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perpendicular height; and of the product will be the solidity: that is, if A be the area of the greater end, and a of the lesser, and h the height. Then Atat Aux jh= the solidity.

in EXAMPLES. (10) How

many

solid feet are there in a tree, whose bases are squares, each side of the one being 15 inches, and each side of the other 6, and the length measures along

the side 24 feet? (U) What is the content of a frustum of a cone 60 feet high,

the diameters of its ends being 20 and 3 feet? (12) How many solid feet are there in a conical frustum, the

circumferences of whose bases are 66 and 56 feet, height is 4 feet?

PROBLEM XVIII.

To find the convex surface of the frustum of a pyramid or right cone

RULE. Multiply the sum of the perimeters or circumferences of the ends by the slant height, and half the product will be the surface required.

EXAMPLES (13) How many square feet are in the surface of a frustum of

à square pyramid, whose slant height is 10 feet, each side of the greater basis being 3 feet 4 inches, and

each side of the less 2 feet 2 inches? (14) How many square feet are in the surface of a frustum of

a cone, whose circumferences of its ends are 32 and s

feet, and slant side 7 feet? (15) If a segment of 6 feet slant height be cut off a cone,

whose slant height is 30 feet, and the circumference of its basę 10 feet, what will be the surface of the frustum?

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PROBLEM XIX:

To find the solidity of a cuneus or wedge.

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RULE.
Multiply the area of the base, ADE, or BCF, by half the
altitude, Dc, of the wedge, and the product will give the so-
lidity.

CD
Thus, ADX DE X = the solidity:

2

H

EXAMPLE (16) What is the solidity of a wedge, whose base measures

30 feet by 16, and whose height is 12!

PROBLEM XX.
To find the solidity of a pavilion roof.

Fig. 20.

A

RULE. To the length of the ridge, add twice the side of the base which is parallel to it: multiply the sum by the other side of the base, and the product which arises by a sixth part of the altitude; and the second product will give the solidity. s,

-alt Thu EF+2AB X BC X

6'

(17) What is the solidity of a pavilion roof, whose base is 36

by 20, ridge parallel to the greatest side 16, and altitude 12 feet?

PROBLEM XXI. To find the solidity of the frustum of a square pyramid made by a section parallel to the base.

RULE. To the area of the ends add the product of their sides; multiply the sum by a third part of the altitude; and the product will give the solidity.

EXAMPLE. (18) What is the solidity of the frustum of a pyramid 60

feet high, whose ends are 16 and 13 feet square?

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RULE. To the area of the ends add the product of the sums of the lengths and breadths; multiply this sum by a sixth part of the altitude; and the product will give the solidity.

alt. FGX

6

EXAMPLE.

(19) What is the solid content of a canal 304 feet by 20 at

top, 300 feet by 16 at bottom, and 5 feet deep

PROBLEM XXIII. To find the surface of a sphere or globe, or of any segment or zone of it.

Fig. 22.

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A GENERAL RULE. Multiply the circumference of the sphere into the diameter or height of the part required; and the product will be the curve surface, whether it be segment, zone, hemisphere, or the whole sphere.

Note.The height of the whole sphere is its diameter.

PARTICULAR RULES. 1. Find the circumference of a great circle upon the globe, by Prob. VI. Rule I. or by multiplying the radius by 6,2832: multiply the circumference by the diameter; the product will give the superficies. Or,

2. Multiply 3,1416 by the square of the diameter, and the product will give the superficies.

EXAMPLES. (20) What is the surface of a globe, whose diameter is 7? (21) What is the surface of a globe, whose semi-diameter is

6 inches? (22) If the diameter, or axis of the earth be 79574 miles,

what is the whole surface, supposing it a perfect

sphere? (23) What is the superficies of a segment 9 feet high, cut

from a globe of 42 feet diameter?

PROBLEM XXIV. To find the solidity or content of a sphere or globe. (See Fig. to Prob. XXIII.)

RULE. 1. Find the superficies by the last Problem; multiply the superficies by of the radius, or by of the diameter; and the product will be the solidity--Or,

2. Multiply the cube of the diameter by ,5236; and the product will give the solidity; that is,

3,1416

=,5236, the solidity.-Or, 6

3. Find the content of a circumscribing cylinder, by Problem XIII. and take of it for the content of the globe.

For a globe is of its circumscribing cylinder, and ,5236 is the content of a globe whose diameter is 1.

EXAMPLES (24) What is the content of a globe whose diameter is 7? (25) Suppose the earth to be spherical, and its diameter

79574 miles, what is its solidity?

PROBLEM XXV. To find the solidity of the segment of a globe. (See Fig. 23.)

RULE. 1. From three times the diameter of the globe, take twice the altitude of the segment; multiply together the remainder, the square of the altitude, and ,5236; and this product will give the solidity. Thus let h=CD the height of the segment, and ds the

diameter. Then 3d-2h ,5236 x hh= the solidity of ACB.--Or,

2. To three times the square of the radius of its base, A B, add the square of its height; multiply the sum by the height, and that product again by ,5236, will give the solidity.

That is, if r=AD, the radius of its bust, h=CD, the

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