,640
470 1,107 1,205
263,466
720 1,519 1,681
546,840
620 861 1,061
266,910
836 1,325 1,565
554,014
585 928 1,097
271,440
780 1,421 1,621
554,190
731 780 1,069
285,090
936 1,127 1,465
556,738
504 1,247 1,345
314,244
1,036 1,173 1,565
607,614
704 903 1,145
317,856
988 1,275 1,613
629,850
660 989 1,189
326,370
880 1,479 1,721
650,760
612 1,075 2,237
329,950
1,113 1,184 1,625
658,896
765 868 1,157
332,010
1,140 1,219 1,669
694,830
705 992 1,217
349,680
1,040 1,431 1,769
744,120
832 855 1,193
355,680
1,248 1,265 1,777
799,480
532 1,395 1,493
371,070
1,148 1,485 1,877
852,390
799 960 1,249
383,520
1,312 1,425 1,937
863,550
Sides.
Area.
4,059 4,060 5,741
23,600 23,661 33,461
207,000 207,151 292,849
159,140,519 159,140,520 225,058,681
150 MENSURATION.
t/l- +'LV t Jo =
3. a2 + 62-
6. Two sides, a = 6-708, b = 5, contain an obtuse angle.
If .7 = 3, find c. 10. Am.
1. If a = 13, 5 = 11, c = 20, find ^ 5. Ans.
(?ro-/61- fZt) ± 21* 1-
8. The three sides of a triangle are 2-5 meters, 4-8
meters, 3-2 meters ; find the projections of the other two
sides on b = 4-8 meters.
1-985 meters and 2-815 meters. Ans.
9. Two sides of a triangle, 3 meters and 8 meters long,
enclose an angle of 60? Find the third side.
/£-^-<~,$-/£,/i 7 meters. Ans.
HINT. Joining the midpoint of 3 with the vertex of the rt. $.
made in projecting 3 on 8 gives two isosceles triangles, and
.%;-H?
10. Two sides of a triangle are 13 meters and 15 meters.
Opposite the first is an angle of 60? Find the third.
7 or 8 meters. Ans.
HINT. Drop _L to 15-meter side. The segment adjacent the third
side is half the third side.
11. Two sides of a triangle are 9-6 meters and 12-8
meters ; the perpendicular from their vertex on the third
side is 7-68 meters; find that side. ^ <v\*r 4. Ans.
6. o» + c»-*&* = 2i'.
12. Given 4, the medial from A to a ; 4> from B to b ;
and ?3, from C to c. Find a, bt and c.
From 6,
2 2 2 2
EXERCISES AND PROBLEMS. 151
Taking the last equation from twice the sum of the two
former gives
f 2 42 — i\> Am.
Symmetrically, find b and c.
13. If ^ = 18, 4 = 24, 4 = 30 ; find a, b, and c.
a = 34-176, £ = 28-844, c = 20. Am.
-n -9 . -2 . -2 1/2. 12 i 9\ O^A- *.GM<*b"*4 1*fd&&^ t*»
14. Prove if -f V + ?s = f (a + o + <^). 5^V i
15. In any right-angled triangle prove $a = ^l±h
15
16. The locus of a point, the sum of the squares of whose
distances from two fixed points is constant, is a circumfer-
ence whose center is the midpoint of the straight line join-
ing the two fixed points.
17. The locus of points, the difference of the squares of
whose distances from two fixed points is constant, is a
straight line perpendicular to that which joins the two
fixed points.
18. The sum of any two sides of a triangle is greater than
twice the concurrent medial.
19. In every quadrilateral the sum of the squares of the
four sides exceeds the sum of the squares of the two diag-
onals by four times the square of the straight line joining
the middle points of the diagonals.
20. The sum of the squares on the four sides of a paral-
lelogram is equivalent to the sum of the squares on the
diagonals.
21. The sum of the squares of the diagonals of a trape-
zoid is equal to the sum of the squares of the non-parallel
sides augmented by twice the product of the parallel
sides.
152 MENSURATION.
22. On the three sides of a triangle squares are described
outward. Prove that the three lines joining the ends of
their outer sides are twice the medials of the triangle and
perpendicular to them.
23. Prove that the medials of a triangle cut each other
into parts which are as 1 to 2.
24. The intersection-point of medials is the center of
mass of the triangle.
25. Prove an ^ in a A, aci*te, rt., or obtuse, according as
the medial through the vertex of that 2£ is >, =, or < half
the opposite side.
7. 8, =
26. The three sides of a triangle are 17-4 meters, 23-4
meters, 31-8 meters. The smallest side of a similar tri-
angle is 5-8 meters. Find the other sides.
7-8 meters and 10-6 meters. Ans.
27. Find the height of an object whose shadow is 37-8
meters, when a rod of 2-75 meters casts 1-4 meters shadow.
74-25 meters. Ans.
28. The perpendicular from any point on a circumfer-
ence to the diameter is a mean proportional between the
two segments of the diameter.
29. Every chord of a circle is a mean proportional be-
tween the diameter drawn from one of its extremities, and
its projection on that diameter.
30. From a hypothenuse of 72-9 meters, a perpendicular
from the right angle cuts a part equal to 6-4 meters ; find
the sides. a — 21-6 meters, b = 69-62 meters. Ans.
EXEKCISES AND PROBLEMS. 153
31. The hypothenuse is 32-5 meters, and the perpendicu-
lar on it 15-6 meters ; find the segments.
11-7 and 20-8. Ans.
32. In a rt. A, c = 36-5 meters; a -f- 6 = 51-1 meters;
find a and b. a = 21-9 meters, b = 29-2 meters. Ans.
33. The sides of a triangle are 4-55, 6-3, and 4-445 ; the
perimeter of a similar triangle is 4-37. Find its sides.
1-3, 1-8, and 1-27. Ans.
34. A lamp-post 3 meters high is 5 meters from a man 2
meters tall ; find the length of his shadow. /O T-VI, .
35. The sides of a pentagon are 12, 20, 11, 15, and 22;
the perimeter of a similar pentagon is 16 meters. Find its
sides. 2-4, 4, 2-2, 3, and 4-4. Ans.
36. Through the point of intersection of the diagonals
of a trapezoid a line is drawn parallel to the parallel sides.
Prove that the parallel sides have the same ratio as the
parts into which this line cuts the non-parallel sides.
7.. 2
8. <*=&
h
10 /-i -
1U. K^ -
37. Find the side of a regular decagon.
If a radius is divided in extreme and mean ratio, the
greater segment is equal to a side of the inscribed regular
Ww. 394; (Eu. IV. 10; Cv. V. 17).
V5-1
= r— — . Ans.
154 MENSURATION.
11. Jsk = V2 r2 - r V4 r* - I2.
38. Prove that the sides of the regular pentagon, hexa-
gon, and decagon will form a right-angled triangle ; or
"'lO T~ "-6 == ^5 '
39,
>. Show that £24 = rA/2 - V2
kM = r \2-\2 + \2+V2 + V3, etc.
40. Show that &4 =
JL - r V2 - V2,
V2,
ka = r \2 - A'2 + V2 + V2, etc.
O 7...
12. *=•
41. If an is the apothegm of a regular inscribed w-gon,
Vr (r 4- c?.,)
~^2
HINT. Use 7 and 2.
42. If also p'n be perimeter of inscribed polygon, pn of
7?' T
circumscribed, prove p'2n = — — •
43. pn : p'n : : r : an.
44. Indicate how to reckon TT by 40, 41, and 42, or by
2p'npn
the following p2n — — \ . , - and p^^= vp'np2n.
13.
45. When a quantity increases continually without be-
coming greater than a fixed quantity, it has a limit equal
or inferior to this constant.
EXERCISES AND PROBLEMS. 155
46. When a quantity decreases continually without be-
coming less than a fixed quantity, it has a limit equal or
superior to this constant.
47. When two variables are always equal, if one of them
has a limit, so has the other also.
14. Lim pn = lim p'n — c.
48. Show that p'2n2 = p'np2n-
I 2pn ZpnP'n
49. Prove p'* = p'n\ — — r, and also p2n = — -r— r-
* J_Jn \ J/ n jfn I f n
50. If 9£n, rn be the radii of the circumscribed and in-
scribed circles of a regular polygon of n sides, and $l2n, r2n
the corresponding radii for a regular 2n-gon of the same
perimeter as the n-gon ; then 9?nr2n= $t2n2, and 9?n+ rn= 2r2n.
51. If %i, %.2, ^!3, etc., 2f2n be the angles of a polygon of
2n sides, inscribed in a circle, then 2£i + ^3 + etc. -{- ^2n-i
52. The greater the number of sides of a regular poly-
gon, the greater is the magnitude of each of its angles.
The limit is /, which can never be reached since the sum
of the exterior angles is always 6.
15. TT = 3-141592653589793238462643383279+.
1 A
AO. C-± '. C2 '. '. T\ '. r2.
17. TT = f = t^ = 3-1416-
d r
18. c = d-rr = 2 rir.
53. Find c when r = 14, taking TT = %f- . 88. ^Iws.
19 d-2r--~
JL t^ • tX/ ^J / *~ — O /\
7T
54. Find the diameter of a wheel, which, in a street 19,635
meters long, makes 3,125 revolutions. 2 meters. Ans.
156
MENSURATION.
55. The hypothenuse is 10, and one side is 8 ; semicircles
are described on the three sides. Find the radius of the
semicircle whose circumference equals the circumferences
of the three semicircles so described. 12.
56. Find the radius of a sphere in which a section 30
centimeters from the center has a circumference of 251-2
centimeters.
r =
EXERCISES AND PROBLEMS ON CHAPTER II.
20. Arc measures "%. at center.
57. Find the third angle of a triangle whose first angle
is 12° 56" and second angle 114° 48". 53° 58' 16". Ans.
58. The angles of a triangle are as 1 : 2 : 3. Find each.
30°, 60°, 90°. Am.
59. Of the three angles of a triangle ^ a is 12° 20' small-
er than £ p, and £ y is 5° 43' smaller than £ p. Find
each. a -53° 41', 0 = 6G°1', y = 60°18'. Ans.
60. The sum of two angles of a triangle is 174°48'24";
the difference of the same two is 48° 24' 50". Find all
three. 110°36'37", 63°11'47", 5°11'36" Ans.
61. Three angles of a quadrilateral are 125° 48' 32",
127° 58' 45", 85° 37' 27"; find the fourth. 20°35f16". Ans.
62. The angles of a quadrilateral are as 2 : 3 : 4 : 7 ; find
each. 45°, 67° 30', 90°, 157° 30'. An*.
63. In what regular polygon is every angle 168° 45'?
A 32-gon. Ans.
64. The vertical angle of an isosceles triangle is 148°47";
find the others. 15°59'36£". Ans.
EXERCISES AND PROBLEMS. 157
65. The sum of two angles adjacent to one side of an
isosceles triangle is/35°23'48"; find the three angles.
44° 36' 12", 44°36'12", and 90° 47' 36". Am.
21. u = L
r
66. Find the circular measure of the angle subtended by
a circular watch-spring 3 millimeters long and radius li
millimeters, }l*/~
67. If a perigon be divided into n equal parts, how many
of them would a radian contain ? n
• — . Ans.
68. Find the arc pertaining to a central angle of 78°
when r = 1-5 meters. I = 2-042 meters. Ans.
69. Find the arc intercepted by a central angle of 36°25f
when r = 8-5 meters. 1= 5-39 meters. Ans.
70. Find an arc of 112° which is 4 meters longer than its
radius. £ = 8-189 meters. Ans.
23. =
71. When c?=. 11-5, find arc 4-6 meters long.
. Ans.
72. Calling TT = %f-, find r when 64° measure 70-4 meters.
r = 63 meters. Ans.
HINT. 2 TIT: 704:: 360: 64.
24, 25, 26, 27, and 28.
73. Find the complement, supplement, and explement of
30°- c
158 MENSURATION.
74. Find the angle between the bisectors of any two ad-
jacent angles. / bQ - 2 ~
75. If a medial equals half its base, its angle is right.
76. If one angle in a right-angled triangle is 30°, one
side is half the hypothenuse.
77. In every isosceles right triangle half the hypothenuse
equals the altitude upon it.
78. One angle in a right triangle is 30° ; into what parts
does the altitude divide the right angle ?
79. How large is the angle between perpendiculars on
two sides of an equilateral triangle ? • ) V" I
80. Find the inscribed angle standing on an arc of
116°27'38". 58°13'49". Ans.
81. Find the inscribed angle cutting out one-tenth cir-
cumference. 18°. Ans.
82. Find the angle of intersection of two secants which
include arcs of 100° 48' and 54° 12'. 23° 18'. Ans.
83. An angle made by two tangents is measured by the
difference between 180° and the smaller intercepted arc.
84. From the same point in a circumference two chords
are drawn which cut off respectively arcs of 120° and 80° ;
find the included angle.
85. The four angle-bisectors of any quadrilateral from
a quadrilateral whose opposite angles are supplemental.
86. Find the circular measure of 42°. -73303. Ans.
87. Find the circular measure of 45°. -785398. Ans.
EXERCISES AND PROBLEMS. 159
88. Find the circular measure of 30°. -523598. Ans.
89. Find the circular measure of 60°. ^. Ans.
7T2
90. Find the circular measure of TT°. — . Ans.
180
91. Express seven-sixteenths of a right angle in circular
measure. JIT
T^
92. Express in circular measure an angle of 240°.
A 'VI O
-
3
93. Find in circular measure the angle made by the
hands of a watch at 5 : 15 o'clock.!''*' £/^ °
Of
94. Find u of %. made by watch-hands at a quarter to 8. £y "
95. Find u of watch ^ at 3 : 30 o'clock. 7? ~ n.
96. Find u of watch %. at 6:05 o'clock. /i>2i
97. The length of an arc of 45° in one circle is equal to
that of 60° in another. Find the circular measure of an
angle which would be subtended at the center of the first
by an arc equal to the radius of the second. f . Ans.
