AN
MENSURATION
RACTICAL GEOMETRY.
I:E WITH
iOUS PROBLEMS OF PRACTICAL IMPORTANCE
MECHANICS.
WILLIAM VOGDES, LL.IX
)F MATHEMATICS W THB CENTRAL HJOH BCBOOL OF F
AT'THOR OF TH? UNITED 8TATK8 AEITHM1IIO.
[PAET FIES1.]
PHILADELPniA:
E. C\ & J. BIDDLE fe CO., No. 508 MINOR ST.
(Betwee* Market, and Chtstttul, and fifth and Sixth Sts.)
1861.
-7
y^-fc?^ vp/z*+
)
JfafivQ
\
AN
ELEMENTARY TREATISE
ON
MENSURATION
PRACTICAL GEOMETKY,
TOGETHER WITH
NUMEROUS PROBLEMS OF PRACTICAL IMPORTANCE
MECHANICS.
WILLIAM VOGDES, LL.D.
LATE PROFESSOR OF MATHEMATICS IN THE CENTRAL HIGH SCHOOL OF PHILA-
DELPHIA, AUTHOR OP THE UNITED STATES ARITHMETIC.
[PAET FIEST.]
PHILADELPHIA:
E. C. & J. BIDDLE, No. 508 MINOK ST.
(Between Market and Chestnut, and Fifth and Sixth Sis.)
1862.
Chamber of the Controllers of Pttblic Schools, \
First School District of Pennsylvania. /
PHILADELPHIA, November 15th, 1849.
AT a meeting of the Controllers of Public Schools, First School
District of Pennsylvania, held at the Controllers' Chamber, on Tues-
day, November 13th, 1849, the following Resolution was adopted:
Resolved, That Vogdes' Mensuration be introduced as a Class Book
into the Grammar Schools of the District.
From the Minutes.
ROBERT J HEMPHILL, Secretary.
ESTEBED according to act of Congress, in the year 1846, by
E. C. & J. DIDDLE,
m the Clerk's Office of the District Court of the Eastern District of Pennsyl-
BTEREOTYPED BY >.. JOHNSOK & CO.
PHILADELPHIA.
COLLINS, PRINT**
PREFACE.
IT has been the design of the author, in the following
pages, to compile a work adapted, by its practical cha-
racter, to the wants of those of the rising generation,
who, not being able to command a collegiate education,
are fitting themselves to fill useful stations in society as
mechanics, merchants, &c. With tiiis end in view, it
was deemed inexpedient to devote any portion of the
book to theoretical demonstrations of the principles in-
volved in the rules given, when those demonstrations are
based upon principles illustrated by more advanced
branches of mathematical science, with which the pupil
is supposed to be unacquainted. By pursuing this
course, room was afforded for the introduction of more
numerous examples illustrating the respective rules, than
could otherwise have been given without the enlarge-
ment of the work beyond expedient limits. The intro-
duction of these examples, it is believed, will enhance
the value of the work in the estimation of teachers gene-
rally, inasmuch as the operation of the rules is more
likely to be permanently impressed on the mind of the
pupil, by long continued practice, than by the solution
of one or two problems only.
In a treatise on Mensuration, little that is new can be
looked for, other than the collection and judicious
arrangement of matters not heretofore presented to the
public in a form adapted to the purposes for which this
3
4 PREFACE.
work is designed. The greater portion of this volume
has been derived from the works of Bonnycastle, Has-
well, Hutton, Gregory, and Grier, To some of these,
special acknowledgment of the obligation has been
made in the subsequent pages.
The application of science to the arts of industry has
oeen one of the most potent operative causes of the
rapid increase of our country in wealth and power. In
the hope that this compilation may be found a useful
assistant to the teacher who is engaged in preparing
pupils for an active participation in these industrial pur-
suits, the author submits it to the inspection of his co-
labourers in this field.
Philadelphia, June 27, 1846.
A KEY to this work has been published.
CONTENTS.
PRACTICAL GEOMETRY.
Definitions 1&
Instruments 23
Geometrical Problems 27
MENSURATION OF SUPERFICIES.
To find the area of a square 57
To find the side of a square 58
The diagonal of a square being given, to find the area 58
The area of a square being given, to find the diagonal 59
The diagonal of a square being given, to find the side 60
To cut off a given area from a square parallel to either side. 60
To find the area of a rectangle 61
To find the side of a rectangle 62
The area and proportion of the two sides of a rectangle being
given, to find the sides 63
The sides of a rectangle being given, to cut off a given area
parallel to either side 64
To find the area of a rhombus 65
To find the area of a rhomboid 66
To find the side of a rhombus or rhomboid 66
To find the area of a triangle, when the base and perpendi-
cular are given 67
The three sides of a triangle being given, to find the area. ... 68
Any two sides of a right-angled triangle being given, to find
the third side 69
The sum of the hypothenuse and perpendicular, and the base
of a right-angled triangle being given, to find the hypothe-
nuse and perpendicular 70
To determine the area of an equilateral triangle 71
The area and the base of any triangle being given, to find the
perpendicular height 72
The proportion of the three sides of a triangle being given, to
find the sides of a triangle corresponding with a given area 73
The base and perpendicular of any triangle being given, to
find the side of an inscribed square 74
The three sides of any triangle being given, to find the length
of the perpendicular which will divide it into two right- -
angled triangles fK
1* 5
D CONTENTS.
Page
The area and base of a triangle being given, to cut off a given
part of the area by a line running from the angle opposite
the base 76
The area and base of a triangle being given, to cut off a tri-
angle containing a given area by a line running parallel to
one of its sides 77
The area and two sides of a triangle being given, to cut off a
triangle containing a given area, by a line running from a
given point in one of the given sides, and falling on the
other 78
To find the area of a trapezium 79
To find the area of a trapezoid 80
To find the area of a regular polygon 81
To find the area of a regular polygon when one of its equal
sides only is given. . . . 82
When the area of any regular polygon is given to find the
side 83
To find the area of an irregular right-lined figure of any num-
ber of sides 83
To find the area of a mixtilineal figure, or one formed by
right lines and curves 85
To find the circumference of a circle when the diameter is
given, or the diameter when the circumference is given. . . 87
To find the area of a circle 88
The area of a circle being given, to find the diameter or cir-
cumference 89
To find the area of a circular ring 89
The diameter or circumference of a circle being given, to find
the side of an equivalent square 91
The diameter or circumference of a circle being given, to find
the side of the inscribed square 9i
To find the diameter of a circle equal in area to any given
superficies 92
The diameter of a circle being given, to find another contain-
ing a proportionate quantity 93
To find the length of any arc of a circle 93
To find the area of a sector of a circle 97
To find the area of a segment of a circle 100
To find the area of a circular zone 105
To find the area of a lune 107
To find the area of a part of a ring, or of the segment of a sector 109
CONIC SECTIONS.
Definitions 110
To describe an ellipse 113
The ellipse and its parts 114
To describe a parabola 120
The parabola and its parts 120
To construct the hyperbola 124
The hyperbola and its parts 124
CONTENTS. 7
MENSURATION OF SOLIDS.
Page
Definitions I'M
To find the area of the surface of a cube. ... 133
The area of the surface of a cube being given, to find the
side 133
To find the solidity of a cube 134
To find the side of a cube the solidity being given 135
To find the solidity of a parallelopipedon 135
To find the solidity of a prism 136
To find the convex surface of a cylinder 137
To find the solidity of a cylinder 138
To find the curve surface of a cylindric ungula 139
To find the solidity of a cylindric ungula 139
To find the convex surface of a cylindric ring 140
To find the solidity of a cylindric ring ]41
The solidity and thickness of a cylindric ring being given, to
find the inner diameter 142
To find the surface of a right cone or pyramid 142
To find the surface of the frustum of a right cone or pyramid 144
To find the solidity of a cone or pyramid 145
To find the solidity of the frustum of a cone or pyramid 146
The solidity and altitude of a cone being given, to find the
diameter of the base 148
The solidity and diameter of the base of a cone being given,
to find the altitude 149
The altitude of a cone or pyramid being given, to divide it
into two or more equal parts, by sections parallel to the
base, to find the perpendicular height of each part 149
To find the solidity of an ungula, when the section passes
through the opposite extremities of the ends of the frustum 150
To find the solidity of a cuneus or wedge 151
To find the solidity of a prismoid 152
To find the convex surface of a sphere 153
To find the solidity of a sphere or globe 153
The convex surface of a globe being given, to find its diameter 154
The solidity of a globe being given to find the diameter 155
To find the solidity of the segment of a sphere 155
To find the solidity of a frustum or zone of a sphere 156
To find the solidity of a circular spindle, its length and middle
diameter being given 157
To find the solidity of the frustum of a circular spindle, its
length, the middle diameter, and that of either of the ends.
being given 158
To find the solidity of a spheroid, its two axes being given . . 159
To find the solidity of the middle frustum of a spheroid, its
length, the middle diameter, and that of either of the ends
being given 1 CO
To find the solidity of the segment of a spheroid 162
To find the solidity of a paraboloid '63
8 CONTENTS.
Page
To find the solidity of the frustum of a parabolo d, when its
ends are perpendicular to the axis of the solid 164
To find the solidity of a parabolic spindle 165
To find the solidity of the middle frustum of a parabolic spindle 165
To find the solidity of a hyperboloid 166
To find the solidity of the frustum of a hyperboloid 167
REGULAR BODIES.
To find the solidity of a tetraedron 16
To find the side of a tetraedron 1 70
To find the solidity of an octaedron 170
To find the side of an octaedron 171
To find the solidity of a dodecaedron 171
To find the side of a dodecaedron 172
To find the solidity of an icosaedron 173
To find the side of an icosaedron 173
To find the surface of any regular body 174
To find the solidity of any regular body ; . 1 75
BALLS AND SHELLS.
Weights and dimensions of balls and shells 176
Piling of balls and shells 182
CARPENTERS' RULE 185
The use of the Sliding Rule 186
TIMBER MEASURE.
To find the content of a board or piece of lumber one inch in
thickness 188
To find the content of any piece of lumber, whose thickness
is more than an inch 189
HEWN TIMBER.
To find the solid content of squared or four-sided timber 190
The breadth of a board being given, to find how much in
length will make a square foot, or any other required
quantity 192
When the board is wider at one end than at the other, and
the quantity is taken from the less end 192
To find how much in length will make a solid foot or any
other required quantity 193
ROUND TIMBER.
To find the solidity of round or unsquared timber 194
To find the number of cord feet in a log 196
To find the number of solid or square feet that a piece of
round timber will hew to when squared 197
CONTENTS.
SAW-LOGS.
Page
To find the Dumber o f feet, board measure, that a log contains 197
MASONS' WORK 198
BRICKLAYERS' WORK 200
CARPENTERS' AND JOINERS' WORK 202
SLATERS' AND TILERS' WORK 203
PLASTERERS' WORK 204
PAINTERS' WORK 206
GLAZIERS' WORK 207
PLUMBERS' WORK 208
PAVERS' WORK 209
VAULTED AND ARCHED ROOFS.
To find the solid content of a circular, elliptical, or Gothic
vaulted roof 211
To find the concave or convex surface of a circular, elliptical,
or Gothic vaulted roof 213
To find the solidity of a dome 214
To find the superficial content of a dome 214
To find the solid content of a saloon 215
To find the superficies of a saloon 216
GAUGING.
The gauging rule 21V
The use of the gauging rule 219
The gauging or diagonal rod 221
To find the content of a cask of the first variety 222
To find the content of a cask of the second variety 223
To find the content of a cask of the third variety 224
To find the content of a cask of the fourth variety 225
To find the content of a cask by four dimensions 226
To find the content of any cask from three dimensions 227
To ullage a standing cask 228
To ullage a lying cask 228
To find the ullage by the gauging rule 229
To find the content of a cask by the mean diameter 229
TONNAGE.
Carpenters' tonnage 2.'U
Government tonnage 232
10 CONTENTS.
CENTRE OF GRAVITY 236
FALLING BODIES 240
GRAVITIES OF BODIES 245
SPECIFIC GRAVITY 246
TABLE OF SPECIFIC GRAVITIES 249
THE PENDULUM 253
In any latitude to find the length of the pendulum which beats
seconds 256
CENTRES OF PERCUSSION AND OSCILLATION 257
CENTRE OF GYRATION 259
PRESSURE OF THE AIR 260
DISTANCES, AS DETERMINED BY THE VELOCITY
OF SOUND 261
MECHANICAL POWERS.
The Lever 263
Wheel and Axle 265
The Pulley 267
The Inclined Plane 270
The Wedge 271
The Screw 272
WHEELS 274
HYDRAULICS 277
PUMPS. 279
WATER-WHEELS 280
MISCELLANEOUS QUESTIONS 284
TABLE CfF AREAS OF CIRCULAR SEGMENTS . . .295
TABLES.
<y
I
3'S
SquarePole
Perches, o
Rods.
3 V
1*
O '
3
i -w O O
C< -<
O O
^ sO
|^. Tfl O O O
O 00 T O
CO rf (M C3D W
^ ry ^
* C5 -H- O O O Q
et 25 Q
^ CO 00 lO ^
(M rf O co OD
*** v c;
|rt* CO -rf Tf* O O
n'WTj* OS O O O Tt< O
-ta-H (^ (JJ _| ^5 ;Q
^ C5 f^ 'X (N Ci
o co ^ yj ^ ao
<O lO (^ Tf
i i CO 'T
11
12
TABLES.
TABLE II. LINEAL MEASURE.
Inches.
Corner's
Link.
Feet.
Yards.
Fa-
thoms.
Rods,
Poles,or
Perches.
Gunter's
Chains.
Fur-
longs.
Mile.
7fi
= 1
12
IIT.
= 1
36
^TT
3
= 1
72
198
Q i
&
6
2
" 2|
= 1
792
100
66
22
11
4
= 1
7920
1000
660
220
110
40
10
= 1
63360 8000
5280
1760
880
320
80
8
= 1
TABLE III. CUBIC OR SOLID MEASURE.
Cubic Inches.
Cubic Feet.
287496000
Cubic Yards.
166|
10648000
147197952000 5451776000 32768000
Cub. Poles,
Rods, or
Perches.
= 1
64000
Cub.
Fur*
longs.
1
612
Cub.
Mile.
TABLE IV. OTHER MEASURES.
40 cubic feet of round timber make
50 cubic feet of hewn timber "
40 cubic feet - "
128 cubic feet, or 8 feet in length, and ~)
4 in breadth, and 4 in height, make 5
24? cubic feet, or 161 feet in length,
1 in breadth and 1 in height make
282 cubic inches -
231 cubic inches
277.274 cubic inches (Eng.)
268 cubic inches
21 50.42 cubic inches
1 ton, T.
- 1 ton, T.
1 ton of shipping.
1 cord of wood.
1 perch of stone.
1 gallon, ale measure.
1 gallon, wine measure.
1 imperial gallon.
1 gallon, dry measure.
1 bushel
EXPLANATION OF THE CHARACTERS
USED IN THIS WORK.
+ denotes plus, or more. The sign of addition, signify-
ing that the numbers between which it is placed are to be
added together. Thus, 8 + 5, denotes that 5 is to be added
to 8. Geometrical lines are generally represented by capi-
tal letters. Thus, A B -f C D, signifies that the line A B is
to be added to the line C D.
denotes minus, or less. The sign of subtraction, sig-
nifying that the latter of the two numbers between which it
is placed is to be taken from the former. Thus, 4 2, de-
notes that 2 is to be taken from 4. In geometrical lines,
also, A B CD, signifies that the line C D is to be sub-
tracted from the line A B.
X denotes into, or by. The sign of multiplication, sig-
nifying that the numbers between which it is placed are to
be multiplied together. Thus, 7x5, denotes that 7 is to
be multiplied by 5. In geometrical lines, also, A B x C D,
signifies that the number of units in the line A B is to be
multiplied by the number of units in the line C D. Instead
of the sign x, a point is sometimes employed. Thus,
A B . C D, is the same as A B x C D.
-4- denotes divided by. The sign of division, signifying
that the former of the two numbers between which it is
placed is to be divided by the latter. Thus, 6 -r- 3, denotes
that 6 is to be divided by 3. This is also expressed by
placing the dividend above a line, and the divisor below it.
/
Thus, .5 denotes that 6 is to be divided by 3. In geometri-
o
cal lines, also, A B -4- C D, signifies that the line A B is to
A B
be divided by the line C D, or thus, T^-TV
O L)
... "| are proportionals, signifying that the numbers
. [between which they are placed are proportional.
\ ' f Thus, as 2 : 4 : : 8 : 16, denotes that the number
J 2 has the same proportion to 4 that 8 has to Hi.
14 EXPLANATIONS.
= denotes equal to. The sign of equality, signifying
that the numbers between which it is placed are equal to
each other, Thus, 2 poles + 2 poles = 4 poles = 22
yards = 1 chain = 100 links.
( ) the parenthesis ; [ ] the crotchet ; and { } the brace,
are signs mad use of to connect two or more quantities
together, and they are synonymous with regard to their
application ; for
is the same as
{(7 + 4) - 5} x 8 = (11 - 5) x 8 = 6 x 8 = 48.
The parenthesis which includes the 7 and 4, serves as a
chain to link them together, and shows that they are to be
added together before the number 5 is subtracted ; and the
brace also shows that the numerals which it includes must
be operated upon ; and the result multiplied by the num-
ber 8.
a This sign is placed above a quantity, signifying that the
quantity is to be squared.
Thus, (5 + 2) 3 = 7 a = 7 x 7 = 49.
8 This sign is placed above a quantity, signifying that the
quantity is to be cubed.
Thus, [(7 + 9) 8] 3 = (16 8) 3 = 8 s = 8x8x8=512.
v/ is a radical sign, signifying that the quantity before
which it is placed is to have the square root extracted.
Thus, _
+ 6) -(4 x 2)]' + {[(12 x 9) -*- 3] - [8 + (9 -2)}
xl2]-[3x(4-2)]} =
-[(108--3)-(8+7)] + [(12xl2)-
= ^ [6* + (36 15) -f (144 -*- 6)] = v/ (36 +21 + 24)
vK is a radical sign, signifying that the quantity before
which it is placed is to have the cube root extracted.
Thus, ^/[(6 x 4 x 3) 8] = # (72 8) = ^64 = 4.
.. denotes therefore.
JL denotes a perpendicular.
< denotes an angle.
A denotes a triangle.
PRACTICAL GEOMETRY.
DEFINITIONS.
1 . PRACTICAL GEOMETRY is a mechanical method of de-
scribing mathematical figures, by means of the scale and
compasses, or other instruments proper for the purpose. It
is founded upon the properties and relations of certain mag-
nitudes, which may be found treated at large in the works
of Euclid and other authors. The definitions of the princi-
pal figures are as follows :
1. A point, considered mathematically, is that which has
no parts or dimensions, but merely position.
2. A line is length without breadth, and its bounds or
extremes are points.
3. A right or straight line is that which lies
evenly between its extreme points ; as A B.
4. A broken line is one which changes fi
its direction at intervals so large that f
they can be perceived ; as A B C D. A /
5. A curved line is one which _-
changes its direction at intervals so A **
small that they cannot be perceived ; as A B.
6. A superficies is that which has A
length and breadth only ; and its bounds
or extremes are lines ; as A B C D.
7. A plane, or plane superficies, is that which is every-
where perfectly flat and even. Or, in other words, it is
that with which a right line will every way coincide.
"71 A
8. A body, or solid, is that which has ~
length, breadth, and thickness, and its bounds,
or extremes, are superficies ; as A B C D.
16
PRACTICAL GEOMETRY.
0. A plane rectilineal angle is the
inclination or opening of two right
lines, which meet in a point without
cutting each other ; as A B C. /
Here it is to be observed that the greater or less length
of the lines makes no alteration in the angle.
10. One right line is said to be perpen- A
dicular to another, when the angles on each
side of it are equal. ^.
Thus A B is perpendicular to C D. An-
gles are of three kinds; being either right, c-
acute, or obtuse.
11. A right angle is that which is formed by two righl
lines, that are perpendicular to each other ; as .
ABC.
Any angle differing from a right angle, whether
it be greater or less, is called an oblique angle,
and the lines that form it are called oblique B
lint>s.
D
12. An acute angle is that which is less
than a right angle ; as A B C.
13. An obtuse angle is that which is
greater than a right angle, as ABC.
14. A figure is a space boxinded by
one or more lines.
15. All plane figures bounded by
three right lines are called triangles,
and receive different denominations according to the nature
of their sides and angles.
16. An equilateral triangle is that
which has all its sides equal ; as A B C.
PRACTICAL GEOMETRY.
17
17. An isosceles triangle is that which
has only two of its sides equal ; as ABC.
18. A scalene triangle is that which
has all its three sides unequal ; as
ABC.
10. A right-angled triangle is that which
has one right angle ; the side opposite to
the right angle is called the hypothenuse,
and the other- two sides the legs ; as A B C,
where A B is the hypothenuse, and B C,
A C the two legs, or base and perpendi-
cular.
Any triangle differing from a right-angled one is frequently
called an oblique-angled triangle.
20. An obtuse-angled triangle is that
which has one obtuse angle ; as A C B,
where C is the obtuse angle.
21. An acute-angled triangle is that which
has all its angles acute ; as A B C.
22. All plane figures bounded by four right lines, are
ailed quadrangles, or quadrilaterals; and receive different
;ames according to the nature of their sides and angles.
23. A square is a quadrilateral, whose
*ides are all equal, and its angles all right
angles ; as A B C D.
2*
18 PRACTICAL GEOMETRY.
A square is also an instrument used by artificers for what
is called squaring their work ; being of various forms, as the
T square, normal square, &c.
A.I
24. A rhombus is a quadrilateral
whose sides are all equal, but its angles
not right angles ; as A B C D.
Bl
This figure, by mechanics, is generally called a lozenge;
and both it and the square belong to the ckss of parallelo-
grams.
25. A parallelogram is a quadri-
lateral whose opposite sides are
parallel ; as A B C D.
1
,/ -In
26. A rectangle is a parallelogram A
whose angles are all right angles ;
as A B C D.
27. A rhomboid is a parallelogram
whose angles are not right angles ;
as A B C D.
B
28. A trapezium is a quadrilateral
which hath not its opposite sides parallel ;
as A B C D.
B
29. A trapezoid is a quadrilateral, having
two of its opposite sides parallel ; as A B
CD.
30. The right line joining any two
opposite angles of a quadrangle, or quad-
rilateral, is called its diagonal ; as B D in
the figure A B C D.
B
c
A/ 7D
PRACTICAL GEOMETRY.
19
31. All plane figures contained under more -than four
sides are called polygons ; and receive different names, ac-
cording to the number of their sides, or angles.
32. Thus, polygons having five sides are called pentagons;
those of six sides, hexagons, those of seven, heptagons ; and
so on.
33. A regular polygon is that which has all its sides as
well as its angles equal to each other, and if the sides or
angles are unequal, it is called an irregular polygon.
An equilateral triangle is also a regular polygon of three
sides, and a square is one of four sides.
34. Parallel right lines are such as are
everywhere at an equal distance from each A u
other ; or which, if infinitely produced,
would never meet
to CD.
thus A B is parallel
35. The base of any figure is that side on which it is sup-
posed to stand ; and its altitude is the perpendicular falling
upon the base from the opposite angle.
36. An angle is usually denoted
by three letters, the one which stands
at the angular point being always to
be read in the middle, as A B C, C A_
B D, D B E, &c.
37. A circle is a plane figure bounded
by a curve line called the circumference or
periphery, which is everywhere equidis-
tant from a point within, called its centre ; c
and is formed by the revolution of a right
line (O A) about one of its extremities (O),
which remains fixed.
38. The centre of a circle is the point (O) about which it
is described ; and the circumference or periphery is the line
or boundary A B C A, by which it is contained.
The circumference itself, as well as the space which is
bounded by it, is also, for the sake of conciseness, sometimes
called a circle.
20
PRACTICAL GEOMETRY.
39. The radius of a circle is a right line
drawn from the centre to the circumference ;
asO A.
A tangent is a line touching a circle, and
which produced, does not cut it, as B A C.
40. The diameter of a circle is a right
line passing through the centre, and A
terminated both ways by the circumfe-
rence ; as A B.
41. An arc of a circle is any part of its
circumference, or periphery ; as A B.
42. A chord is a right line which joins
the extremities of an arc ; as A B.
43. A segment of a circle is the space
contained between an arc and its chord ;
as A B C,
44. A sector is the space contained be-
tween an arc and the two radii drawn to
its extremities ; as A B C.
PRACTICAL GEOMETRY.
21
45. A zone is a part of a circle included A
between two parallel chords and their inter-
cepted arcs ; as A B C D.
46. The versed sine or height of an arc,
is that part of .the diameter contained be- A
tween the middle of the chord and the arc ;
as D B, or D E.
47. A lune is the space included between
the intersecting arcs of two eccentric circles ;
as A B C.
48. A semicircle is a half of a circle ; a quadrant is a
quarter of a circle ; a sextant the sixth part of it, and an
octant the eighth part, where it may be observed that these
names are often applied to instruments used for taking
angles.
49. The circumference of every circle is supposed to be
divided into 360 equal parts called degrees ; each degree
into 60 equal parts, called minutes; and each minute into 60
equal parts, called seconds.
50. The measure of any right-lined
angle is an arc of a circle contained be- /
tween the two lines which form that '
angle, the angular point being the ' x
centre ; thus the angle A O B is mea- \ /
sured by the arc m n. ** ..*''
The angle is estimated by the number of degrees, minutes,
&c. contained in the arc ; whence a right angle is an angle
of 90 degrees or | of the circumference.
AXIOMS.
1. An axiom is an established principle, or self-evident
truth, requiring no other conviction than that which arises
from a proper understanding of the terms in which if is pro-
posed.
22 PRACTICAL GEOMETRY.
2. Things which are equal to the same thing are equal to
other.
3. If equals be added to equals the wholes will be equal.
4. If equals be taken from equals the remainders will be
equal.
5. If equals be added to unequals the wholes will be un-
equal.
6. If equals be taken from unequals the remainders will
be unequal.
7. Things which are half, double, or any number of times
the same thing, are equal.
8. The whole is greater than its part
9. Every whole is equal to all its parts taken together.
10. All right angles are equal to each other.
11. Angles that have equal measures, or arcs, are equal.
12. Things which coincide, or fill the same space, are
identical, or mutually equal in all their parts.
13. Two straight lines cannot enclose a space.
REMARKS.
1 A problem is something proposed to be done.
2. The perimeter of a figure is the sum of all its sides
taken together.
3. The sum of any two sides of a triangle is greater than
the third side.
4. In any triangle the sum of the three angles is equal to
two right angles.
f>. Every triangle is half the parallelogram which has the
same base and the same altitude.
6. An angle inscribed in a semicircle is a right angle.
7. All angles in the same segment of a circle are equal to
each other.
PRACTICAL GEOMWPRY. 23
8. Triangles that have all the three angles of the one
respectively equal to all the three angles of the other, arc
called equiangular triangles, or similar triangles.
9. In similar triangles the like sides, or sides opposite to
the equal angles, are proportional.
10. The areas or spaces of similar triangles are to each
other as the squares of their like sides.
11. The areas of circles are to each other as the squares
of their diameters, radii, or circumference.
12. Similar figures are such as have the same number of
sides, and the angles contained by their sides respectively
equal.
13. The areas of similar figures are to each as the squares
of their like sides.
14. If three quantities are proportional, the middle one is re-
peated, and the first is to the second as the second is to the third.
In such a case the middle quantity is a mean proportional
between the other two, and the last is a third proportional to
the first and second; but, if there are four proportional quan-
tities, the last is called a fourth proportional to the other three
INSTRUMENTS.
2. The principal instruments used in describing or con-
structing geometrical figures are as follows :
THE DIVIDERS OR COMPASSES.
The plain compasses consist of two inflexible rods of brass,
revolving upon an axis at the vertex, and furnished with
steel points.
GEOMETRY.
THE PARALLEL RULER.
The parallel ruler consists of two flat pieces of ebony, con-
nected together with brass bars, having their extremities
equidistant, by which contrivance, when the ruler is opened,
the sides necessarily move in parallel lines.
THE SCALE OF EQUAL PARTS.
The scale of equal parts consists of a certain number of
equal portions of any convenient length, the extreme one or
the left hand being subdivided into ten equal parts, and
is called the unit of the scale, and the rest being numbered
1, 2, 3, &c.
In most scales an inch is taken for a common measure, and
what an inch is divided into is generally set at the end of the
scale.
This scale is used in laying down any distance, as inches,
feet, chains, miles, &c. The several divisions may be con-
sidered as feet, for example, the decimal subdivision would
be tenths of a foot. So also each of the principal divisions
may be regarded as ten inches, ten feet, &c., and in this
case the decimal subdivision will represent inches, feet, &c.,
respectively. This scale is limited to two figures, or any
number less than 100 may be readily taken ; but if the
number should consist of three places of figures, the value
oi the third figure cannot be exactly ascertained, and in this
case it is better to use a diagonal scale, by which any num-
ber consisting of three places of figures may be exactly
found.
Let it be required to take from the scale a line equal to
fire inches and eight-tenths.
PRACTICAL GEOMETRY.
Place one foot of the dividers at 5 on the right, and ex-
tend the other to. 8, which makes the eighth of the small
divisions. The dividers will then embrace the required
distance.
THE DIAGONAL SCALE OF EQUAL PARTS.
