112 
MENSURATION'. 
3. Multiply the fquare of the circumference by -07958, and 
the produft will be the area. 
Ex. 1. What is the area of a circle who fe diamete r is 106 
and the circumference 333 feet?—Here 333X106+4 = 
. j 29 8-f-4=8824g feet, the area required. 
Ex. 2. Required the area of the end or bafe of a roller, 
whofe diameter is2 feet 3 inches.—Here 2-2 5 X a 1 * 5=5-062 5, 
the fquare of the diameter ; and 5-0625 x-7854=3-97608 
feet = 3 ft. 11 in. 8 pa. the area required. 
To find the Area of a Sector of a Circle. —Multiply the 
length of the arc by the radius of the feCtor, and half the 
p>rodu£t will be the area. 
Ex. The radius AB, fig. 10. is 15, the chord B D of 
the whole arc 24, and the verfed fine CE 6; wh at is the 
area of the feCtor ABCD ?—Here 3/ i2 z + 6 2 =y'144+3® 
180=13-4164=00, the chord of half the arc j and 
i3-4i64X8-24 r= i°7-33^- z 4 _83-33_^__, r:>77Afij the 
3 3 3 
---406.6559 
length of the arc 3 then 27-77706x15+2=--— = 
*03-32795, the area required. 
To find the Area of a Segment of a Circle. 
Rule I. Find the area of the feCtor, having the fame 
arc as the fegment; alfo, find the area of the triangle 
formed by the chord of the fegment and the radii of the 
feCtor ; then the difference of thefe areas, when the leg- 
tnent is lefs than a femicircle, or their fum,when it is greater, 
will be the area of the fegment. 
Ex. The radius AD, fig. n, is 25, and the chord AB of 
the whole arc 40 ; what is the area of the fegment ABC? 
—Here 3/ 25*—2o 2 =3/ 625—400=3/225=15=DE, there¬ 
fore the verfed fine CE=io. Again, >v'20 2 + io 2 ~ 
.3/400 + 100 = 3/500 = 22-36068 = A C, the chord of 
22-26068x8—40 178-88544—40 
half the arc ; and ■ -=--= 
*38-88544 
=46-295146, the length of the arc ACB ; then 
46-295146 X25+2=ii57-37865+2=578-689325, the area 
of the feCtor AD B C. Now, fABx DE=2 oXi 5=300, 
the area of the triangle AD B j hence 578-689325—300= 
278-689325, the area of the fegment required. 
Rule II. Divide the height of the fegment by the dia¬ 
meter, and find the quotient in the column of Heights or 
Verfed Sines, in the Table of Areas at the end of this 
article, p. 117 - 
Take out the correfponding area, which multiply by 
the fquare of the diameter, and the product will be the 
area of the fegment. 
Ex. What is the area of a fegment whofe chord is 32, 
the verfed fine 8, and the diameter of the circle 40 ?—■ 
Here-— —2, the quotient, or tabular height; and the 
4 ° 
correfponding Tabular Area is'111823; hence, -111823 
X4 o 2 = i 11823 X 1600=178-9168, the area of the legment 
required. 
To find the Area of an Ellipfe. —Multiply continually 
together the two diameters and the number -7854, and the 
produft will be the area of the ellipfe. 
Ex. What is the area of the ellipfe ABCD, fig. 12. whofe 
tranfverfe diameter AB is 34, and the conjugate CD 25 ?— 
Here 34 X 25 X -7854=850 X-7854=667-59, the area re¬ 
quired. 
To find the Area of tin Elliptical 'Segment, the Bafe of 
which is parallel to cither of the Diameters of the Ellipfe .— 
Divide the height of the legment by that diameter of the 
ellipfe of which it is a part, and find the area anfwering to 
the quotient, in the Table of Areas. Multiply the two 
..diameters of the ellipfe and the area thus found, conti¬ 
nually together, and the product will be the area required. 