7T
98. The angle whose circular measure equals one-half is
28° 38' 52" 24'". fg
f
99. Find the number of degrees in an angle whose u = f .
/ = 1(57-2957795+) = 38-197186+. Ans. T
100. Find % whose u = f. 42°-9718346. Ans. t£
101. Find the number of degrees in an angle whose
u = . 71-61972439. Ans.
160 MENSURATION.
102. Find the number of degrees, minutes, and seconds
in an angle whose u = ±\. 30°0'43"-45. .
103. The circular measure of the sum of two angles is
y|-7r, and their difference is 17°; find the
1;V and 10° 15'. .
104. Express in degrees the angle whose u = $ IT.
120°. Ans.
105. How many degrees, minutes, and seconds are there
in the angle whose circular measure is £ ?
47° 44' 17" nearly, A
106. Express in degrees and circular measure the verti-
cal angle of an isosceles triangle which is half of each of the
angles at the base. 3 & - T-
o
107. How many times is the angle between two consecu-
tive sides of ;i ivmilar hexagon contained (1) in a right
angle? (2) in a radian? (i) 2f (2) A. j
4 L i 7T
108. Two wheels with fixed centers roll up<»n i-adi other,
and the circular measure of the angle through which one
turns gives the number of degrees through which the other
turns in the same time ; find the ratio of the radii of the
. Am.
o
irg
109. The length of an arc of 60° is 36 \ ; find the radius.
r = 35. Ans.
110. Find circumference where ^ 30° is subtended by
arc 4 meters.
111. If 4 be the length of an arc of 45° to radius r1} and
4 the length of arc 60° to r2, prove 3^4 = 4r2 1\.
EXERCISES AND PROBLEMS. 161
EXERCISES AND PROBLEMS ON CHAPTER III.
32. R = aJb.
112. The area of a rectangle is 2,883 square meters ; the
diagonal measures 77-5 meters. Find the sides.
5 = 46-5 meters, a = 62 meters. Am,
HINT. a2 + Z>2 = (77-5)2 ...... I.
a& = 2,883 ...... II.
Substitute in I. the value of a from II. This gives a biquadratic
soluble as a quadratic. Or, to avoid the solution of a quadratic,
multiply II. by 2, then add and subtract from I. This gives
a2 + 2ab + b* = 11,772-25,
a2-2a& + 62 = 240-25.
.-. a + b = 108-5,
a-b = 15-5.
113. Find the perimeter and area of a rectangle whose
altitude a --=. 1,843-02 meters and base b = 845-6 meters.
p = 5,377-24 meters,
R= 1,558,457-712 square meters. Ans.
114. Find the number of boards 4 meters long, 0-5 me-
ters broad, necessary "to floor a rectangular room 16 meters
long and 8 meters broad. ~T7^- 64. Ans.
115. Of two equivalent rectangles, one is 4-87 meters
long and 2-84 meters broad, the other is 4-26 meters long.
How broad is it ? V-^7 X 1-jrfr 3.945 meters. Ans.
116. The perimeter of a rectangle is 24-54 meters ; the
base is double the altitude. Find the area. Wl V %<!%
It = 33-4562 square meters. Ans.
117. The difference of two sides of a rectangle is 1-4 me-
ters ; their sum, 8-2 meters. Find its area.
-E = 16-32 square meters. Ans.
162 MENSURATION.
118. The base of a rectangle of 46-44 square meters is
3-2 meters longer than the altitude ; find these dimensions.
&4.3-KL + 2-Hc V9 l = 8-6 meters, a = 5-4 meters. Ans.
a*A6-7
HINT. <i -fr.t^ *>•*/• & = a + 3-2 meters.
.-. a (a + 3-2) = 46-44, etc.
*- 119- The perimeter of a rectangle is 13 meters longer
o r
than the base; the area is 20-88 square meters. Find the
5-8 and 3-6 meters, or 7-2 and 2-9 meters. An*.
' l' * 120. The perimeter of a rectangle is 3-78 meters; its di-
agonal, 1-35. Find the area.
X0#X-g/ =- R = 0-8748 square meters, ^n*.
121. A rectangular field is 60 meters long by 40 mo 1 ITS
wide. It is surrounded by a ru.-id <>t' uniform width, the
whole area of which is equal to the area of the field. Find
the width of the road. 10 meters. Ans.
122. A rectangular court is 20 meters longer than broad,
and its area is 4,524 square meters; find its .length and
breadth. /V^x -= " 78 and 58 meters. Ans.
33. q = b*.
123. Find the area of a square whose side is 15-4 meters.
237-16 square meters. Ans.
124. The area of a square is f square meter ; find its side.
r? =. 0-79057 meters. Ans.
%
125. The side of a square is a ; find the side when the
area is n times as great. at = a~Vn. Ans.
126. The sum of two squares is 900 square meters, the
difference 252 square meters ; find the sides.
24 and 18 meters. Ans.
HINT. at2 + <z22 = 909,
a2 + - «2 = 252.
EXERCISES AND PROBLEMS. 163
127. The sides of two squares differ by 12 meters ; their
areas by 240 square meters. Find the side and area of
a /\ i
i c)
each. Sides, 4 and 16 meters ; = .
Areas, 16 and 256 square meters. Ans.
HINT. (a + 12)2-«2=240.
128. The perimeter of a square is 48 meters longer than
the diagonal ; find the area.
344-544375 square meters. Ans.
HINT. 4a = aV2 + 48.
129. The sum of the diagonal and side of a square is 100
meters; find the area. 1715-6164 square meters. Ans.
HINT. 2 a2 -(100 -a)2.
34. a = ab.
130. The area of a parallelogram is 120 square meters,
two sides are 12 and 14 meters ; find both diagonals.
24 and 10-2 meters. Ans.
131. The altitudes «j and a2 'of a parallelogram are 5 and
8 meters ; one diagonal is 10 meters. Find the area.
40-285 or 100-65 square meters. Ans.
HINT. &f = H,
&t = 102 + J22-2Zy.
Also, / = 100 - 64 = 36. .-. j = 6.
132. One side of a parallelogram is 8 meters longer than
the corresponding altitude, and O = 384 square meters ;
find this side. 24 meters. Ans.
35. A = lab.
133. The area of a rhombus is half the product of its
diagonals.
164 MENSURATION.
134. From any point in an equilateral triangle the three
perpendiculars on the sides together equal the altitude.
135. Find the area of a right triangle whose two sides
are 248-2 and 160-5 meters.
19,918-05 square meters. Ans.
136. If 18-4 meters is the altitude of a A = 125-36 square
meters, find b. 13-626 meters. Ans.
137. The two diagonals of a rhombus are 8-52 and 6-38
meters; find the area. 27-1788 square meters. Ans.
138. The altitude of a triangle is 8 meters longer than its
base, and area is 44-02 square meters ; find b.
6-2 meters. Ans.
HINT. 44-02
139. The altitude of a right triangle cuts the hypothe-
nuse into two parts, 7-2 and 16-2 meters long ; find the
area. /0.fc , &r/3» A ..126-36 square meters. Ans.
*
36. A = Vs(s- a) (s-b) (s - c).
140. 2 log A = log s -f log (s — a) -f- log (s — b) + log (s — c).
141. If b is the base of an equilateral triangle, find the
area.
Here
.-.A =
142. The altitude of an equilateral triangle is 8-5 meters;
find the area. *-, >,-,
41-71 square meters. Ans.
143. The area of an equilateral triangle is 5-00548 square
meters ; find the side. 3-4. Ans.
EXERCISES AND PROBLEMS. 165
144. The side of an equilateral triangle is 4 meters
longer than the altitude ; find both.
a = 25-856, J = 29-856. Am.
.
62-32a; = -64.
145. The three sides of a triangle are 1O2, 13-6, and 17
meters ; find the area. 69'36 square meters. Ans.
Here
s = 20-4. .-. A = V20-4(20-4 - 10-2) (20-4 - 13-6) (20-4 - 17).
146. By measurement, a = 37*18 meters, b = 48'72 me-
ters, c = 56-46 meters ; find A.
s =71-18; .-. logs =1-85236
s - a = 34-00 ; .-. log (s - a) = 1-53148
s-6 = 22-46; .-. log (s- 6) = 1-35141
s-c = 14-72; .-. log(s- c) = 1-16791
.-. log A* = 5-90316
.-. logA =2-95158
.'. A = 894-50 square meters. Ans.
147. If three arcs, whose radii are 3, 2, 1, at their centers
subtend angles of 60°, 90°, 120°, and intersect each other
at their extremities, prove that the sides of the triangle
formed by their chords are 3, 2V2, V3; and its area
HINT. The perpendicular from vertical 2£ 120° of isosceles A
equals half a side, since joining its foot with midpoint of side makes
an equilateral A.
148. The area of a triangle is 1012; the length of the
side a is to that of b as 4 to 3, and c is to l> as 3 to 2. Re-
quired the length of the sides.
a = 52-470, 5 = 39-353, c = 59-029. Ans.
166
MENSURATION.
Here
and
and
.*. 2s = b + b
s — b = ^ b,
- c =
.'. Area = V .
= 1012.
V885562 = 1012 x 144 = 145,728.
94- 101 62= 145, 728.
149. The area of a triangle is 144, and one of two equal
sides is 24 ; find the third side, or base.
Here
s = 24 + -, and s — a = s-c = %b. s - 6 = 24 — £ &.
.-. 576 = V(2304-62) Z>2.
150. Show that, in terms of its three medials,
A =
- if - i24 - if.
Proof: 4 (?'t2 + i22 + is2) = 3 (a2 + Z>2 + c2),
16 (v8^2 + i*i* + ?:22i32) = 9(a2i2 + a2c2 +
16 (ix4 + i24 + is4) = 9 (a4 H- i4 + c4).
But, by multiplying out 36, we have
A = J \/2(a262 + a»c* + 6V) - (a4 + b4 + c4).
151. Prove that the triangle whose sides equal the me-
dials of a given triangle is three-fourths of the latter.
EXERCISES AND PROBLEMS.
167
TABLE II. — SCALENE TRIANGLES.
Sides.
Area.
Sides.
Area.
4 13 15
24
25 33 52
330
3 25 26
36
11 100 109
330
9 10 17
36
17 39 44
330
7 15 20
42
24 35 53
336
6 25 29
60
25 29 36
360
11 13 20
66
13 68 75
390
5 29 30
72
20 51 65
408
13 14 15
84
25 39 56
420
8 29 35
84
21 85 104
420
10 17 21
84
26 35 51
420
12 17 25
90
21 61 65
420
19 20 37
114
19 60 73
456
16 25 39
120
35 44 75
462
13 20 21
126
25 39 40
468
15 28 41
126
8 123 125
480
11 25 30
132
29 35 48
504
11 90 97
132
51 52 101
510
13 40 51
156
29 60 85
522
15 26 37
156
28 65 89
546
10 35 39
168
25 51 52
624
13 30 37
180
25 52 63
630
12 55 65
198
36 91 125
630
7 65 68
210
26 51 55
660
17 25 28
210
25 92 113
690
9 73 80
216
29 52 69
690
15 41 52
234
17 105 116
714
13 37 40
240
32 53 75
720
9 65 70
252
34 65 93
744
33 34 65
264
25 63 74
756
15 37 44
264
39 41 50
780
25 51 74
300
21 89 100
840
20 37 51
306
35 52 73
840
168
MENSURATION.
TABLE II. — Continued.
Sides.
Area.
Sides.
Area.
25 84 101
840
43 259 300
1,806
14 157 165
924
26 145 153
1,836
35 53 66
!>24
SI 75 78
1,836
33 56 65
924
80 91 165
1,8 is
22 85 91
934
55 84 125
1,8 is
40 51 77
924
45 85 104
1,872
:U 156 185
930
45 91 116
1,890
23 140 159
75 88
1,980
34 61 7">
i.dSo
65 66 109
1,980
57 60 111
1,608
48 85 (-»l
2,01<;
:;i; 61 65
1,080
65 72 119
2,016
31 97 120
1,116
17 260 21 ; 7
2,010
39 62 85
1,116
92 117 205
2,070
25 101 114
1,140
61 69 100
2,070
38 (ir> 87
1,140
65 68 105
2,142
51 98 145
1,176
60 73 !'l
2,184
35 78 97
' 1,260
61 74 87
2,220
16 195 205
1,288
55 136 183
2,244
41 66 85
1,320
19 289 300
2,280
40 111 1-15
1,332
68 75 77
2,310
23 123 130
1,380
58 85 117
2,310
4<> 75 109
1,380
45 133 liil
2,3!) (
51 74 115
1,380
29 182 195
2,436
44 75 97
1,584
87 119 200
2,436
35 100 117
1,638
35 174 197
2,436
39 85 92
1,656
41 169 200
2,460
50 69 73
1,656
85 123 202
2,460
41 84 85
1,680
65 89 132
2,574
56 61 75
1,680
31 193 210
2,604
57 65 68
1,710
39 145 1G4
2,610
39 110 137
1,716
65 87 88
2,640
20 150 169
1,740
61 91 100
2,730
29 125 130
1,740
21 340 353
2,856
52 73 75
1,800
49 200 241
2,940
EXERCISES AND PROBLEMS.
169
TABLE II. — Continued.
Sides.
Area.
Sides.
Area.
27 275 292
2,970
105 124 205
5,208
35 197 216
3,024
75 176 229
5,280
76 85 105
3,196
51 233 260
5,304
37 195 212
3,330
65 173 204
5,304
87 112 185
3,360
45 296 325 .
5,328
45 164 187
3,366
91 125 174
5,460
78 95 97
3,420
104 111 175
5,460
57 122 125
3,420
99 113 140
5,544
65 109 116
3,480
47 250 267
5,640
73 102 145
3,480
55 244 273
6,006
65 126 173
3,484
105 116 143
6,006
65 119 156
3,570
100 217 303
6,510
40 231 257
3,696
91 145 180
6,552
69 113 140
3,864
153 185 328
6,660
65 119 138
3,864
43 520 555
6,708
60 145 161
3,864
119 150 241
7,140
89 99 100
3,960
50 369 401
7,380
57 148 175
3,990
89 170 189
7,560
75 109 136
4,080
65 297 340
7,722
85 99 140
4,158
37 525 548
7,770
91 100 159
4,200
85 234 293
7,956
90 97 119
4,284
123 133 200
7,980
40 291 325
4,290
65 272 303
8,160
87 100 143
4,290
111 200 281
8,880
68 87 145
4,350
140 143 157
9,240
39 280 305
4,368
68 273 275
9,240
89 111 170
4,440
111 175 176
9,240
55 207 244
4,554
89 208 231
9,240
66 175 221
4,620
116 231 325
9,240
143 168 305
4,620
111 175 232
9,324
61 155 156
4,650
74 277 315
9,324
37 411 440
4,884
117 164 175
9,450
41 337 360
4,904
116 181 225
10,440
123 208 325
4,920
91 253 300
10,626
170
MENSURATION.