The construction of this scale is as follows :
Having prepared a ruler of convenient breadth for your
scale, draw near the edges thereof two right lines, a e, c f,
parallel to each other ; divide one of these lines, as a e, into
equal parts, according to the size of your scale, and through
each of these divisions draw right lines perpendicular to
a e, to meet cf; then divide the breadth into ten equal parts,
and through each of these divisions draw right lines paralle.
to a e and cf', divide the lines a b, c d, into ten equal parts,
and from- the point a, to the first division in the line c d,
draw a diagonal line ; then parallel to that line, draw diago-
nal lines through all the other divisions, and the scale is com-
plete. Then, if any number consisting of three places of
figures, as 468, fy required from the larger sca\e,fd, you
must place one foot of the compasses on the figure 4, on the
line/rf, then the extent from 4 to the point d will represent
400. The second figure being 6, count six of the smaller
divisions from d towards c, and the extent from 4 to that
point will be 460. Mo T 'e both points of the compasses down-
wards till they are on the eighth parallel line below/ d, and
open them a little till the one point rests on the vertical line
drawn through 4, and the other on the diagonal line drawn
through 6 ; the extent, then, in the compasses will represent
468. " In the same manner the quantities 46.8, 4.68, 0.468.
&c., are measured.
There are generally two diagonal scales laid down on the
same face of the instrument, the unit of the one being double
that of the other, and commencing on opposite ends of the
scale.
PRACTICAL GEOMETRY.
THE SCALE OF CHORDS.
uTlC
Draw the lines B A, B C, at right angles to each other,
and with any convenient distance, B A, describe the arc
A s C ; divide it into 90 equal parts, and join A C. From
A as a centre, with the distances A 10, A 20, &c., describe
the arcs 10, 10, 20, 20, &c., meeting the line A C. Fill
up the separate degrees, which are not marked in the diagram
to prevent confusion, and the scale is complete. It is evi-
dent, by inspection, that the chord of 60 is equal to the radius,
as shown by the letter r upon the rule ; which distance is
therefore always to be taken in laying down angles.
THE SEMICIRCULAR PROTRACTOR.
The protractor is a semicircular piece of brass divided into
180 degrees, and numbered each way from end to end ; that
is, from A to B, and from B to A. There is a small notch
PRACTICAL GEOMETRY. 27
in the middle of the diameter A B, denoting the centre of
the protractor. In some boxes of mathematical instruments
this is omitted, and the degrees are transferred to the border
of the plain scale.
GUNTER S SCALE.
Gunter's scale, commonly of two feet in length, contains
on one side the lines of the plain scale, already described,
and on the other corresponding logarithmic lines.
PROBLEM I.
3. To describe from a given centre the circumference
of a circle having a given radius.
Let A be the given centre, and A B
the given radius.
Place one foot of the dividers at A, and
extend the other leg until it shall reach B[
to B.
Then turn the dividers around the leg
at A, and the other leg will describe the
required circumference.
PROBLEM II.
Through a given point A, to draw a line parallel to a given
line, D E.
Lay the edge of the parallel ruler B A c
upon D E, and move it upwards till it
reaches the point A, through which
draw B C , and it will be parallel to D E. D ~
28 PRACTICAL GEOMETRY.
PROBLEM III.
To lay off on a given line, as AB, a distance equal
to C D.
EXAMPLE.
Let C D be the distance to be laid off, c D
and A B the given line. A f T,
Place one foot of the dividers at C, and extend the other
leg until the foot reaches D.
Then, raising the dividers, place one foot at A, and mark
with the other the distance AE, this will evidently be
equal to C D.
PROBLEM IV.
To lay down a line of given length, on a scale of a given
number to the inch, to determine how many parts of it are
to be represented on the paper by a distance equal to the
unit of the scale.
EXAMPLE.
If a line 320 feet in length is to be laid down on paper,
on a scale of 40 feet to the inch, what length must be taken
from the scale ?
Divide the length of the line by the number of parts which
is represented by the unit of the scale ; the quotient wiJ.
give the number of parts which is to be taken from the
scale.
Here 320-i-40=8 the number of parts to be taken from
the scale.
PROBLEM V.
The length of the line being given on the paper, to deter
mine the true length of the line which it represents.
EXAMPLE.
The length of the line on the paper is 4.75 inches, aicJ
the scale is one of 20 feet to the inch ; what is the true lenyl
of the line ?
PRACTICAL GEOMETRY. 29
Take the line in your dividers and apply it to the scale,
and note the number of units and parts of a unit to which
it is equal ; then multiply this number by the number of
parts which the unit of the scale represents, and the product
will be the length of the line.
Here 4.75x20=95 feet, the length of the line.
PROBLEM VI.
To make an angle of any proposed number of degrees.
1. Draw any line A B, and having taken first 60 degrees
from the scale of chords, describe with this radius the arc n m.
2. Take in like manner the chord of the proposed num-
ber of degrees from the same scale, and apply it from n to m,
3. Then if the line A C be drawn from the point A
through m, the angle BAG will be that required.
Angles greater than 90 degrees are usually made by first
laying ofF90 degrees upon the arc n m, and then the remain-
ing part.
This problem may be performed by the protractor.
Place the central notch of the instrument upon A, and the
edge along A B ; make a point m against the proposed num-
ber of degrees, and through it draw the line A C.
3*
30
PRACTICAL GEOMETRY.
PROBLEM VH.
Any angle BAG being given, to find the number of de-
grees it contains.
1. From the angular point A, with the chord of 60
degrees, describe the arc n m, cutting the lines A B, A C in
the points n and m.
2. Then take the distance n m, and apply it to the scale
of chords, and it will show the degrees required.
And if the distance n m be greater than 90, it must be
taken at twice, and each part applied separately to the scale.
This problem may be performed by the protractor.
Place the central notch of the instrument upon A, and
the edge along A C, and observe the number of degrees cut
by the line A B, which will show the degrees required.
PROBLEM VIII.
To divide a given line AB into two equal parts.
1. From the points A and B, as
centres, with any distance greater
than half A B, describe arcs cutting
each other in n and m.
2. Through these points, draw
tne line n c ra, and the point c, where
it cuts A B, will be the middle of the
'ine required.
PRACTICAL GEOMETRY.
3J
PROBLEM IX.
To divide a given angle ABC into two equal parts.
1. From the point B, with any radius,
describe the arc A C ; and from A and C,
with the same, or any other radius, describe
arcs cutting each other in n.
2. Then, through the point n draw the
line B n, and it will bisect the angle ABC,
as was required.
PROBLEM X.
From a given point C, in a given right line A B, to erect
a perpendicular.
CASE I. When the point is near the middle of the line.
1. On each side of the point C
take any two equal distances Cn,
Cm.
2. From n and m, with any ra-
dius greater than nC or m C, de-
scribe arcs cutting each other in s.
3. Then through the point s, A
draw the line sC, and it will be n " c m
the perpendicular required.
CASE II. When the point is at, or near, the end of the
line.
Supposing C to be the given point, as before.
1. Take any point 0, and with the
radius or distance, o C, describe the
arc mCn, cutting A B in m and C.
2. Through the centre o, and the
point m, draw the line m o n, cutting
the arc mCn in n.
3. Then from the point n, draw
the line nC, and it will be the per- A i
pendicular required.
32 PRACTICAL GEOMETRY.
The same otherwise.
1. Set one leg of the compasses
on B, and with any extent B m
describe an arc mp ; then set off
the same extent from m to n.
2. Then join m n, and from n
as a centre with the extent m n as
radius, describe an arc q ; pro- /
duce m n to q, and the line join- /
ing q B will be perpendicular to /'
AD A
- m
Another method.
1. From any scale of equal parts
take a distance equal to 3 divisions,
and set it from B to m.
2. And from the points B and
m, with the distances 4 and 5,
taken from the same scale, describe
arcs cutting each other in n.
3. Through the points n, B, draw
the line BC, and it will be the A
perpendicular required.
The same thing may also be readily done, by an instru-
ment in the form of a square, or by the plain scale.
PROBLEM XL
From a given point C, without a given line A B, to let
fall a perpendicular.
CASE I. When the point is nearly opposite to the middle
of the line.
1. From the point C, with any
radius, describe the arc n m, cut-
ting A B in n and m.
2. From the points n, m, with
the same or any other radius, de- A.
scribe two arcs cutting each other
in 8.
3. Through the points C,
draw the line C G s, and C G will \|X
be the peroendicular required ** ' **
PRACTICAL GEOMETRY.
A.
This problem, also, may be performed like the last by
means of a square.
CASE II. When the point is opposite or nearly opposite
to the end of the line.
1. Take any point m, in the
line A B, and from C draw the Xc
Jine C m.
2. Bisect the line C m, or di-
vide it into two equal parts in the
point n.
3. From n, with the radius nm,
or nC, describe the arc CDm, A-
cutting A B in D. N ^ *'' l
4. Then through the point C,
draw the line C D, and it will be the perpendicular required.
This method may also be used in the first case, if the line
A B, when necessary, be produced.
The same otherwise.
1. From A, or any other point in ^\ ^
A B, with the radius A C, describe the
arcs C, D.
2. And from any other point n, in
AB, with the radius nC, describe
another arc cutting the former in C, D.
3. Then through the points C, D, A
draw the line C E D, and C E will be
the perpendicular required.
Perpendiculars may be more easily
raised, and let fall, in practice, by means
of a square, or other instrument proper
for this purpose.
PROBLEM XII.
To draw a perpendicular, from any angle of a triangle
A B C, to its opposite side.
1. Bisect either of the sides con-
taining the angle from which the per-
pendicular is to be drawn, as B C in
the point n
34
PRACTICAL GEOMETRY.
2. Then with the radius nC,
and from the centre n describe an
arc cutting AB, (or AB produced if
necessary, as in the second figure,)
in the point D ; the line joining C
D will be perpendicular to A B or
A B produced.
PROBLEM Xin.
To trisect, or divide a right angle ABC, into three equal
parts.
1. From the point B, with any
radius B A, describe the arc A C, cut-
ting the legs B A, BC, in A, C.
2. From the point A, with the
radius A B, or B C, cross the arc A C
in n ; and with the same radius, from
the point C, cross it in m.
3. Then through the points m, n,
draw the lines Bra, Bn, and they
will trisect the angle as was required.
By this means the circumference of any given circle may
be divided into 12 equal parts ; and thence by bisection into
24, 48, &c.
PROBLEM XIV.
At a given point D, to make an angle equal to a given
angle ABC.
B m AD T E
1. From the point B, with any radius, describe the arc n
, cutting the lines B A, B C, in the points m, n.
PRACTICAL GEOME1RY. 35
2. Draw the line D E, and from the point D, with the
same radius as before, describe the arc r s.
3. Take the distance m n, on the former arc, and apply it
to the arc r s, from r to s.
4. Then through the points D, s t draw the line D F, and
the angle E D F will be equal to the angle ABC, as was
required.
PROBLEM XV.
To draw a line parallel to a given line A B.
I. When the parallel line is to pass through a given
point C.
1. Take any point m in D _ ^ _ f E
the line A B, and from the > ^^^^ \
point C draw the line C m. ' ^^ /'
2. From the point m, ; ^ x /
with the radius m C, de- ^ ^ m i
scribe the arc C n, cutting
A B in n ; and with the same radius, from the point C, de-
scribe the arc m r.
3. Take the distance C n, and apply it to the arc m r, from
m to r; then through the points C, r, draw the line D C r E,
and it will be parallel to A B, as was required.
CASE II. When the parallel line is to be at a given dis
tance from A B.
1. From any two points r, s, D n c
in the line A B, with a radius
equal to the given distance,
describe the arcs, n, m.
2. Then draw the line C D
to touch these arcs without A r
cutting them, and it will be parallel to A B, as was required
36
PRACTICAL GEOMETRY.
s^.--**"" \
PROBLEM XVI.
To divide a given line A B into any proposed number of
equal parts.
1. From one end of the line A, . **
draw A m, making any angle with
A B ; and from the other end B,
draw B n, making an equal angle A<(
A En. \
2. In each of the lines A m, B n, \
beginning at A and B, set off as
many equal parts, of any conve-
nient length, as A B is to be divided into. ,
3. Then join the points A 5, 1, 4, 2, 3, &c., and A B will
be divided as was required.
B n may be drawn parallel to A m, by means of the
parallel ruler, or the same may be done by taking the aic
A n equal to B m.
Another method.
1. Through the point B draw the
indefinite line C E, making an angle
with A B.
2. Take any point E in that line,
through which draw E D parallel to
B A, and set off as many equal parts
E H, H G, &c., from E toward D, as
<he line A B is to be divided into.
3. Through the points D, A, draw the line DA, producing
it till it meets E C in C ; then lines drawn from C through
ihe points F, G, H, will divide the line A B, into the re-
quired number of parts.
PnACTlCAL GEOMETRY.
37
PROBLEM XVII.
To find the centre of a given circle, or of one already
described.
1. Draw any chord, A B, and bisect
it with the perpendicular, C D.
2. Bisect C D, in like manner, with
the chord E F, and the intersection, O, E
will be the centre required ; observing
that the bisection of the chords, and the
raising of the perpendicular, may be per-
formed as in Problems VIII. and X.
PROBLEM XVIII.
To describe the circumference of a circle through any
three given points, A, B, C, provided they are not in a right
line.
1. From the middle point B, draw the
lines, or chords, B A, B C ; and bisect
them perpendicularly with the lines
meeting each other in O.
2. Then from the point of intersec-
tion, O, with the distance O A, O B, or
O C, describe the circle ABC, and it
will be that required.
PROBLEM XLX.
To draw a tangent to a given circle, that shall pass through
a given point A.
CASE I. When the point A is in the circumference of
the circle.
1 . From the given point A, to the
centre of the circle, draw the radius
AO.
2. Then through the point A, draw
B C perpendicular to O A, and it will
be the tangent required.
38
PRACTICAL GEOMETRY.
CASK II. When the given point, A, is without the circle.
1. To the given point A, from the A ,''~"^^L
centre O, draw the line O A, and bisect
it in n.
2. From the point n, with the radius
n A, or n O, describe the semicircle
ABO, cutting the given circle in B.
3. Then through the points A,B, draw
the line A B, and it will be the tangent
required.
PROBLEM XX.
To two given right lines, A,B, to find a third proportional.
1. From any point, C, draw two right lines, making any
angle, F C G.
2. In these lines take C E equal to the first term A, and
C G, C D, each equal to the second term B.
3. Join E D, and draw G F parallel to it ; and C F will
be the third proportional required.
That is C E (A) : C G (B) : : C D (B) : C F.
PROBLEM XXI.
To three given right lines, A,B,C, to find a fourth pro-
portional.
PRACTICAL GEOMETRY. 39
1 . From any point, D, draw two right lines, making any
angle G D H.
2. In these lines take D F equal to the first term A, D E
equal to the second B, and D H equal to the third C.
3. Join F E, and draw H G parallel to it, then D G will be
the fourth proportional required.
That is D F (A) : D E (B) : : D H (C) : D G.
PROBLEM XXII.
Between two given right lines A,B, to find a mean pro-
portional.
1. Draw any right line, in which take C E, equal to A,
and E D, equal to B.
2. Bisect C D in O, and with O D or O C, as radius, de-
scribe the semicircle C F D.
3. From the point E draw E F perpendicular to C D, and
it will be the mean proportional required.
That is C E (A) : E F : : E F : ED (B).
PROBLEM XXIII.
To divide a given line, A B, in the same proportion that
another given line, C, is divided.
1. From the point A draw A D
equal to the given line C, and making
any angle with A B.
2. To A D apply the several divi-
sions of C, and join D,B.
3. Draw the lines 4,4, 3,3, &c.,
each parallel to D B, and the line
A B will be divided as was required.
40
PRACTICAL GEOMETRY.
That is, the parts A, 1 ; 1, 2; 2, 3; 3, 4; 4, B; on the,
Sine A B, will be proportional to the parts 0, 1 ; 1, 2 ; 2, 3 ;
3; 4 ; 4, 5 ; on the line C.
PROBLEM XXIV.
Two angles of a triangle being given to find the third.
1. Draw the indefinite
line ABC; at any point,
as B, make the angle
A B D equal to one of
the given angles, and the
angle D B E, equal to
the other.
2. The remaining an- A B
gle E B C will be the
third angle required, because these three angles are together
<;qual to two right angles.
PROBLEM XXV.
Two sides of a triangle, and the angle which they contain,
being given to construct the triangle.
1. Let A and B be
equal to the given
sides, and C the given
angle.
2. Draw the indefi-
nite line D E ; at the
point D, make the
angle E D F equal to A
the given angle C ;
then take D G=A, B
D H=B, and draw
G H ; then D G H will be the triangle required.
PRACTICAL GEOMETRY.
41
PROBLEM XXVI.
A side and two angles being given to construct the triangle.
1. The two angles will either be
adjacent to the given side, or the one
adjacent and the other opposite ; in
the latter case, find the third angle by
subtracting the sum of the two given
angles from 180 degrees ; then the two
adjacent angles will be known.
2. Draw the line A B equal to the given side ; at the
point A, make an angle, BAD, equal to one of the adjacent
angles, and at B, an angle equal to the other ; the two lines
A D, B E, will cut each other at C ; and ABC will be the
triangle required.
PROBLEM XXVII.
Upon a given right line, A B, to make an equilateral tri-
angle.
1. From the points A and B, with
a radius equal to A B, describe arcs
cutting each other in C.
2. Draw the lines A C, B C, and
the figure A C B will be the triangle
r equired.
An isosceles triangle may be formed A
in the same manner, by taking any
distance for radius.
4
42
PRACTICAL GEOMETRY.
PROBLEM XXVIIL
The three sides of a triangle being given, to describe the
triangle.
Let A, B, and C, be the sides.
1. Draw a line, D E, equal to one of the given lines, C.
2. From the point D, with a radius equal to B, describe
an arc.
3. And from the point E, with a radius equal to A, de-
scribe another arc, cutting the former in F.
4. Then draw the lines D F, E F, and D F E will be the
triangle required.
PROBLEM XXIX.
Upon a given line, A B, to describe a square.
1. From the point B, draw B C per- ri /
pendicular, and equal to A B.
2. From the points A and C, with the
radius A B or C B, describe two arcs
cutting each other in D ; then draw the
lines A D, C D, and the figure A BC D
will be the square required.
A rhombus may be made on the given
line A B, in exactly the same manner, if B C be drawn with
tne proper obliquity, instead of perpendicularly.
PRACTICAL GEOMETRY.
Another method.
1. From the points A, B, as centres,
with the radius A B, describe arcs
crossing each other in ra.
2. Bisect A m in n ; and from the
centre m, with the radius m n, cross
the other two arcs in C and D.
3. Then draw the lines A D, B C,
and C D, and A B C D will be the
square required.
PROBLEM XXX.
To describe a rectangle, whose length and breadth shall
be equal to two given lines, A B, and C.
1. At the point B, in the given
line A B, erect the perpendicular
B D, and make it equal to C.
2. From the points D, A, with
the radii A B and C respectively,
describe two arcs cutting each other
in E ; then join E A and E D, and
A B D E will be the rectangle re-
quired.
A parallelogram may be described in nearly the same
manner.
PROBLEM XXXI.
In a given triangle, A B C, to inscribe a circle.
1 . Bisect any two of the angles,
as A and B, with the lines A O,
and B O.
2. Then from the point of in-
tersection, O, let fall the perpen-
dicular O n, upon either of the
sides, and it will be the radius of
the circle required.
44 PRACTICAL GEOMETRY.
PROBLEM XXXII.
In a given circle to inscribe an equilateral triangle, a hexa-
gon, or a dodecagon.
FOR THE HEXAGON.
1. From any point, A, in the circum-
ference, as a centre, with a distance
equal to the radius A O, describe the
arc, EOF.
2. Join the points A, B, or A, F, and
either of these lines being carried six
times round the circle will form the
hexagon required.
That is, the radius of the circle is equal to the side of the
inscribed hexagon ; and the sides of the hexagon divide the
circumference of the circle into six equal parts, each contain-
ing 60 degrees.
FOR THE EQUILATERAL TRIANGLE.
1. From the point A, to the second and fourth divisions,
or angles of the hexagon, draw the lines A C, A E.
2. Then join the points C, E, and ACE will be the equi-
lateral triangle required ; the arcs, A B C, C D E, and E F A,
being each one-third of the circumference, or 120 degrees.
FOR THE DODECAGON.
Bisect the arc A B, subtending the side of the hexagon in
irhe point n, and the line A n being carried twelve times
lound the circumference, will form the dodecagon required ;
the arc A n being 30 degrees.
Then, if A n be again bisected, a polygon may be formed
of 24 sides ; and by another bisection a polygon of 48 sides ;
and so on.
PRACTICAL GEOMKTKV.
45
PROBLEM XXXm.
To inscribe a square or an octagon, in a given circle.
FOR THE SQUARE.
1. Dra w the diameters BD and AC,
intersecting each other at right angles.
2. Then join the points A,B, B,C, B
C, A, and D, A; and A B G D will be
the square required.
FOR THE OCTAGON.
Bisect the arc A B, subtending the side of the square, in
the point E, and the line A E being carried eight times round
the circumference will form the octagon.
Also, if the arc A E be again bisected, a polygon may be
formed of 16 sides : and by another bisection, a polygon of
32 sides ; and so on.
PROBLEM XXXIV.
To inscribe a pentagon, or decagon, in a given circle.
FOR THE PENTAGON.
1. Draw the diameters A p t n m,
at right angles to each other, and
bisect the radius O n in r.
2. From the point r, with the dis-
tance r A, describe the arc A s, and
from the point A, with the distance
A s, describe the arc s B.
8. Join the points A, B, and the
line A B being carried five times
round the circle, will form the pentagon required.
40
PRACTICAL GEOMETRY.
FOR THE DECAGON.
Bisect the arc A E, subtending the side of the pentagon in
c, and the line A c being carried ten times round the cir-
cumference will form the decagon required.
Also, if the arc A C be again bisected, a polygon of 20
sides may be formed ; and by another bisection a polygon
of 40 sides ; and so on.
PROBLEM XXXV.
In a given circle, to inscribe a regular heptagon.
1. From any point A in the circum-
ference, with the radius A O of the
circle, describe the arc BOG, cutting
the circumference in B and C.
2. Draw the chord B C, cutting O A
m D, and B D, or C D, carried seven
times round the circle from A, will form
the heptagon required.
An exact heptagon, as is well known, cannot be inscribed
in a circle from pure geometrical principles ; but, the above
method of constructing it, which is extremely simple, will
be found sufficiently accurate to answer most practical
purposes ; the approximation for the side being true, as far
as the second place of decimals, inclusively.
PROBLEM XXXVI.
In a given circle, to inscribe any regular polygon.
1. Draw the diameter A B, which ^ c
divide into as many equal parts as the x/X
ligure has sides. /
2. From the points A, B, as centres,
with the radius A B, describe arcs
crossing each other in C.
3. From the point C, through the
second division of the diameter, draw
the line C D ; then if the points A and
D be joined, the line A D will be the
side of the polygon required.
PRACTICAL GEOMETRY.
47
It m?iy be observed, that in the construction here given.
A D is the side of a pentagon.
This rule is not exact, except for the equilateral triangle
and hexagon ; being for polygons in general only a near
approximation.
PROBLEM XXXVII.
On a given line A G, to describe a regular polygon of any
proposed number of sides.
1. From the point A, with
the distance A G, describe the
semicircle G B H, which divide
into as many equal parts as H a,
a B, Be, &c., as the polygon
is to have sides.
2. From A to the second part
of the division draw A B, and
through the other points c, d, e,
f, &c., draw the lines A C, A D, u
A E, A F, &c.
3. Apply the distance A G from G to F, from F to E,
from E to L>, from D to C, &c., and join B C, CD, D E,
E F, &c., and A B C D E F G will be the polygon required.
PROBLEM XXXVIII.
About a given triangle A B C, to circumscribe a circle.
1. Bisect the two sides A B, BC,
with the perpendiculars m O, and n O.
2. From the point of intersection O,
with the distance OA, OB, or OC, de-
scribe the circle A C B, and it will be
that required.
If any two of the angles be bisected,
instead of the sides, the intersection of
these lines will give, as before shown,
the centre of the inscribed circle.
48
PRACTICAL GEOMETRY.
PROBLEM XXXIX.
About a given square, A B C D, to circumscribe a
1. Draw the two diagonals AC and
B D intersecting each other in O.
2. Then from the point O, with the
distance O A, O B, O C or O D, describe
the circle A B C D, and it will be that
required.
PROBLEM XL.
To circumscribe a square about a given circle.
1. Draw any two diameters n o and *
r m at right angles to each other.
2. Then through the points m, o, r, ,
draw the lines AB, BC, CD, and
D A, perpendicular to r m, and n o, and
A. B C D will be the square required.
PROBLEM XLI.
About a given circle to circumscribe a pentagon
1 . Inscribe a pentagon in the circle ;
or which is the same thing, find the
points m,n, v, r,s, as in Prob. XXXIV.
2. From the centre o, to each of
these points, draw the radii o n, o m,
o u, o r, and o s.
3. Through the points n, m, draw
the lines A B, B C, perpendicular to
on, o m ; producing them till they
meet each other at B.
Draw in like manner the lines C D, D E, E A, perpen-
dicular to o v, o r, and o s, and A B C D E will be the
pentagon required.
PRACTICAL GEOMETRY.
49
Any other polygon may be made to circumscribe a circle,
by first inscribing a similar one, and then drawing tangents
to the circle at the angular points.
PROBLEM XLH.
On a given line, A B, to make a regular pentagon.
1. Make Bm perpendicular to AB,
and equal to one half of it.
2. Draw A m, and produce it till
the part m n is equal to B m.
3. From A and B as centres, with
the radius B n, describe arcs cutting
each other in o.
4. Then from the point o, with the
same radius, or with o A, or o B, de-
scribe the circle, ABODE; and the line A B, applied five
times round the circumference of this circle, will form the
pentagon required.
If tangents be drawn through the angular points A, B, C,
D, E, a pentagon circumscribing the circle will be formed ;
and if the arcs be bisected, a circumscribing decagon may be
formed ; and so on.
Vc
Another method.
1. Produce A B towards n, and
at the point B make the perpendi-
cular B m, equal to A B.
2. Bisect A B in r, and from r
as a centre, with the radius r m, de-
scribe the arc m n, cutting the pro-
duced line A B in n.
3. From the points A and B,
with the radius A n, describe arcs cutting each other in D ;
and from the points A, D, and B, with the radius A B, de-
scribe arcs cutting each other in C and E.
4. Then if the lines B C, C D, D E, and E A, be drawn,
A B C D E will be the pentagon required.
50
PRACTICAL GEOMETRY.
PROBLEM XLHI.
On a given line, A B, to make a regular hexagon.
1. From the points A, and B, as cen-
tres, with the radius A B, describe arcs
cutting each other in ; and from the
point O, with the distance O A, or O B,
describe the circle A B C D E F.
2. Then if the line A B be applied
six times round the circumference, it
will form the hexagon required.
PROBLEM XLIV.
On a given line, A B, to form a regular octagon.
1. On the extremities of the given
line, A B, erect the indefinite perpen-
diculars A F, and B E.
2. Produce A B both ways to m and
n, and bisect the angles m A F and
n B E with the lines A H and B C.
3. Make A H and B C each equal to
A B, and draw H G,C D, parallel to A F, m '
or B E, and also each equal to A B.
4. From G, and D, as centres, with a radius equal to A B,
describe arcs crossing A F, B E, in F and E ; then if G F,
FE, and E D be drawn, ABCDEFGH will be the octa-
gon required.
PROBLEM XLV.
To make a figure similar to a given figure, A B
1 . Take A b, equal to one of the sides
of the figure required, and from the angle
A draw the diagonals A c, A d. e
2. From the points b, c, d, draw b c, c d, E
d e, parallel to B C, C D, D E, and f
A b c d e will be similar to A BC D E.
The same thing may also be done by A'
making the angles b, c, d, e, respectively
equal to the angles B, C, D, E.
PRACTICAL GEOMETRY.
PROBLEM XL VI.
51
To make a triangle equal to a given trapezium, A B C D.
1. Draw the diagonal D B, and
make C E parallel to it, meeting the
side A B produced in E.
2. Join the points D, E, and
A D E will be the triangle re-
quired.
PROBLEM XLVII.
To make a triangle equal to any right-lined figure,
ABCDEFGA.
1. Produce the side A B both
ways at pleasure.
2. Draw the diagonals E A,
E B; and by the last problem,
make the triangles A E I, B E K,
equal to the figures A E F G, and
BE DC.
3. Draw I L, K M, parallel to
E A, E B.
4. Then if the points E, L, and E, M, be joined, ELM
will be the triangle required.
And in the same manner may any right-lined figure what-
ever be reduced to a triangle.
PROBLEM XL VIII.
To describe a square that shall be any multiple of a given
square, A B C D.
1. Draw the diagonal B D, and on L
A B, A D, produced, take A G, A E, fT"~
each equal to B.D, then A F, the square G
on A E, will be double the square AC. D
2. Draw in like manner the diagonal
B G, and make A L, A H, each equal to
B G ; then A K, the square on A H, will
be triple the square A C.
52
PRACTICAL GEOMETRY.
3. Proceed in the same manner with the diagonal B L ;
and a square will be formed which is quadruple the 'square
A C ; and so on.
PROBLEM XLIX.
To inscribe a square in a given triangle, ABC.
1 . Through the ver-
tex A, parallel to B C,
draw the straight line
A E, and from C raise
a perpendicular to
meet it in D. B
2. Make D E equal
to D C ; join E B, cutting D C in F.
3. Through F draw F G, parallel to B C, and through H
and G, draw H L, G K, parallel to D C. Then L H G K
will be the square required.
