PROBLEMS.

918. To find the convex surface of a prism or cylinder.

1. Find the area of the convex surface of a prism whose altitude is 7 ft., and its base a pentagon, each side of which is 4 feet.

OPERATION.-4 ft. x 5 = 20 ft., peri

meter.

20 ft. x 7-140 sq. ft., convex surface.

2. Find the area of the convex surface of a triangular prism, whose altitude is 8 feet, and the sides of its base 4, 5, and 6 feet, respectively.

OPERATION. -4 ft. + 5 ft. + 6 ft. = 15 ft., perimeter.

15 ft. x 81=127 sq. ft., convex surface.

3. Find the area of the convex surface of a cylinder whose altitude is 2 ft. 5 in. and the circumference of its base 4 ft. 9 in.

OPERATION.-2 ft. 5 in.-29 in.; 4 ft.

9 in. = 57 in.

57 in. x 291653 sq. in. = 11 sq. ft. 69 sq. inches, convex surface.

RULE.-Multiply the perimeter of the base by the altitude.

To find the entire surface, add the area of the bases or ends.

4. If a gate 8 ft. high and 6 ft. wide revolves upon a point in its center, what is the entire surface of the cylinder described by it? 5. Find the superficial contents, or entire surface of a parallelopipedon 8 ft. 9 in. long, 4 ft. 8 in. wide, and 3 ft. 3 in. high.

6. What is the entire surface of a cylinder formed by the revolution about one of its sides of a rectangle that is 6 ft. 6 in. long and 4 ft. wide?

7. Find the entire surface of a prism whose base is an equilateral triangle, the perimeter being 18 ft., and the altitude 15 ft.

919. To find the volume of any prism or cylinder. 1. Find the volume of a triangular prism, whose altitude is 20 ft., and each side of the base 4 feet.

OPERATION.-The area of the base is 6.928 sq. ft. (882). 6.928 sq. ft. x 20 = 138.56 cu. ft., volume.

2. Find the volume of a cylinder whose altitude is 8 ft. 6 in., and the diameter of its base 3 feet.

OPERATION.-32 x .78547.0686 square feet, area of base (905). 7.0686 sq. ft. x 8.5 = 60.083 cubic feet, volume.

RULE.-Multiply the area of the base by the altitude.

3. Find the solid contents of a cube whose edges are 6 ft. 6 in. 4. Find the cost of a piece of timber 18 in. square and 40 ft. long, at $.30 a cubic foot.

5. Required the solid contents of a cylinder whose altitude is 15 ft. and its radius 1 ft. 3 in.

6. What is the value of a log 24 ft. long, of the average circumference of 7.9 ft., at $.45 a cubic foot?

PYRAMIDS AND CONES.

920. A Pyramid is a body having for its base a polygon, and for its other faces three or more triangles, which terminate in a common point called the vertex.

Pyramids, like prisms, take their names from their bases, and are called triangular, square, or quadrangular, pentagonal, etc.

921. A Cone is a body having a circular base, and whose convex surface tapers uniformly to the vertex.

It is a body conceived to be formed by the revolution of a right angled triangle about one of its sides containing the right angle, as an immovable axis.

922. The Altitude of a pyramid or of a cone is the perpendicular distance from its vertex to the plane of its base.

923. The Slant Height of a pyramid is the perpendicular distance from its vertex to one of the sides of the base; of a cone, is a straight line from the vertex to the circumference of the base.

924. The Frustum of a pyramid or cone is that part which remains after cutting off the top by a plane parallel to the base.

PROBLEMS.

925. To find the convex surface of a pyramid or

cone.

1. Find the convex surface of a triangular pyramid, the slant height being 16 ft., and each side of the base 5 feet.

OPERATION.—(5 ft.+5 ft. +5 ft.) × 16÷2 = 120 sq. ft., conv. surf. 2. Find the convex surface of a cone whose diameter is 17 ft. 6 in., and the slant height 30 feet.

RULE.-Multiply the perimeter or circumference of the base by onehalf the slant height.

To find the entire surface, add to this product the area of the base.

3. Find the entire surface of a square pyramid whose base is 8 ft. 6 in. square, and its slant height 21 feet.

4. Find the entire surface of a cone the diameter of whose base is 6 ft. 9 in. and the slant height 45 ft.

5. Find the cost of painting a church spire, at $.25 a sq. yd., whose base is a hexagon 5 ft. on each side, and the slant height 60 feet.