TABLE II. — Concluded.
Sides.
Area.
Sides.
Area.
148 153 175
10,710
190 231 377
17,556
113 195 238
10,920
175 221 318
18,564
149 15G 175
10,920
175 221 27«i
19,320
66 389 4J5
11,220
125 312 323
19,380
123 187 200
11,220
143 296 375
19,536
85 293 336
11,424
186 221 275
20,460
170 171 305
11,628
I'll! ±T, 217
22,230
75 403 452
12,090
260 287 519
22,386
93 325 388
12,090
12!) 377 440
22,7<M
130 185 231
12,012
205 286 411
27,060
113 225 238
12,600
2J1 346 525
27,300
157 165 184
li',l 11
123 595 676
29,274
87 340 385
13,398
253 260 315
31,878
164 225 349
14,760
277 304
38,304
1 25 253 312
15,180
255 407 596
41,514
225 287 496
15,624
217 404 495
42,966
195 203 356
15,834
1 76 527 600
44,268
144 221 275
15,840
273 425 628
46,410
126 269 325
16,380
37. r=
152. Any right triangle equals the rectangle of the seg-
ments of the hypothenuse made by a perpendicular from
center of inscribed circle.
153. In any triangle ABC, let J/be the mid-point of the
base AC, I the point of contact of the inscribed circle, IT
and K the points where the perpendicular from the vertex
£, and the bisector of the angle B meet AC; prove the
relation MI. HI= Mil. KL
EXERCISES AND PROBLEMS. 171
154. Each tangent from A equals s — a ; from B equals
s — b ; from C equals s — c.
155. In a right triangle, CD is the perpendicular from C
on hypothenuse ; prove that the circles inscribed in triangles
CAD, CBD have the same ratio as these triangles.
156. If hi, h2, hs be the perpendiculars from the angles of
a triangle upon the opposite sides, and r the radius of the
inscribed circle, prove _!_ , 1_ , _!_ __ I
hi h2 hz r
HINT. r = ~ and - = — . etc.
At 2A A2 2A
157. Prove a = (a +
abc
158. If h\, h'2, h's be the perpendiculars from any point
within a triangle, upon the sides, prove
4^ + 7^4- — = !.
159. If TI, r2, r3 be the distances from the angles of a
triangle to the points of contact of the inscribed circle,
prove / TiT2T3
Bv
160. If r4, T6, r6 be the distances from the angles of a
triangle to the center of the inscribed circle, prove
161. Prove afo = ar42 + ^^52 + erf-
162. Prove r42 + r52 + r62 -= a5 + ac + 5c
38. ^ =
4A
163. The radius of a circle is 8 meters. Find the side of
an inscribed equilateral triangle, b = 13-8564 meters. Ans.
172
MENSURATION.
164. Find the radius circumscribing the equilateral tri-
angle whose base equals 8' 66 meters. 5 meters. Ans.
165. In every triangle, the sum of the perpendiculars from
the center of the circumscribed circle on the three sides is
equal to the sum of the radii of the inscribed and circum-
scribed circles.
166. If from the vertices of an equilateral triangle per-
pendiculars be drawn to any diameter of the circle circum-
scribing it, the perpendicular which falls on one side of this
diameter will be equal to the sum of the two which fall on
the other side.
167. If the altitude of an isosceles triangle is equal to its
base, f 1 = 9t
168. If A', .5', C' be the feet of the perpendiculars from
the angles of a triangle upon the sides, prove that the
radius circumscribing ABC is twice the radius circum-
scribing A'3'C1.
169. If a, ft, y be the perpendiculars from the center of
the circumscribing circle upon a, b, c, the sides of a triangle,
FOVe 2^=4^4.* +
afty ^ /?
on A A A
39. n = - - ; r2 = ; rs = - - .
s — a s — o s — c
170. If the sides of a triangle be in arithmetical progres-
sion, the perpendicular on the mean side from the opposite
angle, and the radius of the circle which touches the mean
side and the two other sides produced, are each three times
the radius of the inscribed circle.
171. Each of the common outer tangents to two circles
equals the part of the common inner tangent intercepted
between them.
EXERCISES AND PROBLEMS. 173
172. Each tangent from A to the circle escribed to a
equals s ; from B to circle escribed to a equals s — c; from
C to circle escribed to a equals s — b. Similar theorems
hold for the escribed circles which touch b and c.
173. The area of a triangle of which the centers of the
escribed circles are the angular points is -
174. If a, b, c denote the sides of a triangle ; hi, h2, hs the
three altitudes ; ql} q2, q3 the sides of the three inscribed
squares, prove the relations
qi hi a' q2 h2 b q3 h3 c
175. Prove = - - + — + — ;
1 111
_ = _j_ _|_ . etc.
r2 hi h2 h3
176. Prove |? = i_± = ± _j_ 1 ;
hi r TI r2 r3
I .
fl2 T T2 T3 TI
? =i_l =1+1.
h3 r r3 TI TZ
177. Prove
178. If T, T7, T8, r9 are the distances from the center of the
circumscribed circle to the centers of the inscribed and
escribed circles, prove the relations
ft>2 __ . 2 I
ul — T -f-
HINT. r2 =9l2-2r9?; r?2 = IR2 +
r2 = 2- = 2
174
MENSURATION.
179. Prove rft = —
abc ™ abc abc
4(8- 0)'
180. Prove rl + ra + r8 = r + 43t
181. In any triangle prove s2 — if = n\ ; etc.
40. r=a?
182. The base of a triangle is 20 meters, and its altitude
18 meters. It is required to draw a line parallel to the
base so as to cut off a trapezoid containing 80 square
meters. What is the length of the line of section, and its
distance from the base of the triangle ?
Calling b2 the line of section, and x its distance from 6lf
Now, &2:20::18-=-z:18.
80 = (20 + 20 - -V°- x) x = (20 - f x) x.
720 = 180 x- 5 tf.
.-. ^-36 a: = -144.
/. x - 18 = V324 -44.
/. a? =18 -13-416 = 4-584,
and &2 = 20 - -^(4-584) = 20 - 5-093 = 14-907.
183. In a perpendicular section of a ditch, the breadth at
the top is 26 feet, the slopes of the sides are each 45°, and
the area 140 square feet. Required the breadth at bottom
and the depth of the ditch.
EXERCISES AND PROBLEMS. 175
Here
T= 140 = £ [26 + 26 - 2 x] x = [26 - ar] x.
.-. 140 -26 a; -a".
... x = 13 = V169 - 140 = 13 = V29 = 13 = 5-385 = 7-615.
/. ^=26-15-230 = 1077.
184. The altitude of a trapezoid is 23 meters ; the two
parallel sides are 76 and 36 meters ; it is required to
draw a line parallel to the parallel sides,. so as to cut off
from the smaller end of the trapezoid a part containing 560
square meters. What is the length of the line of section,
and its distance from the shorter of the two parallel sides?
Let x equal altitude of required part.
Tt = l [76 + 36] 23 = 1288,
r2=728 = £[76 + Z] [23+4
76 + I
Also,
=23-
-x
36
1120 1456
36 + I 76 + Z
137,536 + 2576 . 1 = 62928 + 2576 . 1 + 23 .
62,928
74,608 = 23 . P.
/. Z2 = 3243-8.
.-. I =56-95.
= 12-048.
36 + 56-95
185. The two parallel sides of a trapezoid are 83'2 and
110-4 meters; the altitude, 50*4 meters. Find the area.
T= 4878-72 square meters. Ans.
186. The perimeter of a trapezoid is 122 meters. The
non-parallel sides are 36 and 32 meters ; the altitude, 30'4
meters. Find the area. T = 820'8 square meters. Ans.
176
MENSURATION.
187. T= 151-9 square meters, a = 12*4 meters, ^ = 18'6
meters. Find the other parallel side.
b2 = 5*9 meters. Ans.
188. The altitude and two parallel sides of a trapezoid are
2 : 3 : 5, and T= 1270-08 square meters. Find the parallel
sides. bi = 63 meters ; b2 = 37'8 meters. Ans.
189. The triangle formed by joining the mid-point of one
of the non-parallel sides of a trapezoid to the extremities of
the opposite side is equivalent to half the trapezoid.
190. The area of a trapezoid is equal to the product of
one of its non-parallel sides, and the perpendicular from the
mid-point of the other upon the first.
191. The line which joins the mid-points of the diagonals
of a trapezoid is parallel to the bases, and equals half their
difference.
192. Cutting each base of a trapezoid into the same num-
ber of equal parts, and joining the corresponding points,
divides the trapezoid into that number of equivalent
parts.
193. If the mean line of a trapezoid be divided into n
equal parts, and through these points lines, not intersecting
within the trapezoid, be extended to its bases, they cut the
trapezoid into n equal trapezoids.
194. In every trapezoid, the difference of the squares of
the diagonals has to the difference of the squares of the non-
parallel sides the same ratio that the sum of the parallel
sides has to their difference.
195. Let l>i be the longer, b2 the shorter, of the two parallel
sides in any trapezoid, zx and z2 the other two sides, and take
A =
EXERCISES AND PROBLEMS. 177
Prove r
From the intersection point of b2 and z2 draw a line
parallel to zx ; the base of the triangle so formed is (&i— &2),
and its other sides are zt and z2.
/.by 36,
• g=
196. The two parallel sides of a trapezoid are 184 and
68 meters ; the two others, 84 and 72 meters. Find the
area. 6536 square meters. Ans.
197. The diagonal of a symmetric trapezoid is Vz2 -f- b± b2.
198. The altitude of a trapezoid is 80 meters ; the two
diagonals 110 and 100 meters. Find the area.
5419*6 square meters. Ans.
HINT. T=$a(bl + bz} = ^a(Vc^a* + Vcf^a*).
199. In a trapezoid a = 140, bt = 160, b2 = 120 meters ;
if the area is halved by a line parallel to the bases, find its
length and distance below the shorter base.
1= 141-42 meters, d— 74*97 meters. Ans.
HINT I.
224
or, I. 71 -2d= 840,
II. 120 d+ W= 19,600.
Substitute the value of d from I. in II.
200. In a trapezoid bt = 312, b2 = 39, Zj === 350, z2 = 287
meters ; if cut by parallels to b into three similar trapezoids,
178
MENSURATION.
find where the two parallels cut the sides, and find the areas
of the three trapezoids.
If — 2 = n, then lz = nb2.
.'. Zj = nl2 = n?b2.
.-. 6± = nlt = n\ = n\ = 39 n3 = 312.
.'. n = 2.
2.
•'• 28 = 7" ~ 50 meters.
z4 = ,_* = 41 meters, etc.
41. "n
201. Find the distance between the points 1 and 2.
Between two points (x^yi), (.r2y2) the distance
202. Find the sum of the two right trapezoids determined
by the ordinates of the three points (12'3, 45-6), (78-9, 13),
(24, 57).
203. If the cross section of an excavation is a trapezoid,
b breadth of top, h depth, with side slopes m and n in 1,
which means that one side falls m meters vertically for one
meter of horizontal distance ; then show T=bh— ~h2.
2mn
42. If = -i- [>! (yn - ?/2) + x2 fa - 3/3) -f xs (y2 - y4) + .....
+ ^»(yn-i — yi)]-
204. Prove that a polygon may be constructed when all
but three adjacent parts (1 side and 2 ^s, or % sides and
1 ^) are given. What theorem for congruence of polygons
follows from this?
EXERCISES AND PROBLEMS.
179
205. Find the area of a heptagon from the coordinates of
its vertices, measured as follows :
x y
1
0
1-72
2
10-48
16-84
3
16-26
14-36
4
32-54
4-84
5
50-02
10-32
6
50-02
0
7
0
0
j\r=476'21 square meters. Ans.
206. Find the area of an enneagon from the following
measurements.
x y
I
0
16-96
2
26-36
20-04
3
58-02
22-16
4
104-00
11-24
5
97-48
2-48
6
92-22
-11-86
7
61-00
- 2-36
8
35-46
- 4-10
9
9-84
-14-22
Also draw the figure.
2429- 16 square meters. Ans.
207. Find the area of a pentagon, the coordinates of whose
vertices are as follows : (133, 917), (261, 325), (486, 916),
(547, 325), (828, 916).
180
MENSURATION.
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1
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<£)
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EXERCISES AND PROBLEMS.
181
209. Find the area of a hexagon from its coordinates
(719, 313), (512, 852), (719, 454), (513, 116), (720, 242),
(513, 993).
43. 2 Q=xl2—
210. The area of a quadrilateral inscribed in a circle is
/7 - w -- IT?— ^\ *L
= V(s — a)(s—o)(s—c)(s — d) where $=
211. If through the mid-point E of the diagonal BD of a
quadrilateral A BCD, FEG be drawn parallel to the other
diagonal AC, prove that the straight line AG divides the
quadrilateral into two equivalent parts.
212. Show that two quadrilaterals whose diagonals con-
tain the same angle are as the products of their diagonals.
213. A circle of r is inscribed in a kite, and another of rr
in the triangle formed by the axis of the kite and the two
unequal sides ; show that, if 2 1 be the length of the other
kite-diagonal, 111
rf r I
214. To find the area of any quadri-
lateral from one side, and the distances
from that side of the other two ver-
tices, and the intersection-point of the
diagonals.
Given the side AB = b, and the
ordinates from C, D, and E\ namely, y3,
2/4, and y&.