PROBLEM L.
In a given circle to inscribe a polygon of any proposed
number of sides.
1. Divide 360 by the numberof sides
of the figure and make an angle A O B,
at the centre, whose measure shall be
equal to the degrees in the quotient.
2. Then join the points A, B, and
apply the chord A B to the circumfe-
rence the given number of times, and it
will form the polygon required.
PROBLEM LI.
On a given line A B to form a regular polygon of any
proposed number of sides.
1. Divide 360 by the number of sides of the figure, and
subtract the quotient from 180 degrees.
PRACTICAL GEOMETRY.
53
2. Make the angles ABO and BAG
each equal to half the difference last
found ; and from the point of intersec-
tion O, with the distance O A or OB,
describe a circle.
3. Then apply the chord A B to the
circumference the proposed number of
times, and it will form the polygon
required.
PROBLEM LIT.
To describe a circle within or without a regular polygon.
1. Bisect any two angles, A ED,
EDO, by the lines EG, DG, and
from G let fall GF, perpendicular to
the side E D.
2. Then with the radius GE de-
scribe the outer circle, and with the
radius G F describe the inner circle.
PROBLEM LIII.
To make a square that shall be nearly equal to a given
circle.
1. In the given circle A C B D, draw
the two diameters A B, CD, cutting
each other perpendicularly in the
centre O.
2. Bisect the radius O C, in E, and A
/hrough the points A, E, draw the chord
A E F, which will be the side of a
square that is nearly equal in area to
ihe circle.
If F G be drawn parallel to C D, it will be nearly equal
to $ of the circumference of the circle.
5*
64
PRACTICAL GEOMETRY.
PROBLEM LIV.
To find a right line that shall be nearly equal to any given
arc A D B of a circle.
1. Divide the chord AB into four
equal parts, and set one of the parts A C,
on the arc from B to D.
2. Draw C D, and the double of this
line will be nearly equal to the arc A D B.
If a right line be made equal to 3|
times the diameter of a circle, it will
be nearly equal to the circumference.
PROBLEM LV.
To divide a given circle into any proposed number of
parts, that shall be mutually equal to each other, both in area
and perimeter.
1. Divide the diameter A B into the
proposed number of equal parts, at the
points a, b, c, d.
2. On Aa, Aft, Ac, Ad, &c., as
diameters, describe semicircles on one
side of the diameter A B ; and on Br/,
Be, B6, Ba, &c., describe semicircles
on the other side of the diameter.
3. Then the corresponding semicircles, joining each other
as above, will divide the circle in the manner proposed.
PROBLEM LVI.
In a given circle, to describe three equal circles which
shall touch one another, and also the circumference of the
given circle.
From the centre O, let the right-
lines O H, O D. and O E be drawn,
dividing the circumference into
three equal parts in the points H,
D, and E ; join D E, and in O E
produced take E F. equal to on**
PRACTICAL GF.OMETHY.
55
half of DE; draw DF, and parallel thereto draw E A meeting
O D in A ; make H C and E 1? each equal D A, and upon
the centres A, B and C, through the points D, H, and E, let
the circles A r D, C m H and B n E be described.
PROBLEM LVII.
To divide a given circle into any number of equal parts
by means of concentric circles.
Let it be required to divide the circle A K L into three
equal parts.
Divide the radius A B into three
equal parts ; and from the points of
section c, d, draw the perpendiculars
a b, d c, meeting the circumference of A|
a semicircle described on A B, in b
and c ; and join B c, B b. Then, if
circles be described from B, as a cen-
tre, with the radii B c, B b, the circle L
A K L will be divided into three equal parts, as required.
PROBLEM LVIII.
To describe the circumference of a circle through two given
points, A, B, which shall touch a right line, C D, whose
position is given.
1. Draw the right line A B,
joining the two given points,
which bisect in E, by the perpen-
dicular E F, meeting the given
line C D in F.
2. Join B F, and from any
point, H, in F E, let fall the per-
pendicular H G ; and having made H I equal to H G, draw
B K, parallel to I H, meeting F E in K.
3. Then, if a circle be described from the point K, as a
centre, with the radius K B, it will pass through the points
A and B. and touch the line C D, as was required.
56
PRACTICAL GEOMETRY.
PROBLEM LIX.
Upon a given line, A B, to describe an oval, or a figure
resembling an ellipse.
1. Divide A B into three equal parts,
A C, C D, D B ; and from the points
C, D, with the radii C A, D B, describe A
che circles A G D E, and C H B F.
2. Through the intersections m, n,
and centres C, D, draw the lines m H,
n E ; and from the points n, m, with the radii n E, m H, de-
scribe the arcs E F, H G, and A G H B F E will be the
oval required.
Another method.
c
1. On A B, as a common base,
describe the two equal isosceles
triangles A C B, A D B, produc-
ing their sides C A, C B, D A, and
DB.
2. From D and C, as centres,
with any convenient radius, de-
scribe the arcs E G, F H ; and
from A and B, with the radius A E,
B G, describe the arcs G H, E F, and the figure E F H G
will be the oval required.
PROBLEM LX.
To divide a given straight line A B in extreme and mean
ratio ; that is so that the whole line shall be to the greater
part as the greater part is to the less.
1. From A, draw AC at right
angles to A B, and make A C
= A B ; produce B A to D,
making A D = half A B ; and
about D, at the distance D C,
describe the arc C E. Then
A B is divided in extreme and
mean ratio at the point E; and
AB AE::AE:EB.
MENSURATION OF SUPERFICIES.
4- THE area of any figure is the measure of its surface,
or of the space contained within the bounds of that surface,
without any regard to the thick-
ness. A square whose side is 1
inch, 1 foot, or 1 yard, &c., is
called the measuring unit ; and the
area, or content of any figure, is
estimated by the number of squares
of this kind that are contained in it, A
as in the rectangle, ABCD.
THE SQUARE.
PROBLEM I.
5- To find the area of a square.
RULE. Multiply the side by itself, and the product will
be the area.
EXAMPLES.
1. What is the area of the square
ABCD, whose side is 21 inches ?
Here, 21 x 21 = 441 inches : the area
required.
2. What is the area of a square whose
side is 4 ft. 2 in. ? Jlns. 17 ft. 4 in. 4".
67
58 MENSURATION OF SUPERFICIES.
3. What is the area of a square field whose side is 50
perches ? Ans. 15 a. 2 r. 20 p.
4- What te the area of a square meadow whose side is
35.25 chains? Ans. 124 a. 1 r. 1 p.
5. How many yards are contained in a square whose side
is 30 feet? Ana. 144yds.
(3. How many men can stand on acres of land, each
occupying a space of 3 feet square ? Ans. 29040 men.
PROBLEM II.
The area of a square being given, to find the length of the
side.
RULE. Extract the square root of the area.
EXAMPLES.
1 . The area of a square is 2025 feet ; what is the side ?
Here </ 2025 = 45 ft. the side required.
2. What is the side of a square floor containing 729 square
feet ? tins. 27 ft.
3. What is the side of a square whose area is 8 acres,
roods, 16 perches ? Jins. 36 p.
4. What is the side of a square field whose area is 7 acres ?
tins. 8.3666 chains.
5. Required the side of a square floor containing 1734
square feet ? Am. 41.6413 ft.
6. Required the side of a square meadow whose area is
6 acres, 2 roods, 14 perches 1 Ans. 32.4653 p.
PROBLEM HI.
The diagonal of a square Hing given, to find the area.
RULE. Divide the square of the diagonal by 2, and the
quotient will be the area.
MENSURATION OF SUPERFICIES. 59
EXAMPLES.
1. Thediagonalofthesquare,ABCD, D
is 8 chains ; what is the area ?
Here (8x8)-*-2=64-*-2=32 square
chains, then 32-=- 10=3 a. r.32 p.
2. The diagonal of a square is 12
yards ; what is the area ? Jins. 72 yds. A
3. The diagonal of a square is 34 perches ; what is the
area ? Jlns. 3 a. 2 r. 18 p.
4. The diagonal of a square is 16 chains ; what is the
area ? Ans. 12 a. 3 r. 8 p.
5. The diagonal of a square is 324 feet ; what is the area ?
tins. 52488 ft.
C. How many acres are contained in a square field whose
diagonal is 29 chains ? Ans. 42 a. r. 8 p.
PROBLEM IV.
The area of a square being given, to find the diagonal.
RULE. Extract the square root of double the area.
EXAMPLES.
1. The area of a square piece of land is 64 acres, 3 roods,
8 perches ; what is the diagonal ?
Here 64 a. 3 r. 8 p. = 10368 perches; then </ (10368
X2)== v/ 20736= 144 perches, the diagonal.
2. The area of a square is 578 feet ; what is the diago-
nal ? Am. 34 ft.
3. The area of a square is 128 yards ; what is the diago-
nal ? Ans. 16 yds.
4. The area of a square field ia 28.8 acres ; what is the
diagonal in chains ? Ans. 24 chs.
60 MENSURATION OF SUPERFICIES.
6. The area of a square meadow is 16.2 acres ; what is
the diagonal in chains 1 Jins. 18 chains.
6. The area of a square is 4 acres and 8 perches ; what is
the diagonal ? Jlns. 36 perches.
PROBLEM V.
The diagonal of a square being given, to find the side.
RULE. Extract the square root of half the square of the
diagonal.
EXAMPLES.
1. The diagonal of a square is 24 yards ; what is the side ?
HeKV[(24 X 24) -r-2] = <v/(576-r-2) = v/288= 16.9705 yds.
2. The diagonal of a square is 18 feet; what is the side?
Jlns. 12.7279 feet.
3. The diagonal of a square is 36 chains ; what is the
side ? Jlns. 25.4558 chains.
4. What is the side of a square whose diagonal is 48
perches ? Jins. 33.941 1 perches.
6. What is the side of a square whose diagonal is 12 feet ?
Jlns. 8.4852 feet.
6. What is the side of a square piece of land whose
diagonal is 58 perches 1 Jlns. 41.0121 perches.
PROBLEM VL
To cut off" a given area from a square, parallel to either
side.
RULE. Divide the given area by the length of the side,
the quotient will be the length of the other side to be cut off.
MENSURATION OF SUPERFICIES.
61
EXAMPLES.
1. What length must be cut from the
square A B C D, whose sides are 25 D
chains, to have an area B C E F, of 40
acres, at the end ?
Here 400 ch.-j- 25 = 16 chains, length
required.
2. The sides of a square are 17 feet ;
what must be the length of another side A
to give an area of 153 square feet ?
Jlns. 9 feet.
3. The side of a square being 40 rods, what must be the
length of the other side, to give 4 acres ?
Jlns. 18 rods.
4. The sides of a square are 15 chains ; what must be the
length of the other side to have an area of 7 acres ?
Jlns. 5 chains.
5. What length must be cut off from a square field whose
sides are 125 perches to have an area of 50 acres ?
Jlns. 64 perches.
6. The area of a square piece of land is 29 acres, 3 r. 1 p.;
what length must be cut off from the same to give an area
of 10 acres and .8 perches ? Jlns. 23.2 perches.
THE RECTANGLE.
PROBLEM I.
G. The length and breadth of a rectangle being given
w find the area.
RULE. -.Multiply the length by the breadth, and the
product will be the area.
6
62 MENSURATION OF SUPERFICIES.
EXAMPLES.
1. What is the area of the rect-
angle ABCD, whose length AB D
is 14.5 feet, and breadth B C 11.6
feet ?
Here, 14.5x11.6=168.2 feet,
area required.
2. What is the area of a rectangle
whose length is 14 feet 6 inches, and breadth 4 feet 9
inches ? Ans. 68 ft. 10 in.
3. How many acres are contained in a rectangular piece
of land whose sides are 46 and 58 chains ?
Ans. 266 acres, 3 r. 8 p.
4. How many square feet are contained in 28 boards, each
18 feet long and 16 inches wide ? tins. 672 feet.
5. How many squares of 100 feet each are contained in
a floor 48 feet long and 21 feet wide ?
Ans. 10.08 squares.
6. What is the area of a rectangular piece of land whose
length is 204.7 and breadth 117.8 yards ?
Ans. 241 13.66 yards.
PROBLEM II.
The area and either side of a rectangle being given, to find
the other side.
RULE. Divide the area by the given side, and the
quotient will be the other side.
EXAMPLES.
1. The area of a rectangle is 456 feet, and the length
30 feet ; what is the breadth ?
Here 456 -j- 30 = 15.2 feet, the breadth.
2. The area of a rectangle is 846 chains, and its length
42 chains ; what is the breadth ? Ans. 20j chains.
MENSURATION OF SUPERFICIES. 63
3. The area of a rectangular piece of land is 6 a. 2 r. 16 p.,
and breadth 29^ p.; what is the length ?
Jlns. 36 perches.
4. The area of a rectangle is 392 yards, and the shortest
side 12 yards ; what is the longest side ?
Jlns. 32f yards.
5. The area of a rectangular meadow is 73 acres, 3 roods
20 perches, and the length 120 perches ; what is the breadth ?
Jlns. 98s p.
6. The area of a rectangle is 1728 feet, and the breadth
36 feet ; what is the length ? Jlns. 48 ft.
PROBLEM III.
The area and the proportion of the two sides of a rectangle
being given, to find the sides.
RULE. Multiply the area by the greater number of the
proportion, and divide the product by the less ; the square
root of the quotient will be the length : then multiply the
length by the less number of the proportion, and divide the
product by the greater ; the quotient will give the breadth.
EXAMPLES.
1. The area of a rectangular piece of land is 432 acres,
and the length is to the breadth as 5 to 3 ; what are the
sides ?
Here 432 acres = 69120 perches.
Then^/ [(69120x5)-r-3]=v/ (345600-4-3) = v/ 115200
= 339.41125 perches, the length ;
And (339.41 125x3)^-5=203.64675 perches, the breadth.
2. The area of a rectangle is 1472 yards, and the breadth
is to the length as 3 to 4 ; required the sides.
Jlns. 33.2264, and 44.3019 yds.
3. The area of a rectangle is 24 acres, and the length is
to the breadth as 3 to 2 ; what are the sides ?
Jlns. 18.9736, and 12.6491 chains.
4. The area of a rectangle is 27 acres, 3 roods, 20 perches,
and the length is to the breadth as 9 to 7 ; required the sidex
Jlns. 75.725, and 58.897 p.
64 MENSURATION OF SUPERFICIES.
6. The area of a rectangle is 28 acres, and the breadth i
to the length as 4 to 7 ; required the sides.
Ans. 12.6491, and 22.1359 chains.
6. It is required to lay out 1 acre in a rectangular form,
having the length to the breadth in the ratio of 8 to 5.
tfns. 88, and 55 yds.
PROBLEM IV.
The sides of a rectangle being given, to cut off a given area
parallel to either side,
RULE. Divide the area by the side which is to retain its
length or breadth, and the quotient will be the length 01
breadth of the other side.
EXAMPLES.
1. The sides of the rectangle,
A B C D, are 18.16, and 12.15
chains ; what must be the length
to leave an area, BCEF,of 12
acres adjoining the breadth ?
Here 120 chs. -4- 12.15 =
9.8765 chains, the length.
E
F
2. The sides of a rectangle are 24 and 1 6 chains ; what
must be the breadth to leave 18 acres adjoining the length ?
Ans. 7s chs.
3. The sides of a rectangle are 216 and 124 feet; what
must be the length to leave 10106 square feet adjoining the
breadth? tins. 81 ft.
4. The sides of a rectangle are 180 and 75 perches ; what
must be the breadth so as to leave 22 acres adjoining the
length ? Ans. 20 p.
5. The sides of a rectangle are 8 and 15 yards ; what
must be the length to leave 46 square yards at the shortest
side ? Am. 5| yds.
6. The length of a rectangle is 14.5 chains, and the breadth
6.4 chains ; what must be the breadth, the length being the
same, to contain 5.8 acres ? Am. 4 chs.
MENSURATION OF SUPERFICIES.
65
. THE RHOMBUS.
PROBLEM I.
7. To find the area of a rhombus.
RULE. Multiply the length by the perpendicular height,
and the product will be the area.
EXAMPLES.
1. The length of a rhombus, A B,
is 185 feet, and the perpendicular
height, D E, 7f feet ; required the
area.
Here, 18.5x7.75=143.375 feet.
2. Required the area of a rhombus
whose length is 14 feet 4 inches, and
its height 12 feet 2 inches.
rfns. 174 ft. 4 in. 8".
3. What is the area of a rhombus whose length is 12 feet
3 inches, and height 9 feet 4 inches ? Ans. 114 ft. 4 in,
4. Required the area of a rhombus whose length is 38
perches, and height 17 perches. Ans. 4 a. r. 6 p.
5. What is the area of a rhombus whose length is 19
chains, and height 15 chains ? Jlns. 28 a. 2 r.
6. Required the area of a rhombus whose length is 27
yards, and height 21.5 yards. Ans. 580.5 yds.
G
66 MENSURATION OF SUPERFICIES.
THE RHOMBOID.
PROBLEM I.
8. To find the area of a rhomboid.
RULE. Multiply the length by the perpendicular height,
and the product will be the area.
EXAMPLES.
1. The length of a rhomboid, A B,
is 28 perches and height, D E, 12 ;
required the area.
Here 28 x 12=336 perches =2 a.
r. 16 p.
2. The length of a rhomboid is
22 feet 9 inches, and height 14 feet
3 inches ; how many square yards does it contain ?
Ans. 36.0208 yds.
3. The length of a rhomboid is 16.2 yards, and height
9.6 yards ; how many square perches does it contain ?
jlns. 5.1411 sq. p.
4. The length of a rhomboid is 21 chains, and the height
11.5 chains ; what is the area ? Jlns. 24 a. r. 24 p.
5. How many acres are contained in a rhomboid whose
length is 130 perches and height 57 perches?
Jlns. 46 a. 1 r. 10 p.
6. How many square yards are contained in a rhomboid
whose length is 271 feet and height 107 feet?
Ans. 3221 1 sq. yds.
PROBLEM II.
The area of a rhombus or rhomboid, and the length of the
side being given, to find the perpendicular height ; or the
area and the height being given, to find the length of the
side.
ROLE. Divide the area bv the length of the side, and the
quotient will be the perpenaieular height ; or divide by the
height, and the quotient will be the length of the side.
MENSURATION OF SUPERFICIES. G7
EXAMPLES.
1 . The area of a rhombus is 27 perches, and the length
of the side 6.75 perches ; what is the perpendicular height ?
Here 27 -f- 6.75 = 4 perches.
2. The area of a rhombus is 4 acres, and the height
5 chains ; what is the length of the side ?
Jlns. 8 chains.
3. Required the length of a rhombus whose area is 17
acres, and height 35 perches. Jlns. 77-y perches.
4. The area of a rhomboid is 4 acres, 3 roods, 18 perches,
and the length of the side 38 perches ; what is the height ?
Am. 20 T 9 g- perches.
5. The area of a rhomboid is 1776 square feet, and the
height 24 feet ; what is the length ? Ans. 74 feet.
6. The area of a rhomboid is 36 acres, and the length of
the sides 24 chains ; what is the height ?
Ans. 15 chains.
THE TRIANGLE.
PROBLEM I.
9. To find the area of a triangle, when the base and
perpendicular height are given.
RULE. Multiply the base by the perpendicular height,
and half the product will be the area.
EXAMPLES.
1. What is the area of a triangle
ABC, whose base, AB, is 46 feet,
and height, DC, 23 feet?
Here (46 x 23) -4-2 = 1058-j-2=
529 feet, the area.
2. What is the area of a triangle
whose base is 67 yards, and height
is 14.5 yards ? Ans. 485.75 yds.
68 MENSURATION OF SUPERFICIES.
3. What is the area of a triangle whose base is 207.5 feet,
and height is 59.5 feet ? dns. 6173.125 feet.
4. The base of a triangle is 72 perches, and the height 12|
perches ; how many acres does it contain ?
Ans. 2 a. 3 r. 10 p.
5. What is the area of a triangle whose base is 12.25
chains, and the height 8.5 chains 1 rfns. 5 a. r. 33 p.
6. What is the area of a triangular field whose base is 24 2
chains, and the height 18 chains 1 Ans. 22 a. r. 8 p.
PROBLEM II.
The three sides of a triangle being given, to find the area.
RULE. From half the sum of the three sides, subtract
each side severally. Then multiply the half sum, and the
three remainders continually together, and the square root
of the product will be the area required.
EXAMPLES.
1. What is the area of a triangle
ABC, whose three sides, B C, C A,
AB, are 23.7, 29.25 and 40.1
yards ?
Here (23.7 + 29.25 + 40.1) -f- 2
= 93.05 -5- 2 = 46.525 = half the
sum of the sides. A B
46.525 23.7 = 22.825 = first diffejence.
46.525 29.25 = 17.275 = second difference.
46.525 40.1 = 6.425 = third difference.
Whence ^ (46.525 x 22.825 x 17.275 x 6.425)
1=^/117865.94866835 = 343.3161 yards, the area.
2. What is the area of a triangular field whose sides are
26, 29 and 30 chains ? Jlns. 33 a. 2 r. 16 p.
3 Required the area of an equilateral triangle whose side
is 22 perches. Jlns. 1 a. 1 r. 9.5781 p.
4. Required the area of an isosceles triangle whose base is
HO. and each of its eaual sides 45 feet ?
rfns. 636.3961 feet.
MENSURATION OF SUPERFICIES. 69
5. What is the area of a triangle whose sides are 22.2, 38,
and 40.1 feet ? tins. 413.7114 feet.
6. What is the area of a triangular field whose sides are
27.35, 31.15, and 38 chains ? Ans. 42 a. r. 6.6955 p.
PROBLEM HI.
Any two sides of a right-angled triangle being given, to
find the third side.
RULE 1st. To the square of the perpendicular add the
square of the base, and the square root of the sum will give
the hypothenuse. x
2d. The square root of the difference of the squares of the
hypothenuse, and either side will give the other.
3d. Or multiply the sum of the hypothenuse and either
side by their difference, and the square root of the product
will give the other.
EXAMPLES.
1. The base of a right-angled tri-
angle, A B, is 27 yards, and the per-
pendicular, B C, 36 yards; what is
the hypothenuse?
Here </ (36 s -f 27 a ) = </ (1296
+ 729) = </ 2025 = 45 yards, the
hypothenuse. A B
2. The hypothenuse of a right-angled triangle is 242 feet,
and the perpendicular 182 feet ; what is the base ?
Here ^ (242 2 182 a ) = ^/ (5856433124) = v / 25440
= 159.4989 feet, the base.
Or, v/ [ (242 -f 182) x (242 182) ] = v/ (424 x 60)
= v/ 25440 = 159.4989 feet, the base as before.
3. The base of a right-angled triangle is 38 chains, and
the perpendicular 41 chains ; required the hypothenuse.
Ans. 55.9016 chs
4. The hypothenuse of a right-angled triangle is 68
perches, and the perpendicular 33 perches ; what is the base ?
Ans. 59.4558 p.
70 MENSURATION OF SUPERFICIES.
5. The hypothenuse of a right-angled triangle is 315 feet,
and the base 299 feet ; required the perpendicular.
Ans. 99. 1160 feet.
6. The top of a May pole, being broken off by a blast of
wind, struck the ground at 9 feet distance from the foot of
the pole ; what was the height of the whole May pole, sup-
posing the length of the broken piece to be 41 feet ?
dns. 81 ft.
7. Two ships sail from the same port, one east 60 miles,
and the other north 80 miles ; how far are they apart ?
tins. 100 m.
8. A line 78 yards long will reach from the top of a fort,
on the opposite bank of a river, to the water edge on this side
of the river; what is the height of the fort, the river being
76 yards across 1 Jlns. 17.5499 yds.
9. A ladder of 100 feet in length was placed against a
building of 100 feet high, in such a manner that the top of
it reached within six inches of the top of the building ; what
was the distance of the foot of the ladder from the base of the
edifice ? Am. 9.9874 ft.
10. A ladder 30 feet long, placed near the middle of a
street, reached the buildings at one side 24 feet from the
ground ; and the opposite side, without moving the foot, 18
feet : what was the breadth of the street ? Jlns. 42 ft.
PROBLEM IV.
The sum of the hypothenuse and perpendicular, and the
base of a right-angled triangle being given, to find the hypo-
thenuse and the perpendicular.
RULE. To the square of the sum add the square of the
base, and divide the number by twice the sum of the hypo-
thenuse and perpendicular, and the quotient will be the
hypothenuse.
Subtract the hypothenuse from the sum of the hypothenuse
and perpendicular, and the remainder will be the perpen-
dicular.
MENSURATION OF SUPERFICIES. 71
EXAMPLES.
1. The sum of the hypothenuse
and perpendicular is 100 feet, and
the base 40 feet ; required the hypo-
thenuse and the perpendicular.
Here (100 3 + 40*) -r- (100 x 2)
= (10000 + 1600) -4- 200 =
1 1000 -T- 200 = 58 feet, the hypo-
thenuse. And 100 58= 42 feet, the perpendicular.
2. The height of a tree, standing perpendicularly on a
plane, is 110 feet. At what height must it break off, so that
the top may rest on the ground 50 feet from the base, and
the place broken on the upright part ? fins. 43 T 7 T ft.
3. The base of a right-angled triangle is 36 chains, and
the sum of the hypothenuse and perpendicular is 84 chains ;
what is the hypothenuse ? jins. 49f ch.
4. A tree 90 feet high, growing perpendicularly on a
plane, was broken off by the wind ; the broken part resting
on the upright, and the top on the ground 30 feet from the
base ; what was the length of the broken part ?
Ans. 50 ft.
5. The sum of the hypothenuse and perpendicular is 240
yards, and the base 80 yards; required the perpendicular.
Jlns. 106f yds.
6. A May pole, whose height was 84 feet, standing on a
horizontal plane, was broken by the wind, and the ex-
tremity of the top part struck the ground at the distance ot
24 feet from the bottom of the pole ; required the length of
each part.
Ans. 45^ feet, the hypoth. and 38^ the perp.
PROBLEM V.
To determine the area of an equilateral triangle.
RULE. Multiply the square of the side by .433013, and
the product will be the area.
72 MENSURATION OF SUPERFICIES.
EXAMPLES.
1. What is the area of an equilate-
ral triangle, ABC, whose side is 20
feet?
Here (20) 2 x .433013 = 400
X .433013 = 173.2052 feet, the
area. A
2. What is the area of an equilateral triangle whose side is
40 feet ? Ans. 692.8208 feet.
3. What is the area of an equilateral triangle whose side is
80 perches ? Ans. 17.3205 acres.
4. What is the area of an equilateral triangle whose side
is 24.4 yards ? Ans. 257.7986 yards.
5. How many acres are contained in an equilateral triangle
whose side is 16 chains ? Ans. 11 a. r. 13.6212 p.
6. How many acres are contained in an equilateral triangle
whose side is 32 perches ? Ans. 2 a. 3 r. 3.4053 p.
PROBLEM VI.
The area and the base of any triangle being given, to find
the perpendicular height.
Or, the area and height being given, to find the base.
KULE. Divide double the area by the base, and the quo-
tient will be the perpendicular height; or divide double the
area by the height, and the quotient will be the base.
EXAMPLES.
1. The area of a triangle, ABC,
is 4 a. 3 r. 32 p., and the base, A B,
24 perches; what is the perpen-
dicular height, DC?
Here 4 a. 3 r. 32 p. = 792 p.
and (792 x 2) -r- 24 = 1584 -r- 24
= 66 perches, the height.
MENSURATION OF SUPERFICIES. 73
2. The area of a triangJe is 806.3125 yards, and the per-
pendicular 33.25 yards ; what is the base ?
(806.3125 x 2) -T- 33.25 = 1612.625 -7- 33.25 = 48.5
yards, the base.
3. The area of a triangle is 102| feet, and the base 20
feet ; what is the perpendicular 1 Ans. 10? feet.
4. The area of a triangle is 5 a. r. 33 p. and the per-
pendicular 28.5 p. ; what is the base ? Ans. 58.4561 perches.
5. If the area of a triangle be 802 jf yards, and its base
17} yards ; what is its perpendicular height ?
Ans. lOOf yards.
6. If the area of a triangular field be 4.39775 acres, and
its perpendicular 7.18 chains ; what is its base ?
Jlns. 12.25 chains.
PROBLEM VII
The proportion of the three sides of a triangle being given,
to find the sides of a triangle corresponding with a given area.
RULE. Find the area of the triangle according to the
given proportion, by Problem II. page 68 ; then as that
area is to the area given, so is the square of either of its
sides to the square of the similar side ; the square root of
which will be the required side. The other sides of the
triangle will be proportional to the corresponding given sides.
EXAMPLES.
1. A person wishes to enclose
6 acres, 1 rood, 12 perches, in a
triangle, similar to a small triangle
whose sides are 6, 8, and 9 perches
respectively ; required the sides of
the triangle.
Here 23.5252 perches = the area A
of the triangle, whose sides are 6, 8, and 9.
6 a. 1 r. 12 p. = 1012 perches.
As 23.5252 : 1012 :: 64 : 2753.1328, and ^ 2753.1328
e= 52.47 perches, one of the sides.
Now as 8 : 9 : : 52.47 : 59.029 p. 7 ,, . ,
"8:6:: 52.47 : 39.353 p. 5 the Other tWO Slde8 '
7
74 MENSURATION OF SUPERFluiES.
2. The area of a triangle is 24 acres ; what must be the
length of the sides, in the proportion 3, 4 and 5 chains ?
dns. 18.973, 25.298, and 31.622 chains.