Ex. What is the area of the elliptical fegment ABC, 
fig. 13. cut off by the double ordinate A B ; the height CG 
being 12, the diameter CD of the whole ellipfe 40, and 
I 2> 
EF 25 ?— Here—=- 3 , the tabular height; and the cor¬ 
refponding Area in the Table, is, -198168; then -19.8168 
X 2 5X4 °= i 9^' i ®8 , the area of the fegment required. 
To find the area of the Segment of a Circle, or any other 
cur vilineal figure, hy means of eqnidifiant ordinates. —If a 
right line AN, fig. 14. be divided into any even number 
of equal parts AC, CE, EG, See. and at the points of di- 
vifion be erefted perpendicular ordinates AB, CD, EF, &c. 
terminated by any curve BDF, See. and if A be put for 
the fum of the extreme or firft and laft ordinates AB, NO 
B for the fum of the even ordinates CD, GH, LM, See. 
viz. the fecond, fourth, forth, &c. and C for the fum of all 
the reft, EF, IK, &c. viz. the third, fifth, &c. or the odd 
ordinates, wanting the firft and laft; then the common 
diftance AC, or CE, &c. of the ordinates, being multi¬ 
plied by the fum arifing from the addition of A, four 
times B, and two times C ; one-third of the product will 
be the area ABON, very nearly; that is, A "l~4-B+2C 
3 
j<D= the area, putting D=AC, the common diftance of 
the ordinates. 
Ex. Required to findthearea ofthecurved fpaceABCD, 
fig. 15. the lengths of the five equidiftant ordinates being 
as follow : viz. the firft or AD=8, the fecond =10, the 
third =10, the fourth =14, and the fifth or laft B C=i 5; 
and the length of the bafe AB=*4.—Here A=8+i5=23, 
£=10+14=24, C=i2, and D=—=6; then A+4B+2C 
__ 4 
X iD = 23-p96+24X-§=i43X2=236, the area required. 
Mensuration of Solids. 
To find the folid or cubic content of a Prifm or Cylinder .—• 
Multiply the area of the bafe by the height, and the pro¬ 
duct will be the folid or cubic content. 
Ex. 1. How many cubic inches are contained in the 
rectangular prifm or parallelopiped AB, tig. 16. the 
length AD being 3 inches, breadth AC=i, and height 
C 0=2 ? 
2 X 3=6, area of the bafe 
and 6x2=12 inches, the cubic content. 
This is called cubic meafure, becaufe the capacity OP 
magnitude is eftimated in cubic integers, as cubic yards, 
cubic feet, or cubic inches. Thus in the prefent exam¬ 
ple a cubic inch is the meafuring integer or unit, the 
whole prifm containing 12 of thefe units or inch cubes. 
Ex 2. How many gallons of water will a cubic ciftera 
contain, its depth being 4 feet ? 
4X4 X 4=64 cubic feet, the capacity. 
_ 64X1728 . . 
and-=47 gallons, wine meafure. 
231 
To find the folicl content of a Pyramid or Cone. —Multiply 
the bale by the perpendicular height, and | of the pro¬ 
duct will be the area. Or, multiply the bafe by 5 of the 
height. 
Ex. 1. How many cubic feet in a triangular pyramid, 
the fides of the bafe being 7, 8, and 9, feet, and the per¬ 
pendicular height 17 ?—Anfwer, 152-4. 
Ex. 2. Required the number of cubic yards in an up¬ 
right pyramid, the bafe being a regular heptagon, whofe 
fide is 10 feet, and the llant height from the middle of.the 
fide of the bafe =30 feet?—Anfwer, 126-27. 
To find the content of a Cuneus, or Wedge. —A wedge is 
a folid, having one of its ends flat, and the other an edge 
made by the concourfe of two oppoiite plane fides. Thus 
the trapezoid ARBO, fig. 17, is the flat end; and GG, 
the concourfe of the planes AOCG and RBCG, the other 
end or edge of the wedge EG. When the planes AOCG 
and RBCG are reCtangular and equal, tlfo end ARBO 
will alfo be a reCtangle, and the wedge is of the common 
form, or a rectangular pfifin, or half a parallelopiped, hav- 
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