Parallel to BD draw CM to intersection with AD pro-
longed, and drop y6, the ordinate of M.
c
Then
But
ABCD = A AMB = J by6
2/5
n
w =
182 MENSURATION.
215. The area of a triangle equals 3259*6 square meters;
one side equals 112'4 meters. Find the area of a similar
triangle whose corresponding side equals 28' 1 meters.
203' 725 square meters. Ans.
216. The sides of a triangle are 389'2, 486-5, and 291-9
meters. The area of a similar triangle is 2098' 14 square
meters. Find its sides. 74'8, 56*1, 93*5 meters. Ans.
217. The areas of two similar triangles are 24'36 and
182'7 square meters. One side of the first is 8-5 imMcrs
shorter than the homologue of the second. Find these
sides. 4'88 and 13'38 meters. Ans.
HINT. 24-36: 182-7 ::(x-t
218. Two triangles are 21'66 and 43' 74 square meters,
and have an equal angle whose including sides in the first
are 7'6 and 5'7 meters. The corresponding sides in second
differ by 2- 7 meters. Find them.
10*8 and 8*1 meters. Ans.
219. The areas of two similar polygons are 46'37 and
185'48 square meters. A side of tin- first is 15 meters
smaller than the corresponding side of the other. Find
these sides. 15 and 30 meters. Ans.
HINT. 46-37 : 18548 : x2 : (x + 15)2.
220. The sum of perpendiculars dropped from any point
within a regular polygon upon all the sides is constant.
221. In area, an inscribed 2n-gon is a mean proportional
between the inscribed and circumscribed w-gons.
EXEECISES AND PROBLEMS. 183
222. With what regular polygons can a vestibule be
paved ?
223. If a regular n-gon is revolved about its center through
A
the X -, it coincides with itself.
6
224. In a regular w-gon, each %=f
46. N^liNi.
225. A hexagon is inscribed in a circle, and the alternate
angles are joined, forming another hexagon. Find its area.
V3 5
— • r. Ans.
z>
226. What is the area of a regular dodecagon whose side
is 54 feet ?
(54)2 = 2916, and 2916 X 11*1961524 = 32647-980+. Ans.
184 MENSURATION.
47. 0
227. There are three circles whose radii are 20, 28, and
29 meters respectively. Kequired the radius of a fourth
circle whose area is equal to the sum of the areas of the
other three.
O2 =78471-,
/. r = = V2025 == 45. Am.
228. If a circle equals 34'36 square meters, find its radius.
3-3 meters. Ans.
229. Two O's together equal 740-4232 square meters, and
differ by 683*8744 square meters. Find radii.
r = 15'056 meters and r1 = 3 meters. Ans.
II. 7T/-2 -7rV = 683-87 U.
230. If O be the area of the inscribed circle of a triangle,
©i, O2, O3 the areas of the three escribed circles, prove
V©i Vo2 v©8 Vd
231. If from any point in a semicircumference a per-
pendicular be dropped to the diameter, and semicircles
described on these segments, the area between the thn^
semicircumferences equals the circle on the perpendicular
as diameter.
232. The perimeters of a circle, a square, and an equi-
lateral triangle are each of them 12 meters. Find the area
of each of these figures to the nearest hundredth of a square
meter. 11-46, 9, 6.93 square meters. Ans.
EXERCISES AND PROBLEMS. 185
233. Find the side of a square inscribed in a semicircle.
|r Vo. Ans.
234. An equilateral triangle and a regular hexagon have
the same perimeter ; show that the .areas of their inscribed
circles are as 4 to 9.
235. How far must the diameter of a circle be prolonged,
in order that the tangent to the circle from the end of the
prolongation may be m long? £(Vc?2-|-4m2— d~). Ans.
48. 8=Mr =
236. Find the area of a sector of 68° 36' when r = 7-2.
31-03398 square meters. Ans.
237. When circle equals 432 square meters, find sector of
84° 12'. 100-8 square meters. Am.
HINT. 432 : 8 : : 360 : 84£.
238. Find the number of degrees in the arc of a sector
equivalent to the square of its radius.
239. In different circles, sectors are equivalent whose
angles have a ratio inverse to that of the squared radii.
240. Find radius when sector of 7° 12' is 2 square centi-
meters.
241. Find sector whose radius equals 25, and the circular
measure of whose angle equals f . 234-375. Ans.
242. The length of the arc of a sector of a given circle is
16 meters ; the angle of the sector at center is i of a right
angle. Find sector. 488-9 square meters. Ans.
186 MENSURATION.
4A
243. AB is a chord of a given circle ; if on the radius
CA, which passes through one of its extremities, taken as
diameter, a circle be described, the segments cut off from
the two circles by the chord AB are in the ratio of
4 to 1.
244. Show that, if a is the angle or arc of a segment, for
a= 60°, G=^
LJj
a=120°, £=
\.
a= 90°, £=-
7T
jj
a== 36°, £=^
jj
a= 72°, £=|
245. In a segment of 60°, to how many places of decimals
is our approximation correct ?
246. Prove that there can be no segment with Jc = 120,
247. In a circle, given two parallel chords ^ and Jc2, and
their distance apart r ; find the diameter.
d =
HINT.
then
EXERCISES AND PROBLEMS. 187
50.
248. What is the area of a circular zone, one side of which
is 96 and the other 60, and the distance between them
26 (r = 50), when the area of the larger sector is 3217-484,
and of the smaller 1608-736 ? 2136-75. Ans.
51.
249. HIPPOCEATES'S THEOREM.
The two crescents made by de-
scribing semicircles outward on
the two sides of a right triangle,
and a semicircle toward them on
the hypothenuse, are equivalent
to the right triangle.
250. The crescent made by describing a semicircle on the
chord of a quadrant equals the right triangle.
52. -4==(r1
251. A circle of 60 meters diameter is divided into seven
equal parts by concentric circles ; find the parts of the
iiarneter.
r27r = 900 X 3-14159 = 2827'431.
/. outer annulus = 2827'431 = 403-9186 = (900 - r22)7r.
... r22 = 900- °3'9186 = 900 - 128-575 = 771-425.
O-14159
.-. r2 = 27-77+.
.-. dz= 55-55.
.-. 1st part = 60 - 55-55 = 4-450+.
In the same way, 2d part = 4-840,
3d part = 5-353+,
4th part = 6-076+, etc.
188
MENSURATION.
252. Find the annulus between the concentric circumfer-
ences, <?! = 21-98 meters and c2 == 18-84 meters, taking
TT — 3-14. A = 10-205 square meters. Ans.
53. S.A=
253. To trisect a sector of an annulus by concentric circles.
54. J=$hlc.
254. What is the area of a parabola whose base is 18
meters and height 5 meters ? 60 square meters. Ans.
255. What is the area of a parabola whose base is 525
meters and height 350 meters ?
122,500 square meters. Ans.
55. E —- abir.
256. The area of an ellipse is to the area of the circum-
scribed circle as the minor axis is to the major axis.
257. The area of an ellipse is to the area of the inscribed
circle as the major axis is to the minor axis.
258. The area of an ellipse is a mean proportional between
the inscribed and circumscribed circles.
259. What is the area of an ellipse whose major axis is
70 meters, and minor axis 60 meters?
3298-67 square meters. Ans.
260. What is the area of an ellipse whose axes are 340
and 310? 82,780-896. Ans.
EXERCISES AND PROBLEMS. 189
EXERCISES AND PROBLEMS ON CHAPTER IV.
POLYHEDKONS.
56.
261. The number of plane angles in the surface of any
polyhedron is twice the number of its edges.
HINT. Each face has as many plane angles as sides. Each edge
pertains, as side, to two faces.
262. The number of plane angles on the surface of a
polyhedron is always an even number.
263. If a polyhedron has for faces only polygons with an
odd number of sides, e.g., trigons, pentagons, heptagons, etc.,
it must have an even number of faces.
264. If the faces of a polyhedron are partly of an even,
partly of an odd number of sides, there must be an even
number of odd-sided faces.
265. In every polyhedron •§ g ^ (§.
HINT. The number of plane angles on a polyhedron can never be
less than thrice the number of faces.
266. In every polyhedron -|(S ^ (£.
267. In any polyhedron ($ -[- 6 ^ 3 <3.
268. In any polyhedron (g -f 6 ^ 3 ft.
269. In every polyhedron (£ < 3 @, and @< 3 g.
270. In a polyhedron not all the summits are more than
five-sided ; nor have all the faces more than five sides.
271. There is no seven-edged polyhedron.
190
MENSURATION.
272. For every polyhedron s = 0((g — J), that is, the sum
of the plane angles is as many perigons as the difference
between the number of edges and faces.
273. For every polyhedron s = 6 ( © — 2), just as for every
polygon s =f(n — 2).
274. How many regular convex polyhedrons are possible ?
275. In no polyhedron can triangles and three-faced sum-
mits both be absent ; together are present at least eight.
276. A polyhedron without triangular and quadrangular
faces has at least twelve pentagons ; a polyhedron without
three-faced and four-faced summits has at least twelve
five-faced.
57. P = lp.
277. In a right prism of 9 meters altitude, the base is a
right triangle whose legs are 3 and 4 meters. Find the
mantel.
EXERCISES AND PROBLEMS. 191
278. The base of a right prism 12 meters tall is a triangle
whose sides are 12, 14, and 15 meters. Find its surface.
279. To find the mantel of a truncated prism.
Rule : Multiply each side of the perimeter of the right
section by the sum. of the two edges in which it terminates.
The sum of these products will be twice the area.
280. The mantel of a truncated prism equals the axis
multiplied by perimeter of a right section.
281. A right prism 4 meters tall has for base a regular
hexagon whose side is 1-2 meters. Find its surface.
282. In a right triangular prism the lateral edges equal
the radius of the circle inscribed in the base. Show that
the mantel equals the sum of the bases.
58. C=cl=2irrl=27rra.
283. In a right circular cylinder,
(1) Given a and C; find r. '—. Ans.
(2) Given B and C] find a.
(3) Given C and a — 2r; find surface.
(4) Given surface and a = r\ find C.
(5) Given a and B + C; find r.
284. The mantel of a right cylinder is equal to a circle
whose radius is a mean proportional between the altitude
of the cylinder and the diameter of its base.
285. The bases of a right circular cylinder together are to
the mantel as radius to altitude.
286. If the altitude of a right circular cylinder is equal
to the diameter of its base, the mantel is four times the base.
192 MENSURATION.
287. Find a cylinder equivalent to a sum of right circular
cylinders of the same height.
HINT. Find a radius whose square equals squares of the n radii.
288. How muoh must the altitude of a right circular
cylinder be prolonged to make its mantel equal its previous
surface ?
289. A plane perpendicular to the base of a right cylinder
cuts it in a chord whose angle at the center is a ; find the
ratio of the curved surfaces of the two parts of the cylinder.
59. Y=lhp.
290. The surface of any regular tetrahedron is to that of
the cube on its edge as 1 to 2 V3.
291. Each edge of a regular tetrahedron is 2 meters.
Find mantel.
292. Each edge of a regular square pyramid is 2 meters.
Find surface.
293. From the altitude a and basal edge b of a regular
hexagonal pyramid, find its surface.
Ans.
294. In a regular square pyramid, given p, the perimeter
of the base, and the area A of the triangle made by a basal
diagonal and the two opposite lateral edges ; find the sur-
face of the pyramid. -&p* + i V32A- 4-^jt?4. Ans.
60. K=\CJl = TTTll.
295. The convex surface of a right cone is twice the area
of the base ; find the vertical angle.
Here
EXERCISES AND PROBLEMS. 193
Thus the section containing the axis is an equilateral
triangle; so the angle equals 60°. Ans.
296. Find the ratio of the mantels of a cone and cylinder
whose axis-sections are equilateral.
297. Find the locus of the point equally distant from three
given points.
298. In a right cone
(1) Given a and r ; find K. TTT
(2) Given a and h ; find K.
(3) Given K and h ; find r.
299. In an oblique circular cone, given hi, the longest
slant height, A2, the shortest, and a, the axis ; find r, the
radius of the base. Vi (h? -+- A22) — a2. Ans.
300. How many square meters of canvas are required to
make a conical tent which is 20 meters in diameter and 12
meters high ?
TT PT*P
= 314159 X 10 x V144 + 100.
K= 31-4159 X 15-6205
= 490' 7320+ square meters. Ans.
61. F=
301. Given a basal edge 5b and a top edge b2, of the frus-
tum of a regular tetrahedron ; also a, the altitude of the
frustum. Find A, its slant height, and F, its mantel.
h =
F = I (b, + £2) VK*i~^)2 + 4a2. Ans.
302. Same for a regular four-sided pyramid.
194 MENSURATION.
62. 2??=*
303. In the frustum of a right circular cone, given rb r2,
and a ; find h.
304. In the frustum of a right circular cone, on each base
stands a cone with its vertex in the center of the other base ;
from the basal radii ri arid r2 find the radius of the circle in
which the two cones cut. r^r* ^
305. Given 5b b2) the basal edges, and I, the lateral edge
of a frustum of a regular square pyramid ; the frustum of a
cone is so constructed that its upper base circumscribes the
upper base of this pyramid-frustum, while its lower base is
inscribed in its lower base. Find the slant height of the
cone-frustum. v72 - \b? + (1 - * V2)&i b» Ans.
306. How far from the vertex is the cross-section which
halves the mantel of a right cone ? J a v2. Ans.
307. Reckon the mantel from the two radii when the in-
clination of a slant height to one base is 45°.
(rf-rf)ir^/2. Ans.
308. If in the frustum of a right cone the diameter of the
upper base equals the slant height, reckon the mantel from
the altitude a and the perimeter p of an axial section.
T- (p* + 12 a2 +p Vp2 - 12 a2). Ans.
oO
63. F=
309. In the frustum of a cone of revolution, given rlf r2, h ;
find a.
310. Find the altitude of the frustum of revolution from
the mantel k and the bases B\ and J52.
•V;
EXERCISES AND PROBLEMS. 195
311. A right-angled triangle is revolved about an axis
parallel to, and at the distance r from its side a ; the areas
of the circles described by its base are as m to n. Find the
whole surface described by the triangle.