3. The area of a triangle is 10 acres ; what must be the
length of the sides, in the proportion of 5, 11 and 13 chains ?
rfns. 9.64, 21.208 and 25.064 chains.
4. What are the sides of a triangle containing one acre,
in the proportion of 3, 4 and 6 chains?
tfns. 4.1082, 5.4776 and 8.2164 chains.
5. What are the sides of a triangle containing 33.6 acres,
in the proportion of 13, 14 and 15 chains ?
Ans. 26, 28 and 30 chains.
6. What are the sides of a triangle containing 24 acres, in
the proportion of 5, 12 and 13 chains ?
dns. 14.1421, 33.9411 and 36.7695 chains.
PROBLEM VIIL
The base and perpendicular of any plane triangle being
given, to find the side of its inscribed square.
RULE. Divide the product of the base and perpendicular
by their sum, and the quotient will be the side of the in-
scribed square
EXAMPLES.
1. The base A B of a triangle is
12 feet^and the perpendicular, C D,
18 feet; what is the side, EF, of
the inscribed square ?
Here (12 x 18) -r- (12 -f- 18)
= 216 -T- 30 = 7.2 feet, = EF,
the side of the inscribed square
2. The base of an isosceles triangle is 24 yards, and the
perpendicular 16 yards ; what is the side of the inscribed
square ? rfns. 9.6 yards.
3. The hypothenuse of a right-angled triangle is 16
perches, and the base 14 perches ; what is the side of the
inscribed square ? *4ns. 4.9808 p.
MENSURATION OF SUPERFICIES. 75
4. The base of a scalene triangle is 20 chains, and the
perpendicular 15 chains; what is the side of the inscribed
square ? Am. 8$ chs.
ft. The area of a scalene triangle is 32 acres, and the base
25 chains ; required the side of the inscribed square.
Ana. 12.6482 chs.
6, The area of an isosceles triangle is 324 feet, and the
perpendicular 18 feet ; required the side of the inscribed
squa/e. Jins. 12 ft.
PROBLEM IX.
Tiie three sides of any triangle being given, to find the
lengih of a perpendicular which will divide it into two right-
angled triangles.
RULE. Upon the base, let fall a perpendicular from the
opposite angle ; this perpendicular will divide the base into
two parts called segments, and the whole triangle into two
right-angled triangles.
Then, as the base or sum of the segments is to the sum of
the other two sides, so is the difference of these sides to the
difference of the segments of the base.
To half the base add half the difference of the segments,
and the sum will be the greater segment ; also from half the
base subtract half the difference of the segments, and the re-
mainder will be the less segment.
When the perpendicular falls without the triangle, the
base is to the sum of the sides as the difference of the sidos
is to the sum of the segments of the base.
Then in each of the two right-angled triangles, ihere will
be known the hypothenuse and base ; consequently, the per-
pendicular vv.ay be found by Problem HI., page 69.
EXAMPLES.
1. In tho triangle, ABC, are
riven, A B 2's A C 18, and B C 14
yards ; required the perpendicular,
C D.
Here, As 20: (18+14) : : (18
14) : 6.4, the difference of the seg-
ments of the ba&e : then, 6.4 -f- 2 =
3.2 yards, half tieir difference, and
76 MENSURATION OF SUPERFICIES.
20 -r- 2 = 10 yards, half the base ; now 10 + 3.2 = 13.3
yards, the segment A D, and 10 3.2 = 6.8 yards, the seg-
ment B D. Therefore, v/ (B C 3 B D 2 ) = ^/ ( 14 3 6.8 a )
= v' (196. 46.24) = v/ 149.76 = 12.2376 yards, the
perpendicular, D C.
2. The base of a triangle is 426 feet, and the other two
sides 365 .and 230 feet ; required the length of the perpen-
dicular, rfns. 196.9904ft.
3. The base of a triangle is 80 chains, and the other two
sides 60 and 40 chains ; what is the length of the perpen-
dicular which divides it into two right-angled triangles ?
Am. 29.0473 chs.
4. The base of a triangle is 128 perches, and the other
two sides 94 and 68 perches ; required the length of the
perpendicular. Jlns. 48.6137 p.
5. The base of a triangle is 324 yards, and the other two
sides 264 and 162 yards ; required the length of the perpen-
dicular. Ans. 131.2613yds.
6. The base of a triangle is 30 perches, and the other two
sides 24 and 18 perches ; required the length of the perpen-
dicular. Jlns. 14.4 p.
PROBLEM X.
The area and base of a triangle being given, to cut off a
given part of the area by a line running from the angle oppo-
site the base.
RULE. As the given area of the triangle is to the area of
the part to be cut off, so is the given base to the base corre-
sponding to that area.
EXAMPLES.
1. Given the area of a triangle,
A B C, 12 acres, and the length of
the base, A B, 24 chains ; it is re-
quired to cut off 5 acres towards the
angle A, by a line running from the
angle C to the base.
Here 12 acres = 120 chains, and
6 acres = 50 chains.
MENSURATION OF SUPERFICIES. 77
Then, as A B C (120) : A D C (50) : : A B (24) : A D
(10 chains.)
2. In the triangle ABC, there are given the area 54
acres, 2 roods, 30 perches, and the base, A B, 70 perches,
to cut off 20 acres towards the angle B, by a line, C D. run-
ning from the angle C to the base ; required the part B D of
the base. . Jlns. 25.6 p.
3. In the triangle ABC, there are given the area 7
acres, and the base, A B, 8 chains, to cut off lg acres towards
the angle A, by a line, C D, running from the angle C to the
base ; required the part A D of the base. Jlns. 2 chs.
PROBLEM XI.
The area and base of a triangle being given, to cut off a
triangle containing a given area, by a line running parallel
to one of its sides.
RULE. As the given area of the triangle is to the area
of the triangle to be cut off, so is the square of the given base
to the square of the required base. The square root of the
result will be the base of the required triangle.
EXAMPLES.
1. Given the area of the triangle
ABC, 250 square chains, and the
base, A B, 20 chains ; it is required
to cut off (50 square chains towards
the angle A, by a line, D E, running
parallel to B C.
As A B C (250) : A D E (60) : :
A B 2 (400) : A D a (96). And A D
= v' A D* (96) = 9.7979 chains.
2. Given the area of a triangle, A B C, 20 acres, and the
base, A B, 50 chains, to find D B, a part of the base, so that
a line, D E, running from the point D, parallel to the side.
A C, may cut off a triangle, B D E, containing 9 acres.
. B D = 33.541 chs.
3. Given the area of a triangle, A B C, 5 acres, and the
7*
78 MENSURATION OF SUPLRFICIES.
base, A B, 12| perches, to find A D, a part of the base, so
that a line, D E, running from the point D, parallel to the
side B C, may cut off a triangle, A D E, containing 2| acres.
Ana. A D = 8.3852 p.
PROBLEM XII.
The area and two sides of a triangle being given, to cut
off a triangle containing a given area, by a line running from
a given point in one of the given sides, and falling on the
other.
RULE. As the given area of a triangle is to the area of
the part to be cut off, so is the rectangle of the given sides to
a fourth term.
Divide this fourth term by the distance of the given point
from the angular point of the two given sides ; the quotient
will be the distance of the required point from the same
angle.
EXAMPLES.
1. Given the area of a triangle,
A B C, 2| acres ; the side A B, 25
perches ; the side A C, 20 perches ;
and the distance of a point D, from
the angle A, 18 perches ; it is re-
quired to find a point, E, to which,
if a line be drawn from the point D, A
it shall cut off a triangle, A D E,
containing 1 acre, 2 roods, 10 perches.
Here, as the area of the triangle ABC, 400 sq. p. : the
area of the triangle A D E, 250 sq. p. : : A B x A C (25 x
20) : A D x A E (312.5). Then A D x A E (312.5) -f-
A D (18) = A E = 312.5 -4- 18 = 17.36 per.
2. Given the area of a triangle, ABC, 24.7875 acres ;
the side A B, 40 chains ; the side A C, 32.5 chains ; and
the distance of a point E, in the side A B, from the angle A.
17 chains ; it is required to find the distance A D, in the
line A C, so that a line drawn from E to D may cut off &
triangle A E D, containing 6 acres. Jlns. 18.51 chs.
MENSURATION OF SUPERFICIES. 7?
THE TRAPEZIUM.
PROBLEM L
1O. To find the area of a trapezium.
RULE. Multiply the sum of the two perpendiculars by
the diagonal upon which they fall, from the opposite angles
and half the product will be the area.
EXAMPLES.
1. Required the area of the trape-
zium, A C B D, whose diagonal, A B,
is 84 yards, the perpendicular C E, A
28 yards, and the perpendicular D F,
21 yards.
Here [(28 + 21) x 84] -s- 2 =
(49 x 84) -=- 2 = 4116 -T- 2 = 2058 yards, the area of the
trapezium A C B D.
2. Required the area of a trapezium whose diagonal is 33
perches, and the perpendiculars 11 perches and 13 perches.
Am. 2 a. 1 r. 36 p.
3. How many acres are there in the trapezium whose
diagonal is 80.5 chains, and the perpendiculars 22.4 and
28.3 chains ? Am. 204 a. r. 10.8 p.
4. What is the area of a trapezium whose diagonal is 108
feet 6 inches, and the perpendiculars 56 feet 3 inches and
60 feet 9 inches ? Arts. 6347.25 ft.
5. How many square yards of paving are there in a trape-
zium whose diagonal is 65 feet, and the two perpendiculars
let fall on it, from its opposite angles, 28 and 33 feet re-
spectively ? Am. 222.083 yds.
6. How many acres are there in the trapezium whose
diagonal is 4.75 chains, and the two perpendiculars falling
on it, from its opposite angles, 2.25 and 3.6 chains respect-
ively ? Am. 1 a. 1 r. 22.3 p.
80 MENSURATION OF SUPERFICIES.
THE TRAPEZOID.
PROBLEM I.
11. To find the area of a trapezoid.
KncE. Multiply the sum of the two parallel sides by the
perpendicular distance between them, and half the product
will be the area.
EXAMPLES.
1. Required the area of the trape-
zoid, A B C D, whose parallel sides,
D C and A B, are 14 feet 6 inches
and 24 feet 9 inches, and the perpen-
dicular distance, D E, 8 feet 3 inches.
Here [(A B + D C) x D E] -f- 2
= [(24.75 + 14.5) x 8.25] -f- 2 =
(39.25 x 8.25) -*- 2 = 323.8125 -r- 2 == 161.90625 feet, the
area.
2. How many square feet are there in a plank 1 foot 6
inches broad at one end, and 1 foot 3 inches at the other, the
length being 20 feet ? Am. 27| ft.
3. Required the area of a trapezoid whose parallel sides are
24.46 chains, and 38.4 chains, and the perpendicular dis-
tance 16.2 chains. Am. 50 a. 3 r. 20.6 p.
4. Required the area of a trapezoid whose two parallel
sides are 25 feet 6 inches, and 18 feet 9 inches, and the per-
pendicular distance between them 10 feet 5 inches.
Jlns. 230i| ft.
5. The two parallel side's of a trapezoid arb 12.41 and
8.22 chains, respectively, and the perpendicular distance
between them 5.15'chains ; required the area.
Ana. 5 a. 1 r. 9.956 p.
6. Required the area of a trapezoid whose two parallel
sides are 750 and 1225 links, and the perpendicular distance
between them 1540 links. Ans. 15 a. r. 33.2 p
MENSURATION OF SUPERFICIES 81
POLYGONS.
PROBLEM I.
12. To find the area of a regular polygon.
RULE. Multiply the perimeter, or sum of all the sides of
the figure, by the perpendicular falling from its centre upon
one of the sides, and half the product will be the area.
EXAMPLES.
1. Required the area of the regular
pentagon, ABODE, one of whose
equal sides, AB, or BC, &., is 25
yards, and the perpendicular O P from
its centre, 17.2 yards.
Here (25 x 5 x 17.2) -j- 3 2150
-f- 2 = 1075 yards, the area.
2. The side of a regular hexaron is 15 feet, and the per-
pendicular 13 feet ; what is the area ? Ans. 585 feet.
3. Required the area of a regular hexagon whose side is
14.6 yards, and perpendicular 12.04 yards ?
Ans. 553.632 yds.
4. How many acres are contained in a regular heptagon
whose sides are each 19.38 chains, &.nc' wrpendicular from
the centre 20 chains ? Am. 135.66 acres.
5. The side of a regular pentagon is TO yards, and the
perpendicular is 6.882 yards ; what is the aroa ?
tins. 172.% yards.
6. What is the area of a regular octagon whose sides are
each 19.882 feet, and the perpendicular from the crctre 24
feet ? Am. 1908.672 fe'
MENSURATION OF SUPERKICIE?.
PROBLEM II.
To find the area of a regular polygon when one of its
equal sides only is given.
RULE. Multiply the square of the side of the polygon,
by the number standing opposite to its name in the following
table, and the product will be the area.
Table, when the side of the polygon is 1.
No. of
sides.
Names.
Areas, or
Multipliers.
Radius of
inscribed circle
3
Trigon or equil. A
0.433013
0.288675
4
Tetragon or square
1.000000
0.500000
5
Pentagon
1.720477
0.688191
6
Hexagon
2.598076
0.866025
7
Heptagon
3.633912
1.038262
8
Octagon
4.828427
1.207107
9
Nonagon
6.181824
1.373739
10
Decagon
7.694209
1.538842
11
Hendecagon
9.365640
1.702844
12
Duodecagon
11.196152
1.866025
EXAMPLES.
1. The side of a regular pentagon is 12 feet ; what is the
area?
Here 12 a x 1.720477 = 144 x 1.720177 = 247.748688
feet, the area.
2. What is the area of a regular octagon, the side being
20 feet? rfns. 1931.3708 ft.
3. The side of a regular hexagon is 21 feet ; what is its
area? Ans. 1496.4917 ft.
4. What is the area of a regular heptagon whose side is
16 yards ? Ans. 930.2814 yds.
5. What is the area of a regular nonagon whose side is
36 inches ? Ans. 801 1.6439 in.
6. fik>w many pieces, each 4 inches square, may be cut
from {^fegular decagon whose side is 12 inches ?
fins. 69.2178 pieces.
MENSURATION OF SUPERFICIES. 83
PROBLEM HI.
When the area of any regular polygon is given, to find
the side.
RULE. Divide the area by the number in the table cor-
responding with the figure, and the square root of the
quotient will be the length of the side.
EXAMPLES.
1. The area of a regular pentagon is 4 a res ; how many
perches are contained in the side ?
Here 160 x 4 = 640 perches,
Then V (640 -j- 1.720477) = ,/ 371.9898 = 19.2870
perches, the length of the side.
2. Required the length of the side of a regular hexagon
containing one acre. rfns. 7.8475 perches.
3. The area of an octagonal floor is 560 feet ; what is the
length of the side ? Ans. 10.7693 feet.
4. The area of a regular hexagon is 73.9 feet ; what is
the side ? Jins. 5i feet.
5. The area of a regular heptagon is 1356.6 yards; what
is the length of the side ? Jlns. 19.3214 yards.
6. The area of a regular decagon is 3233.4912 feet ; what
is the length of the side ? rfns. 20$ fret.
IRREGULAR FIGURES.
PROBLEM I.
13. To find the area of an irregular right-lined figure
of any number of sides.
RULE 1. Divide the figure into triangles and trapeziums,
and find the areas of each of them separately by Prob. I. p. 67,
and Prob. I. p. 79 ; then add these areas together, and their
sum will give the area of the whole figure.
84
MENSURATION OF SUPERFICIES.
EXAMPLES.
1 . Reqi lired the area of the i rregular
right-lined figure, A B C D E F, the
dimensions of which are as follows :
F B = 20.75,F C = 27.48. E C =
18.5, B n = 14.25, E m = 9.35,
D r = 12.8, and A s = 8.6 perches
respectively.
Here (F B x A *) -=- 2 = (20.75 A
X 8.6) -f- 2 = 178.45 -~ 2 = 89.225 perches, the area of the
triangle, A B F.
And (E C x D r) -r- 2 = (18.5 x 12.8) ~- 2 = 236.8 -r-
2=1 18.4 perches, the area of the triangle, DEC.
Also, [(Bra -f Em) x F C] -r- 2 = [(14.25 -f 9.35) x
27.48] -7- 2 = (23.6 x 27.48) -r- 2= 648.528 -~ 2 = 324.264
perches, the area of the trapezium, F B C E.
Whence 324.264 -f 89.225 -f 118.4 = 531.889 perches
= 3 a. 1 r. 11.889 p., the area of the whole figure.
2. Required the area of an irregular hexagon, like that in
the last example, supposing the dimensions of the different
lines to be the halves of those before given.
Ans. 3 r. 12.9722 p.
RULE 2. The area of any irregular right-lined figure may
also be determined, by drawing perpendiculars from all its
angles, to one of its diagonals, considered as a base ; and
then adding the areas of all the triangles and trapezoids to-
gether for the content.
EXAMPLES.
1. Required the area of the irregular right-lined figure,
A B C D E F, the dimensions of which are as follows :
E n <= 4.54, E m = 8.26, E r =20.01, E s = 26.22, E B
= 30.15, D m = 10.56, C r = 12.24, F n = 8.56, and A s
= 9.26 chains respectively.
Here m r = E r E m = 20.01
8.26 =11.75, n* = Es En==
26.22 4.54 = 21 .68, B r = E B
E r -= 30.15 20.01 = 10.14, and
B*=EB E*= 30.15 26.22
<= 3.93.
MENSURATION OF SUPERFICIES. 85
Whence (F n x E n) -4-2= (8.56 x 4.54) -s- 2 = 38.8624
-i- 2 = 19.4312 chains, area of the triangle E F n.
And ^D m x E m) -r- 2 = (10.56 x 8.26) -4- 2 = 87.2256
.4- 2 =43.6 128 chains, area of the triangle, E D m.
Also (C r x B r) -4- 2 = (12.24 x 10. 14) -- 2 = 124.1 136
-i. 2 = 62.0568 chains, area of the triangle C B r.
And (A x B s] -4- 2 = (9.26 x 3.93) -5- 2 = 36.3918 -r-
2 = 18.1959 chains, area of the triangle A B s.
Then [(D ra + C r) x m r] -f- 2 = [(10.56 + 12.24) x
11.75] -4- 2= (22.8 x 11.75) -r- 2 = 267.9 -4-2= 133.95
chains, area of the trapezoid D C r m.
And [(F n + A s) x n s] -=- 2 = [(8.56 -f 9.26) x 21.68]
~ 2 = (17.82 x 21.68) -=- 2 = 386.3376 -4- 2 = 193.1688
chains, area of the trapezoid F A s n.
Hence 19.4312 + 43.6128 + 62.0568 + 18.1959 + 133.95
+ 193.1688= 470.4155 chains = 47 a. r. 6.648 p., the
area of the whole figure.
2. Required the area of an irregular figure, like that in the
last example, only doubling the dimensions of the diagonal,
and the several perpendiculars, rfns. 188 a. Or. 26.592 p.
PROBLEM II.
To find the area of a mixtilineal figure, or one formed by
right lines and curves.
RULE 1. Take the perpendicular breadths of the figure
in several places, at equal distances from each other, and
divide their sum by their number, for the mean breadth ; and
this quotient, multiplied by the length, will give nearly the
true area of the figure.
2d. Or, if greater accuracy be required, take half the sum
of the two extreme breadths, for one of the said breadths,
and add it to the others, as before ; then divide this sum
by the number of parts in the base (instead of by the number
of breadths), and the result multiplied by the length will
give the area, with a sufficient degree of correctness to
answer most questions of this kind that can occur. When
the curved, or mixti'ineal, boundary meets the base, as is
8
86
MENSURATION OF SUPERFICIES.
frequently the case in surveying, the area is found by
dividing the sum of all the breadths by the number of parts
in the base, and then multiplying the result by the length,
as before ; observing, in each of these cases, that the greater
the number of parts into which the base is divided, the
nearer will the approximation be to the exact area.
It may likewise be further remarked, that if the perpen-
diculars, or breadths, be not at equal distances from each
other, the parts should be computed separately, as so nutny
trapezoids, and then added together, for the area.
EXAMPLES.
1. The perpendicular breadths
of the irregular mixtilineal figure,
A B C D, at 5 equidistant places,
AEGIB, being 9.2, 10.5, 8.3,
9.4, and 10.7 yards, and its length, A
I 1 ' IT
A B, 20 yards ; what is its area ?
Here (9.2 -f 10.5 + 8.3 + 9.4 + 10.7) -i- 5 = 48. i -,- 5
= 9.62,
And 9.62 X 20 = 192.4 yards, area by the first part of the
rule.
Or,
(AD + BC) -v- 2 = (9.2 + 10.7) -+ 2 = 19.9 -=- 2
= 9.95.
Then (9.95+10.5+8.3 + 9.4)-=- 4=38.15 -r- 4 =9.5375,
and 9.5375 x 20 = 190.75 yards, the area by the second
part of the rule.
2. Required the area of the figure,
A B C D, of which the part A C is D
a rectangle, whose sides are 20.}
and 10^ chains respectively; and
the perpendicular breadths of the
curvilineal spaces, reckoning from A
DC at 4 equidistant places, are 10.2,
8.7, 10.9, and 8. 5 chains respectively.
Here (10.2 + 8.7 + 10.9 + 8.5) -r- 5 =38.3 -4- 5 = 7.06,
And 7.66 x 20.5 = 157.03 chains, the area of the curved
space.
Then 20.5 x 10.5 = 215.25 chains, the area of the rect'
angle.
Whence 215.25 + 157.03 = 372.28 chains = 37 a Or
36.48 p., the area of the whole figure.
MENSURATION OF SUPERFICIES. 87
3. The length of an irregular mixtilineal figure is 47
chains, and Us breadth, at 6 equidistant places, beginning at
the left hand extremity of the base, 5.7, 4.8, 7.5, 5.1, 8.4,
and 6.5 chains respectively ; what is its area?
Am. 29 a. 3 r. 2 p.. by the first rule.
4. The length of an irregular mixtilineal figure, of which
the curvilinear boundary meets the base, is 37i chains, and
its breadth, at 7 equidistant places, is 4.9, 5.6, 4.5, 8.2, 7.3,
5.9, and 8.5 chains respectively ; what is its area ?
Jlns. 24 a. r. 8.571 p. by the first rule.
And 23 a. 3 r. 20 p. by the second rule.
5. The length of an irregular field is 39 rods, and its
breadths, at five equidistant places, are 2.4, 2.6, 2.05, 3.05
3.6 rods respectively ; what is the area ?
Jlns. 111.54 rods by the first rule.
6. The length of an irregular piece of land being 42
chains, and the breadths, at six equidistant points, being 8.7,
10.3, 7.1, 8.34, 10.04, 12.2 chains respectively ; what is the
area ? Jlns. 38 a. 2 r. 39.872 p., by the second rule.
THE CIRCLE.
PROBLEM I.
14. To find the circumference of a circle, when the
diameter is given, or the diameter, when the circumference
is given.
RULE. Multiply the diameter by 3.1416, and the product
will be the circumference ; or divide the circumference by
3.1416, or multiply the circumference by .31831, and the
result will be the diameter.
EXAMPLES.
1. What is the circumference of the
circle, A C B D, whose diameter, A B, is 7
fi'Pt ? A
Here 3.1410 x 7 = 21.9912 feet, the cir-
cumference.
88 MENSURATION OF SUPERFICIES.
2. What is the diameter of a circle whose circumference
is 100 yards?
Here 100 -j- 3.1416 = 31.831 yards, the diameter.
Or,
100 X .31831 = 31.831 yards, the diameter as before.
3. If the diameter of a circle be 17 chains ; what is the
circumference ? Jlns. 53.4072 chains.
4. If the circumference of a circle be 354 perches ; what
is the diameter ? Jlns. 112.0817 perches.
5. What is the circumference of the earth, supposing its
diameter to be 7935 miles, which it is very nearly ?
Jlns. 24928.590 miles.
6. If the circumference of a carriage-wheel be 16 feet 6
inches; what is its diameter? Jlns. 5.2521 feet.
PROBLEM II.
To find the area of a circle.
RULE. Multiply the square of the diameter by .7854 ;
or the square of the circumference by .07958, and the pro-
duct in either case will be the area.
EXAMPLES.
1. How many square feet are there in a circle whose
diameter is 5 feet 6 inches ?
Here (5.5) 3 X .7854 = 30.25 x .7854 = 23.75835 sq. feet.
2. Required the area of a circle, the circumference of
which is 9| yards.
Here (9.2) 2 x .07958 = 84.64 X .Q7958 = 6.7356512
sq. yds.
3. How many square yards are there in a circle whose
radius is 15* feet? Jlns. 81.1798 sq. yards.
4. Hov many square feet are there in a circle whose
circumference is 10* yards ? Jlns. 82.7081 sq. feet.
5. The diameter of a circle is 16 chains; how many acres
does it contain ? Jlns. 20 a. r. 16.U9M p.
6. What is the value of a circular garden whose diameler
is 6 perches, at the rate of 75 cents per square yard ?
Jlns. 9641.471*
MENSURATION OF SUPERFICIES. 89
PROBLEM III.
The area of a circle being given, to find the diameter or
circumference.
RULE. Divide the area by .7854, and the square root of
the quotient will be the diameter. Or, divide the area by
.07958, and the square root of the quotient will be the cir-
cumference.
EXAMPLES.
1. The area of a circle is 5 acres, 3 roods, and 26 perches ;
what is the diameter ?
Here 5 a. 3 r. 26 p. = 946 perches ; and ^7(946 -f- .7854)
= v/ 1204.48179271 = 34.7056 perches, the diameter.
2. The area of a circle being 2 acres, 3 roods, and 12
perches, what is the circumference ?
Here 2 a. 3 r. 12 p. = 452 perches ; and ^/ (452-*- .07958)
= v/ 5(579.09 = 75.3637 perches, the circumference.
3. The area of a circle is 5028f square yards ; what is its
diameter ? Ans, 80.0160 yds.
4. The area of a circular garden being 1 acre, what is the
length of a stone wall which will enclose it ?
Ans. 44.8392 p.
5. It is required to find the radius of a circle whose area
is an acre. Jlns. 39.2507 yds.
6. The value of a circular piece of ground, at $4 per
square perch, is $46.50. How many dollars will encircle it
if the diameter of a dollar be 1 inches?
Ans. $1595.3916.
PROBLEM IV.
To find the area of a circular ring, or the space included
between two concentric circles.
RULE. Find the areas of the two circles separately.
Then the difference of these areas will be the area of the
rug.
Or, multiply the sum of the diameters by their difference.
90 MENSURATION OF SUPERFICIES.
and this product again by .7854, and it will give the area
required.
EXAMPLES.
1. The diameters of the two circles
are, A B 20, and D C 12 yards ; re-
quired the area of the ring.
Here 400 x .7854 = 314.16, area of A(
the outer circle.
144 x .7854= 113.0976, area of the
inner circle.
And 314.16113.0976 = 201.0624 yards, area of the
ring.
Or, 20 + 12 = 32, sum of diameters.
20 12 = 8, difference of diameters.
And 32 x 8 X .7854 = 201.0024 yards, area as before.
2. What is the area of a circular ring, the diameters of
the concentric circles being 20 and 30 feet 1 Jlns. 392.7 ft.
3. The diameters of two concentric circles are 8 and 12
yards ; what is the area of the ring contained between their
circumferences ? Jlns. 62.832 yds.
4. The diameters of two circles are 21| and 9 feet; re-
quired the area of the ring. Jlns. 300.6009 ft.
5. The diameter of the inner circle is 6, and the outer 10
chains ; what is the area of the ring 1 Ans. 50.2656 chs.
6. The area of the outer circle contains 100 acres, and
the diameter of the inner is equal to f of the diameter of the
greater ; what is the area of the ring ? Jins. 555.55 chs.
PROBLEM V.
The diameter or circumference of a circle being given, to
find the side of an equivalent square.
RULE. Multiply the diameter by .8862, "or the circum-
ference by .2821, the product in either case will be the side
of an equivalent square.
MENSURATION OF SUPERFICIES.
91
EXAMPLES.
1. The diameter of a circle is 200 yards ; what is the side
of a square of equal area ?
Here 200 x .8862 = 177.24 yards.
2. The circumference of a circle is 316 yards ; what is
he side of a square of equal area ?
Here 310. x .2821 = 89.1436 yards.
3. The diameter of a circle is 1 142 feet ; what is the side
of a square of equal area ? Ans. 1012.0-104 ft.
4. The circumference of a circle is 18.8 chains ; what is
the side of a square of equal area ? rfns. 5.3034 chs.
5. The diameter of a circular fish-pond is 22 perches ;
what would be the side of a square fish-pond of an equal
area ? dns. 19.4964 p.
6. The circumference of a circular walk is 64 rods ; what
!s the side of a square containing the same area ?
fins. 18.0544 r.
PROBLEM VI.
The diameter or circumference of a circle being given, 19
find the side of the inscribed square.
RULE. Multiply the diameter by .7071, or the circum-
ference by .2251, and the product in either case will be the
side of the inscribed square.
EXAMPLES.
1. The diameter, A B, of a circle is
614 feet ; what is the side, A C, of the
inscribed square ?
Here 614 x .7071 = 434.1594 feet
=AC.
2. The circumference of a circle is
1804 feet; what is the side of the in-
scribed square? Ans. 406.0804 ft.
3. The diameter of a circle is 239 feet ; what is the side
of the inscribed square ? tfns. 168.9909 fl.
92 MENSURATION OF SUPERFICIES.
4. The circumference of a circle is 98 chains ; what is
the side of the inscribed square ? Jlns. 22.0598 chs.
