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To find the area of the convex surface of a frustum of a right pyramid.

111. From the principle demonstrated in Book XII., Prop. IV., C., we may write the following

RULE.

Multiply the half sum of the perimeters of the two bases by the slant height; the product will be the area required.

EXAMPLES.

1. How many square feet are there in the convex surface of the frustum of a square pyramid, whose slant height is 10 feet, each side of the lower base 3 feet 4 inches, and each side of the upper base 2 feet 2 inches? Ans. 110 sq. ft. 2. What is the convex surface of the frustum of a heptagonal pyramid, whose slant height is 55 feet, each side of the lower base 8 feet, and each side of the upper base 4 feet? Ans. 2310 sq. ft.

112. Since a cylinder may be regarded as a prism whose base has an infinite number of sides, and a cone as a pyramid whose base has an infinite number of sides, the rules just given, may be applied to find the areas of the surfaces of right cylinders, cones, and frustums of cones, by simply changing the term perimeter, to circumference.

EXAMPLES.

1. What is the convex sur e of a cylinder, the diamete. of whose base is 20, and whose altitude 50 ? Ans. 3141.6.

2. What is the entire surface of a cylinder, the altitude being 20, and diameter of the base 2 feet? 131.9472 sq. ft. 3. Required the convex surface of a cone, whose slant height is 50 feet, and the diameter of its base 8 feet. Ans. 667.59 sq. ft.

4. Required the entire surface of a cone, whose slant height is 36, and the diameter of its base 18 feet.

Ans. 1272.348 sq. ft.

5. Find the convex surface of the frustum of a cone, the slant height of the frustum being 12 feet, and the circumferences of the bases 8.4 feet and 6 feet. Ans. 90 sq. ft.

6. Find the entire surface of the frustum of a cone, the slant height being 16 feet, and the radii of the bases 3 feet, and 2 feet. Ans. 292.1688 sq. ft.

To find the area of the surface of a sphere.

113. From the principle demonstrated in Book VIII, Prop. X., C. 1, we may write the following

Find the area of one

it by 4; the product will

RULE.

of its great circles, and multiply be the area required.

EXAMPLES.

1. What is the area of the surface of a sphere, whose radius is 16 ?

Ans. 3216.9984.

2. What is the area of the surface of a sphere, whose radius is 27.25 Ans. 9331.3374.

To find the area of a zone.

M

114. From the principle demonstrated in Book VIII Prop. X., C. 2, we may write the following

RULE.

Find the circumference of a great circle of the sphere, and multiply it by the altitude of the zone; the product will be the area required.

EXAMPLES.

1. The diameter of a sphere being 42 inches, what is the area of the surface of a zone whose altitude is 9 inches. Ans. 1187.5248 sq. in.

2. If the diameter of a sphere is 12 feet, what will be the surface of a zone whose altitude is 2 feet?

To find the area of a spherical polygon.

78.54 sq. ft.

115. From the principle demonstrated in Book IX., Prop. XIX., we may write the following

RULE.

From the sum of the angles of the polygon, subtract 180° taken as many times as the polygon has sides, less two, and divide the remainder by 90°; the quotient will be the spherical excess. Find the area of a great circle of the sphere, and divide it by 2; the quotient will be the area of a tri-rectangular triangle. Multiply the area of the trirectangular triangle by the spherical excess, and the product will be the area required.

This rule applies to the spherical triangle, as well as to any other spherical polygon.

EXAMPLES.

1. Required the area of a triangle described on a sphere, whose diameter is 30 feet, the angles being 140°, 92°, and Ans. 471.24 sq. ft.

68°.

2. What is the area of a polygon of seven sides, described on a sphere whose diameter is 17 feet, the sum of

the angles being 1080° ?

Ans. 226.98.

of eight sides,

3. What is the area of a regular polygon described on a sphere whose diameter is 30 yards, each an

gle of the polygon being 140° ?

Ans. 157.08 sq. yds.

MENSURATION OF VOLUMES.

To find the volume of a prism.

116. From the principle demonstrated in

Prop. XIV., we may write the following

RULE.

Book VII,

Multiply the area of the base by the altitude; the product will be the volume required.

EXAMPLES.

1. What is the volume of a cube, whose side is 24 inches? Ans. 13824 cu. in.

2. How many cubic feet in a block of marble, of which the length is 3 feet 2 inches, breadth 2 feet 8 inches, and height or thickness 2 feet 6 inches ? Ans. 21 cu. ft.

3. Required the volume of a triangular prism, whose height is 10 feet, and the three sides of its triangular base 3, 4, and 5 feet. Ans. 60.

To find the volume of a pyramid.

117. From the principle demonstrated in Book VII., Prop. XVII., we may write the following

RULE.

Multiply the area of the base by one-third of the altitude; the product will be the volume required.

EXAMPLES.

1. Required the volume of a square pyramid, each side of its base being 30, and the altitude 25.

Ans. 7500.

2. Find the volume of a triangular pyramid, whose altitude is 30, and each side of the base 3 feet. 38.9711 cu. ft.

3. What is the volume of a pentagonal pyramid, its altitude being 12 feet, and each side of its base feet.

Ans. 27.5276 cu. ft.

4. What is the volume of an hexagonal pyramid, whose altitude is 6.4 feet, and each side of its base 6 inches? Ans. 1.38564 cu. ft.

To find the volume of a frustum of a pyramid.

118. From the principle demonstrated in Book VII., Prop., XVIII., C., we may write the following

RULE.

Find the sum of the upper base, the lower base, and a mean proportional between them; multiply the result by onethird of the altitude; the product will be the volume required.

EXAMPLES.

1. Find the number of cubic feet in a piece of timber, whose bases are squares, each side of the lower base being 15 inches, and each side of the upper base 6 inches, the altitude being 24 feet. Ans. 19.5.

2. Required the volume of a pentagonal frustum, whose altitude is 5 feet, each side of the lower base 18 inches, and

each side of the upper base 6 inches.

Ans. 9.31925 cu. ft.

119. Since cylinders and cones are limiting cases of prisms and pyramids, the three preceding rules are equally applicable to them.

EXAMPLES.

1. Required the volume of a cylinder whose altitude is 12 feet, and the diameter of its base 15 feet.

Ans. 2120.58 cu. ft.

2. Required the volume of a cylinder whose altitude is 20 feet, and the circumference of whose base is 5 feet

6 inches.

Ans. 48.144 cu. ft.

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