TIT
[fm 1 \ I O I / !m , 1 Y* I 2 . 2 / [^ 1 \n A
-1 -f2a+ A — -fl \a*+r2 A 1 . Am.
\n J \\ n J \\ n J
fi4- 77 —
vFTt» JLL —
312. Find the surface of a cube inscribed in a sphere
whose surface is H.
313. A sphere is to the entire surface of its circumscrib-
ing cylinder as 2 is to 3.
314. Given TI and r2, the radii of two section-circles of a
sphere, and the ratio (m : n) of their distances from its
center. Find its radius. \m2r? — n<>r%
r= - '
m2 — n2
315. Find the sphere whose radius is 12-6156 meters.
2000. Ans.
316. Find the sphere whose radius is 19-j-f j meters.
5000. Ans.
317. A sphere is 50-265 square meters ; find its radius.
2 meters. Ans.
318. Find a sphere from a section-circle c whose distance
from the center is r. (c , *\ A A
^
319. What will it cost to gild a sphere of 22-6 centimeters
radius, if 100 square centimeters cost 87 2 cents ?
$56-16. Ans.
320. Find the ratios of the mantel of the cone, described
by rotating an equilateral triangle about its altitude, to the
sphere generated by the circle inscribed in this triangle.
3 : 2. Ans.
196
MENSURATIOH.
65. Z=
321. Cut a sphere into n equal parts by parallel circles.
322. In a calot,
(1) Given r and a ; find r^.
(2) Given r and TI ; find Z±.
(3) Given a and TI ; find Z±.
(4) Given a and J?"; find ^i. a^/ HTT. Ans.
323. In a segment 6 centimeters high, the radii of base
and top are 9 and 3 centimeters. Find area of the zone.
36 TT VlO square centimeters. Ans.
324. In a segment of altitude a, and congruent bases
calling the top and base radii r1? find the zone.
2 + a2. Ans.
325. How far above the surface of the earth must a person
be raised to see one-third of its surface ?
Here a-id = fr;
and, by similar triangles,
x + r:r = r: r — a.
.'. x = 2/ = d, Ans.
326. A luminous point is distant r from a sphere of radius
r ; how large is the lighted surface ? 2rrir^
327. Find a zone from the radii of its bases rlt r2, and the
radius of the sphere r. 2 rir [ V?-2 — r22 + Vv2 — r^]. Am.
328. How far from the center must a plane be passed to
divide a hemisphere into equal zone and calot ?
EXERCISES AND PROBLEMS. 197
66. THEOEEM OF PAPPUS.
329. An acute-angled triangle is revolved about each side
as axis ; express the ratio of the surfaces of the three double-
cones in terms of a, 5, <?, the sides of the triangle.
a + bma + ctb + c *
, . . ^x/t-o.
c o a
330. The sides of a symmetric trapezoid are bi, b2, and z.
Express the surface generated by rotating the trapezoid
about one of the non-parallel sides.
(b? + bl + ^ z + b2 z) Vz2-i(6i-&2)2. Ans.
z
67. 0 =
331. An equilateral triangle rotates about an axis with-
out it, parallel to, and at a distance a from one of its sides b.
Find the surface thus generated. b-n-(b V3 -f- 6 a). Ans.
332. A rectangle with sides a and b is revolved about an
axis through one of its vertices, and parallel to a diagonal.
Find the generated surface.
§ (L). SPHEEICS AND SOLID ANGLES.
68. h =
333. Find the area of a lime whose angle is 36°.
-frV. Ans.
334. Find lune of 36° when r = 1-26156. 2. Ans.
69.
335. A conical sector is one-fourth of a globe ; find its
solid angle. 90°. Ans.
Find the vertex-angle of an axial section. 120°. Ans.
HINT. By 65, Cor. 2, generating arc = 60°.
198
MENSURATION.
70. A-er2.
336. If two angles of a spherical triangle be right, its
area varies as the third angle.
337. In a cube each solid angle is one-eighth of a stere-
gon. (For eight cubes may be placed together, touching at
a point.)
338. Find the ratio of the solid angle of a regular right
triangular prism to the solid angle of a quader. 2 : 3. Ans.
339. Find the ratio of the trihedral angles of two regular
right prisms of m and n sides. (ra — 2) n A
(n — 2) m
340. Find the area of a spherical triangle from the
radius r, and the angles a = 20° 9' 30", ft = 55° 53' 32",
y = 114° 20' 14". 018137-2. Ans.
341. Given r, and a = 73° 12' 8", /? = 85° 3' 14", y =
32° 9' 16"; find A. O-lSlGr2. Ans.
342. Given r, and a = 114° 20' 5"-92, /? = 30° 57' 18"-41,
y = 90° 9' 41"-67 ; find A. 0-9678^. Ans.
343. Spherical triangles on the same base are equivalent
if their vertices lie in a circumference passing through the
opposite extremities of sphere-diameters from the ends of
the base.
344. All trihedral angles having two edges common, and,
on the same side of these, their third edges prolongations of
elements of a right cone containing the two common edges,
are equivalent.
Proof: On the edges of a trihedral angle take SA, SB,
SO equal ; and pass through the three extremities a circle
ABC of center 0. Join SO, and suppose three planes to
start from SO and to pass one through each edge of the
trihedral angle. These planes form three new trihedrals
EXERCISES AND PROBLEMS.
199
having a common summit /S, and one common edge SO. In
each of these are a pair of equal dihedral angles, since each
stands on an isosceles tri-
angle with vertex at O.
Thus,
CB.O = BC80 . (2)
ACsO=CA80 . (3)
Therefore,
AB.C+CA.B-BC.A
+CA80+£A8O
-BC80-ACaO
==2BA80 .
(4)
Now, 2J3AaO remains
constant as long as the
summit jS of the trihe-
dral, the two edges J3/S
and AS, and the center
0 of the circle are unchanged ; and equation (4) holds as
long as the edge SO passes through the arc ACB.
But
BC,A = BOSA
(5)
(6)
(7)
Making the substitutions (5), (6), (7), equation (4) be-
comes
AB8C* + BA.O + BO, A = Q - 2 BA80.
The second member is constant ; therefore, in the trihe-
dral SABC1, the sum of the three interior angles, and
consequently the area of its intercepted spherical triangle,
is constant.
200
MENSURATION.
345. Equivalent spherical triangles upon the same base,
and on the same side of it, are between the same parallel
and equal lesser circles of the sphere.
346. The locus of B, the vertex of a spherical triangle of
given base and area, is on a lesser circle equal to a parallel
lesser circle passed through A and C, the extremities of the
given base.
347. Find the spherical excess of a triangle in degrees
from its area and the radius. e _ A -^0° ^^
348. Find the ratio of the spherical excesses of two equiv-
alent triangles on different spheres. ^ : c3 = r* : r?. Ans.
349. A spherical triangle whose
a -91° 12' 17", p = 120° 9' 41", y = 100° 42' 2",
contains 3,962 square meters. Find the sphere.
21,600 square meters. Ans.
71. jf = [« — (n — 2)7r]?*2.
350. Find the ratio of the vertical solid angles of two
regular pyramids of m and n sides, having the inclinations
of two contiguous faces respectively, a and (3.
2 TT — ??? (TT — a) A
j ^. Ans.
351. What is the area of a spherical pentegon on a sphere
of radius 5 meters, supposing the sum of the angles 640° ?
43-633 square meters. Ans.
EXERCISES AND PROBLEMS. 201
EXERCISES AND PROBLEMS ON CHAPTER V.
72. U=abL
352. The diagonal of a cube is n ; find its volume.
353. Find the volume of a cube whose surface is 3-9402
square meters.
354. The edge of a cube is n ; approximate to the edge
of a cube twice as large.
355. Find the edge of a cube equal to three whose edges
are a, b, I.
356. Find the cube whose volume equals its superficial
area.
357. If a cubical block of marble, of which the edge is 1
meter, costs 1 dollar, what costs a cubical block whose edge
is equal to the diagonal of the first block.
3 V3 dollars. Ans.
358. In any quader,
(1) Given a, 5, and mantel ; find U.
(2) Given a, b, U-, find I.
(3) Given U, £, and (ah) ; find I and b.
(4) Given U, ff\ (j\ ; find a and b.
\bj \IJ
(5) Given (o5), (al), (bl) ; find a and b.
359. If 97 centimeters is the diagonal of a quader with
square base of 43 centimeters side, find its volume.
HINT. a2 = (97)2-2(43)2.
202
MENSURATION.
360. What weight will keep under water a cork quader
55 centimeters long, 43 centimeters broad, and 97 centi-
meters thick, density 0-24 ?
229-405 - 55-0572 kilograms. Ans.
361. The volume of a quader whose basal edges are 12
and 4 meters is equal to the superficial area. Find its
altitude.
362. In a quader of 360 square meters superficial area
the base is a square of 6 meters edge. Find the volume.
363. A quader of 864 cubic centimeters volume has a
square base equal to the area of two adjacent sides. Find
its three dimensions.
364. In a quader of 8 meters altitude and 160 square
meters surface the base is square. Find the volume.
365. The volume of a quader is 144 cubic centimeters ; its
diagonal 13 centimeters ; the diagonal of its base 5 centi-
meters. Find its three dimensions.
366. In a quader of 108 square meters surface, the base, a
square, equals the mantel. Find volume.
367. If in the three edges of a quader, which meet in an
angle, the distances of three points A, B, and 0 from that
angle be a, b, c ; then triangle A£C= % ^/a?
368. How many square meters of metal will be required
to construct a rectangular tank (open at top) 12 meters
long, 10 meters broad, and 8 meters deep. 472. Ans.
369. The three external edges of a box are 3, 2-52, and
1-523 meters. It is constructed of a material 0-1 meters in
thickness. Find the cubic space inside.
8-594208 cubic meters. Ans.
EXERCISES AND PROBLEMS. 203
70 $$ _ «>*__ _ wkg
- yccm - y i '
370. A brick 11 centimeters long, 3 centimeters broad, 2
centimeters thick, weighs 45 grams ; find its density.
371. A cube of pine wood of 12 centimeters edge weighs
1 kilogram ; find the density of pine. 0-57. Ans.
372. If a mass of ice containing 270 cubic meters weighs
229,000 kilograms, find the density of ice. ? 0-84. Ans.
373. If a cubic centimeter of metal weighs 6-9 grams,
what is its density ?
74. V. P = M.
374. If the base of a parallelepiped is a square, can you
find the altitude a and basal edge b from the volume and
mantel?
75. V. P = aJB.
375. The base of a prism 10 meters tall is an isosceles
right triangle of 6 meters hypothenuse ; find volume.
376. In a prism whose base is 210 square meters, the
three sides are rectangles of 336, 300, 204 square meters ;
find volume.
377. Find altitude of a right prism of 480 cubic centi-
meters volume, standing upon an isosceles triangle whose
base is 10 centimeters and side 13 centimeters.
378. In a right prism of 54 cubic centimeters volume, the
mantel is four times the base, an equilateral triangle ; find
basal edge.
204
MENSURATION.
379. The vertical ends of a hollow trough are parallel
equilateral triangles, with 1 meter in each side, the bases
of the triangles being horizontal. If the distance between
the triangular ends be 6 meters, find the number of cubic
meters of water the trough will contain.
2-598 cubic meters. Ans.
76. V. C = a/V
380. In a right circular cylinder,
(1) Given a and c ; find V. C. — . Ans.
(2) Given a and (7; find V. C.
(3) Given (V. C) and C; find r.
v . \j — ui /i , . . a = — - — .
C a2rTr • a C"
2V. C
2rir
V. C 0
rV 2nr'
C '
Ans.
381. If (7=91-84 square meters, and V. 0 = 145 cubic
meters, find a. a = 4-628986 meters. Ans.
382. A right cylinder of 50 cubic centimeters volume has
a circumference of 9 centimeters ; find mantel.
383. In a right cylinder of 8 cubic centimeters the man-
tel equals the sum of the bases ; find altitude.
384. If, in three cylinders of the same height, one radius
is the sum of the other two, then one curved surface is the
sum of the others, but contains a greater volume.
385. Find the ratio of two cylinders when the radius of
one equals the altitude of the other.
386. Find the ratio of two cylinders whose mantels are
equivalent.
EXERCISES AND PROBLEMS. 205
387. If 1728 cubic meters of brass were to be drawn into
wire of one-thirtieth of a meter in diameter, determine the
length of the wire.
\60J 3600
1728 X 3600
T nom/iK
= 1,980,145 meters. Am.
IT
388. What must be the ratio of the radius of a right
cylinder to its altitude, in order that the axis-section may
equal the base ? 2 : TT. Am.
389. A cylindric glass of 5 meters diameter holds half a
liter; find its height. iZn^iro "'
390. A rectangle whose sides are 3 meters and 6 meters
is turned about the 6-meter side as axis ; find the volume
of the generated cylinder.
391. The diagonal of the axis-section of a right cylinder
O v
is 5 centimeters ; the diameter of its base is three-fourths
its height. Find its volume.
392. In a right cylinder, from A, the area of the axis-
section, reckon the area of that section which halves the
basal radius normal to it. % A-\/3. Ans.
393. The longest side of a truncated circular cylinder of
1-5 meters radius is 2 meters; the shortest, 1-75 meters.
Find volume.
394. If a room be 40 meters long by 20 meters broad,
what addition will be made to its cubic contents by throw-
ing out a semicircular bow at one end ?
2513.28 cubic meters. Ans.
395. The French and German liquid measures must be
cylinders of altitude twice diameter. Find the altitude for
measures holding 2 liters, 1 liter, and \ liter.
216-7, 172-1, and 136-5 millimeters. Ans.
206
MENSURATION.
396. The German dry measures must be cylinders of alti-
tude two-thirds diameter. Find diameter of a measure
containing 100 liters. 575-9 millimeters. Ans.
397. In the French grain measure the altitude equals
diameter. Find for hectoliter. 503-7 millimeters. Ans.
77. V. d - V. C2 = «7r (n + r,)(/i - r2).
398. How many cubic meters of iron are there in a roller
which is half a meter thick, with an outer circumference of
61 meters, and a width of 37 meters? (7r = -^-2-).
1353 cubic meters. Ans.