5. The diameter of a circle is 65 rods ; what is the side
of the inscribed square ? Ans. 45.9015 r.
6. The circumference of a circular walk is 721 perches ;
what is the side of an inscribed square ?
tins. 162.2971 p.
PROBLEM VII.
To find the diameter of a circle equal in area to any given
superficies.
RULE. Divide the area by .7854, and the square root of
the quotient will be the diameter.
EXAMPLES.
1. The length and breadth of a rectangle are 24 and 10
chains ; what is the diameter of a circle which contains the
same area?
Here 24 x 16 = 384, the area of the rectangle.
Then v/(384 -=- .7854) = v/488.9228 = 22. 1 1 16 chains,
the diameter.
2. The side of a square is 16 perches ; what is the diame-
ter of a circle containing the same area ?
Jlns. 18.054 p.
3. The base and perpendicular of a right-angled triangle
are 16 and 20 feet ; what will be the diameter of a circle
which contains the same area ? rfns. 14.2/29 ft.
4. The three sides of a scalene triangle are 14, 18, and 24
yards ; what is the diameter of a circle containing the sam
area? -Ans. 12.6267 yds.
5. The three sides of a triangle are 18, 20, and 26 feet ;
what is the diameter of a circle containing three times as
much? dns. 26.1919 ft.
6. The length and breadth of a parallelogram are 32 and
18 yards : what is the diameter of a circle that contains the
same area ? -fins. 27.0810 \ds.
MENSURATION OF SUPERFICIES. 93
PROBLEM VIII.
The diameter of a circle being given, to find another con-
taining a proportionate quantity.
RULE. Multiply the square of the given diameter by the
given proportion, and the square root of the product will be
the diameter required.
EXAMPLES.
1. The diameter of a circle is 24 chains ; what is the
diameter of one containing one-fourth of the area ?
Here x /(24 2 x *) = v/(576 x |) = v/144 = 12 chains.
2. The diameter of a circle is 16 perches ; what is the
diameter of one containing nine times as much ?
rfns. 48 p.
3. The diameter of a circle is 36 yards; what is the
diameter of one containing four limes as much ?
rfns. 72 yds.
4. The diameter of a circle is 81 feet ; what is the diame-
ter of one containing five times as much ?
Ans. 181.1215 ft.
5. A gentleman has a circular grass-plat in his yard, the
diameter of which is 25 yards; required the length of the string
that would describe a circle to contain sixteen times as much.
Jlns. 50 yds.
6. The diameter of a circle is 9 rods ; what is the diame-
ter of one containing six times as much? Jlns. 22.0454 r.
PROBLEM IX.
To find the length of any arc of a circle.
RULE. 1 . From eight times the chord of half the arc, s
tract the chord of the whole arc, and one-third of the
mainder will be the length of the arc nearly.
94 MENSURATION OF SUPERFICIES.
The chord of the whole arc, or sim-
ply the chord, is a right line which joins
the extremities of an arc. Thus in the
figure, D C is the chord of the whole
arc, D F C ; D F the chord of half the A[
arc, D F C ; and D E, or E C, half the
chord of the arc D F C. The height of
an arc, called its versed sine, is that part
of the diameter contained between the
middle of the chord and the arc. Thus, in the case of the
arc D F C, the versed sine, or the height of the arc, is the
line E F.
The chord of half the arc may be found by adding together
the square of half the chord and the square of the versed
sine, and extracting the square root of the sum.
Or, by taking the square root of the product of the diame-
ter and the versed sine.
Half the chord of the whole arc may be found by sub-
tracting the versed sine from the diameter, multiplying the
remainder by the versed sine, and taking the square root of
the product. By doubling this we get the chord of the
whole arc.
To find the versed ine, or height of the arc, subtract the
square of the chord from the square of the diameter, and ex-
tract the square root of the remainder ; subtract this root from
the diameter, and one-half of the remainder will be the versed
sine.
Or, from the square of the chord of half the arc, subtract
the square of half the chord of the arc, and the square root
of the remainder will give the versed sine.
Again, to obtain the versed sine, divide the square of the
chord of half the arc by the diameter.
The diameter may be known by adding together the square
of the versed sine and the square of half the chord of the arc,
and dividing the sum by the versed sine. The diameter
may likewise be obtained by dividing the square of the chord
of half the arc by the versed sine.
It may here likewise be observed, as another rule for the
parne purpose, that if the number of degrees in the arc be
multiplied by radius, and that product again by .01745, the
result will give the length of the arc nearly.
MENSURATION OF SUPERFICIES.
95
EXAMPLES.
1. The chord A B of the whole arc
A C B, is 48.74 feet, and the chord A C
of half the arc 30.25 feet ; what is the
length of the arc ?
Here [ (30.25 x 8) 48.74) ] -r- 3 =
(242 48.74) -j- 3 = 193.26 -=- 3 =
64.42 ft. the length of the arc A C B nearly.
2. If the chord A B, of the arc A C B, be 30 yards, and
the versed sine, or height C D, 8 yards, what is the length
of the arc ?
Here */ (15 9 + 8*} = </ (225 -f 64) = </ 289 = 17 =
A C, the chord of half the arc.
And [(17x8) 30)] -=- 3 =(136 30) ~ 3= 106 -r- 3
= 35| yards, the length of the arc nearly.
3. If the versed sine, or height of half the arc, be 4 feet,
and the diameter of the circle 30 feet, what is the length of
the arc ?
Here ^7(30 x 4) = v/120 = 10.95445 => A C, the chord
of half the arc.
And ^[(30 4) x 4] x2 = ^/(26x4) x2 = v/(104)
X 2= 10.19803 x 2 =20.39606 = A B, the chord of the
whole arc.
Then [(10.95445 x 8) 20.39606] +- 3 = (87.6356
20.39606) -4-3 = 67.23954 -7- 3 = 22.41318 feet, the
length of the arc nearly.
4. Required the length of an arc of 30 degrees, the radius
of the circle being 14 feet.
(30 x 14 X .01745) = 7.329 feet, the length of the arc.
5. The chord of the whole arc is 50f yards, and the chord
of half the arc is 33| yards ; required the. length of the arc.
Ans, 71.6 yards.
6. The length of the chord of the whole arc is 36| feet,
and the length of the chord of half the arc is 231 feet ; what
is the length of the arc ? Jlns. 49.6166 feet.
7. The chord. of the whole arc is 48k feet, and its versed
sine, or height of the arc, 18| feet ; what is the length of the
arc ? ns. 64.7667 fee-..
96 MENSURATION OF SUPERFICIES.
8. If the versed sine or height of the arc be 2 yards, and
the diameter of the circle 36 yards ; what is the length of
the arc? Ans. 17.1299 yards.
9. If the chord of the whole arc be 16 chains, and the
radius of the circle 10 chains ; what is the length of the arc ?
Jlns. 18.518 chains.
10. The chord of half the arc is 10 perches, and the
diameter of the circle 16 perches ; what is the length of
the arc? rfns. 21 5 perches.
11. Required the length of an arc of 57 17' 44*", the
diameter of the circle being 50 feet.
Jlns. 25 feet, which equals the radius.
12. Required the length of a degree of a great circle of
the earth, supposing its circumference to be 25000 miles.
Ans. 69j miles nearly.
RULE 2. Let d = the diameter C E of the circle, and
t; = the versed sine or height C D of half the arc, then will
the length of the arc be expressed by the following series :
6_, , v." I r^ f\ i "-'7 vwv '
X 7rf 8 x9e?
Where A, B, C, &c. represent the terms immediately pre-
ceding those in which they first occur.
Which rule is to be used, by substituting in the above
series the numerical values of the given parts, in the place
of the letters by which they are denoted, and then finding the
sum of such a number of its terms as may be thought suffi-
cient for determining the length of the arc, to a degree of
accuracy required.
EXAMPLES.
1. Required the length of the arc,
A C B, whose versed sine or height,
DC, is 5 feet, and the diameter, CE, of
>he circle 25 feet.
Here d . 25, and v = 5.
MENSURATION OF SUPERFICIES. 97
Then 2 v/ dv = 2 v/(25 x 5) = 2 v/ 125 = 22.3606798 == A
, " , X A = ^-^ X 22.3606798 = .7453559= B
2x3 d 6 x 25
*' v x B = J! X JL x .7453559 = .0670820 = C
x 5 d 20 x 25
f^i 71
6x7<* xc =4ra x 0670820= OOT9859=
7 3 u 49 x 5
8 9~d xD= 72 ox x -0079859 = .0010869= E
The sum = 23.1821905"
fret, = the length of the arc, A C B.
2. Required the length of the arc, A C B, whose versed
sine or height, DC, is 2 feet, and the diameter, C E, of the
circle 52 feet. Ans. 20.5291 feet.
3. Required the length of the arc whose versed sine is 9
yards, and the diameter of the circle 100 yards.
Ans. 60.9385 yards.
4. It is required to find the length of the arc whose chord
is 16 perches, and its height 4 perches.
Ans. 18.5459 perches.
5. It is required to find the length of the arc whose height
is 6 inches, and the chord of half the arc 1 foot.
Ans. 2.0943 feet.
6. It is required to find the length of the arc whose chord
is 48 chains, and the radius of the circle 25 chains.
Ans. 64.3501 chains.
PROBLEM X.
To find the area of a sector of a circle.
ROLE 1. Multiply the radius or half the diameter of the
circle by half the length of the arc of the sector, as found by
the last Problem, and the product will be the area.
To this we may add, that as 360 is to the number of de-
grees in the arc of the sector, so is the area of the circle, to
the area of the sector.
9
MENSURATION OF SUPERFICIES.
EXAMPLES.
1 The chord, A B, of the whole arc,
A C B, is 24 feet, and the chord, A C,
of half the arc 13 feet ; what is the area
of the sector, OBC AO?
Here v/ (13 3 12 3 ) = v/ (169 - 144)
= ^/ 25 = 5 =C D, the versed sine.
And (13 a -=- 5) =169 -T- 5 = 33.8
= C E, the diameter. E
[713 x 8) 24] -4- 3 = (104 24) -=- 3 = 80 -v-3 =20
= the length of arc A C B.
Then 13 x 16.9 = 225! feet, the area of the sector.
2. Required the area of the sector, the arc of which is 30,
and the diameter 3 yards.
Here .7854 x 9 = 7.0686 = the area of the circle.
As 360 : 30 : : 7.0686 : .58905 yards, the area of the
sector.
3. The chord of the whole arc is 8 yards, and its height
3 yards ; what is the area of the sector ?
Ans. 22.2222 yards.
4. What is the area of a sector whose chord is 18| chains,
and the diameter of the circle 20 chains ?
Ans. 118.954 chains.
5. Required the area of the sector whose height is 4
perches, and the radius of the circle 8 perches.
Ans. 66.8581 perches.
6. Required the area of the sector whose arc is 17 15',
and the diameter of the circle 19 feet.
Ans. 13.5857 feet.
RULE 2. Let d = the diameter C E of the circle, and
v = the versed sine or height, C D, of half the arc, then
will the area of the sector be expressed by the following
series.
D, &c.
MENSURATION OF SUPERFICIES.
99
When A, B, C, &c., represent the terms immediately
preceding those where they first occur. This Rule is to bo
used in the same manner as directed in regard to Rule 2,
Problem IX. p. 96.
EXAMPLES.
1. Required the area of the sector,
A C B O, whose versed sine, or height,
C D, is 7 feet, and the diameter, C E, of
the circle 28 feet.
Here d = 28, and v = 7.
Then J
v=14 v /(28x7)=
7
XA =
xB =
C =
XD =
6x28
= 196.000000 = A
= 8.1 60666 = B
9X ~X8. 166666 =
20 x 28
25x7
42x28
49x7
72x28
.918750 =
X .918750= . 136718 = D
X .136718 =
= E
9 a u 81 x 7
10 X lid X " 110x23 X
IPu 121x7
v T^ v
.023261= .004282 =F
.004282= .000830 = G
The sum = 205.250507
feet = the area of the sector O A C B O.
2. Required the area of the sector O A C B O, whose
versed sine, C D, is 2 yards, and the diameter, C E, of th
circle 52 yards. Ans. 266.8787 yds.
3. Required the area of the sector, the radius of the cir-
cle being 10 feet, and the chord of the arc 12 feet.
tins. 64.3501 ft.
4. Required the area of the sector of a circle whose diame-
ter is 20 perches, and the chcrd of its arc 16 perches
An* 92.7295 p
100 MENSURATION OF SUPERFICIES.
5. Required the area of the sector of a circle whose versed
sine is 9 feet, and the chord of half the arc 30 feet.
tins. 1523.4632ft.
6. Required the area of the sector of a circle whose chord
is 48 feet, and versed sine 18 feet. rfns. 804.370 ft.
PROBLEM XT.
To find the area of a segment of a circle.
RULE 1. Find the area of the sector, having the same
arc as the segment, by the last problem.
Also find the area of the triangle formed by the chord of
the segment, and the two radii of the sector, by Problem I.
page 67.
The sum, or difference of these areas, according as the
segment is greater or less than a semicircle, will be the area
of the segment required.
EXAMPLES.
1 . The chord A B is 24 feet, and the
versed sine or height C D, of half the
arc A C B, is 5 feet ; what is the area
of the segment A B C A ?
Here ^ (A D a + C D a ) = v/ (12 a -f
5 3 ) = v' (144 + 25) = ^ 169 = 13 =
A C, the chord of half the arc.
Again, (A C 3 -7- C D) = (169 -4- 5) = 33.8 = C E, the
diameter; therefore 16.9 = C O, the radius.
Then [ (13 x 8) 24] -7- 3 = 26f , the length of the arc
AC B.
And (13 x 16.9) =225.3333 feet, the area of the sec-
tor O A C B O.
But (C O C D) = (10.9 5) = 11.9 = perpendicular
OD.
Whence (11.9 x 12) = 142.8, area of the A A O B.
And, consequently, (225.3333142.8) =82.5333 feet
area of the segment A B C A.
2. What is the area of a segment of a circle whose arc is
00, and the diameter of the circle 10 feet ?
Here .7854 x 100 = 78.54, area of the whole circle.
MENSURATION OF SUPERFICIES. 101
Then, as 360 : 60 : : 78.54 : 13.09, area of the sector
<> A C B O.
And since the chord A B (to an arc of 60) is = radius
O A or O C, which = 5, then by Problem II., page 82, the
area of the A A O B = 10.8253.
Whence (13.09 10.8253) = 2.2647 feet, the area of
the segment required.
3. Required the area of the segment of a circle, whose
versed sine, or height, of half the arc is 5 yards, and the
diameter of the circle 20 yards. Am. 61.1(345 yds.
4. Required the area of the segment of a circle, whose
chord is 16 chains, and diameter of the circle 16| chains.
Am. 70.2222 chs.
5. What is the area of the segment of a circle, whose arc
is a quadrant, the diameter being 24 perches ?
Am. 41. 0976 p.
6. Required the area of a segment of a circle, whose
chord is 18.9 feet, and height 2.4 feet. Am. 30.601 ft.
RULE 2. Let d equal the diameter C E of the circle, and
v equal the versed sine C D, of half the arc, or height of tne
segment ; then will the area of the segment be expressed by
the following series.
Area of the segment A C B A =
x9cT 8xll<T
Where A, B, C, &c., are the first, second, third, &c., terms
of the series.
This rule is to be used in a manner similar to that directed
in regard to Rule 2d, Problem IX., page 96.
EXAMPLES.
1. Required the area of the segment
A B C A of a circle, whose chord A B
is 16 feet, and the diameter C E 20
feet.
Here ^/(C E' A B a )= x /(20 3 16 a )
= v/ (400 256) = v/144 = 12 ; then
(20 12) -=- 2 = 4 = C D, the versed
sine.
<J'
102 JNSURATION OF SUPERFICIES.
4v^/dv 16./(20X4) 16^/80 Atinotai A
S3 ^S -+- %f / U<vf OO =^ /\
333
3g xA= X4 X +47.702783 = 2.862 167 = B
2x5d 10x20
_ -!L. X B= JL* A. X 2.862167 = .102220 =
4 X 7 d 28 X 20
3 X 7p X C= 21 x 4 X .102220 = .007950 = D
6 X 9 d 54 X 20
5 x 9 x ^^ 45 X 4 X .007950 = .000813 = B
8 X 1 1 d 88 X 20
7 X 11 v xE __ 77 X 4 .000813= .00009 6 = F
10 X 13d 130X20
ThesumofBCDEF= 2.973246
The difference = 44.729537, ft.
= the area of the segment A B C A.
2. Required the area of the segment of a circle whose
height is 24 feet, and the radius 26 feet. Am. 957.9609 ft.
3. Required the area of the segment of a circle whose
versed sine is 3 yards, and the diameter 60 yards.
Ans. 52.8533 yds.
4. Required the area of the segment of a circle whose
height is 4 feet, and the radius 50 feet.
Am. 105.3773 ft.
5. Required the area of the segment of a circle whobe
chord is 20 chains, and the diameter 52 chains.
Am. 26.8787 chs.
6. Required the area of the segment of a circle whose
height is 2 feet, and chord 12 feet. Ans. 16.35 ft.
RULE 3. 1. From seven times the diameter subtract five
times the versed sine ; multiply the remainder by seven
times the versed sine, and extract the square root of the
product.
2. Multiply the diameter by the versed sine, and extract
the square root of the product.
3. To the first root add four thirds of the second, and
multiply the sum by four twenty-fifths of the versed sine,
and the product will be the area of the segment nearly.
MENSURATION OF SUPERFICIES.
103
EXAMPLES.
1. Required the area of the segment,
A B C A, of a circle whose versed sine,
or height, C D, is 4 feet, and the diame-
ter, C E, 20 feet.
Here N /{[(20 X 7)-(4 X 5)]x(4x7)}
= v/ [(14020) X 28] = v/(120 x 28)
= v/ 3360 = 57.96550 the first root.
And ^ (20 x 4) = ^80 = 8.94427,
the second root.
Then 57.96550 + (8.94427 X |)= (57. 96550 +11. 92569)
= 69.89119.
Whence 69.89 1 19 x (4 X^)= (69.891 19 x. 64) =44.73036
2o
feet, the area of the segment A B C A.
2. What is the area of a segment of a circle whose versed
sine, or height, is 4 feet, and the diameter of the circle is
40 feet ? dm. 65.4005 feet.
3. Required the area of the segment of a circle whosa
versed sine is 4 yards, and the diameter 100 yards.
dns. 105.3773 yards.
4. Required the area of the segment of a circle whose
height is 3 yards, and the diameter 60 yards.
Jlns. 52.8533 yards.
5. What is the area of a segment of a circle whose height,
or versed sine, is 5 feet, and the diameter of the circle 25
feet? An. 69.8911 ft.
6. What is the area of the segment of a circle whose
height, or versed sine, is 3 feet, and the diameter of the
circle 8 feet? Ans. 17.2198 ft.
RULE 4. To two-thirds of the product of the chord and
versed sine of the segment, add the cube of the versed sine
divided by twice the chord, and the sum will give the area of
the segment nearly.
104
MENSURATION OF SUPERFICIES.
EXAMPLES.
1 . What is the area of the segment,
A BC A, of a circle whose versed sine,
C D, is 4 feet, and the chord, A B, is 16
feet?
Here [(16 x4) x|] + [4 3 -(16x2)]
2 128
(64 x S + (64 32) = |p + 2)
= (42 + 2) = 44 feet, the area of the segment A B C A.
2. What is the area of the segment of a circle whose
versed sine is 2 feet, and the chord 20 feet ?
Ans. 26.8 ft.
3. Required the area of the segment of a circle whose
height is 2 feet, and the chord 12 feet. Jlns. 16? ft.
4. Required the area of the segment of a circle whose
versed sine is 15 feet, and the chord 40 feet.
Ans. 442.1875ft.
5. Required the area of the segment of a circle whose
versed sine is 2 feet, and the chord 7 feet. Jlns. 9^ ft.
6. Required the area of the segment of a circle whose
versed sine is 18 feet, and the chord 48 feet.
rfns. 6361 ft.
RULE 5. Divide the versed sine, or height of the seg-
ment by the diameter of the circle, and find the quotient in
the table of verged sines, p. 295.
Then multiply the tabular area on the right hand of the
versed sine so found, (which is the tabular segment,) by the
square of the diameter, and the product will be the area.
When the quotient arising from dividing the versed sin
by the diameter has a remainder, or fraction, after the third
place of decimals, subtract the tabular area, answering to the
first three figures, from the next following area ; then if the
remainder be multiplied by the fractional part, and the result
be added to the frst area, it will give the tabular area for the
whole quotient ; which must be multiplied by the square of
the diameter, as before.
MENSURATION OF SUPERFICIES. 105
EXAMPLES.
1. What is the area of the segment
of a circle whose versed sine, C D, is
2 yards, and the .diameter of the circle,
C E, 52 yards ?
Here (2 -4- 52) = .038^, the tabular
versed sine.
And .009763 the tab. segment to .038.
.010148 the next do. .039.
.000385 difference.
Then .000385 x -^ = .000178
.009763 tab. segment to .038.
.009941 tab. segment to .038 r V
Now (.009941 x 52 a ) = (.009941 x 2704) = 26.880464
yards, the area of the segment.
2. What is the area of the segment of a circle whose
height is 10 feet, and the diameter of the circle 50 feet?
Here (10 -f- 50) = .2 the tabular versed sine.
And .111823, the tabular segment to .2.
Now (.111823 x 50 3 ) = (.111823 x 2500) = 279.5575
feet, the area required.
3. What is the area of the segment of a circle whose
versed sine is 3 feet, and the diameter of the circle 8 feet ?
Ans. 17.2168 ft.
4. What is the area of the segment of a circle whose
height is 15 yards, and the diameter of the circle 75 yards ?
Ans. 629.0043 yds.
5. What is the area of the segment of a circle whose
height is 6 feet, and the diameter of the circle 21 feet ?
Ans. 81.6582 ft.
6. What is the area of the segment of a circle whose
versed sine is 7 feet, and the diameter of the circle 38 feet *
Ans. 143.5104 ft.
PROBLEM Xn.
To find the area of a circular zone, or the space included
between two parallel chords and their intercepted arcs.
106
MENSURATION OF SUPERFICIES.
RULE. Find the area of that part of the zone A BCD,
which forms a trapezoid, by Prob. I. p. 80, and the area of
the small segment Bn C B by rule 5, in the last Problem.
Then add the area of the trapezoid to twice the area of
the segment, and it will give the area of the zone.
1. Let d = the diameter E F ; C, c = the two parallel
chords A B, DC; v = the versed sine m n, and b = G H,
the breadth cf the zone, we shall have
ET7
J? >^
And,
2. When the two parallel chords are equal to each other,
the diameter d or E F will be = ^/ (6 a + c 2 ) ; and the
versed sine mn = 5 (d c).
And when one of the parallel chords is the diameter, the
versed sine mn - d ^ </ i 6 3 + ( n ) S '
which expressions will be found of considerable use in facili-
tating the computation of the segment B n C B.
EXAMPLES.
1. The greater chord A B is 8 yards,
the less D C 6 yards, and their perpen-
dicular distance G H is 7 yards ; re-
quired the area of the zone.
Here A B + D C = 8 + 6 = 14, and
Whence (14 x 7) -r- 2 = 49, the area
of the trapezoid A B C D
And
5 -
-f 50 -f- 1) = -v/100 = 10 the diameter.
10 1 /8
G Xc.
H f*
== 53.535533 = 1.464467, the versed sine.
MENSURATION OF SUPERFICIES. 107
Therefore 1.464467 -r- 10 = .1464467, the tabular versed
sine.
Answering to which, when the work is performed as in
the last problem, we shall have .071349.
Hence .071319 x 10 s = .071349 x 100 =
7.1349, the area of the segment B n C B.
7.1349 x 2 = 14.2098, twice do.
49.0000, area of the trapezoid A B C D.
63.2698 yards, area of the zone.
2. One of the parallel chords of a circular zone is 48 feet,
and the other 30 feet, and its breadth 13 feet ; what is the
area of the zone 1 Ana. 534.19 ft.
3. The greater chord of a circular zone is 16 yards, and
the less chord 12 yards, and their perpendicular distance 2
yards ; what is the area of the zone ? Am. 28.379 yds.
4. Supposing the greater chord of a circular zone to be
20 feet, and the less 15 feet, and their distance 17 feet;
what is the area of the zone ? Am. 395.4362 ft.
5. Required the area of a circular zone, each of whose
parallel chords are 50 feet, and their perpendicular distance
30 feet. Ans. 1668.7093 ft.
6. It is required to find the area of a circular zone, the
greater chord of which, being equal to the diameter of the
circle, is 40 feet, and the less 20 feet. Am. 592.086 ft.
PROBLEM XIII.
To find the area of a lune, or the space included between
.the intersecting arcs of the two eccentric circles.
RULE. Find the areas of the two segments, from which
the lune is formed by Problem XL, rule 5, page 104, and
their difference will be the area of the lune.
If semicircles be described on the three sides of a right-
angled triangle as diameters, the two lunes formed on the
base and perpendicular, taken together, will be equal to the
area of the triangle.
108
MENSURATION OF SUPERFICIES.
If A B C, or H, be a right-angled
triangle, and semicircles be de-
scribed on the three sides as di-
ameters, then will the triangle H
be equal to the two lunes D and E
taken together.
For, if from the greater semicircle ABC there be taken
the two segments F and G, there will remain the triangle
H ; and if the same segments be taken from the other two
semicircles, there will remain the lunes D and E : hence,
since the greater semicircle is equal to the sum of the other
two, the triangle H must be equal to the sum of the lunes
DandE.
EXAMPLES.
1. The length of the chord A B is 40
feet, the height D C 10 feet, and D E
4 feet ; required the area of the lune.
Here (A D a + C D a ) -4- C D =
(20 3 + 10 2 ) -f- 10 = (400 + 100) -f- A(/!_.
10 = 500 -=- 10 = 50, the diameter be-
longing to the circle of which A C B is a part.
Again (A D 9 + D E a ) -f- DE = (20 3 + 4 2 ) -r- 4 = (400
+ 16) -r 4 = 416 -r- 4 = 104, the diameter belonging to the
circle of which A E B is a part.
Whence 10 -;- 50 = .2 the first tabular versed sine.
Answering to which is .111823, the first tabular segment.
Therefore .111823 x 50'=. 11 1823 x 2500 = 279.5575,
the area of the segment A B C A.
Also 4 -j- 104 =.038-A-, the second tabular versed sine.
Answering to which is .009941, the second tabular seg-
ment.
Consequently, .009941 x 104 2 = .009941 x 10816 =
107.521856, the area of the segment A E B A.
Whence 279.5575 107.521856 = 172.035644 feet,
area of tne ;une A C B E A.
2. The chord is 48 feet, and the height of the segments
18 and 12 feet ; required the area of the lune.
Jlns. 233.8122 ft.
MENSURATION OF SUPERFICIES.
109
3. The chord is 40 feet, and the height of the segments
20 and 4 feet ; required the area of the lune.
dns. 520.7965 ft.
4. Supposing the length of the chord to be 96 feet, and
the height of the segments to be 36 and 14 feet ; what is the
area of the lune ? Ana. 1634.4350 ft.
5. The length of the chord is 50 feet, and the heights of
\he segments 18 and 15 feet ; required the area of the lune.
rfns. 123.8785 ft.
6. The length of the chord is 40 feet, and the heights of
.he segments 10 and 8 feet ; what is the area of the lune?
Ans. 59.5485 ft.
PROBLEM XIV.
To find the area of a part of a ring, or of the segment of a
sector.
RULE. Multiply half the sum of the bounding arcs by
their distance asunder, and the product will give the area.
EXAMPLES.
1. Let A B be 50 inches, and D C
30 inches, and the distance D A 10
inches ; what is the area of the space
A.BCD?
Here [(50 + 30) -=- 2] X 10 = (80
-T- 2) x 10 = 40 x 10 = 400 inches.
2. Let A B = 60 inches, D C =
40 inches, and D A = 2 ; required
the area of the space A B C D.
Ans. 100 in.
3. Let A B = 25 feet, D C == 1 5 feet, and D A = 6 feet
required the area of the segment of the sector.
Ana. 120 ft.
10
CONIC SECTIONS.
DEFINITIONS.
15. 1- The conic sections are certain plane figures
foi ned by the cutting of a cone, which are of great use in
some of the higher branches of mathematics.
2. A cone is a solid described by the
revolution of a right-angled triangle about
one of its legs, which remains fixed ; as
ABC.
3. The axis of the cone is the right line
about which the triangle revolves. c
4. The base of a cone is the circle which is described by
the revolving leg of the triangle ; and its altitude is a per*
pendicular drawn from the vertex to the base; as A n.
5. If a cone be cut through the vertex by
a plane, perpendicular or oblique to that of
the base, the section will be a triangle ; as
ABC.
0. If a cone be cut into two parts, by a
plane parallel to. the base, the section will
be a circle, as A C B D.
7. If a cone be cut by a plane which
passes obliquely through its two slant sides,
the section will be an ellipse, as A C B D.
110
CONIC SECTIONS.
Ill
8. If a cone be cut ty a plane, which is
parallel to either of its slant sides, the sec-
tion will be a parabola ; as A B C A.
9. If a cone be cut into two parts, by a
plane, which, being continued, would meet
the opposite cone, the section is called an
hyperbola ; as A B C.
The two opposite cones, in this definition,
are supposed to be generated together, by
the revolution of the same line.
It may here also be observed, that all
figures which can possibly be formed by the
cutting of a cone, are mentioned in these definitions, and are
the five following : viz., a triangle, a circle, an ellipsis, a
parabola, and an hyperbola; but the last three only are
usually called the conic sections.
