Reason: A cylinder can be converted into an approximate rectangular solid with half the circumference of the cylinder for its base, with the radius for its altitude, and with the altitude of the cylinder for its length. 275. A pyramid is a solid whose base is a polygon, and whose sides are triangles meeting in a point called the vertex. The altitude is the shortest distance from the vertex (called apex) to the plane of the base; the slant height is the shortest distance from the vertex to the middle point of one of the sides of the base. 276. RULE. To find the convex surface of a regular pyramid: Multiply the perimeter of the base by half the slant height. Reason: The convex surface of a pyramid can be converted into a rectangle with the perimeter of the base for its base, and half the slant height for its altitude. 277. RULE. To find the solid contents of a pyramid: Multiply the area of the base by one third of the altitude. Reason: It is proved in geometry that the solid contents of a pyramid is equal to one-third of the solid contents of a prism with an equal base and the same altitude. 278. A cone is a solid whose base is a circle, and whose surface tapers uniformly to a point called the vertex. 279. RULE. To find the convex surface of a right cone: Multiply the circumference of the base by half the slant height, Reason: The convex surface of a cone is a sector of a circle, and can be converted into an approximate rectangle, with the circumference of the base for its base, and half the slant height for its altitude. 280. RULE.-To find the solid contents of a cone: Multiply the area of the base by one-third of the altitude. Reason: It is proved in geometry that the volume of a cone is equal to one-third of the volume of a cylinder with an equal base and the same altitude. 281. The frustum of a pyramid or cone is a solid which remains when a portion having the vertex is cut off by a plane parallel to the base. 282. RULE. To find the convex surface of a frustum of a pyramid or cone: Multiply half the sum of the perimeters of the bases by the slant height. 283. RULE. To find the solid contents of a frustum of a pyramid or cone: To the sum of the two bases add the square root of their product, and multiply the amount by one-third of the altitude. Multiply the circumference by the diameter. NOTE.-The surface of a sphere is 4 times as great as the surface of a circle whose diameter equals the diameter of the sphere. Since the surface of a circle equals Tr2, the surface of a sphere may be represented by either 4r2, or d2. 286. RULE.-To find the solid contents of a sphere: Multiply the area of the surface by one-third of the radius. Reason: A sphere may be regarded as composed of pyramids whose bases form the surface of the sphere, and whose common altitude is the radius of the sphere. NOTE. Since the surface of a sphere may be represented by 4r2, the solid contents may be represented by 287. Similar figures are those that have the same shape. The parts similarly placed are said to be homologous. 288. Fundamental principles of similar figures: 1. Homologous lines are proportional. 2. Two similar surfaces are to each other as the squares of any two homologous lines. 3. Two similar solids are to each other as the cubes of any two homologous lines. Exercise XXXII 1. Find the area of a rectangle 9 ft. 3 in. long and 3 ft. 9 in. wide. 2. Find the area of a parallelogram whose base is 42 rd. and altitude 17 rd. 3. How many tiles 6 in. square in a floor 36 ft. long and 20 ft. wide? 4. Find the area of a triangular field whose base is 72 rd. and altitude 16 rd. 5. Find the area of a triangular garden whose base is 7 ch. and whose altitude is 5 ch. 6. How many bricks 8 in. long and 4 in. wide will it take to pave a cellar floor 36 ft. long and 24 ft. wide? 7. Find the area of a walk 12 ft. wide at one end, 6 ft. wide at the other end, and 120 ft. long. 8. Find the area of a circle whose radius is 6 ft.; whose diameter is 8 ft. 9. How long a rope will it take to fasten a horse to a post so that he may graze over 2 acres? 11. Show that the area of an inscribed circle is equal to .7854 of the area of the square. Cir.=.7854 of Square 12. Show that the area of an inscribed square is equal to .6366 of the area of the circle. Sq.=.6366 13. Show that the area of a square inscribed in a circle is equal to half the area of the square circumscribed about the same circle. 14. Show that the area of a circle inscribed in a square is equal to half the area of the circle circumscribed about the same square. 15. In a square field containing 100 acres, a circle is inscribed; how many acres in one corner cut off by the circle? 16. What is the thickness of the largest square stick of timber that can be cut out of a cylindrical log 17 inches in diameter? 17. Around a circular grass plot containing 3490 sq. yd. is a concrete walk 10 ft. wide. Find the number of sq. yd. in the walk. 18. Find the cost, at 1 cent a sq. ft., of painting a church spire whose base is 10 ft. sq. and whose slant height is 50 ft. 19. What is the convex surface of a cone whose base is 10 ft. in diameter, and whose slant height is 20 ft.? 20. What is the convex surface of a vat whose top is 8 ft. sq., whose base is 9 ft. sq., and whose slant height is 10 ft.? 21. Find the number of sq. ft. in a triangular lot whose sides are 40, 50, and 60 yd. |