399. Find the amount of metal in a pipe 3-1831 meters
long, with TI= 12 meters and ra = 8 meters.
800 cubic meters. Ans.
400. The amount of metal in a pipe is 175-9292 cubic
meters, its length is 3-5 meters, and its greater radius is
5 meters. Find its thickness. 2 meters. Ans.
78. SECTIONS SIMILAE.
401. A regular square pyramid, whose basal edge is b,
is so cut parallel to the base that the altitude is halved ;
find the area of this cross-section.
402. A section parallel to the base of a cone (base-radius
r), cuts its altitude in the ratio of ra to n. Find the area of
this section. wVn-
(m -f nf
403. On each of the bases of a right cylinder, radius r,
stands a cone whose vertex is the center of the other base.
Find the circumference in which the cone-mantels cut.
r-rr. Ans.
EXERCISES AND PROBLEMS. 207
79. EQUIVALENT TETRAHEDRA.
404. If a plane be drawn through the points of bisection
of two opposite edges of a tetrahedron, it will bisect the
tetrahedron.
80. V.Y==*GLB.
405. A pyramid of 9 decimeters altitude contains 15|
cubic meters ; find its base. 52.5 square meters. Ans.
406. The pyramid of Memphis has an altitude of 73
Toises ; the base is a square whose side is 116 Toises. If
a Toise is 1-95 meters, find the volume of this pyramid.
About 2,427,847-578 cubic meters. Ans.
407. A goldsmith uses up a triangular pyramid of gold,
density 19-325, and charges $900 a kilogram. What is his
bill if the altitude of the pyramid is 4 centimeters, the alti-
tude of its base 4 millimeters, and the base of its base 1-5
centimeters. $6-957. Ans.
408. Find the volume of a pyramid of 30 meters altitude,
having for base a right triangle of 25 meters hypothenuse
and 7 meters altitude.
81. V. K =
409. In a right circular cone,
(1) Given r and h ; find V. K. JrVVA2 — r2. Ans.
(2) Given a and h ; find V. K. %a7r(h? — a2). Ans.
(3) Given r and K- find V. K. jrV^'^V. Ans.
(4) Given h and K] find V. K.
P ~K •* 2~1
(5) Given a and K; find V. K. ^-n-al *\/— *+——-• . Ans.
L\7r 4 2J
208 MENSURATION.
410. A cone and cylinder have equal surfaces, and their
axis-sections are equilateral ; find the ratio of their volumes.
Surface of cylinder — — - + d2 TT = - ^".
2 2
& f f
Surface of cone
411. In a triangular prism of 9 meters altitude, whose
base has 4 square meters area and 8-85437 meters perimeter,
a cylinder is inscribed. Find the base and altitude of an
equivalent cone whose axial section is equihitcml.
B = 17-1236 square meters, a = 4-043738 meters. Ans.
412. Find the edge of an equilateral cone holding a liter.
16-4 centimeters. Ans.
413. Halve an equilateral cone by a plane parallel to the
base.
414. Find the ratio of the volumes of the cones inscribed
and circumscribed to a regular tetrahedron whose edge is n.
1 : 4. Ans.
82. PRISMOIDAL FORMULA: D = la(£1-}-4 M +-B2).
415. Find the volume of a, rectangular prismoid of 12
meters altitude, whose top is 5 meters long and 2 meters
broad, and base 7 meters long arid 4 meters broad.
220 cubic meters. Ans.
416. In a prismoid 15 meters tall, whose base is 36 square
meters, each basal edge is to the top edge as 3 to 2. Find
the volume. 380 cubic meters. Ans.
EXERCISES AND PROBLEMS.
209
417. Every regular octahedron is a prismatoid whose
bases and lateral faces are all congruent
equilateral triangles. Find its volume in
terms of an edge b. 53iV2. Ans.
418. The bases of a prismatoid are con-
gruent squares of side b, whose sides are
not parallel; the lateral faces are eight
isosceles triangles. Find the volume.
419. If} from a regular icosahedron, we
take off two five-sided pyramids whose
vertices are opposite summits, there re-
mains a solid bounded by two congruent
regular pentagons and ten equilateral tri-
angles. Find its volume from an edge b.
Ans.
420. Both bases of a prismatoid of altitude a are squares ;
the lateral faces isosceles triangles; the sides of the upper
base are parallel to the diagonals of the lower base, and
half as long as these diagonals ; b is a side of the lower base.
Find the volume. £ abz. Ans.
421. The upper base of a prismatoid of altitude a = 6 is
a square of side b2 = 7-07107 ; the lower base is a square of
side bi = 10, with its diagonals parallel to sides of the upper
base ; the lateral faces are isosceles triangles. Find volume.
i a (l.f + fci b2 V2 -f b?) = 500. Ans.
422. Every prismatoid is equivalent to three pyramids of
the same altitude with it, of which one has for base half the
sum of the prismatoid's bases, and each of the others its
midcross section.
210 MENSURATION.
423. Every prismoid is equivalent to a prism plus a
pyramid, both of the same altitude with it, whose bases
have the same angles as the bases of the prismoid ; but the
basal edges of the prism are half the sum, and of the pyramid
half the difference, of the corresponding sides of both the
prismoid's bases.
424. If the bases of a prismoid are trapezoids whose mid-
lines are bi and b2, and whose altitudes are «j and «2,
83. V. F==
425. A side of the base of a frustum of a square pyramid
is 25 meters, a side of the top is 9 meters, arid the height is
240 meters. Required the volume of the frustum.
Here V. F = * 240 (625 + 225 -f 81)
= 80 X 931 = 74,480 cubic meters. Ans.
426. The sides of the square bases of a frustum are 50
and 40 centimeters ; each lateral edge is 30 centimeters.
Find the volume. 59-28 liters. Ans.
427. In the frustum of a pyramid whose base is 50 square
meters, and altitude 6 meters, the basal edge is to the cor-
responding top edge as 5 to 3. Find volume.
196 cubic meters. Ans.
428. Near Memphis stands a frustum whose height is
142-85 meters, and bases are squares on edges of 185-5
and 3-714 meters. Find its volume.
429. In the frustum of a regular pyramid, volume is 327
cubic meters, altitude 9 meters, and sum of basal and top
edge 12 meters. Find these. 7 meters and 5 meters. Ans.
EXERCISES AND PROBLEMS. 211
430. In the frustrum of a regular tetrahedron, given a
basal edge, a top edge, and the volume. Find the altitude.
84. V. F = * air (rf + n r2 + r22).
431. Divide a cone whose altitude is 20 into three equiv-
alent parts by planes parallel to the base.
Volume of whole cone = J r2n- 20.
Volume of midcone = £rt27r(20 — a).
.'. ^(20-0) = $^ 20.
But r : 20 = rt : 20 - a.
20 - a
(20 - of _ 40
400 3'
r
.-. (20-a)3 = - - = 5333-333+.
O
.-. 20 -a =\/5333-333+ = 17-471+.
.-. a = 2-528+.
In the same way, a'= 3-604+.
THEOREM OF CLAVIUS.
432. The frustum of a cone equals the sum of a cylindo;
and cone of frustral altitude whose radii are respectively
the half-sum and half-difference of the frustral radii.
2
This is a formula convenient for computation.
433. A frustum of 8 meters altitude, with TI = 4 and
r2 = 2, is halved by a plane parallel to the base. Find
radius of section and its distance from top of frustum.
6-~8. Ans.
212 MENSURATION.
434. In a frustum, of 3 meters altitude and 63 cubic me-
ters volume, TI = 2 r2 ; find r2. o II A
435. In the frustum where a = 8 meters, i\ = 10 meters,
r2 — 6 meters, the altitude is cut into four equal parts by
planes parallel to the base. Find the radii of these sections.
HINT. The altitude of the completed cone is 8 + 12 = 20, and of
the others, 18, 16, 14. .'. by similar triangles,
20:18:16:14::10:9:8:7.
7, 8, 9 meters. Ans.
436. The frustum of an equilateral cone contains 2 hec-
toliters, and is 40 centimeters in altitude. Find the radii.
27-785 and 50-879 centimeters. Ans.
85. PRISMOIDAL FORMULA: D = fca(-B1 + 4 M +JB9).
437. A solid is bounded by
the triangles ABC, CBD, the
parallelogram ACDE, and the
skew quadrilateral B A E D
whose elements are parallel to
the plane BCD. Find its vol-
ume. \ a. ABC. Ans.
The skew quadrilateral is part of a warped surface called the
hyperbolic paraboloid.
438. A tetrahedron is bisected by the hyperbolic parabo-
loid whose directrices are two opposite edges, and whose
plane directer is parallel to another pair of opposite edges.
439. A solid is bounded by a parallelogram, two skew
quadrilaterals, and two parallel triangles ; find its volume.
^(Aj-f A2). Ans.
EXERCISES AND PROBLEMS. 213
440. Twice the volume of the segment of a ruled surface
between parallel planes is equivalent to the sum of the
cylinders on its bases, diminished by the cone whose ver-
tex is in one of the parallel planes, and whose elements are
respectively parallel to the lines of the ruled surface.
86. W=*aw(2b1 + ba).
441. A wedge of 10 centimeters altitude, 4 centimeters
edge, has a square base of 36 centimeters perimeter. Find
volume.
442. The three parallel sides of a truncated prism are 8,
9, and 11 meters. The section at right angles to them is a
right-angled triangle, with hypothenuse 17 meters, and one
side 15 meters. Find volume.
443. The volume of a truncated regular prism is equal to
the area of a right section multiplied by the axis or mean
length of all the lateral edges.
n
444. To find the volume of any truncated prism.
Rule : Multiply the length of each edge by the sum of
the areas of all the triangles in the right section which have
an angular point in that edge. The sum of the products
will be three times the volume.
Formula: 3F=2A£.
87. X=laM.
445. Given V, the volume of a parallelepiped ; in each
of two parallel faces draw a diagonal, so that the two diag-
onals cross. Take the ends of these as summits of a tetra-
hedron, and find its volume. i V. Ans.
214 MENSURATION.
88. V. H-
446. Find 'the volume of a sphere whose superficial area
is 20 meters.
447. Find the radius of a globe equal to the sum of two
globes whose radii are 3 and 6 centimeters.
3 / —
3S/9 centimeters. ATIS.
448. Find the radius of a golden globe, density 19-35,
weighing a kilogram.
449. A solid metal globe 6 meters in diameter is formed
into a tube 10 meters in external diameter and 4 meters in
length. Find the thickness of the tube. 1 meter. Ans.
450. If a cone and half-globe of equal bases and altitudes
be placed with their axes parallel, and the vertex of the
cone in the plane of the base of the half-globe, and be cut
by a plane normal to their axes, the sum of the sections will
be a constant.
ARCHIMEDES' THKOKKM.
451. Cone, half-globe, and cylinder, of same base and
altitude, are as 1:2:3.
452. The surfaces and the volumes of a sphere, a circum-
scribed right cylinder, and a circumscribed right cone whose
axial section is an equilateral triangle, are as 4:6:9. There-
fore, the cylinder is a geometric mean between the sphere
and cone.
HINT. H = 4 rV. V. H = f r3-.
C + 2J3 = 6r27r. V. C -ff*r.
K + J3=9r*7r. V. K = |r3-.
453. A quader, having a square base of 5 centimeters
edge, is partly filled with water. Into it is put an iron ball
EXERCISES AND PROBLEMS. 215
going fully under the water, which rises 1*33972 centi-
meters. Find the diameter of the sphere.
— = 5ax T33972.
89. V. G = * air [3 Of -f r22) + a2].
454. If a heavy globe, whose diameter is 4 meters, be let
fall into a conical glass, full of water, whose diameter is 5
meters and altitude 6 meters, it 'is required to determine
how much water will run over.
The slant height of the cone
h = V36 + 6-25 = V&25 = 6-5.
If a is the altitude of the dry calot,
6-5 : 2-5 = 6 - (2 - a) : 2.
13 = 10 + 2-5a.
.'. a = 1-2.
But dry segment equals
aV(r-£a) = l-447r(2-0-4) = 1-44 7rl-6 = 2- 304 TT = 7-2382233+.
But V. H=£d37r= 33-5104.
.'. volume of segment immersed is
26-272+ cubic meters. Ans.
455. A section parallel to the base of a half-globe .bisects
its altitude ; find the ratio of the parts of the half-globe.
5 : 11. Ans.
456. A sphere is divided by a plane in the ratio 5 : 7.
In what ratio is the globe cut ? 325 : 539. Ans.
457. A calot 8 centimeters high contains 1200 cubic
centimeters ; find radius of the sphere.
— cubic centimeters. Ans.
7T
216 MENSURATION.
458. Find the volume of a segment of 12 centimeters
altitude, the radius of whose single base is 24 centimeters.
r = 30 centimeters ;
V. G = -014976 TT cubic meters. Ans.
459. In terms of sphere-radius, find the altitude of a
calot n times as large as its base. (n — 1\ 0
a=[ }Zr. Ans.
\ n J
460. Find the ratio of the volume of a sphere to the vol-
ume of its segment whose calot is n times its base.
V. H:to V. G::n3:(n-l)2(tt4-2). Ans.
461. Find volume of a segment whose calot is 15-085
square meters, and base 2 meters from sphere-center.
V. G = 5-737025 cubic meters. Ans.
462. In a sphere of 10 centimeters radius, find the radii
TI and r2 of the base and top of a segment whose altitude is
6 centimeters, and base 2 centimeters from the sphere-
center. TI = 4 V6 centimeters, ?*2 = 6 centimeters. Ans.
463. Out of a globe of 12 centimeters radius is cut a seg-
ment whose volume is one-third the globe, and whose bases
are congruent ; find the radius of bases.
90. V. 8 =
464. In a spherical sector,
(1) Given r, rlt r2 ; find V. S.
(2) Given a, rit r2 ; find V. S.
465. In a sphere of radius r, find the altitude of a seg-
ment which is to its sector as n to m.
/3 , S 2w\ ,
= n-d=AT — -). -4w«.
\2 \4 mj
EXERCISES AND PROBLEMS. 217
466. A sector is - of its globe, whose diameter is d ; find
the volume of its segment. V. Gr = - r [ - - 1 Ans.