10. If two lines be drawn through the
centre of an ellipse, perpendicular to
each other, and terminated both ways by
the circumference, the longest of them,
A B, is called the transverse diameter,
or axis, and the shortest, C D, the con-
jugate.
11. An ordinate of an ellipse is a
right line E F, or E G, drawn from
any point E, in the curve, perpendicu- A
lar to either of the diameters.
12. An abscissa is that part, A F, or
C G, of the diameter which is contain-
ed between either of the extremities of that diameter and the
ordinate.
An abscissa may be more generally considered as any
part of the diameter or axis of a curve comprised between
any fixed point, from which all the abscissas are supposed to
take their origin, and another line called the ordinate, drawn
so as to make a given angle with the former, and terminated
in the curve ; but when not otherwise specified, they an*
commonly taken as above. The abscissa and its correspond-
112 CONIC SECTIONS.
ing ordinates, when considered together, are also frequently
called co-ordinates.
13. The axis of a parabola, BAG,
is a right line, A D, drawn from the
vertex, so as to divide the figure into two
equal parts.
14. An ordinate of a parabola is a right B/
line E F, drawn from any point in the
curve, perpendicular to the axis.
15. An abscissa is that part of the axis, A F, which is con-
tained between the vertex of the curve and the ordinate.
16. The transverse diameter of an hyperbola is that part
of the axis which is intercepted between the two opposite
cones, as a B, in the figure accompanying definition ninth.
17. The conjugate diameter is a line drawn through the
centre, perpendicular to the transverse.
18. An ordinate of an hyperbola is a line drawn from any
point in the curve, perpendicular to either of the diameters ;
and an abscissa is that part of the diameter which is con-
tained between either of the extremities of that diameter and
the ordinate.
Hence, in the ellipse and hyperbola, every ordinate has
two abscissas, but in the parabola only one, the other vertex
of the diameter being at an infinite distance.
19. The parameter of any diameter is a third proportional
to that diameter and its conjugate.
20. The focus of a conic section is the point in the axis
where the ordinate is equal to half the parameter, as F, in
the figure accompanying definition eleventh, where E F is
equal to the semi-parameter of the section.
The ellipse and hyperbola have each two foci, as F,/: in
the figure accompanying the following problem, the parabola
has but one.
21. The vertices of a conic section are the points where
the cutting plane meets the opposite sides of the cone, or the
sides of the vertical triangular section. Hence the ellipse
and the opposite hyperbolas have each two vertices, as A, B,
and B,a, in the figures accompanying definitions seventh and
ninth ; but the parabola only one, as A, in the figure accom-
panying definition eighth.
CONIC SECTIONS.
113
THE ELLIPSE:
PROBLEM I.
16. To describe an ellipse, the transverse and conjugate
diameters being given.
Construction. 1. Draw the trans-
verse and conjugate diameters, A B,
C D, bisecting each other perpendicu- A
larly in the centre o.
2. With the radius A o, and centre
C, describe an arc cutting A B in F,f;
and these two points will be the foci of the ellipse.
3. Take any number of points n, n, &c., in the transverse
diameter AB, and with the radii A n, n B, and centres F,/,
describe arcs intersecting each other in s, s, &c.
4. Through the points s, s,&c., draw the curve AsCsBD,
and it will be the circumference of the ellipse required.
It is a well-known property of the ellipse, that the sum
of two lines drawn from the foci, to meet in any point in
the curve, is equal to the transverse diameter ; and from this
the truth of the construction is evident.
From the same principle is
derived another method of describ-
ing an ellipse, by means of a string A
and two pins.
Having found the foci F,y, as
before, take a thread of the length
of the transverse diameter, and
fasten its ends with two pins in the points F,y"; then stretch
the thread Fsyto its greatest extent, and it will reach to the
point s in the curve ; and by moving a pencil round within
the thread, keeping it always stretched, it will trace out the
curve required.
10*
114 CONIC SECTIONS.
PROBLEM II.
In an ellipse, any three of the four following terms being
given, viz., the transverse and conjugate diameters, an ordi-
cate and its abscissa, to find the fourth.
CASE 1. When the transverse, conjugate, and abscissa
are given, to find the ordinate.
RULE. As the transverse diameter is to the conjugate, so
is the square root of the product of the two abscissas, to the
ordinate which divides them.
It may be remarked that if but one abscissa is given, by
subtracting it from the transverse diameter, we obtain the
other.
EXAMPLES.
1 . In the ellipse A D B C, the trans-
verse diameter A B is 50, the conjugate
diameter C D is 30, and the abscissa A(
B E 18 ; what is the length of the or-
dinate E F ?
Here A B = 50, C D = 30, B E =
18, and A E = 50 18 = 32.
Whence 50: 30 :: >/ (18 x 32) : E F. Or, E F =
nn o yo
v/576 = - X 24 = = 14.4, the ordinate required.
50 5 &
2. If the transverse diameter be 40, the conjugate 30, and
the abscissa 24, what is the ordinate ? rfns. 14.6969.
3. If the transverse diameter be 120, the conjugate 40,
and the abscissa 24, what is the ordinate ? Jlns. 16.
4. If the transverse diameter be 35, the conjugate 25, and
he abscissa 28, what is the ordinate ? jins. 10.
CASE 2. When the transverse, conjugate, and ordinate
are given, to find the abscissa.
RULE. As the conjugate diameter is to the transverse, so
is the square root of the difference of the squares of the ordi-
nate and semi-conjugate to the distance between the ordinate
and centre.
CONIC SECTIONS. 115
And this distance being added to, and subtracted from, the
semi-transverse, will give the two abscissas required.
EXAMPLES.
1. The transverse diameter A B is 60, the conjugate
diameter C D 40, and the ordinate F E 12; what is the
length of each of the two abscissas B E and A E ?
Here A B = 60, C D = 40, and F E = 12.
Whence 40 : 60 :: v /(20 2 12 2 ) : O E.
Or O E = v/(400- 144)=| ^ 256 = | x 16 = I 8
= 24.
And 30 24 = 6 = B E, and 30 + 24 = 54 =A E.
2. What are the two abscissas to the ordinate 10, the
diameters being 35 and 25 ? Jlns. 7 and 28.
3. What are the t\vo abscissas to the ordinate 16, the
diameters being 120 and 40 ? Jlns. 24 and 96.
4. What are the two abscissas to the ordinate 14.4, the
diameters being 50 and 30 1 Jlns. 18 and 32.
CASE 3. When the conjugate, ordinate, and abscissa are
given, to find the transverse diameter.
RULE. To, or from the semi-conjugate, according as the
less or greater abscissa is used, add or subtract the squaio
root of the difference of the squares of the ordinate and semi-
conjugate.
Then, as the square of the ordinate is to the product of the
conjugate and abscissa, so is the sum or difference above
found to the transverse diameter required.
EXAMPLES.
1. The conjugate diameter C D is 60, the ordinate E F
24, and the less abscissa B E 36 ; required the transverse
diameter A B.
Hete O C = 30, E F = 24, and B E = 36.
Whence 30 + ^/ (30 2 24 2 ) = 30 +^(900 576) =^0
-f v/ 324 = 30 + 18 = 48.
And 24 2 : (00 x 36) : : 48: (60 x 36 x 48) -4- 24 2 =
103080 -i- 576 = 180, the transverse diameter reauired.
116 CONIC SECTIONS.
2. The conjugate diameter is 60, the ordinate 18, and the
less abscissa 24 ; required the transverse diameter.
. 240.
3. The conjugate diameter is 40, the ordinate 16, and the
frreater abscissa 96 ; required the transverse diameter.
AM. 120.
4. The conjugate diameter is 25, the ordinate 10, and the
greater abscissa 28 ; required the transverse diameter.
Am. 35.
CASE 4. The transverse, ordinate, and abscissa being
given, to find the conjugate diameter.
RULE. As the square root of the product of the two
abscissas is to the ordinate, so is the transverse diameter to
the conjugate.
EXAMPLES.
1. The transverse AB is 82, the ordinate EF 8, and the
abscissa BE 32; required the conjugate diameter C D.
Here A B = 82, EF = 8, BE = 32, and AE = 82
32 = 50.
Hence ^/ (32 x 50) : 8 : : 82 : (8 x 82) -j- v/ (32 x 50)
= 656 -r- ^/1600 == 656 -r- 40 = 16.4, the conjugate diame-
ter required.
2. The transverse diameter is 35, the ordinate 10, and the
abscissa 28 ; what is the conjugate ? Am. 25.
3. The transverse diameter is 120, the ordinate 16, and
its abscissa 24 ; what is the conjugate ? Jlns. 40.
4. The transverse diameter is 50, the ordinate 14.4, and
the abscissa 32 ; what is the conjugate ? Ans. 30.
PROBLEM III.
The transverse and conjugate diameters being given, to
find the circumference.
RULE. Multiply the square root of half the sum of the
squares of the two diameters by 3.1416, and the product
will be the circumference nearly.
CONIC SECTIONS. 117
EXAMPLES.
1. The transverse is 40, and the
conjugate 80;' required the circum-
ference of the ellipse.
Here ^ [(40 3 + 30 3 ) -4- 2] =
/ v' [(1600 + UOO) -4- 2] =
v' (2500 -4- 2) = v/ 1250
= 35.3553.
Whence 35.3553 x 3.1416 = 111.0722 = the circum
ference.
2. The transverse diameter is 24, and the conjugate 20;
required the circumference of the ellipse. Jim. 69.4001.
3. The transverse diameter is 60, and the conjugate 40 ;
what is the circumference? Jlns. 160.1907.
4. The transverse diameter is 24, and the conjugate 18 ;
what is the circumference 1 Jlns. 66.6433.
PROBLEM IV.
The transverse and conjugate diameters being given, to
find the area.
RULE. Multiply the transverse diameter by the conjugate,
and this product again by .7854, and the resiilt will be the
area.
EXAMPLES.
1. Required the area of an ellipse whose transverse di-
ameter is 24 perches, and the conjugate 18 perches.
Here 24 x 18 X .7854 = 339.2928 perches = 2 a. r.
19.2928 p., the area required.
2. Required the area of an ellipse whose two diameters
are 60 and 40 rods. Jlns. 11 a. 3 r. 4.96 p.
3. Required the area of an ellipse whose two diameters
are 40 and 36 chains. Jlns. 113 a. r. 15.616 p.
4. The transverse and conjugate diameters are 66 and 22
yards ; what is the area 1 Jlns. 1 140.4 yds.
118 CONIC SECTIONS,
PROBLEM V.
The transverse and conjugate diameters of an ellipse being
given, to find the diameter of a circle containing the same
area.
RULE. Multiply the transverse and conjugate diameters
together, and extract the square root of their product.
EXAMPLES.
1. The transverse and conjugate diameters are 70 and 50
chains ; what is the diameter of a circle containing the same
area?
Here ^/ (70 x 50) = -v/3500 = 59.16079 chains, the
diameter.
2. The transverse and conjugate diameters are 36 and 24
chains ; what is the diameter of a circle containing the same
area ? rfns. 29.3938 chs.
3. The transverse and conjugate diameters are 49 and 16
rods ; what is the diameter of a circle containing the same
area ? jlns. 28 rods.
4. The transverse and conjugate diameters are 87 and 52
feet ; what is the diameter of a circle containing the same
area ? rfns. 67.2606 ft.
PROBLEM VI.
To find the area of an elliptic segment whose base is
parallel to either of the axes of the ellipse.
RULE. Divide the height of the segment by that axis of
the ellipse of which it is a part, and find in the table, p. 295,
a circular segment whose versed sine is equal to the quotient.
Then, multiply the segment thus found and the two axes
of the ellipse together, and the product will be the area
required.
iG
CONIC SECTIONS. 119
EXAMPLES.
1. Required the area of the elliptic
segment whose height A G is 20 chains,
and the axes of the ellipse, 2 A B and
C D, 70 and 50 chains respectively.
Here 20 -r- 70 = .285f , the tabular
versed sine. And the tabular segment ;
belonging to this, as found byProb.XI., c l - : D
Rule 5, p. 104, is .185166.
Whence .185166 x 70 (2AB) x 50 (CD) == 648.081
chains = 64 a. 3 r. 9.296 p., the area of the segment E A F.
2. What is the area of an elliptic segment cut off by an
ordinate parallel to the transverse diameter whose height is
20 feet, the axes being 80 and 50 feet ?
fins. 1173.476 ft.
3. What is the area of an elliptic segment cut off by an
ordinate parallel to the transverse diameter whose height is
10 chains, the axes being 35 and 25 chains ?
Ana. 25 a. 2 r. 27.166 p.
4. What is the area of an elliptical segment cut off by an
ordinate parallel to the conjugate diameter whose height is
10 feet, the axes of the ellipse being 35 and 25 feet ?
Ans. 162.0202 ft.
5. What is the area of an elliptic segment cut off by an
ordinate parallel to the transverse diameter whose height is
5 yards, the axes being 35 and 25 yards ?
dns. 97.8451 yds.
6. What is the area of an elliptic segment cut off by an
ordinate parallel to the conjugate diameter, the axes of the
ellipse being 60 and 40 feet, and the height of the segment
15 feet? dns. 368.51 ft.
7. What is the area of an elliptic segment cut off by a
double ordinate parallel to the conjugate axis, at the distance
of 36 yards from the centre, the axes being 120 and 40 yards ?
tins. 536.7504 yds.
120
CONIC SECTIONS.
THE PARABOLA.
PROBLEM L
17. To describe a parabola, any ordinate to the axis
and its abscissa being given.
Construction. 1 . Let V R and R S be
the given abscissa and ordinate ; bi-
sect the latter in m, join Vra, and draw
m n perpendicular to it, meeting the axis
in n.
2. Make V C and V F each equal to
Rn, and F will be the focus of the curve.
3. Take any number of points r, r, &c.
in the axis, through which draw the
double ordinates-S r S, &c., of an indefi-
nite length.
4. With the radii C F, Cr, &c., and centre F, describe
arcs cutting the corresponding ordinates in the points s, s,
&c., and the curve S V S drawn through all the points of in-
tersection will be the parabola required.
The line s F s passing through the focus F is called the
parameter.
PROBLEM H.
In a parabola, any three of the four following terms being
given, viz., any two ordinates and their two abscissas, to find
the fourth.
RULE. As any abscissa is to the square of its ordinate, so
is any other abscissa to the square of its ordinate. Or as the
square root of any abscissa is to its ordinate, so is the square
root of any other abscissa to its ordinate, and conversely.
CONIC SECTIONS. 121
EXAMPLES.
The abscissa V F is 18, and its or-
dinate E F 12 ; required the ordinate
G H, the abscissa of which, V H, is 32.
Here VF = 18, EF = 12, and
V H = 32.
Whence 18: 12 2 :: 32: 12 * * 32^
18
144x32 OK _ , , , A
= 256 = the square oi the
18
ordinate GH, and ^/ 256 = 16 = the ordinate required.
2. The abscissa VF is 25 and its ordinate EF 16;
required the ordinate G H, the abscissa of which, V H.
is 49.
._ 16^/49 16x7 112
Here </25 : 16:: ^/ 49: - - = - =- =
22.4 = the ordinate G H.
3. The abscissa V F is 9, and the ordinates E F 6 and
GH 8 ; required the abscissa VH.
Here 6 2 : 8 2 : : 9 * = - * =64 -7- 4 = 16 = the
abscissa V H.
4. The abscissa V H is 49, and its ordinate G H 22.4 ; re-
quired the ordinate E F, the abscissa of which, V F, is 25.
Ans. 16, the ordinate E F.
PROBLEM III.
To find the length of any arc of a parabola cut off by
double ordinate.
RULE. To the square of the ordinate add 4 of the square
of the abscissa, and twice the square root of the sum will be
the length of the arc nearly.
11
122
CONIC SECTIONS.
\
EXAMPLES.
1. The abscissa VH is 3, and its
ordinate O H 6 ; what is the length of
the arc G V K ?
Here 6 2 +| X 3 2 =36 +| X 9 =
36 + 12 = 48. And v' (48) x 2 =
0.9282 x 2 = 13.8564 = the length
ofthearcGVK.
2. The abscissa VH is 6, and its ordinate GH 12; re-
quired the length of the arc G V K. Am. 27.7128.
3. The abscissa is 15, and its ordinate 8 ; what is the
length of the arc ? tins. 38.1575.
4. The abscissa is 9, and its ordinate 6 ; what is the
length of the arc ? Ans. 24.
PROBLEM IV.
To find the area of a parabola, its base and height being
given.
RULE. Multiply the base by the height, and two-thirds
of the product will be the area required.
EXAMPLES.
1 . What is the area of the parabola
BAG, whose height A D is 12, and
the base, or double ordinate, BC, 30?
Here B C = 30, and A D = 12.
Whence (30 x 12) x f = 360 x |
o o
= 240 = area required.
2. The abscissa is 18, and the base, or double ordinate,
42 ; what is the area ? Jins. 504.
3. What is the area of a parabola whose abscissa is 11,
and its double ordinate 21 ? Ans. 154.
4. What is the area of a parabola whose height is 9, and
the base, or double ordinate, 24 ? Ans. 144.
CONIC SECTIONS.
123
PROBLEM V.
To find the area of a frustum of a parabola.
R ULE . Divide the difference of the cubes of the two ends
of the frustum by the difference of their squares, and this
quotient, multiplied by two-thirds of the altitude, will give
the area required.
EXAMPLES.
A.
1. In the parabolic frustumBEGD,
the two parallel ends, B D and E G,
are 6 and 10, and the altitude, or part
of the abscissa, C F, is 3 ; what is the
area?
Here B D = 6, E G = 10, and C F
= 3. E'-
Whence (10 s 6 J ) -f- (10 2 6 2 ) =
(lOO'O 210) -4- (100 36) = 784 -7- 64 = 12.25.
Then 12.25 x fe of 3) = 12.25 x 2 = 24.50, the area.
3
2. Required the area of a parabolic frustum, the greater
end of which is 10, the less 6, and the height 4.2.
Ans. 34.3.
3. The greater end of the frustum is 12, the less 10, and
the height 2.4 ; what is the area ? Am. 26f f .
4. The greater end of the frustum is 24, and the less 20,
and the altitude 6 ; what is the area ? Am. 132.3636.
5. Required the area of the parabolic frustum, the greater
end of which is 10, the less 6, and the height 5.
Am. 40.8|.
6. The greater. end of the frustum is 24, the less end 20,
and the height 5| ; what is the area ? Am. 121 j.
7. Required the area of the parabolic frustum, the greater
end of which is 10, the less 6, and the height 4.
Am. 32|.
124 . CONIC SECTIONS.
THE HYPERBOLA.
PROBLEM I.
18. To construct an hyperbola, the transverse and con-
jugate diameters being given.
Construction. 1. Make A B the
transverse diameter, and C D perpen-
dicular to it, the conjugate.
2. Bisect A B in O, and from O, with /
ir- * :
the radius O C or O D, describe the cir-
cle D / C F, cutting A B produced in
F and/, which points will be the two
foci.
3. In A B produced, take any num-
ber of points, n, n, &c., and from F and/, as centres, with
the distances B n, A n, as radii, describe arcs cutting each
other in s, s, &e.
4. Through the several points s, s, &c., draw the curve
s B s, and it will be the hyperbola required.
If straight lines be drawn from the point O, through the
extremities C, D of the conjugate diameter C D, they will be
the asymptotes of the hyperbola, whose property it is to ap-
proach continually to the curve, without ever meeting it.
PROBLEM II.
In an hyperbola, any three of the four following terms
being given, viz., the transverse and conjugate diameters, an
ordinate, and its abscissa, to find the fourth.
CASE 1. The transverse and conjugate diameters, and the
two abscissas being given, to find the ordinate.
RULE. As the transverse diameter is to the conjugate, so
is the square root of the product of the two abscissas to the
ordinate required.
It may be observed that in the hyperbola the less abscissa
added to the transverse diameter gives the greater, and the
transverse diameter subtracted from the greater abscissa
gives the less.
CONIC SECTIONS.
125
EXAMPLES.
1. In the hyperbola B A C,
the transverse diameter is 50,
the conjugate 30, and the less
abscissa AD, 12; required the
ordinate D C.
Here transverse diameter =
50, conjugate = 30, A D == 12, and 12 -f 50 = 62,= the
greater abscissa.
When;e 50 : 30 :: v/(62 x 12) : |^ </ (62 x 12) = \
v/744 = - X 27.27636 = 16.3658
5
50
'the ordinate.
2. The transverse diameter is 24, the conjugate 21, and
the less abscissa 8; what is the ordinate ? Ans. 14.
3. The transverse diameter is 36, and the conjugate 24,
and the less abscissa 12; what is the ordinate? Jlns. 16.
4. The transverse diameter is 120, the conjugate 72, and
greater abscissa 160 ; what is the ordinate ? Jlns. 48.
CASE 2. The transverse and conjugate diameters, and an
ordinate being given, to find the two abscissas.
RULE. As the conjugate diameter is to the transverse, so
is the square root of the sum of the squares of the ordinate
and semi-conjugate to the distance between the ordinate and
centre, or half the sum of the abscissas.
Then the sum of this distance and the semi-transverse
will give the greater abscissa, and their difference will give
the less.
EXAMPLES.
1. The transverse diameter is
36, the conjugate 24, and the or-
dinate, B D, 16 ; what are the
two abscissas ?
Here ^/(le 2 + 12 2 ) = v/(256
+ 144) = x/ 400 = 20.
Then 24 : 36 . : 20 : 30 = sum of the abscissas.
ie
And 30 -(- 18 = 48 = the greater abscissa.
Also 30 18 = 12 = the less abscissa.
11*
126 CONIC SECTIONS.
2. The transverse diameter is 120, the conjugate 72, and
the ordinate 48 ; what are the two abscissas ?
tins. 160 and 40.
3. The transverse and conjugate diameters are 24 and
21 ; required the two abscissas to the ordinate 14 ?
tins. 32 and 8.
4. The transverse being 60, and the conjugate 36 ; re-
quired the two abscissas to the ordinate 24.
Ans. 80 and 20.
CASE 3. The transverse diameter, the two abscissas, and
the ordinate being given, to find the conjugate.
RULE. As the square root of the product of the two
abscissas is to the ordinate, so is the transverse diameter to
the conjugate.
EXAMPLES.
1. The transverse diameter is 36, the ordinate 16, and the
two abscissas are 48 and 12 ; required the conjugate.
Here x/(48 x 12) = ^7576 =24.
Whence 24 : 16 : : 36 : 24 = the conjugate required.
2. The transverse diameter is 60, the ordinate 24, and the
two abscissas are 80 and 20 ; required the conjugate.
Jlns. 36.
3. The transverse diameter is 36, the ordinate 21, and the
abscissas 12 and 48; required the conjugate. Am. 31.5.
4. The transverse diameter is 24, the ordinate 14, and the
abscissas 8 and 32; required the conjugate. Ans. 21.
CASE 4. The conjugate diameter, the ordinate, and the
two abscissas being given, to find the transverse.
RULE 1. Add the square of the ordinate to the square of
the semi-conjugate, and find the square root of their sum.
2. Take the sum or difference of the semi-conjugate and
this root, according as the less or greater abscissa is used, and
then say : As the square of the ordinate is to the product of the
abscissa and conjugate, so is the sum or difference above
found to the transverse required.
CONIC SECTIONS. 127
EXAMPLES.
1. The conjugate diameter is 21, the ordinate 14, and the
less abscissa 8 ; required the transverse.
Here ^/(H 2 + 10.5 2 ) = v/(196 -f 1 10.25) = v/306.25 =.
17.5, and 17.5 + 10.5 = 28.
Also 21 x8 = 168.
Whence 196 (14 2 ) : 168 :: 28: 24= the transverse re-
quired.
2. The conjugate diameter is 72, the ordinate 48, and the
less abscissa 40 ; what is the transverse ? Jlns. 120.
3. The conjugate diameter is 36, the less abscissa 20, and
the ordinate 24 ; required the transverse. Jlns. 60.
4. The conjugate diameter is 31.5, the ordinate 21, and
the greater abscissa 48 ; required the transverse.
dns. 36.
PROBLEM III.
To find the length of any arc of an hyperbola, beginning at
the vertex.
RULE. To 19 times the square of the transverse add 21
times the square of the conjugate; also to 9 times the square
of the transverse add as before 21 times the square of the
conjugate ; and multiply each of these sums by the abscissa.
2. To each of the two products thus found add 15 times
the product of the transverse and the square of the conju-
gate.
3. Then as the less of these results is to the greater, so is
the ordinate to the length of the arc nearly.
EXAMPLES.
1. In the hyperbola BAG,
the transverse diameter is 80, the
conjugate 60, the ordinate B D,
10, and the abscissa A D, 2. 1637;
required the length of the arc
BAG.
Here 2.1637 x [(19 x 80 2 ) + (21 x 60 2 )] = 2.1637 x
(121600 + 75600) = 2.1637 x 197200 = 426681.64.
128 CONIC SECTIONS.
And 2.1637 x [(9 x 80 2 )-f(21 x 60 2 )] =2.1637 x (57600
-f- 75000)== 2. 1637 x 133200= 288204.84.
Whence (15 x 80 x 60 2 ) + 426681.64 = 4320000 +
126681.64 = 4746681 .64.
And (15x80 x60 2 )+ 288204.84 =4320000 +288204.84
= 4608204.84.
Then 4608204.84 : 4746681.64 : : 10 : 10.3005 = the
length of the arc A C. Therefore, 10.3005 x 2 = 20.001 =
the length of the whole arc B A C.
2. The transverse diameter of an hyperbola is 120, the
conjugate 72, the ordinate 48, and the abscissa 40 ; required
the whole length of the curve. rfns. 125.304.
3. The transverse diameter is 80, the conjugate 60, and
the ordinate 16 ; required the length of the arc.
Ans. 17.0856.
4. Required the whole length of the curve of an hyperbola,
to the ordinate 20; the transverse and conjugate axis being
80 and 70. An*. 42.267.
PROBLEM IV.
To find the area of an hyperbola, the transverse, conjugate,
and abscissa being given.
RULE. 1. To the product of the transverse and abscissa,
add 4- of the square of the abscissa, and multiply the square
root of the sum by 21.
2. Add 4 times the square root of the product of the trans-
verse and abscissa to the product last found, and divide the
sum by 75.
3. Then if 4 times the product of the conjugate and ab-
scissa be divided by the transverse, this last quotient multi
plied by the former will give the area required nearly.
EXAMPLES.
1. In the hyperbola BAG,
the transverse axis is 100, the
conjugate 60, and the abscissa, or
height A D, is 50 ; required the
urea.
CONIC SECTIONS. 129
Here 21 v/ [(100 x 50) + (| x 50 2 )]
== 21 X 82.37544710 = 1729.8844.
And [4 </ (100 x 50) -f 1729.8844] -r- 75
= [(4 x 70.7107) + 1729.8844] -4- 75
= (282.8428 + 1729.8844) -=- 75
= 2012.7272 -r- 75 = 26.836362.
Whence [(4 x 50 x 60) -=- 100] X 26.836362 =
(12000 -f- 100) x 26.836362 = 120 X 26.836362 =
3220.3634 = the area required.
2. Required the area of the hyperbola to the abscissa 25,
the two axes being 50 and 30. ' tins. 805.0908.
PROBLEM V.
To find the area of a space A N O B, bounded on one
side by the curve of an hyperbola, by means of equidistant
ordinates.
Let A N be divided into a given
number of equal parts, A C, C E,
&c., and let perpendicular ordi-
nates A B, C D, &c., be erected,
and terminated by any hyperbolic
curve, B D F, &c. ; and let A =
AB + NO, B=CD+GH+
L M, &c., and C = E F -f I K, &c. ; then the common dis-
tance A C of the ordinates, being multiplied by the sum
arising from the addition of A, 4 B, and 2 C, and one-third
of the product taken will be the area very nearly.
EXAMPLES.
1. Given the lengths of 9 equidistant ordinates, 14, 15,
16, 17, 18, 20, 22, 23, and 25 feet, and the common distance
2 feet ; what is the area ?
Here [A C x (A + 4 B + 2 C)] -f- 3 = [2 X (39 -f 300
-f- 112)] -T- 3 = (2 x 451) -r- 3 = 902 -r- 3 = 300 feet.
2. Given the lengths of 3 equidistant ordinates, A B =
5 feet, C D = 7, and E F = 8, the length of the base A E =
10 ; what is the area of the figure A B F E ? *2ns. 68? ft.
MENSURATION OF SOLIDS.
DEFINITIONS.
1O. I- The measure of any solid body is the whole
capacity or content of that body, when considered under the
triple dimensions of length, breadth, and thickness.
2. A cube whose side is one inch, one foot, or one yard,
&c., is called the measuring unit; and the content or solidity
of any figure is estimated by the number of cubes of this
kind which are contained in it.
A
3. A cube is a solid contained by six equal
square sides, or faces, as A B C D E F.
4. A parallelopipedon is a solid con- BJ
tained by six rectangular plane faces,
every opposite two of which are equal
and parallel, as A B C D E F. c
5. A prism is a solid whose ends are two
equal, parallel, and similar plane figures, and its
sides parallelograms ; as A B C D E F.
It is called a triangular prism when its ends
are triangles ; a square prism when its ends are
squares ; a pentagonal prism, when its ends are
pentagons, and so on.
6. A cylinder is a solid described by the revo-
lution of a rectangle about one of its sides as an
axis, which remains fixed ; as A B C D.
13C
MENSURATION OF SOLIDS.