926. To find the volume of a pyramid or of a cone. 1. What is the volume, or solid contents, of a square pyramid whose base is 6 feet on each side, and its altitude 12 feet.

OPERATION.-6 × 6 × 12 ÷ 3 = 144 cu. ft., volume.

2. Find the volume of a cone, the diameter of whose base is 5 ft. and its altitude 10 feet.

OPERATION.—(5o ft. x .7854) × 10 ÷ 3 = 68.721 cu. ft., volume. RULE.-Multiply the area of the base by one-third the altitude. 3. Find the solid contents of a cone whose altitude is 24 ft., and the diameter of its base 30 inches.

4. What is the cost of a triangular pyramid of marble, whose altitude is 9 ft., each side of the base being 3 ft., at $2 per cu. foot?

5. Find the volume and the entire surface of a pyramid whose base is a rectangle 80 feet by 60 feet, and the edges which meet at the vertex are 130 feet.

927. To find the convex surface of a frustum of a pyramid or cone.

1. What is the convex surface of a frustum of a square pyramid, whose slant height is 7 feet, each side of the greater base 4 feet, and of the less base 18 inches?

OPERATION.-The perimeter of the greater base is 16 ft., of the less

6 feet.

16 ft.+6 ft. x 7÷2 =77 sq. ft., convex surface.

2. Find the convex surface of a frustum of a cone whose slant height is 15 feet, the circumference of the lower base 30 feet, and of the upper base 16 feet.

RULE.-Multiply the sum of the perimeters, or circumferences, by one-half the slant height.

To find the entire surface, add to this product the area of both ends, or bases.

3. How many square yards in the convex surface of a frustum of a pyramid, whose bases are heptagons, each side of the lower base being 8 feet, and of the upper base 4 feet, and the slant height 55 feet?

928. To find the volume of a frustum of a pyramid

or cone.

1. Find the volume of the frustum of a square pyramid whose altitude is 10 feet, each side of the lower base 12 feet, and of the upper base 9 feet.

OPERATION.—122 +92 = 225; (225+ √/144 × 81) × 10÷3=1000 cu. feet, volume.

2. How many cubic feet in the frustum of a cone whose altitude is 6 ft. and the diameters of its bases 4 ft. and 3 feet?

RULE.-To the sum of the areas of both bases add the square root of the product, and multiply this sum by one-third of the altitude.

3. How many cubic feet in a piece of timber 30 ft. long, the greater end being 15 inches square, and that of the less 12 inches?

4. How many cubic feet in the mast of a ship, its height being 50 ft., the circumference at one end 5 feet and at the other 3 feet?

929. A Sphere is a body bounded by a uniformly curved surface, all the points of which are equally distant from a point within called the center.

930. The Diameter of a sphere is a straight line passing through the center of the sphere, and terminated at both ends by its surface.

931. The Radius of a sphere is a straight line drawn from the center to any point in the surface.

932. To find the surface of a sphere.

1. Find the surface of a sphere whose diameter is 9 in. OPERATION.-9 in. x 3.141628.2744 in., circumference. 28.2744 in. x9= 254.4696 sq. in., surface. RULE.-Multiply the diameter by the circumference of a great circle

of the sphere.

2. What is the surface of a globe 3 feet in diameter? 3. Find the surface of a globe whose radius is 1 foot.

933. To find the volume of a sphere.

1. Find the volume of a sphere whose diameter is 18 inches. OPERATION.-18 in. x 3.1416 = 56.5488 in., circumference. 56.5488 in. x 18 = 1017.8784 sq. in., surface.

1017.8784 sq. in. x 18÷6-3053.6352 cu. in., volume. RULE.-Multiply the surface by of the diameter, or of the radius. 2. Find the volume of a globe whose diameter is 30 in. 3. Find the solid contents of a globe whose radius is 5 yards.

934. To find the three dimensions of a rectangular solid, the volume and the ratio of the dimensions being given.

1. What are the dimensions of a rectangular solid, whose volume is 4480 cu. ft., and its dimensions are to each other as 2, 5, and 7?* OPERATION.-4480 ÷ (2 × 5 × 7)

=

=

4; 4 ft. x 28 ft., height;

4 ft. x 520 ft., width; 4 ft. x 7 28 ft., length.