6 \ w J
467. A sphere of given volume V is cut into two seg-
ments whose altitudes are as m to n ; find both calots Z±
and Z* and the segments.
n
m-\-n
(m + njr (m -f-
91. v =
468. Find the volume of a spherical ungula whose radius
is 7-6 and £ 18° 12'. 92-958.
469. £ = 26° 6', r=13-2. Find v. 69845.
470. A lune of 192 square meters has radius 15 meters ;
find volume of the ungula. 960 cubic meters. Ans.
Lr
471. Given L and r ; find v. 9 = —- . Ans.
o
472. Given 9 and r ; find ^. X — - — . J.TIS.
? 7T
473. Given L and ^ ; find v. v = i-\ — -7. Ans.
\ TT.
92. f^
474. In a spherical pyramid given the angles of its tri-
angular base, a -78° 15', /? = 144°30', y = 108°15', and
given r=10-8; find Y. 1106-61.
— ra — I
475. Given a, /8, y, and A ; find Y. 2 A A
\
218 MENSURATION.
476. A = 486, a = 84° 13', /? = 96° 27', y = 112° 20' ;
find Y. 2543-06. Ans.
477. Given r = 8-8, a = 106°30', /3=120°10', y = 150°15',
8 = 112° 5' ; find the four-faced Y. 511-433. Ans.
93. THEOREM OF PAPPUS.
478. If an equilateral triangle whose sides are halved
by a straight line rotates about its base, the two volumes
generated are equivalent.
479. A trapezoid rotates first about the longer, then
about the shorter, of its parallel sides ; the volumes of the
solids generated are as ra to n. Find the ratio of the
parallel sides. 2 n — m
—. Ans.
2 7n — n
94. V. 0 =
480. Find the volume of a solid generated by rotating a
parallelogram about an axis exterior to it ; given the area
of the parallelogram £17, and the distance r of the inter-
section-point of its diagonals from the axis. 2^x0. Ans.
481. The volume of a spiral spring, whose cross-section
is a circle, equals the product of this generating circle by
the length of the helix along which its center mov-'s.
The helix is the curve traced upon the surface of a cir-
cular cylinder by a point, the direction of whose motion
makes a constant angle with the generating line of the
cylinder.
482. A regular hexagon rotates about I, one of its sides ;
find the volume generated. -| ISTT. Ans.
EXEECISES AND PROBLEMS. 219
483. Any two similar solids may be so placed that all
the lines joining pairs of homologous points intersect in a
point. Every two homologous lines or surfaces in the two
solids are then parallel.
484. Any two symmetric solids may be so placed that
all the lines joining pairs of homologous points intersect
in a point. This point bisects each sect. Every two ho-
mologous lines are then parallel;
485. Three persons having bought a sugar-loaf, would
divide it equally among them by sections parallel to the
base. It is required to find the altitude of each person's
share, supposing the loaf to be a cone whose height is 20.
13-8672, 3-6044, and 2.5284. Am.
Let altitude of upper cone equal x, and its volume equal 1.
Now,
1 : 3 - a,-3 : 203.
x = V2666-666 = 13-867+.
96. IRREGULAR SOLIDS.
486. When a solid is placed in a square quader of basal
edge 6 meters, the liquid, rising 3-97 meters, covers it;
find its volume.
97. Vccm-.
o
487. How much mercury, density 13-60, will weigh 7-59
grams ?
488. If the density of zinc is 7'19, find how much weighs
3-83 kilograms.
220 MENSURATION.
98. 7= i OsGBi - B$ + 0:3(^2 - £4) + etc.
-f 1 [xsHfj. + Os - ^2)^2 + (#4 - ^'3)^3 + etc.
-f (ffn+i —
489. If the areas of six parallel planes 2 meters apart
are 1, 3, 5, 7, 9, 11 square meters, and of the five mid-sec-
tions 2, 4, 6, 8, 10 square meters, find the whole volume.
99. Ax = q -f mx -f nor* -f/a8.
490. Find an expression for the volume of a semicubic
paraboloid generated by the revolution of a semicubic
parabola round its axis. In this curve y2 oc 3?, the revolv-
ing ordinate being y,
491. A paraboloid and a semicubic paraboloid have a
common base and vertex ; show that their volumes are as
2:1.
492. A vessel, whose interior surface has the form of a
prolate spheroid, is placed with its axis vertical, and filled
with a fluid to a depth h; find the depth of the fluid when
the axis is horizontal.
493. A square-threaded screw, with double thread, is
formed upon a solid cylinder 3 meters in diameter; the
thread projects from the cylinder -£$ of meter, and the
screw rises 3 meters in four turns. Find the volume, if
the screw be 9 meters in length.
494. Find the volume of a square groin, the base of
which is 15 meters square, and the guiding curve a semi-
circle.
495. The prismoidal formula applies to any shape con-
tained by two parallel bases, and a lateral surface generated
EXERCISES AND PROBLEMS. 221
by the motion of a parabola or cubic parabola whose plane
is always parallel to a given plane, but whose curvature
may pass through any series of changes in amount, direc-
tion, and position.
496. No equation of finite degree, representing a bounding
surface, can define the limits of applicability of the pris-
moidal formula, because surfaces of higher degrees enclose
prismoidal spaces.
100. V=-(B^
497. Show how existing rules for the estimation of rail-
road excavation may be improved.
498. If a parabolic spindle is equal in volume to one-fifth
of the sphere on its axis as diameter, show that its greatest
diameter is equal to half its length.
499. A parabolic spindle is placed in a cylinder half-full
of water, the greatest diameter of the spindle being equal to
that of the interior of the cylinder ; find the height of the
cylinder so that the water may just rise to the top.
500. A vessel, laden with a cargo, floats at rest in still
water, and the line of flotation is marked. Upon the re-
moval of the cargo every part of the vessel rises 3 meters,
when the line of flotation is again marked. From the
known lines of the vessel the areas of the two planes of
flotation and of five intermediate equidistant sections are
calculated and found to be as follows, the areas being ex-
pressed in square meters: 3918, 3794, 3661, 3517, 3361,
3191t 3004. Find the weight of the cargo removed.
222 MENSURATION.
EXERCISES AND PROBLEMS ON CHAPTER VIII.
501. A square on the line b is divided into four equal
triangles by its diagonals which intersect in C; if one tri-
angle be removed, find the PC of the figure formed by the
three remaining triangles. Q^ __ £_
HINT. For such problems let L be the PC of the part left, and O
of the part cut out ; then
CL X area left = CO X area cut out.
502. If a heavy triangular slab be supported at its an-
gles, the pressure on each prop will be one-third the weight
of the slab.
503. A weight <o is placed at any point 0 upon a trian-
gular table ABO (supposed without weight).
Show that the pressures on the three props
(viz., A, B, C) are proportional to the areas
of the triangles BOG, AOC, AOB respec-
tively.
Draw the straight lines A Of1, BO II,
COE; and let A', B1, C' be the pressures at
E £ A, B, C respectively. Then
C'=AOB
" u ABC'
t
Similarly, for A' and B'.
504. The mid-point of one side of a square is joined with
the mid-points of the adjacent sides, and the triangles thus
formed are cut off; find the PC of the remainder.
EXERCISES AND PROBLEMS. 223
505. If two triangles stand on the same base, the line
joining their PC's is parallel to the line joining their ver-
tices.
506. Find the distance from the base of the PC of four
uniform rods forming a trapezoid, the two parallel sides of
which are respectively 12 meters and 30 meters long, and
the other sides each 15 meters long. 5^- meters. Ans.
507. The altitude of the segment of a globe is a; find
height of ^0 of its zone. £ a. Ans.
508. Find <"C"of a hemisphere.
509. Find PC of cylinder-mantel.
510. Find PC of cone-mantel.
511. If a body of density 8 weighs w, express the distance
of its PC from its midcross-section. a\£2 — J3i) 8
12(0
512. Find the PC of a portion of a parabola cut off by a
line perpendicular to the axis at a distance h from the ver-
tex. -|A. Ans.
513. Find the PC of the segment of a globe at a distance
b from the center. 3 (r -j- b}2 A
-. Ans.
514. Find the distance from vertex of the PC of half a
prolate spheroid.
515. A right circular cone, whose vertical angle is 60°,
is constructed on the base of a half-globe ; find the PC of
the whole body.
516. Show that the compound body of the last exercise
will rest in any position on its convex spherical surface.
517. Every body or system of particles has a PC, and
cannot have more than one.
224 MENSURATION.
518. Find the ^C of any polygon by dividing it into tri-
angles.
519. If the sides of a triangle be 3, 4, and 5 meters, find
the distance of PC from each side. f, 1, -J meter. Am.
MISCELLANEOUS.
520. Find both sides of a rectangle from their ratio ra : n,
and its area R.
ra
521. If two triangles have one angle of the one equal to
one angle of the other, and the sides about a second angle
in each equal, then the third angles will be either equal or
supplemental.
522. Two triangles are congruent, if two sides and a
medial in the one are respectively equal to two sides and a
corresponding medial in the other.
523. Two triangles are congruent, if three medials in one
equal those in the other.
524. On a plane lie three tangent spheres of radius r ;
upon these lies a fourth of radius r'. How high is its cen-
ter above the plane, and how large at least is r', since the
sphere does not fall through ?
525. Calling the solid angle whose faces fall into the
same plane, aflat angle, show that a flat angle contains 2ir
steradians.
526. A steregon contains 4?r steradians.
LOGARITHMS.
133, The logarithm a of a number n to a given base b
is the index of the power to which the base must be raised
to give the number :
So, if ba = n, then blogn — a, or the 5-logarithm of n
is a.
134, Mog5 =1. blogl = 0.
135, b log ran =
136. blog— =blogra — blogw.
w
137. blogn* --=pX
138, Mogw* = -X
139. b'log7i = blogri X -.
r- — — is called the modulus or multiplier for transform-
blog6'
ing the log of a number to base b to the log of same number
to base b1.
140. The base of the common system of logarithms is 10.
10log(nX 10') =
141. 10log(W 10*) = 10logn — p.
142. The mantissa is the decimal part of a logarithm.
The characteristic is the integral part of a logarithm.
226 MENSURATION.
The 10logs of all numbers consisting of the same digits in
the same order have the same mantissa.
143. The characteristic of the 10log of a number is one less
than the number of digits in the integral part.
144. When the number has no integral figures, the
characteristic of its 10log is negative, and is one more than
the number of cyphers which precede the first significant
digit; that is, the number of cyphers (zeros) immediately
after the decimal point.
LOGARITHMS.
227
N
O 1 2 3 4
56789
PP
10
11
12
13
14
0000 0043 0086 0128 0170
0414 0453 0492 0531 0569
0792 0828 0864 0899 0934
1139 1173 1206 1239 1271
1461 1492 1523 1553 1584
0212 0253 0294 0334 0374
0607 0645 0682 0719 9755
0969 1004 1038 1072 1106
1303 1335 1367 1399 1430
1614 1644 1673 1703 1732
4.1
3.8
3.5
3.2
3.0
15
16
17
18
19
1761 1790 1818 1847 1875
2041 2068 2095 2122 2148
2304 2330 2355 2380 2405
2553 2577 2601 2625 2648
2788 2810 2833 2856 2878
1903 1931 1959 1987 2014
2175 2201 2227 2253 2279
2430 2455 2480 2504 2529
2672 2695 2718 2742 2765
2900 2923 2945 2967 2989
2.8
2.6
2.5
2.3
2.2
20
21
22
23
24
3010 3032 3054 3075 3096
3222 3243 3263 3284 3304
3121 3444 3464 3483 3502
3617 3636 3655 3674 3692
3802 3820 3838 3856 3874
3118 3139 3160 3181 3201
3324 3345 3365 3385 3404
3522 3541 3560 3579 3598
3711 3729 3747 3766 3784
3892 3909 3927 3945 3962
2.1
2.0
1.9
1.8
1.8
25
26
27
28
29
3979 3997 4014 4031 4048
4150 4166 4183 4200 4216
4314 4330 4346 4362 4378
4472 4487 4502 4518 4533
4624 4639 4654 4669 4683
4065 4082 4099 4116 4133
4232 4249 4265 4281 4298
4393 4409 4425 4440 4456
4548 4564 4579 4594 4609
4698 4713 4728 4742 4757
1.7
1.6
1.6
1.5
1.5
30
31
32
33
34
4771 4786 4800 4814 4829
4914 4928 4942 4955 4969
5051 5065 5079 5092 5105
5185 5198 5211 5224 5237
5315 5328 5340 5353 5366
4843 4857 4871 4886 4900
4983 4997 5011 5024 5038
5119 5132 5145 5159 5172
5250 5263 5276 5289 5302
5378 5391 5403 5416 5428
1.4
1.4
1.3
1.3
1.3
35
36
37
38
39
5441 5453 5465 5478 5490
5563 5575 5587 5599 5611
5682 5694 5705 5717 5729
5798 5809 5821 5832 5843
5911 5922 5933 5944 5955
5502 5514 5527 5539 5551
5623 5635 5647 5658 5670
5740 5752 5763 5775 5786
5855 5866 5877 5888 5899
5966 5977 5988 5999 6010
1.2
1.2
1.2
1.1
1.1
4O
41
42
43
44
6021 6031 6042 6053 6064
6128 6138 6149 6160 6170
6232 6243 6253 6263 6274
6335 6345 6355 6365 6375
6435 6444 6454 6464 6474
6075 6085 6096 6107 6117
6180 6191 6201 6212 6222
6284 6294 6304 6314 6325
6385 6395 6405 6415 6425
6484 6493 6503 6513 6522
1.1
1.0
1.0
1.0
1.0
228
MENSURATION.