7. A cone is a solid described by the
revolution of a right-angled triangle about
one of its legs, which remains fixed ; as
ABC.
8. A pyramid is a solid whose sides are all
triangles meeting in a point at the vertex, and
the base any plane figure ; as A B G D E, which
is a pentagonal pyramid.
When the base is a triangle, it is called a fri-
angular pyramid; when a square, it is called
a square or quadrangular pyramid ; when a
pentagon, it is called a pentagonal pyramid, &c.
9. A sphere is a solid described by the
revolution of a semicircle about its diameter,
which remains fixed ; as A B C D.
10. The centre of a sphere is a point
within the figure, equally distant from every
point of its convex surface.
1 1 . A diameter of the sphere is a straight line passing
through its centre, and terminated both ways by the convex
surface.
And if it be the diameter about which the generating
semicircle revolves, it is called the axis of the sphere.
12. A circular spindle is a solid
generated by the revolution of a seg-
ment of a circle about its chord, which
remains fixed ; as A B D C.
Elliptic, parabolic, and hyperbolic
spindles are generated in the same
manner as circular spindles, the double ordinate of the sec
tion being always considered as fixed.
13. A spheroid or ellipsoid is a solid
generated by the revolution of a semi-
ellipsis about one of its axes, which re- B
mains fixed ; as A B D C.
The spheroid is called prolate, when
the revolution is made about the trans-
verse axis, and oblate when it is; made
about the conjugate axis.
132 MENSURATION OF SOLIDS.
14. Parabolic and hyperbolic conoids
are solids formed by the revolution of a
semi-parabola or semi-hyperbola about its
transverse axis, which is considered as
quiescent ; as A B D ; and the same for I
any other solid of this kind.
15. A segment of a pyramid, sphere, or of any other solid,
is a part cut off from the top of it by a plane parallel to the
base of the figure.
16. A frustum or trunk, is the part that remains at the
bottom, after the segment is cut off.
17. The zone of a sphere, is that part which is intercepted
between two parallel planes ; and when those planes are
equally distant from the centre, it is called the middle zone
of the sphere.
18. The height of a solid is a perpendicular drawn from
its vertex to the base, or to the plane on which it is supposed
to stand.
19. A wedge is a solid, having a rectangular base, and
two of its opposite sides meeting in an edge.
20. A prismoid is a solid, having on its ends two rect
angles parallel to each other, and its upright sides are foui
trapezoids.
21. An ungula, or hoof, is a part cut off from a solid by a
plane oblique to the base.
The surfaces of all similar solids are to each other as the
squares of their like dimensions ; such as diameters, circum
ferences, like linear sides, &c., &c., and their solidities, as
the cubes of those dimensions.
The solidity of cylinders, prisms, parallelopipedons, &c.,
\vhich have their altitudes equal, are to each other as the
squares of their diameters or like sides. The same remark
is applicable to frustums of a cone or pyramid when the
altitude is the same, and the ends proportional.
MENSURATION OF SOLIDS.
PROBLEM I.
To find the area of the surface of a cube.
. Multiply the square of the length of one side by
the number of sides, and the product will be the area of the
kurface.
EXAMPLES.
1. The side of a cube is 18 inches ; "|
what is the area of its surface ?
Here (18* X 6) = '324 X 6 = 1944 in.
= 13.5 sq. ft.
2. The side of a cube is 25 inches; E
what is the area of its surface ?
tins. 26^ ft.
3. The side of a cube is 12 feet ; what is the area of its
surface ? Am. 864 ft.
4. The side of a cube is 16 feet ; what is the area of its
surface ? tins. 1536 ft.
5. The side of a cube is 19 feet ; what is the area of its
surface ? Ans. 2166 ft.
6. The side of a cube is 10 inches ; what is the area of
its surface ? Ans. 4^f ft.
PROBLEM H.
The area of the surface of a cube being given, to find the
length of the side.
RULE. Divide the area by 6, and extract the square root
of the quotient.
EXAMPLES.
1 . The area of a cube is 2400 square inches ; what is th
length of the side ?
Here v/(2400 -~ 6) = v/400 = 20 inches.
2. The area of a cube is 24 square feet ; what is the length
of the side ? Ans. 2 ft.
12
134 MENSURATION OF SOLIDS.
3. The area of a cube is 216 square feet; what is the
length of the side ? rfns. 6 ft.
4 The area of a cube is 96 feet ; what is the length of
the side ? rfns. 4 ft.
5. The area of a cube is 600 square inches ; what is the
length of the side ? fins. 10 in.
6. The area of a cube is 5400 square inches ; what is the
length of the side ? Ans. 2 ft.
PROBLEM III.
To find the solidity of a cube, the length of one of its
sides being given.
RULE. Multiply the side by itself, and that product again
by the side, or cube the given side, and it will give the
solidity required.
EXAMPLES.
1. The side A B or B C of the cube A B C "
G H E, is 25.5 inches ; what is the solidity ?
Here (ABxBC)xAE= (25.5 x
25.5) x 25.5 = 650.25 x 25.5= 16581.375
cubic inches.
Or, (25.5) 3 =25.5x25.5x25.5=16581.375
cubic inches.
2. What is the solidity of a cube whose side is 5 feet f
dns. 125 ft.
3. The side of a cube is 15 inches ; what is its solidity?
Ans. 1.1)531 ft.
4. What is the solidity of a cube whose side is 5 &', 3
inches ? Jlns. 144^| ft.
5. How many solid feet are contained in a cubic box whose
depth is 32 inches ? Ans. 18| f ft.
6. How many cubic inches are contained in a cubic piece
of timber whose length is 42 inches ? Jlns. 74088 in.
MENSURATION OF SOLIDS. 135
PROBLEM IV.
To find the side of a cube, the solidity being given.
RULE. Extract the cube root of the solidity.
EXAMPLES.
1. What is the length of the side of a cube containing 36
s -lid feet ?
Here ^/36 = 3.3019, the side required.
2. What is the length of the side of a cube containing
274 cubic inches ? Am. 6.4950 in.
3. What is the length of the side of a cube whose solidity
is 1800 cubic inches ? Am. 12.1644 in.
4. What is the length of the side of a cube whose solidity
is 789 cubic feet 1 Am. 9.2404 ft.
x
PROBLEM V.
N.
To find the solidity of a parallelopipedon.
RULE. Multiply the length by the breadth, and that pro-
duct again by the depth, or Altitude, and it will give the
solidity required.
EXAMPLES.
1. Required the solidity of the paral-
lelopipedon A B C G H E, whose
length A B is 9 feet, its breadth A E
65 feet, and its depth, or altitude, E
A D 7| feet.
Here (ABxAExAD) =
(9 x 5.5) x 7.75 = 49.5 x 7.75 =
3S:U>25 solid feet.
2. The length of a parallelopipedon is 8 feet, Us breadth
3 feet, and thickness 2 feet ; how many solid feet does it
contain 1 J) ns . 48 ft.
136 MENSURATION OF SOLIDS.
3. The length of a parallelopipedon is 36 inches, the
width 20 inches, and depth 18 inches ; how many solid feet
does it contain ? Ana. 7.5 feet.
4. The length of a parallelopipedon is 15 feet, and each
side of its square base 21 inches ; what is the solidity ?
Ana. 45.9375 ft.
5. What is the solidity of a block of marble, whose length
is 12 feet, breadth 5? feet, and depth 2 feet ?
Am. 172 ft.
PROBLEM VI.
To find the solidity of a prism.
RULE. Multiply the area of the base by the perpen-
dicular height of a prism, and the product will be the
solidity.
EXAMPLES.
1. What is the solidity of the triangular
prism A B C D E F, whose length A B is
20 feet, and either of the equal sides B C,
C F, or F B, of one of its equilateral ends
B C F, 5 feet ?
Here, by Problem II., page 82, the area of
the base B C F is = 5 a x .433013 = 25 X
.433013 = 10.825325.
And consequently, 10.825325 x 20 =
216.5065 feet, the solidity required.
2. What is the solidity of a triangular prism whose length
is 18 feet, and one side of the equilateral end H feet?
Am. 17.5370 ft.
3. What is the solidity of a triangular prism whose length
is 40 feet, and one side of the equilateral end 18 inches ?
Ana. 38.9711 ft.
4. What is the solidity of a triangular prism whose lenglh
is 24 feet, and one side of the equilateral end 16 inches ?
Ana. 18.4752 ft.
5. What is the value of a prism whose height is 32 fe. %
and each side of the equilateral end 14 inches, at 20 CGI i
per solid foot ? Ana. $3.772.
MENSURATION OF SOLIDS.
137
6. ^Required the solidity of a prism whose base is a hexa-
gon, supposing each of the equal sides to be 1 foot 6 inches,
and the length of the prism 10 feet. rfns. 93.5307 ft.
PROBLEM VII.
To find the convex surface of a cylinder.
RULE. Multiply the circumference or periphery of the
base, by the height of the cylinder, and the product will be
the convex surface required ; to which add the area of each
end, and the sum will be the whole surface of the cylinder
EXAMPLES.
1. What is the convex surface of the
right cylinder A BC D, whose length, BC,
is 24 feet, and the diameter of its base, A B,
16 feet ?
Here 3.1416 x 16 = 50.2650, the circum-
ference of the base.
And 50.2656 x 24 = 1200.3744 square
feet, the convex surface required.
2. What is the whole surface of a right cylinder, the
diameter of whose base is 21 feet, and the height 5 feet ?
Here 3.1410 x 2.5= 7.854, the circumference of the base.
And 7.854 x 5 == 39.27 square feet, the convex surface.
Then (2.5 2 x .7854) x2 = (6.25 x .7854) x 2 = 4.90875 x.2
= 9.8175 square feet, the area of the ends.
Whence 39.27+9.8175 = 49.0875 square feet, the whole
surface.
3. Required the convex surface of a right cylinder, whose
circumference is 8 feet 4 inches, and its length 18 feet.
Ana. 150 ft.
4. What is the convex surface of a right cylinder, the
diameter of whose base is 2 feet, and its length 30 feet?
Am. 188.490ft.
12*
138
MENSURATION OF SOLIDS.
5. What is the whole surface of a right cylinder, the
diameter of whose base is lo inches and its height 20 feet ?
Jlns. 80.5685 ft.
6. How many square yards are contained in the whole
surface of a cylinder, the diameter of whose base is 4 feet
and its length 10 feet ?
,. Jlns. 16.7552yds.
PROBLEM Vin.
To find the solidity of a cylinder.
RULE. Multiply the area of the base by the perpendicu
lar height of the cylinder, and the product will be the
solidity.
EXAMPLES.
1 . What is the solidity of the cylinder
A B C D, the diameter of whose base, A B,
is 30 inches, and the height, B C, 55
inches?
Here (30 2 X .7854) x 55 = (900 X-785 1)
X 55=706.86x55 =38877.3 cubic inches.
2. What is the solidity of a cylinder, whose height is
30 feet, and the circumference of its base 20 feet ?
Here (20 3 x .07958) x 30 = (400 x .07958) x 30
= 31.832 x 30 = 954.90 cubic feet.
3. What is the solidity of a cylinder whose height is>
4 feet, and the diameter of its base 10 inches ?
Ans. 2.1816 ft.
4 The diameter of a cylinder is 16 inches, and the lengtl
20 feet ; what is the solidity ?
rfns. 27.9253ft.
5. The circumference of a cylinder is 2 feet, and the
length 5 feet ; what is the solidity ?
dns. 1.5910 ft.
MENSURATION OF SOLIDS.
139
C. The circumference of a cylinder is 20 feet, and the
height 19.318 feet ; what is the solidity ?
fins. 614.9305ft.
PROBLEM IX. i
To find the curve surface of a cylindric ungula, when
the section passes obliquely through the sides of the cylinder.
RULE. Multiply the circumference of the base by half
the sum of the greatest and the least heights of the ungula,
and the product will be the curve surface.
EXAMPLES.
1. What is the curve surface of a cylin-
dric ungula, the diameter of whose base is
A B, 16 feet, and the greatest and least
heights are B F, 17, and A E, 14 feet ?
Here 3.1416 x 16 = 50.2856, the cir-
cumference of the base.
Then 50.2656 X (17 + 14) -4- 2 =
50.2656 x (31 -s- 2) =50.2656 x 15.5 =
779.1168 feet, the curve surface required.
2. What is the curve surface of a cylindric ungula, the
circumference of whose base is 21 feet, and the greatest and
least heights are 13 and 8 feet? rfns. 220.5 ft.
3. What is the curve surface of a cylindric ungula, the
diameter of whose base is 19 feet, and the greatest and least
heights are 13| and 11| feet? rfns. 746.13 ft.
PROBLEM X.
To find the solidity of a cylindric ungula, when the sec-
tion passes obliquely through the opposite sides of the cylin
der.
RULE. Multiply the area of the base of the cylinder by
half the sum of the greatest and least heights of the unguia,
and the product will be the solidity.
140 MENSURATION OF SOLIDS.
EXAMPLES.
1. What is the solidity of a cylindric ungula, the diame-
ter of whose base, A B, is 12 feet, and the greatest and least
heights are BF, 6, and AE, 4 feet?
Here (12 2 x.7854) = 144 x. 7854 = 113.0976 square feet,
the area of the base.
Then 113.0976 x (6 + 4) -*- 3 = 113.0976 x (10 -4- 2)
= 113.0976 x 5 = 565.488 feet, the solidity required.
2. What is the solidity of a cylindric ungula, the diameter
of the base of which is 10 feet, and the greatest and least
heights are 4 and 3 feet ? Am. 274-89 ft.
3. What is the solidity of a cylindric ungula, the circum-
ference of the base of which is 24 feet, and the greatest and
least heights 18 and 12 feet? Am. 687.5712 feet.
PROBLEM XI.
To find the convex superficies of a cylindric ring.
RULE. To the thickness of the ring add the inner dia-
meter, and this sum being multiplied by the thickness, and
the product again by 9.8696 (or the square of 3.1416) will-
give the superficies required.
EXAMPLES.
1. The thickness, A c, of a cylindric
ring is 3 inches, and the inner diameter,
cd, 12 inches ; what is the convex su-
perficies ?
Here [(12 + 3) x 3] x 9.8696 =
(15 x 3) x 9.8696 = 45 x 9.8696 =
444.132 square inches.
2. The thickness of a cylindric ring is 4 inches, and the
inner diameter 18 inches ; what is the convex superficies ?
fins. 868.5248 inches.
3. The thickness of a cylindric ring is 2 inches, and the
inner diameter 1 loot 6 inches; what is the convex super-
ficies? Ans. 394.784 inches
MENSURATION OF SOLIDS. 141
4. The thickness of a cylindric ring is 3 inches, and its
inner diameter 9 inches ; what is the convex superficies ?
JIns. 355.3056 inches.
5. The thickness of a cylindric ring is 2 inches, and the
inner diameter 12 inches ; what is the convex superficies?
rfns. 276.3488 inches.
6. The thickness of a cylindric ring is 3.5 inches, and its
inner diameter 18.765 feet; what is its convex superficies?
Ana. 7899.4304 inches.
PROBLEM XII.
To find the solidity of a cyiindric ring.
RULE. To the thickness of the ring add the inner diame-
ter, and this sum being multiplied by the square of half the
thickness, and the product again by 9.8096, will give the
solidity.
EXAMPLES.
1. What is the solidity of an anchor ring, whose inner
diameter is 8 inches, and thickness in metal 3 inches ?
Here [(8 + 3) x 1.5 s ] X 9.8696 =(11 X 2.25) x 9.8696
= 24.75 x 9.8690 = 244.2726 cubic inches.
2. What is the solidity of an anchor ring whose inner dia-
meter is 9 inches, and the thickness of metal 3 inches ?
dns. 266.4792 inches.
3. The inner diameter of a cylindric ring is 12 inches,
and its thickness 4 inches ; what is its solidity ?
dns. 631.6544 inchts.
4. Required the solidity of a cylindric ring whose thick-
ness is 2 inches, and its inner diameter 16 inches.
rfns. 177.6528 inches.
5. Required the solidity of a cylindric ring whose inner
diameter is 12 inches, and thickness 5 inches.
Jlns. 1048.645 inches.
6. What is the solidity of a cylindric ring whose thick-
ness is 4 inches, and inner diameter 16 inches ?
Jlns. 789.568 inches.
142 MENSURATION OF SOLIDS.
PROBLEM XIH.
Thi solidity and thickness of a cylindric ring being given,
to find the inner diameter.
RULE. Divide the solidity by 9.8696, and that quotient
by the square of half the thickness, from which subtract the
thickness, and the remainder will be the inner diameter of
the ring.
EXAMPLES.
1. The thickness of a cylindric ring is 4 inches, and its
solidity 789.568 solid inches. What is its inner diameter?
Here 789.568 -r-9.8696 = 80, and SO-s-2 2 = 80-i-4=20,
then 20 4 = 16 inches the diameter.
2. Required the inner diameter of a cylindric ring whose
solidity is 138. 1744 inches, and thickness 2 inches.
JLns. 12 inches.
3. What is the inner diameter of a cylindric ring whose
solidity is 1 solid foot, and thickness 4 inches ?
fins. 39.77 inches.
4. If the solidity of a cylindric ring be 4 solid feet, and
the thickness 3.5 inches, what is the inner diameter ?
rfns. 18.765 feet.
5. What must be the inner diameter of a cylindric ring
whose solidity is 1 solid inch, and thickness J- of an inch ?
rfns. 25.8132 inches.
6. What is the inner diameter of a cylindric ring whose
wlidity is 244.2726 inches, and the thickness 3 inches ?
Ans. 8 inches.
PROBLEM XIV.
To find the surface of a right cone or pyramid.
RULE. Multiply the circumference or perimeter of the
i ase by the slant height or length of the side of the cone or
pyramid, and half the product will be the surface required.
And if this be added to the area of the base, it will give the
whole surface.
MENSURATION OF SOLIDS.
EXAMPLES.
1. The diameter of the base A B, of a
right cone C A B, is 6 feet, and the slant
height A C or B C, 21 feet. Required
the convex surface of the cone.
Here 3.1416 x 6 = 18.8496 = the cir-
cumference of the base.
And (18.8496x21)-*- 2=395.8416-=-2
=197.9208 square feet, the convex sur-
face required.
2. The circumference of a right cone is 10 feet, and the
slant height 12 feet. What is the whole surface of the
cone ?
Here (10 x 12) -r- 2 = 120 -j- 2 = 60 square feet, the con-
vex surface.
And (10 2 x .07958) = 100 X .07958 = 7.958 square feet,
the area of base.
Whence 60 + 7.958 = 67.958 square feet, the whole sur-
face required.
3. Required the whole surface of a triangular pyramid,
each side of its base being 5 feet, and its slant height
171 feet.
Here 5.5 X 3 = 16.5 = the perimeter of the base.
And (16.5 x 17.5) -h 2 = 288.75 -=- 2 = 144.375 square
feet, the outward surface of the pyramid.
Also (5.5 2 x. 433013) = 30.25 x. 433013 = 13.0986 square
feet, the area of 'he base.
Whence 144.375 -f 13.0986 = 157.4736 square feet, the
whole surface required.
4. The slant height of a right cone is 20 feet, and the dia
meter of the base 8 feet ; required the convex surface.
Ans. 251.328 ft.
5. The circumference of a right cone is 27.5 feet, and the
slant height 11 feet; required the convex surface.
rfns. 151.25 ft.
6. The slant height of a right cone is 20 feet, and the
diameter 3 feet ; what is the whole surface ?
Am. 101 3166 ft
J44 MENSURATION OF SOLIDS.
7. Required the outward surface of a triangular pyramid,
each side of its base being 85 feet, and its slant height 14
feet. tins. 73.5 ft.
8. The circumference of a right cone is 10 feet, and the
perpendicular height 12 feet ; required the convex surface.
Ans. 00.525 ft.
PROBLEM XV.
To find the surface of the frustum of a right cone or
pyramid.
RULE. Multiply the sum of the perimeters of the two
ends by the slant height of the frustum, and half the pro-
duct will be the surface required.
EXAMPLES.
1. In the frustum of the cone A B D E,
the circumferences of the two ends A B
and ED, are 22.75 and 15.5 feet respect-
ively, and the slant height, A E, is 26 feet ;
required the convex surface.
Here [(22.75 + 15.5) x 26] -f- 2 =
(38.25 x 26) -^ 2 = 994.5 -^ 2 = 497.25
square feet, the convex surface required.
2. Required the surface of the frustum of a square
pyramid, one side of the base being 12 feet, and of the
upper end 5| feet, and its slant height 40* feet.
Here 12 X 4 = 50 = the perimeter of the base, and
5? x 4 = 23 = the perimeter of the upper end.
Then [(50 + 23) x 40.25] -f- 2 = (73 x 40.25) ~ 2 =
2938.25 -T- 2 = 1469.125 square feet, the surface required.
3. What is the convex surface of the frustum of a right
cone, the circumference of the greater end being 23| feet,
and that of the less end 16| feet, and the length of the slant
side 12 feet ? Ans. 240 ft.
4. What is the convex surface of the frustum of a right
cone, the diameters of the ends being 8 and 4 feet, and the
length of the slant side 20 feet ? Ans. 376.992 ft.
MENSURATION OF SOLIDS.
145
5. Required the surface of a hexagonal pyramid, one side
of the base being 85 feet, and of the upper end 3^ feet, and
the slant height 20 feet. Ans. 733.5 ft.
6. What is the convex surface of the frustum of a right
cone, the diameters of the ends being 5 and 4 feet, and the
slant height 6 feet ? tins. 84.8232 ft.
PROBLEM XVI.
To find the solidity of a cone or pyramid.
RULE. Multiply the area of the base by the perpendicu-
lar height of the cone or pyramid, and one-third of the pro-
duct will be the solidity.
EXAMPLES.
1. Required the solidity of a cone CAB,
whose diameter, A B, is 30 feet, and its per-
pendicular height, G C, 36 feet.
Here (.7854 X 30 2 ) = .7854 X 900 =
706.86 = the area of the base.
And (706.86 x 36) -=- 3 = 25446.96 -^ 3
= 8482.32 feet, the solidity. A
2. Required the solidity of the hex-
agonal pyramid H A D H, each of the
equal sides of its base being 20 feet, and
the perpendicular height, H G, 50 feet.
Here, by the table, page 82, for poly-
gons we have 2.598076 (multiplier when
the side is 1) x 20 3 = 2.598076 x 400 =
1039.2304 = the area of the base.
And (1039.2304 x 50) -j- 3 = 51961.52
-*- 3 = 17320.5066 feet, the solidity.
3. What is the solidity of a cone, the diameter of whose
base is 18 inches, and its altitude 24 feet ?
Ans. 14.1372ft.
13
1-iO MENSURATION OF SOLIDS.
4. If the circumference of the base of a cone be 60 feet
and its height 72 feet ; what is the solidity ?
rfns. 6875.712 ft.
5. What is the solidity of a triangular pyramid, whose
height is 30 feet, and each side of the base 3 ft. ?
dns. 38.9711 ft.
6. What is the solidity of a pentagonal pyramid, its height
being 12 feet, and each side of its base 2 feet ?
Ans. 27.5276 ft.
7. What is the solidity of a cone, whose diameter of the
base is 14 feet, and the slant side being 25 feet ?
Ans. 1231.5072ft.
PROBLEM XVII.
To find the solidity of the frustum of a cone or pyramid.
1. For the frustum of a cone, the diameters of the two
ends and the height being given.
RULE. Divide the difference of the cubes of the diameters
of the two ends by the difference of the diameters, and this
quotient being multiplied by .7854, and again by one-third
of the height, will give the solidity.
Or divide the difference of the cubes of the circumferences
of the two ends, by the difference of the circumferences, and
the quotient being multiplied by .07958, and again by one-
third of the height, will give the solidity.
Or to the product of the diameters add one-third of the
square of their difference, and that sum being multiplied by
.7854, and again by the height, will give the solidity.
Or to the product of the circumferences add one-third of
the square of their difference, and this sum being multiplied
by .07958, and again by the height, will give the solidity.
2. For the frustum of a pyramid, the sides of the base
ond the height being given.
RULE. To the areas of the two ends of the frustum add
the square root of their product, and this sum being multi-
plied by one-third of the height, will give the solidity.
MENSURATION OF SOLIDS.
147
EXAMPLES.
1. What is the solidity of the frustum
of the cone A B D E, the diameter of whose
greater end, A B, is 6 feet, that of the less
end, E D, 4 feet, and the perpendicular
height, F G, 9 feet ?
Here (6 1 4 s ) -f- (6 4) = (216 64)
-4-2 = 152 -4-2 = 76.
And (76 x .7854) x (9 -j- 3) = 59.6904
X3 = 179 .0712 feet, the solidity required.
Or [(6 x 4) + (6 - 4) a + 3] = 24 + (2'-4- 3) = (24+-*)
=25i. And (25 x.7854) x 9 = 19.8968 x 9=179.0712
feet, the solidity, as hefore.
2. What is the solidity of the frustum of a cone, the cir-
cumference of the greater end being 40 feet, and that of the
less 20 feet, and the length or height 51 feet ?
Here (40 3 20 5 ) -f- (40 20) = (64000 8000) ~ 20
= 56000 -4- 20 = 2800.
And (2800 x .07958) x (51 -4- 3) = 222.824 x 17
= 3783.008 feet, the solidity required.
Or [(40 x 20) + (40 20) 2 -4-3] = 800 + (20 a -^3)=800
+ 133^ = 933|, and (9331 x .07958) x 51 = 74.274666
X 51 =3788.008 feet, the solidity, as before.
3. What is the solidity of the frustum aADd of an
hexagonal pyramid, the side A B of whose greater end is
4 feet, that ab of the less end 3, and the height G g 9 feet .'
Here 2.598076 (the tabular multiplier)
x 3 2 = 2.598076 x 9 = 23.382684 the
area of the less end.
And 2.598076 x 4 C = 2.598076 x 16
= 41.5(39216 the area of the greater
end.
Whence ^(23.382684x41.569216)
= v/971.999341 = 31.176912.
And (23.3S2G84 + 41.569216 +31.
176912) x (9 -4- 3) = 96.128812 x 3 = 288.3864556 feet.
4. What is the solidity of the frustum of a cone, the dia-
meter of the greater end being 4 feet, that of the less end 2,
und the altitude 9 feet ? Jlns. 65.9736 feet
148 MENSURATION OF SOLIDS.
5. What is the solidity of the frustum of a cone, the dia-
meter of the greater end of which is 5 feet, that of the less
end 3 feet, and the altitude 4 feet ? Ans. 51.3128 feet.
6. What is the solidity of the frustum of a cone, the cir-
cumference of the greater end being 20 feet, that of the less
end 10 feet, and the length or height 21 feet ?
tins. 389.942 feet.
7. What is the solidity of the frustum of a cone, the cir-
cumference of the greater end being 12 feet, that of the less
end 8 feet, and the height 5 feet ? Am. 40.3205 feet.
8. What is the solidity of the frustum of a square pyra-
mid, one side of the greater end being 18 inches, that of the
less end 15 inches, and the altitude 60 inches ?
Ans. 9.4791 feet.
9. What is the solidity of the frustum of an equilateral
triangular pyramid, one side of the greater end being 14
inches, that of the less end 8 inches, and the height 10 feet?
Ans. 3.7287 feet.
10. What is the solidity of the frustum of a pentagonal
pyramid, the side of whose greater end is 18 inches, that of
the less end 12 inches, and the height 4 feet 6 inches ?
Ann. 12.2583 feet
PROBLEM XVm.
The solidity and altitude of a cone being given, to I nd
the diameter.
RULE. Divide the solidity by the product of .7854, nd
one-third of the altitude, and the square root of the quol snt
will be the diameter.
EXAMPLES.
1. The solidity of a cone is 16 feet, and the altitui 9
feet ; what is the diameter ?
Here x/{16 -- [.7854 x (9 + 3)]} = v /[16-f-(.7854x-<)]
feet, the diame>r.
2. The, altitude of a cone is 15 feet, and the solidity 30
feet ; what is the diameter ? Ans. 2.7639 ft. *.
MENSURATION OF SOLIDS. 149
3. The solidity of a cone is 18 feet, and the altitude 9
feet; what is the diameter? Ans. 2.9316 feet.
PROBLEM XDC.
The solidity and diameter of a cone being given, to find
the altitude.
RULE. Divide the solidity hy the product of .7854 and
the square of the diameter, and the quotient, being multiplied
by 3, will give the altitude.
EXAMPLES.
1. The solidity of a cone is 30 feet, and the diameter
2 feet ; what is the altitude ?
Here [30 -4- (.7854 x 2*)] X 3 = [30 - (.7854 x 4)] x 3
= (30 -=- 3.1410) x 3 = 9.5492 x 3 = 28.6476 feet, the al-
titude.
2. The diameter of a cone is 20 inches ; what must be
the altitude, to make 20 solid feet ? Ans. 27.5019 ft.
3. The solidity of a cone is 2513.28 feet, and the diame-
ter 20 feet ; what is the altitude 1 Ans. 24 ft.
PROBLEM XX.
The altitude of a cone or pyramid being given, to divide
it into two or more equal parts, by sections parallel to the
base, to find the perpendicular height of each part.
RULE. Multiply the cube of the altitude by the numera-
tor of the proportion left at the vertex, and divide the pro
duct by the denominator; the cube root of the quotient will
be the altitude of the cone or pyramid left at the vertex.
EXAMPLES.
1 . The altitude of the cone A B C is
10 feet, to be divided into three equal
parts by sections parallel to the base ;
required the perpendicular height of each
part.
