in the bottom layer of B. As there are three layers, A is contained a total of twelve times in B, i.e., the volume of B is 12 cubic feet. But the area of the base of B is 4 sq. ft. and the altitude is 3 ft. The product of these two gives 12 cu. ft., the same as obtained by the use of the unit A. Rule: The volume of any prism is equal to the product of the area of the base by the altitude. PROBLEMS FOR PRACTICE 1. What is the weight of a cast-iron block 6 inches long, 5 inches wide, and 3 inches high? The block is a rectangular parallelopiped and weighs .26 lbs. per cubic inch. 2. How many square feet of sheet copper will be required to make an open rectangular tank 7 feet long, 3 feet wide, and 11⁄2 feet deep, allowing 10% extra for waste? 3. How many cubic feet of concrete will be required to construct a column 15 feet high having a hexagonal cross-section. The crosssection is 1 feet long on each side of the hexagon and the perpendicular distance from the center to each side is 1.3 feet. 4. What will it cost to paint the above column, exclusive of the base and top, with waterproofing at 35 cents per square yard? 5. Find the volume of a triangular prism whose altitude is 20 feet, and each side of the base is 4 feet. 6. Find the total area of the prism given in Problem 5. CYLINDERS 90. A cylinder is a solid having as bases two equal parallel curved surfaces and as its lateral face the continuous surface generated by a straight line connecting the bases and moving along their circumferences. The bases are usually circles and such a cylinder is called a circular cylinder. A right cylinder, Fig. 48, is one whose side is perpendicular to the bases. Fig. 48. Right Cylinder. 91. Areas of Cylinders. As a cylinder may be considered a prism with an infinite number of sides, its lateral area is equal to the product of the circumference of the base by the altitude. The total area is evidently equal to the sum of the areas of the bases and the lateral area. 92. Volumes of Cylinders. The volume of a cylinder is equal to the product of the area of one base by the altitude. Examples. 1. How much tin will be required to make a cylindrical can, 3 inches in diameter and 4 inches high, allowing nothing for seams and waste? SOLUTION. The can is a right cylinder the circumference of whose base is 9.43 inches, nearly. Now, the radius of the 22 1 inches, and the area of the base × (1)2 7.08 sq. in., nearly. The lateral area is 9.43 X 4 = 37.72 sq. in. The total area equals 37.72 + (2 × 7.08) square inches. = 7 2. A piece of iron weighing 30 pounds has 4 holes 6 inches deep and 1 inch in diameter bored into it. What is the weight of the piece after the boring has been completed, iron weighing .26 lbs. per cubic inch? = There are 4 holes making 4 X 4.7 18.8 cu. in. The weight of the material cut out is = 18.8 X .26 4.9 lbs. Then the final weight is 30 4.9 lbs. 25.1 lbs. = PROBLEMS FOR PRACTICE 1. What is the capacity of a cylindrical tank 9 feet long and 5 feet in diameter, inside measurements? 2. If a gallon contains 231 cubic inches, what must be the diameter of a cylindrical tank 10 feet high which will hold 1,000 gallons? 3. Find the number of square feet of material necessary for a straight piece of copper pipe 10 feet long and 12 inches in diameter. PYRAMIDS 93. A pyramid is a solid whose base is a polygon and whose sides are triangles. The vertices of the triangles meet to form the vertex of the pyramid. The altitude of the pyramid is the perpendicular distance from the vertex to the base. Pyramids are named according to the kind of polygon forming the base, viz, triangular, quadrangular, Fig. 49, pentagonal, Fig. 50, hexagonal, Fig. 51. Fig. 49. Pyramid. Fig. 50. Regular Pyramid. Fig. 51. Hexagonal Pyramid. A regular pyramid is one whose base is a regular polygon and whose vertex lies in a perpendicular erected at the center of the base, Fig. 50. The slant height of a regular pyramid is a line drawn from the vertex perpendicular to a side of the base. (See the line O F, Fig. 52.) In other words, it is the altitude of one of the triangles which form the sides. The lateral edges of a pyramid are the intersections of the triangular sides, called the faces. 94. Areas of Pyramids. (a) The lateral area is the combined area of all the triangles forming the sides. The area of each triangle is equal to the product of the base by one-half the altitude. Therefore the lateral area of a pyramid is found by adding the products of each side of the base by one-half the same altitude; i.e., it is equal to the perimeter of the base multiplied by one-half the slant height. If the slant height is not given it can usually be found by means of the law of the right triangle. (b) The total area of a pyramid is Fig. 52. Pyramid Showing equal to the sum of the lateral area and the area of the base. Example. Find the lateral area and total area of the pyramid shown in Fig. 52, if it has an altitude O E of 12 feet and each side of the square base is 8 feet. SOLUTION. EF 4 feet. Since the angle O E F is a right angle = 266.4 sq. ft. Ans. = 64 sq. ft., making the total 95. Volumes of Pyramids. It may be shown that any triangular prism may be divided into three equivalent triangular pyramids, having bases and altitudes equal to those of the prism; and therefore the volume of each pyramid must be one-third of the volume of the prism. But the volume of a prism is equal to the product of the area of the base by the altitude; therefore, the volume of a pyramid is equal to the product of the area of the base by one-third of the altitude. PROBLEMS FOR PRACTICE 1. A triangular pyramid is 9 inches in height and each side of the base is 4 inches long. Find the volume. 2. What is the volume of a square pyramid one side of whose base is 4 inches and whose height is 4 inches? 3. Find the total area of a square pyramid whose slant height is 28 inches and one side of whose base is 8 inches. 4. A regular hexagonal pyramid has an altitude of 18 inches, and each side of the hexagon is 6 inches long. Find the total area and the volume of the pyramid. CONES 96. A cone is a solid bounded by a conical surface and a plane which cuts the conical surface. It may be considered as a pyramid with an infinite number of sides. C The conical surface is called the lateral area, and the plane is called the base. The conical surface tapers to a point called the vertex. The altitude of a cone is the perpendicular distance from the vertex to the base. An element of the cone is any straight line from the vertex to the perimeter of the base. Fig. 53. Circular Cone. Fig. 54. Cone of Revolution. A circular cone is a cone whose base is a circle, Fig. 53. A right circular cone, or cone of revolution, Fig. 54, is a cone whose axis is perpendicular to the base. It may be generated by the revolution of a right triangle about one of the sides as an axis. 97. Areas of Cones. The lateral area of a cone is found in the same way as in the case of a pyramid. Multiply the perimeter of the base by one-half the slant height. Example. The base of a cone is 12 inches in diameter and the slant height is 16 inches. What is the total area? To find the total area, add to this the area of the base. In the above example, the area of the base is 98. Volumes of Cones. As a cone may be considered a pyramid with an infinite number of sides, it follows from Section 95 that, |