N
O 1 2 3 4
56789
PP
45
46
47
48
49
6532 6542 6551 6561 6571
6628 6637 6646 6656 6665
6721 6730 6739 6749 6758
6812 6821 6830 6839 6848
6902 6911 6920 6928 6937
6580 6590 6599 6609 6618
6675 6684 6693 6702 6712
6767 6776 6785 6794 6803
6857 6866 6875 6884 6893
6946 6955 6964 6972 6981
1.0
0.9
0.9
0.9
0.9
50
51
52
53
54
6990 6998 7007 7016 7024
7076 7084 7093 7101 7110
7160 7168 7177 7185 7193
7243 7251 7259 7267 7275
7324 7332 7340 7348 7356
7033 7042 7050 7059 7067
7118 7126 7135 7143 7152
7202 7210 Ti'l* 7±.v.
7284 7292 7300 7308 7316
7364 7372 7380 7388 7396
0.9
0.8
0.8
0.8
0.8
55
56
57
58
59
7404 7412 7419 7427 7435
7482 7490 7497 7505 7513
71 7582 7589
7o:U TC 12 7019 7057 7664
7709 7716 7723 7731 7738
7443 7451 7459 7466 7474
7520 7528 7536 751.".
1 7612 7619
7079 7686 70! M 7701
7745 7752 7760 7767 7774
0.8
0.8
0.8
0.7
0.7
60
61
62
63
64
7782 7789 7796 7803 7810
7853 7860 7868 7875 7882
7UJ1 7931 7938 7945 7952
7993 8000 8007 8014 8021
8062 8069 8075 8082 8089
7818 7825 7832 7839 7846
7889 7896 7903 7910 7'.' 17
7959 7966 7973 7980 7987
8028 8035 8041 8048 8055
SOOO 8102 8109 8116 8122
0.7
0.7
0.7
0.7
0.7
65
66
67
68
69
8129 8136 8142 8149 8156
8195 8202 8209 82 ir.
8261 8267 8274 8280 8287
8325 8331 8338 8344 8351
8388 8395 8401 8407 8414
8162 8169 8176 8182 8189
8228 8235 8241 8248 8254
8293 8299 8306 8312 8319
8357 8363 8370 8376 8382
8420 8426 8432 8439 8445
0.7
0.7
0.6
0.6
0.6
70
71
72
73
74
8451 8457 8463 8470 8476
8513 8519 8525 8531 8537
8573 8579 8585 8591 8597
8633 8639 8645 8651 8657
8692 8698 8704 8710 8716
8482 8488 8494 8500 8506
8543 8549 8555 8561 8567
8603 8609 8615 8621 8627
8663 8669 8675 8681 8686
8722 8727 8733 8739 8745
0.6
0.6
0.6
0.6
0.6
75
76
77
78
79
8751 8756 8762 8768 8774
8808 8814 8820 8825 8831
8865 8871 8876 8882 8887
8921 8927 8932 8938 8943
8976 8982 8987 8993 8998
8779 8785 8791 8797 8802
8837 8842 8848 8854 8859
8893 8899 8904 8910 8915
8949 8954 8960 8965 8971
9004 9009 9015 9020 9025
0.6
0.6
0.6
0.6
0.5
LOGARITHMS.
229
N
01234
56789
PP
80
81
82
83
84
9031 9036 9042 9047 9053
9085 9090 9096 9101 9106
9138 9143 9149 9154 9159
9191 9196 9201 9206 9212
9243 9248 9253 9258 9263
9058 9063 9069 9074 9079
9112 9117 9122 9128 9133
9165 9170 9175 9180 9186
9217 9222 9227 9232 9238
9269 9274 9279 9284 9289
0.5
0.5
0.5
0.5
0.5
85
86
87
88
89
9294 9299 9304 9309 9315
9345 9350 9355 9360 9365
9395 9400 9405 9410 9415
9445 9450 9455 9460 9465
9494 9499 9504 9509 9513
9320 9325 9330 9335 9340
9370 9375 9380 9385 9390
9420 9425 9430 9435 9440
9469 9474 9479 9484 9489
9518 9523 9528 9533 9538
0.5
0.5
0.5
0.5
0.5
90
91
92
93
94
9542 9547 9552 9557 9562
9590 9595 9600 9605 9609
9638 9643 9647 9652 9657
9685 9689 9694 9699 9703
9731 9736 9741 9745 9750
9566 9571 9576 9581 9586
9614 9619 9624 9628 9633
9661 9666 9671 9675 9680
9708 9713 9717 9722 9727
9754 9759 9763 9768 9773
0.5
0.5
0.5
0.5
0.5
95
96
97
98
99
9777 9782 9786 9791 9795
9823 9827 9832 9836 9841
9868 9872 9877 9881 9886
9912 9917 9921 9926 9930
9956 9961 9965 9969 9974
9800 9805 9809 9814 9818
9845 9850 9854 9859 9863
9890 9894 9899 9903 9908
9934 9939 9943 9948 9952
9978 9983 9987 9991 9996
0.5
0.5
0.4
0.4
0.4
N
O 1 2 3 4
56789
1OO
101
102
103
104
0000 0004 0009 0013 0017
0043 0048 0052 0056 0060
0086 0090 0095 0099 0103
0128 0133 0137 0141 0145
0170 0175 0179 0183 0187
0022 0026 0030 0035 0039
0065 0069 0073 0077 0082
0107 0111 0116 0120 0124
0149 0154 0158 0162 0166
0191 0195 0199 0204 0208
105
106
107
108
109
0212 0216 0220 0221 0228
0253 0257 0261 0265 0269
0294 0298 0302 0306 0310
0334 0338 0342 0346 0350
0374 0378 0382 0386 0390
0233 0237 0241 0245 0249
0273 0278 0282 0286 0290
0314 0318 0322 0326 0330
0354 0358 0362 0366 0370
0394 0398 0402 0406 0410
230
MENSURATION.
N
01234
56789
110
111
112
113
114
0414 0418 0422 0426 0430
0453 0457 0461 0465 0469
0492 0496 0500 0504 0508
0531 0535 0538 0542 0546
0569 0573 0577 0580 0584
0434 0438 0441 0445 0449
0473 0477 0481 0484 0488
0512 0515 0519 0523 0527
0550 0554 0558 0561 0565
0588 0592 0596 0599 0603
115
116
117
118
119
0607 0611 0615 0618 0622
0645 0648 0652 0656 0660
0682 0686 0689 0693 0697
0719 0722 0726 0730 0734
0755 O7.r,'.> 07('.:'> 0766 <>77<>
0626 0630 0633 0637 0641
0663 0667 0671 0674 0678
0700 0704 0708 0711 0715
0737 0741 0745 0748 0752
0774 0777 0781 0785 0788
12O
121
122
123
124
0792 0795 0799 0803 0806
0828 0831 0835 0839 0842
0864 0867 0871 0874 0878
0899 0903 0906 0910 O'-.M:;
0934 0938 0941 0945 0948
0810 0813 0817 0821 0824
0846 0849 0853 0856 0860
0881 0885 0888 0892 0896
0917 0920 0924 0927 0931
~>2 0955 0959 0962 09<;<;
125
126
127
128
129
0969 0973 0976 0980 0983
1004 1007 1011 1014 1017
1038 1041 1045 1048 1052
1072 1075 1079 1082 1086
1106 1109 1113 1116 1119
0986 0990 0993 0997 1000
1021 1024 1028 1031 10:;5
1055 1059 1062 1065 1069
1089 1092 1096 1099 1103
1123 1126 1129 1133 11::.;
130
131
132
133
134
1139 1143 1146 1149 1153
1173 1176 1179 1183 1186
1206 1209 1212 1216 1219
1239 12-12 1245 1218 1252
1271 1274 1278 1281 1284
1156 1159 1163 1166 1169
1189 1193 1196 1199 1202
1222 1225 1229 1232 1235
12.55 1258 1261 1265 1268
1287 1290 1294 1297 1300
135
136
137
138
139
1303 1307 1310 1313 1316
1335 1339 1342 1345 1348
1367 1370 1374 1377 1380
1399 1402 1405 1408 1411
1430 1433 1436 1440 1443
1319 1323 1326 1329 1332
1351 1355 1358 1361 1364
1383 1386 1389 1392 1396
1414 1418 1421 1424 1427
1446 1449 1452 1455 1458
140
141
142
143
144
1461 1464 1467 1471 1474
1492 1495 1498 1501 1504
1523 1526 1529 1532 1535
1553 1556 1559 1562 1565
1584 1587 1590 1593 1596
1477 1480 1483 1486 1489
1508 1511 1514 1517 1520
1538 1541 1544 1547 1550
1569 1572 1575 1578 1581
1599 1602 1605 1608 1611
LOGARITHMS.
231
N
01234
56780
145
146
147
148
149
1614 1617 1620 1623 1626
1644 1647 1649 1652 1655
1673 1676 1679 1682 1685
1703 1706 1708 1711 1714
1732 1735 1738 1741 1744
1629 1632 1635 1638 1641
1658 1661 1664 1667 1670
1688 1691 1694 1697 1700
1717 1720 1723 1726 1729
1746 1749 1752 1755 1758
150
151
152
153
154
1761 1764 1767 1770 1772
1790 1793 1796 1798 1801
1818 1821 1824 1827 1830
1847 1850 1853 1855 1858
1875 1878 1881 1884 1886
1775 1778 1781 1784 1787
1804 1807 1810 1813 1816
1833 1836 1838 1841 1844
1861 1864 1867 1870 1872
1889 1892 1895 1898 1901
155
156
157
158
159
1903 1906 1909 1912 1915
1931 1934 1937 1940 1942
1959 1962 1965 1967 1970
1987 1989 1992 1995 1998
2014 2017 2019 2022 2025
1917 1920 1923 1926 1928
1945 1948 1951 1953 1956
1973 1976 1978 1981 1984
2000 2003 2006 2009 2011
2028 2030 2033 2036 2038
16O
161
162
163
164
2041 2044 2047 2049 2052
2068 2071 2074 2076 2079
2095 2098 2101 2103 2106
2122 2125 2127 2130 2133
2148 2151 2154 2156 2159
2055 2057 2060 2063 2066
2082 2084 2087 2090 2092
2109 2111 2114 2117 2119
2135 2138 2140 2143 2146
2162 2164 2167 2170 2172
165
166
167
168
169
2175 2177 2180 2183 2185
2201 2204 2206 2209 2212
2227 2230 2232 2235 2238
2253 2256 2258 2261 2263
2279 2281 2284 2287 2289
2188 2191 2193 2196 2198
2214 2217 2219 2222 2225
2240 2243 2245 2248 2251
2266 2269 2271 2274 2276
2292 2294 2297 2299 2302
17O
171
172
173
174
2304 2307 2310 2312 2315
2330 2333 2335 2338 2340
2355 2358 2360 2363 2365
2380 2383 2385 2388 2390
2405 2408 2410 2413 2415
2317 2320 2322 2325 2327
2343 2345 2348 2350 2353
2368 2370 2373 2375 2378
2393 2395 2398 2400 2403
2418 2420 2423 2425 2428
175
176
177
178
179
2430 2433 2435 2438 2440
2455 2458 2460 2463 2465
2480 2482 2485 2487 2490
2504 2507 2509 2512 2514
2529 2531 2533 2536 2538
2443 2445 2448 2450 2453
2467 2470 2472 2475 2477
2492 2494 2497 2499 2502
2516 2519 2521 2524 2526
2541 2543 2545 2548 2550
232
MENSURATION.
N
O 1 2 3 4
56789
180
181
182
183
184
2553 2555 2558 2560 2562
2577 2579 2582 2584 2586
2601 2603 2605 2608 2610
2625 2627 2629 2632 2634
2648 2651 2653 2655 2658
2565 2567 2570 2572 2574
2589 2591 2594 2596 2598
2613 2615 2617 2620 2622
2636 2639 2641 2643 2646
2660 2662 2665 2667 2669
185
186
187
188
189
2672 2674 2676 2679 2681
2695 2697 2700 2702 270 i
2718 2721 2723 2725 2728
2742 2744 2746 2749 2751
2765 2767 2769 2772 2774
2683 2686 2688 2690 2693
2707 2709 2711 2714 2716
2730 2732 2735 2737 2739
2753 2755 2758 2760 2762
2776 2778 2781 2783 2785
19O
191
192
193
194
2788 2790 2792 2794 2797
2810 2813 2815 2817 2819
2833 2835 2838 2840 2842
2856 2858 2860 2862 2865
2878 2880 2882 2885 2887
2709 2801 2804 2806 2808
2822 2824 2826 2828 2831
2844 2847 2849 2851 2853
2867 2869 2871 2874 2876
2889 2891 2894 2896 2898
195
196
197
198
199
2900 2903 2905 2907 2909
2923 2925 2927 2!>29 2931
i".i 15 2947 2949 2951 2953
i".ir,7 2969 2!>71 L".i7:J L)(.»7'>
2989 2991 2993 299." 2<i-.i7
2911 2914 2916 2918 2920
2934 2936 2938 2940 2912
->G 2958 2960 2962 2964
2978 2980 2982 2984 2986
2999 3002 3004 3006 3008
NOTE. 233
Proof of Cor. 2, page 123 :
Where r is a positive finite integer, from the equation
a — o
if a > b, we get
a — o
Putting a = b + 1, this inequality gives
Consequently, by addition,
(r + 1) (I' + 2T + & + • • • + O < (n + 1 X+1 - 1.
Subtracting nr+l leaves
(r+1) (I' + 2*1 + 3" + • • • + nr) - n'+l<(n + l)H-i - n'+l -I.
Dividing by (r + l)nr+1 gives
...+nr 1
By taking n sufficiently great we can make the right-hand mem-
ber as small as we please ; therefore, for n = oo, the limit of
lr + 2r + 3r + • •• + wr
Jg
Now, since in Cor. 2 we are given
Q) I OC i ^= tv(\ ~T" / t-i 3C ~\~ i t'o ™ "T" /to t>C ~p Vt-4. w ~f" * "T" f^^X J
therefore, 0(0) = n0,
^la^- ian +la2w -a3w
\n J n n2 ?i3
/2 \=^ 2^ 2_2a2^ ^
\n J n ri* 2 n3
To get the sum of all the prisms of like height add the columns,
and multiply by -.
Thus, 2[,(0H,|Ia) + ,(?,) + ...+
The limit of which, as n = oo, we have just proved to be
anQ + | a2 nj + i a3 n, H H = — am+l nm.
m + 1
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