Here ^[(10x 1) -5-3] = N V[(1000xl)
-5-3]= v 3/( 1000 -r- 3) = x 3/333 = 6.9336
feet, the altitude of the first section = C H.
13*
150 MENSURATION OF SOLIDS.
And # [(10 3 x2)-=-8] = ^[(1000 x 2) -r-3] = ^(2000
-f. 3) = # 666f = 8.7358 feet, the altitude C I. Now
C I C H == 8.7358 6.9336 = 1.8022 feet, the altitude
of the second section I H.
Then C K - C I = 10 - 8.7358 = 1.2642 feet, the alti-
tude of the third section I K.
2. The altitude of a pyramid is 12 feet, to be divided into
three equal parts by sections parallel to the base ; required
the perpendicular height of each part.
Ans. 8.3203, 2.1626, and
1.5171 feet, the altitudes.
3. The altitude of a cone is 20 feet, to be divided into
four equal parts by sections parallel to the base ; required
the perpendicular height of each part.
Ann. 12.5992, 3.2748, 2.2972, and
1.8288 feet, the altitudes.
PROBLEM XXI.
To find the solidity of an ungula when the section passes
through the opposite extremities of the ends of the frustum.
RULE. From the square of the greater diameter subtract
the square root of the product of the two diameters, multi-
plied by the less diameter.
This difference being divided by the difference of the
diameters, and the quotient, multiplied by the greater diame-
ter, that product by the height, and the last product by .2618
will give the solidity.
EXAMPLES.
1. Required the solidity of a conical
ungula, the diameter of the greater end
being 5 feet, that of the less end 1 .8 feet,
and the height 12 feet.
Here [5 1.8 */ (5 X 1.8)] -~ (5
1.8) = [(25 1.8^9) -T- 3.2] = [25
(1.8 x 3) H- 3.2] = [(25 -5.4) -5-3.2]=
(19 6-1-3.2) = 6.125.
MENSURATION OF SOLIDS. 151
And (6.125 x 5 x 12 x .2618) = 96.2115 feet, the so-
lidity.
2. Required the solidity of a conical ungula, the diametei
of the greater end being 10 feet, that of the less end 2| feet,
and the height 15 feet. tfns. 458.15 ft.
3. Required the solidity of a conical ungula the diameter of
the greater end being 4.23 inches, that of the less end 3.7
inches, and the height 5.7 inches.
rfns. 38.7692 in.
PROBLEM XXII.
To find the solidity of a cuneus or wedge.
RULE. Add twice the length of the base to the length of
the edge, then multiply this sum by the height of the wedge,
and again by the breadth of the base, and one sixth of the
last product, will be the solidity.
EXAMPLES.
1. How many solid feet are there in
a wedge whose base is 5 feet 4 inches
long, and 9 inches broad, the length of
the edge being 3 feet 6 inches, and the
perpendicular height*2 feet 4 inches ?
Here 5 ft. 4 in. = 64 in., 3 feet 6 in.
= 42 in. and 2 ft. 4 in. = 28 in.
Then (64 x2+42)x28 =(128+42) A
X 28 = 170 x 28 = 4760.
And (47(50 x 9) -j- 6 = 42840 -- 6 = 7140 solid inches.
Whence 7140 -i- 1728 = 4.1319 solid feet.
2. The length and breadth of the base of a wedge are
35 and 15 inches, the length of the edge 55 inches,
and the perpendicular height 18 inches; what is the
solidity ? rfns. 3.2552 ft.
3. The length and breadth of the base of a wedge are
27 and 8 inches, the length of the edge 36 inches, and
the perpendicular height 3 feet 6 inches ; what is the
solidity? rfns. 2.9166 ft.
162 MENSURATION OF SOLIDS.
PROBLEM XXIII.
To find the solidity of a prismoid.
RULE. Multiply the sum of the lengths of the two ends
by the sum of the breadths, and add this product to the sum
of the areas of the two ends ; multiply the result by one
sixth of the height, and the product will be the solidity.
EXAMPLES.
1. What is the solidity of a rectan-
gular prismoid, the length and breadth
of one end being 14 and 12 inches, and
the corresponding sides of the other 6
and 4 inches ; and the perpendicular
30-a feet?
Here (by Prob. I, p. 61) E F x E H
= 6 x 4 = 24 square inches, the area
of the less end, E F G H.
And, by the same problem ABxAD = 14xl2 = 168
square inches, the area of the greater end, A B C D.
Also 30<|ft. = 366 inches, calling the height h,
[(A B + E F) x (A D + E H) + (E F G H +
A B.C D)] x = [(14 + 6) x (12 + 4) + (168 -f 24)] '^.
= [(20 X 16) -f 192] X 61 = (320+192) X 61 =512 x 61
= 31232 cubic inches, the solidity required.
2. What is the solidity of a rectangular prismoid, the
length and breadth of one end being 12 and 8 inches, and
the corresponding sides of the other 8 and 6 inches, and the
perpendicular height 5 feet? fins. 2.4537 ft.
3. What is the solidity of a stick of hewn timber, whose
ends are respectively 30 by 27 inches, and 24 by 18 inches,
and whose length is 48 feet ? Ans. 204 ft.
4. What is the capacity of a coal wagon, whose inside
dimensions are as follows : at the top the length is 7 feet,
and breadth 6 feet, at the bottom the length is 5 feet, and
breadth 3 feet, and the perpendicular depth is 4 feet?
tins. 110ft
MENSURATION OF SOLIDS. 153
PROBLEM XXIV.
To find the convex surface of a sphere.
RULE. Multiply the diameter of the sphere by its cir-
cumference, and the product will be the convex superficies
required.
The curve surface of any zone or segment will also be found
by multiplying its height by the whole circumference of the
sphere.
EXAMPLES.
1. What is the convex surface of a
globe A D B C, whose diameter, A B,
is 16 inches ?
Here (3.1416x16) x 16 =50.2650
X 16 = 804.2496 square inches, the
surface required.
2. What is the convex surface of a
sphere whose diameter is 10 feet ?
Ans. 314.16 ft.
3. What is the convex surface of a sphere whose diameter
is 4 feet ? Ana. 50.2656 ft.
4. The diameter of a globe is 21 inches ; what is the con-
vex surface of that segment of it whose height is 4% inches?
Jlns. 296.8812 inches.
5. What is the convex surface of a sphere whose diame-
ter is 6 feet ? Ana. 1 13.0976 ft.
6. If the diameter of the globe we inhabit be 7935 rniles;
what is the convex surface ? Jlns. 197808409.26 miles.
PROBLEM XXV.
To find the solidity of a sphere or globe.
RULE. Multiply the cube of the diameter by .5236, and
the product will be the solidity.
154 MENSURATION OF SOLIDS.
EXAMPLES.
1. What is the solidity of a globe whose diameter is 4.5
feet?
Here (4.5 s x .5236) = 4.5 x 4.5x4.5x.5236=91.125x
.5236 =47.71305 solid feet.
2. What is the solidity of a globe whose diameter is 3$
feet? Ans. 22.4493ft.
3. What is the solidity of a. globe whose diameter is 17
inches? Ans. 1.4886ft.
4. What is the solidity of a globe whose diameter is 3
feet 4 inches ? Ans. 19.3925 ft.
5. How many cubic miles are contained in the solidity of
the earth, if its diameter be 7935 miles ?
Ans. 261601621246.35 miles.
PROBLEM XXVI.
The convex surface of a globe being given, to find its dia-
meter.
RULE. Multiply the given area by .31831, and the square
root of the product will be the diameter.
EXAMPLES.
1. What is the diameter of that globe, the area of whose
convex surface is 14 square feet ?
Here ^(14 x .31831) = ^4.45634 = 2.1110 feet, the
diameter required.
2. The convex surface of a sphere is one square rood ;
what is its diameter ? Ans. 3.5682 rods.
3. The expense of gilding a ball at $1.80 per square foot
is thirty-four dollars ; what is its diameter ?
Ans. 2.452ft.
MENSURATION OF SOLIDS.
155
PROBLEM XXVII.
The solidity of a globe being given, to find the diameter.
RULE. Divide the solidity by .5236, and extract the cube
toot of the quotient.
EXAMPLES.
1. The solidity of a globe is 2000 solid inches ; what is
its diameter ?
Here & (2000-=-. 5236) == ^/3819.7097 = 15.631 inches
the diameter.
2. The solidity of a globe is 10 solid feet ; what is its dia-
meter 1 j3ns. 2.67 ft.
3. What is the circumference of a globe whose solidity
is 8 solid feet ? rfns. 7.7952 ft. '
PROBLEM XXVIII.
To find the solidity of the segment of a sphere.
RULE. To three times the square of the radius of its base
add the square of its height ; and this sum multiplied by the
height, and the product again by .5236, will give the solidity.
Or, from three times the diameter of the sphere subtract
twice the height of the segment, multiply by the square of
the height, and that product by .5230 ; the last product will
be the solidity.
EXAMPLES.
1. The radius An of the base of the
segment CAB is 7 inches, and the
height Cn 4 inches ; what is the soli-
dity?
Here [(7 a x 3) + 4 2 ] x4= [(49 x 3)
+ 16] x 4 = (147 + 16) x 4 = 163
X 4 == 652.
Then 652 x .5236 = 341.3872 solid
inches
156 MENSURATION OF SOLIDS.
2. The diameter of a sphere is 6 inches. What is the
solidity of the segment whose height is 2 inches ?
Here Q6 x 3) - (2 x 2)] x 2 9 = (18 4) x 4 = 14 x 4
= 56. Then 56 x .5236 = 29.3216 solid inches.
3. What is the solidity of a spherical segment, the dia-
meter of its base being 40 inches, and the height 10 inches I
3.9391 feet.
4. The diameter of a sphere is 18 inches ; what is the soli-
dity of a segment cut from it, the height being 3 inches ?
Ans. 226. 1952 inches.
5. The diameter of a spherical segment is 20 inches, and
the height 6 inches ; how many gallons of water will it
hold, each gallon containing 282 cubic inches ?
Ans. 3.743 gallons.
PROBLEM XXIX.
To find the solidity of a frustum or zone of a sphere.
RULE. To the sum of the squares of the radii of the two
ends, add one-third of the square of their distance, or of the
breadth of the zone, and this sum multiplied by the said
breadth, and the product again by 1.5708, will give the so-
lidity.
EXAMPLES.
1. What is the solidity of the zone
A BC D, whose greater diameter, A B,
is 1 foot 8 inches, the less diameter,
D C, 1 foot 3 inches, and the distance
nm of the two ends 10 inches ?
Here [(Am a + Dn 9 ) + $ (nm) 8 ]
X nm x 1.5708 = [(10 3 + 7 9 ) + (10 8
-^-3)]xlO x 1.5708=(100 + 561+331) x 15.7(58=189 T \
X 15.708 = 2977.975 cubic inches the solidity of the
zone required.
2. What is the solidity of a zone whose greater diameter
is 9 feet 3 inches, less diameter 6 feet 9 inches, and height
5 feet 6 inches. Ans. 370.3242 feet.
3 What is the solidity of a zone, whose greater diameter
MENSURATION OF SOLIDS. 157
is 2 feet, the less diameter 1 foot 8 inches, and the distance
of the ends 4 inches ? rfns. 1560.0112 inches.
4. Required the solidity of the middle zone of a sphere,
whose top and bottom diameters are each 3 feet, and the
breadth of the zone 4 feet? Ans. 61.7848 feet.
PROBLEM XXX.
To find the solidity of a circular spindle, its length and
middle diameter being given.
RULE. To the square of half the length of the spindle, or
longest diameter, add the square of half the middle diameter,
and this sum divided by the middle diameter will give the
radius of the circle.
2. Take half the middle diameter from the radius thus
found, and it will give the central distance ; or that part of
the radius that lies between the centre of the circle, and that
of the spindle.
3. Find the area of the generating circular segments, by
Problem XL, rule 5, page 104.
4. From one-third of the cube of half the length of the
spindle, subtract the product of the central distance, and
half the area last mentioned, and the remainder multiplied
by 12.5064, will give the solidity of the spindle.
EXAMPLES.
I. The longest diameter, AB,
of the circular spindle A DBG,
is 48 inches, and the middle dia-
meter, C D, 36 inches ; what is
the solidity of the spindle ? F' _ j ! G
First (Ae 2 +Ce 2 ) -=- CD =(24 a
-f IS 2 ) ~S6 = (576 + 324) -=- 36 = 900 -=- 36 = 25, the
radius OC.
Second, OC Ce = 25 18 = 7, O e, the central dis-
tance.
Third, (by Problem XL, rule 5, page 104,) Ce-r-2OC
-= 18 -i- 50 = .36, the tabular versed sine, against which
.stands .254550 the tabular segment.
14
MENSURATION OF SOLIDS
Here .254550 x 50 a =. 254550 x 2500 = 636.375, the
area of the segment A B C A.
Fourth, AAe 3 - (i A EGA x Oe)x 12.5664= [(24 3 -i-3)
- (636.375-r-2)x7]x 12.5664 = [(13824-5-3) (318.1875
X7)] x 12.5664= (4608-2227.3125) x 12.5664=2380.6875
X 12.5664 = 29916.6714, solidity of the spindle.
2. If the length of a circular spindle be 40 inches, and its
middle diameter 30 inches, what is its solidity ?
tins. 17312.8886 inches.
PROBLEM XXXI.
To find the solidity of the middle frustum of a circular
spindle, its length, the middle diameter, and that of either
of the ends, being given.
RULE 1. Divide the square of half the length of the frus-
tum by half the difference of the middle diameter, and that of
either of the two ends ; and half this quotient, added to one-
half of the said difference, will give the radius of the circle.
2. Find the central distance, and the revolving area, as
in the last problem.
3. From the square of the radius take the square of the
central distance, and the square root of the remainder will
give half the length of the spindle.
4. From the square of half the length of the spindle take
one-third of the square of half the length of the frustum, and
multiply the remainder by the said half length.
5. From this product take that of the generating area and
central distance, and the remainder multiplied by 6.2832
will give the solidity of the frustum.
EXAMPLES.
1. What is the solidity of the
middle frustum, A B C D, of a
circular spindle, whose middle
diameter, nm, is 36 inches, the
diameter D A or C B, of the end
i6 inches, and its length or 40
inches ?
MENSURATION OF SOLIDS. 159
First, [Oe 9 ~ (we Do)] + | (ne Do) = % [20* -=-(18
8)]+ (18 8) = i (400 -H 10) + (10 -T- j =20+5
= 25, the radius of the circle.
Second, 25 ne = 25 18 = 7, the central distance.
And ne Do = 18 8 = 10, the versed sine of the
arc On.
Then, by Problem XL, rule 5, page 104, 10 -4- (25 X 2)
= 10 -T- 50 = .2, the tabular versed sine; against which
stands .111823 the tabular segment.
Hence .11 1828 x 50 a =.l 118*23 X 2500 = 279.5575, the
area of the revolving segment DCwD.
Again, by Problem 1, page 01, or x Do = 40 x 8 = 320,
the area of the rectangle DorCD.
And DCD -f D or CD = 279.5575 + 320 = 599.5575,
the generating area OrCnDo.
Third, ,/(26 7 s ) = >/(625 49) = ^7576 = 24 =
Ee, half the length of the spindle.
Fourth, {[(Ee 8 | Oe 2 ) x Oe] (O/nDox7)}X 6.2832
= {[(24 2 133 j) x 20] (599.5575 X 7)} X 6.2832 =
[(442f x 20) 4196.9025] x 6.2832 = (8853.3333^
4196.9025) x 6.2832 = 4656.4308J x6.2832 = 29257.2862
cubic inches, the solidity of the middle frustum AmBCnDA
required.
2. The middle diameter of the frustum of a circular spin-
dle is 2 feet 8 inches, the diameter at the end is 2 feet, and
the length 3 feet 4 inches ; what is the solidity ?
Ans. 27285.0882 inches.
PROBLEM XXXII.
To find the solidity of a spheroid, its two axes being given.
RULE. Multiply the square of the revolving axis by the
fixed axis, and this product again by .523(5, or one-sixth of
3.141(5, and it will give the solidity required.
EXAMPLES.
1. In the prolate spheroid A BCD,
the transverse or fixed axis, AC, is 3 ft,
and the conjugate or revolving axis,
DB, is" 2 feet ; what is the solidity ?
Here (2- x 3) x .523(5 = (4 X 3)
X.5-.J30 = 12 X .523G = :i.2b32 feet,
the solidity required.
1GO MENSURATION OF SOLIDS.
3. What is the solidity of a prolate spheroid whose trans-
verse or fixed axis is 4 feet 2 inches, and conjugate or re-
volving axis 3 feet 4 inches? rfns. 24.2407 ft.
3. What is the solidity of a prolate spheroid whose fixed
axis is 8 feet 4 inches, and its revolving axis 5 feet ?
Am. 109.0833ft.
4. What is the solidity of an oblate spheroid whose con-
jugate or fixed axis is 5 feet, and its transverse or revolving
axis 8 feet 4 inches ? Ans. 181.8055 ft.
5. What is the solidity of an oblate spheroid whose fixed
axis is 30 inches, and its revolving axis 40 inches ?
Ans. 14.5444ft.
PROBLEM XXXIII.
To find the solidity of the middle frustum of a spheroid,
its length, the middle diameter, and that of either of the ends,
being given.
CASE 1. When the ends are circular, or perpendicular to
the fixed axis.
RULE. To twice the square of the middle diameter add
the square of the diameter of either of the ends, arid this sum
multiplied by the length of the frustum, and the product
again by .2618 (or one-twelfth of 3.141(5), will give th
solidity.
EXAMPLES.
1. In the middle frustum of a
prolate spheroid E F G H, the mid-
dle diameter, B D, is 50 inches, and /
that of either of the ends E F or
G H, is 40 inches, and its length,
nm, 18 inches ; what is its solidity?
Here [(50 2 x2 + 40-)xl8] X .2618 = [(2500x2+1600)
Xl8]x.2618= [(5000+ 1600) x 18] x. 2618= (6600 x 18)
X .2618 = 11&800 x .2618 = 31101.84 cubic inches, the
solidity required.
2. What is the solidity of the middle frustum of a prolate
spheroid, the middle diameter being 5 feet, that of either of
the two ends 3 feet, and the distance of the ends 6 feet 8
inches Ans. 102.9746 feet.
MENSURATION OF SOLIDS. 161
3. What is the solidity of the middle frustum of an oblate
spheroid, the middle diameter being 100 inches, that of
either of the ends 80 inches, and the distance of the ends
36 inches ? Ans. 248814.72 inches.
4. What is the solidity of the middle frustum of a prolate
spheroid, the middle diameter being 5 feet, that of either oi
the ends 3 feet, and the distance of the ends G feet ?
Ans. 92.6772 ft.
CASE 2. When the ends are elliptical or perpendicular
to the revolving axis.
RULE 1. Multiply twice the transverse diameter of the
middle section by its conjugate diameter, and to this product
add the product of the transverse and conjugate diameters
of either of the ends.
2. Multiply the sum thus found, by the distance of the
ends or the height of the frustum, and the product again by
.2018, and it will give the solidity required.
EXAMPLES.
1. In the middle frustum, A BC D,
of a prolate spheroid, the diameters of
the middle section are 50 and 30 inches;
those of the end 40 and 24 inches ; and
its height, 2 on, 18 inches ; what is the
solidity ?
Here {[(50x2 x 30) + (40x24)]x 18} x .2618=[(3000
+ 9(50) x 18] x .2618 = (3060 x 18) x .2618 = 71280 X
.2618 = 18661.104 cubic inches, the solidity required.
2. In the middle frustum of a prolate spheroid, the diame-
ters of the middle section are 100 and 60 inches ; those of
the end 80 and 48 inches ; and the length 36 inches ; what
is the solidity ? tfns. 86.3940 ft.
3. In the middle frustum of an oblate spheroid, the
diameters of the middle section are 100 and 60 inches ;
those of the end 60 and 36 inches ; and the length 80
inches ; what is the solidity of the frustum ?
171.6244ft.
14'
A TABLE OF THE AREAS OF THE SEGMENTS
OF A CIRCLE,
Whose diameter is Unity, and supposed to be divided into
1000 equal Parts.
Versed
Sine.
Seg. Area.
Versed
Sine.
Seg. Area.
Versed
Sine.
Seg. Area.
.001
.000042
.034
.008273
.007
.022652
.002
.000119
.035
.008638
.008
.023154
.003-
.000219
.036
.009008
.069
.023659
.004
.000337
.037
.009383
.070
.024168
.005
.000470
.038
.009763
.071
.024080
.006
.000618
.039
.010148
.072
.025195
.007
.000779
.040
.010537
.073
.025714
.008
.000951
.041
.010931
.074
.020236
.009
.001135
.042
.011330
.075
.026761
.010
.001329
.043
.011734
.076
.027289
.011
.001533
.044
.012142
.077
.027821
.012
.001740
.045
.012554
.078
.028356
.013
.001968
.046
.012971
.079
.028894
.014
.002199
.047
.013392
.080
.029435
.015
.002438
.048
.013818
.081
.029979
.010
.002685
.049
.014247
.082
.030526
.017
.002940
.050
.014681
.083
.031076
.018
.003202
.051
.015119
.084
.031629
.019
.003471
.052
.015561
.085
.032186
.020
.003748
.053
.016007
.086
.032745
.021
.004031
.051
.016457
.087
.033307
.022
.004322
.055
.016911
.088
.033872
.023
.004618
.056
.017369
.089
.034441
.024
.004921
.057
.017831
.090
.035011
.025
.005230
.058
.018296
.091
.035585
.026
.005546
.059
.018766
.092
.036162
.027
.005867
.060
.019239
.093
.036741
.028
.006194
.061
.019716
.094
.037323
.029
.006527
.062
.020190
.095
.037909
.030
.006865
.063
.020680
.096
.038490
.0531
.007209
.064
.021108
.097
.039087
1 .032
.007558
.065
.021659
.098
.039680
.033
.007913
.006
.022154
.099
.040276
205
96
THE AREAS OF THE SEGMENTS OF A CIRCLE.
Versed
Sine.
Seg. Area.
Versed
Sine.
Seg. Area. 1
Versed
Sine.
Seg. Aifa.
.100
.040875
.141
.067528
.182
.097674
.101
.041476
.142
.068225
.183
.098447
.102
.042080
.143
.068924
.184
.099221
.10*
.042687
.144
.069625
.185
.099997
.104
.043296
.145
.070328
.186
.100774
.105
.043908
.146
.071033
.187
.101553
.106
.044522
.147
.071741
.188
.102334
.107
.045139
.148
.072450
.189
.103116
.108
.045759
.149
.073161
.190
.103900 1
.109
.046381
.150
.073874
.191
.10^685
.110
.047005
.151
.074589
.192
.105472
.111
.047632
.152
.075306
.193
.106261
.112
.048262
.153
.076026
.194
.107051
.113
.048894
.154
.076747
.195
.107842
.114
.049528
.155
.077469
.196
.108636
.115
.050165
.156
.078194
.197
.109430
.116
.050804
.157
.078921
.198
.110226
.117
.051446
.158
.079649
.199
.111024
.118
.052090
.159
.080380 I
.200
.111823
.119
.052736
.160
.081112
.201
.112624 ;
.120
.053385
.161
.081846
.202
.113426
.121
.054036
.162
.082582
.203
.1142530
.122
.054689
.163
.083320
.204
.115035
.123
.055345
.164
.084059
.205
.115842
.124
.056003
.165
.084801
.206
.116650
.125
.056663
.166
.085544
.207
.117460
.126
.057326
.167
.086289
.208
.118271
.127
.057991
.168
.087036
.209
.119083
.128
.058658
.169
.087785
.210
.119897
.129
.059327
.170
.088535
.211
.120712
.130
.059999
.171
.089287
.212
.121529
.131
.060872
.172
.090041
.213
.122347
.132
.061348
.173
.090797
.214
.123167
.133
.06202o
.174
.091554
.215
.123988
.134
.062707
.175
.092313
.216
.124810
.135
.063389
.176
.09:1074
.217
.125634
.136
.004074
.177
.093836
.218
.126459
.137
.064760
.178
.094601
.219
.127285
.138
.005449
.179
.095366
.220
.128113
.139
.066140
.180
.096134
.221
.128942
.140
.0668*3
.181
.096903
.222
.129773
THE AREAS OF THE SEGMENTS OF A CIRCLE. 297
f Versed
Sine.
Seg. Area.
Versed
Sine.
Seg. Area.
Versed
Sine.
Seg. Area.
.223
.130005
.264
.165780
.305
.202761
.224
.131438
.265
.166663
.306
.203683
.225
.132272
.266
.167546
.307
.204605
.226
.133108
.267
.168430
.308
.205527
.227
.133945
.268
.169315
.309
.206451
.228
.134784
.269
.170202
.310
.207376
.229
.135624
.270
.171089
.311
.208301
.230
.136465
.271
.171978
.312
.209227
.231
.137307
.272
.172867
.313
.210154
.232
.138150
.273
.173758
.314
.211082
.233
.138995
.274
.174649
.315
.212011
.234
.139841
.275
.175542
.316
.212940'
.235
.140688
.276
.176435
.317
.213871
.236
.141537
.277
.177330
.318
.214802
.237
.142387
.278
.178225
.319
.215733
.238
143238
.279
.179122
.320
.216666
.239
.144091
.280
.180019
.321
.217599
.240
.144944
.281
.180918
.322
.218533
.241
.145799
.282
.181817
.323
.219468
.242
.146655
.283
.182718
.324
.220404
.243
.147512
.284
.183619
.325
.221340
.244
.148371
.285
.184521
.326
.222277
.245
.149230
.286
.185425
.327
.223215
.246
.1500!)!
.287
.186329
.328
.224154
.247
.150953
.288
.187234
.329
.225093
.248
.151816
.289
.188140
.330
.226033
.249
.152680
.290
.189047
.331
.226974
.250
.153546
.291
.189955
.332
.227915
.251
.154412
.292
.190864
.333
.228858
.252
.155280
.293
.191775
.334
.229801
.253
.156149
.294
.192684
.335
.230745
.254
.157019
.295
.193596
.336
.231689
.255
.157890
.296
.194509
.337
.232634
.256
.158762
.297
.195422
.338
.233580
.257
.159636
.298
.196337
.339
.234526
.258
.160510
.299
.197252
.340
.235473
.259
.161386
.300
.198168
.341
.236421
.260
.162263
.301
.199085
.342
.237369
.261
.103140
.302
.200003
.343
.238318
.2(52
.164019
.303
.200922
.344
.239268
.263
.164899
.304 .201841
.345
.2402 is
298 THE AREAS OF THE SEGMENTS OF A CIRCLE.
Versed
Sine.
Seg. Area.
Versed
Sine.
Seg. Area.
Versed
Sine.
Seg. Area.
.346
.241169
.387
.280668
.428
.320948
.347
.242121
.388
.281642
.429
.321938
.348
.243074
.389
.282617
.430
.322928
.349
.244026
.390
.283592
.431
.323918
.350
.244980
.391
.284568
.432
.324909
.351
.245934
.392
.285544
.433
.325900
.352
.246889
.393
.286521
.434
.326892
.353
.247845
394
.287498
.435
.327882
.354
.248801
.395
.288476
.436
.328874
.355
.249757
.396
.289453
.437
.329866
.356
.250715
.397
.290432
.438
.330858
.357
.251673
.398
.291411
.439
.331850
.358
.252631
.399
.292390
.440
.332843
.359
.253590
400
.293369
.441
.333836
.360
.254550
401
.294349
.442
.334829
.361
.255510
.402
.295330
.443
.335822
.362
.256471
403
.296311
.444
.336816
.363
.257433
404
.297292
.445
.337810
.364
.258395
405
.298273
.446
.338804
.365
.259357
.406
.299255
.447
.339798
.366
.260320
.407
.300238
.448
.340703
.367
.261284
408
.301220
.449
.341787
.368
.262248
409
.302203
.450
.342782
.369
.263213
410
.303187
.451
.343777
.370
.264178
.411
.304171
.452
.344772
.371
.265144
.412
.305155
.453
.345768
.372
.266111
.413
.306140
.454
.346764
.373
.267078
.414
.307125
.455
.347759
.374
.268045
.415
.308110
.456
.348755
.375
.269013
.416
.309095
.457
.349752
.376
.269982
.417
.310081
.458
.350748
.377
.270951
.418
.311068
.459
.351745
.378
.271920
.419
.312054
.460
.352741
.379
.272890
.420
.313041
.461
.353739
.380
.273861
.421
.314029
.462
.354736
.381
.274832
.422
.315016
.463
.355732
382
.275803
.423
.316004
.464
.356730
.383
.276775
.424
.316992
.465
.357727
.384
.277748
.425
.317981
.466
.358725
.385
.278721
.426
.318970
.467
.359723 '
.386
.279694
.427
.319959
.468
.360721
THE AR-EAS OF THE SEGMENTS OF A CIRCLE. 299
Versed
Baa.
Seg. Area.
Versed
Sine.
Seg. Area.
Versed
Sine.
Seg. 4 r ea.
.469
.361719
.480
.372764
.491
.383699
.470
.362717
.481
.373703
.492
384699
.471
.363715
.482
.374702
.493
.385699
.472
.364713
.483
.375702
.494
.386699
.473
.365712
.484
.376702
.495
.387699
.474
.366710
.485
.377701
.496
.388699
.475
.367709
.486
.378701
.497
.389699
.476
.368708
.487
.379700
.498
.390699
.477
.369707
.488
.380700
.499
.391699
.478
.370706
.489
.381699
.500
.39269U
.479
.371705
.490
.382699
THE END.
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