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PROBLEM III.

To find the Surface of the Frustum of a Cone or

Pyramid. RULE.-Multiply the sum of the perimeters of the two ends by the slant height, and half the product will be the slant surface; to which add the areas of the two ends, and the product will be the whole surface.

EXAMPLE.—Required the convex surface of the frustum of a cone A B C D, whose base A B = = 20 inches, the slant height B C = 19, and top end CD= 11.

F

3.1416 x 20 + 3.1416 x 11 x 19

2
=925.2012 square inches, and
divided by 144 = 6.425 feet
nearly.

B

PROBLEM IV.

To find the Solid Content of the Frustum of a Cone.

Rule.—To the product of the diameters of the two ends, add the sum of their squares ; multiply this sum by the perpendicular height and by .2618, the product is the solid content.

Example 1.-Required the solid content of the frustum in problem 3, whose perpendicular E F= 18 inches.

20 X 11 = 220 and 220 + 202 + 112 x 18 x .2618 = 8491.8884 cubic inches, and divided "by 1728 = 2.0208 cubic feet nearly. EXAMPLE 2.-Required the content in imperial gallons, of the inverted frustum of a cone A B CD,

D

whose inner dimensions are 31 feet deep, 18 inches diameter at bottom, and 22 inches diameter at top, 22 x 18=396, and 396 +222 +182 x 42 x .2618= 13238.7024

277.274 47.745 gallons nearly.

Or, 13238.7024 x .003607 = 47.75 gallons nearly as before.

PROBLEM V.

To find the Solid Content of the Frustum of a Pyramid.

RULE.—To the sum of the areas of the two ends, add the square root of their product, multiply this sum by the perpendicular height; and f of the product is the solid content.

Example.—Required the solid content of the frustum of a pyramid A B C D, whose perpendicular height = 24 inches, the area of the base = 144 inches, and area of the top end

144 + 64 = 208 and ✓ 144 X 64 = 96 then 208 + 96 x 24

= 2432

3
cubic inches, and = 1728=1.4074
cubic feet nearly.

= 64.

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A

PROBLEM VI.

To find the Solidity of a Wedge. RULE.-To the length of the edge add twice the length of the base; multiply that sum by the height, and by the breadth of the base, and of the product will be the solidity.

EXAMPLE.—Required the content in cubic inches of the wedge A B C D E, whose basė A B C = 12 inches long and 4 inches broad, the length of the edge D E = 10 inches, and perpendicular height r E= 20 inches.

10 + 24 x 20 x 4

= 453.33
6
cubic inches.

A

B

PROBLEM VII.

To find the Convex Surface and Solid Content of a

Sphere or Globe. Rule 1.–Multiply the square of the diameter by 3.1416, the product will be the convex superficies.

RULE 2.-Multiply the cube of the diameter by .5236, and the product is the solid content.

Example 1.-Required the convex surface of a sphere, whose diameter A B = 254 inches.

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EXAMPLE 2.-Required the solid content of a sphere, whose diameter A B = 254 inches.

25.53 x .5236 = 8682.00795 cubic inches • 1728 = 5.0243 cubic feet nearly.

PROBLEM VIII.

To find the Convex Surface and Solid Content of the

Segment of a Sphere. Rule 1.-Multiply the height of the segment by the whole circumference of the sphere, and the product is the curve surface.

RULE 2.-Add the square of the height to three times the square of the radius of the base ; multiply that sum by the height, and by .6236, and the product is the solid content.

EXAMPLE 1.-The diameter A B of the sphere A B CD = 20 inches; what is the convex surface of that segment of it, whose height E D = 8 inches ?

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EXAMPLE 2.--The base F G of the segment F DG = 18 inches, and perpendicular ED= 8, what is the solid content?

82 = 64., and 92 X 3 = 243, then 243 + 64 x 8 x .5236 = 1285.9616 cubic inches = 1728

= .7441 cubic feet nearly. EXAMPLE 3.-Suppose A B C D to be a sugar pan, and that the diameter of the mouth A B is 4 feet, the depth D C being 25 inches, how many imperial gallons will it contain ?

252 = 625, and 242 X 3= 1728, then 1728 +625 x 25 x .5236=30800.77

277.274 =111.084 gallons nearly.

PROBLEM IX.

To find the Solidity of a Spheroid. Rule.-Multiply the square of the revolving axis by the fixed axis, and by 5236, and the product will be the solidity.

ExamPLE 1.-Required the solid content of the prolate spheroid ABCD, whose fixed axis A C is 50, and revolving axis B D 30.

B

302 X 50 X 5236 = 23562 the solidity.

EXAMPLE 2.-What is the solid content of an oblate spheroid, the fixed axis being 30, and revolving axis 50 ?

502 x 30 x .5236 = 39270 the solid content.

PROBLEM X.

To find the Solidity of the Segment of a Spheroid when the base is circular or parallel to the revolving axis.

Rule.-From triple the fixed axis, take double the height of the segment; multiply the difference by the square of the height, and by .5236; then say, as the square of the fixed axis is to the square of the revolving axis, so is the former product to the solidity.

EXAMPLE 1.-. Required the solid content of the segment A B C; whose height B r is 10; the revolving axis E F being 40, and fixed axis B D 25.

B

25 x 3 10 x 2 = 55 and 55 x 102 x .5236 = 2879.8; then, as 252 : 402 2879.8 : 7372.3 nearly.

!

D

EXAMPLE 2.- What is the solid content of the segment of a spheroid, whose height = 20 inches, the revolving axis being 25, and fixed axis 50.

50 x 3 20 x 2 = 110, and 110 x 202 X .5236 = 23038.4; then, as 502 : 252 :: 23038.4 : 5759.6 inches, the solid content.

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