(13) The slant height of an hexagonal pyramid is 16 ft., and each side of its base 5 ft.; find the cost of tooling the side faces, at 1s. 6d. per sq. ft. (14) The base of a pyramid is a regular octagon, each side being 18 ft., and each of the other edges of the pyramid 41 ft.; find the upright surface. THE RIGHT CONE. 166. A Cone is a solid of which the base is a circle, and whose curved surface tapers to a point, called, as in the pyramid, the vertex of the solid. Thus A B C represents a cone, of which A is the vertex. The straight line A D drawn from the vertex A to the centre D of the base is called the axis of the cone. When the axis is perpendicular to the base of a cone, it is said to be a right cone. The cones mentioned in this book are all right cones. A D is the perpendicular height, and AB the slant height. E Suppose a cylindrical vessel ABCD to be filled with water, and to contain D three pints. If now a right cone of the same base and height be carefully dropped into it, as in the case of the pyramid one-third of the contents will be found to overflow, thus showing that the volume of a right cone is equal to onethird of the volume of a cylinder whose base and height are the same. Hence A the rule. 167. To find the volume of a right cone. A D RULE. Multiply the area of the base by the perpendicular height, and one-third of the product will be the volume. Ex. 1. The diameter of the base of a right cone is 28 in., and the perpendicular height 15 in.; required the volume. Area of base = 14 × 28 × 28=616 sq. in. (Art. 102.) Volume of cone=× 616 × 15=3080 c. in. Ex. 2. The radius of the base of a right cone is 21 ft., and the slant height 35 ft.; find the volume. As in the case of the pyramid (Art. 163, worked example 2), the perpendicular height, the radius of the base, and the slant height, form respectively the perpendicular, the base, and the hypotenuse of a right-angled triangle. (Art. 70.) Hence, the square of AD=the square of AB-the square of B D, =1225-441=784, and A D=the square root of 784, or 28 ft. Area of base1 × 42 × 42=1386 sq. ft. (Art. 102.) .. Volume of conex 1386 x 28=12936 c. ft. Ex. LXXII. Find the volumes of right cones having the following dimensions: (1) Diameter of base, 42 ft.; perpendicular height, 9 ft. (2) Radius of base, 28 ft.; perpendicular height, 6 ft. (3) Circumference of base, 8.8 ft.; perpendicular height, 4.5 ft. (4) Diameter of base, 1 ft. 2 in.; slant height, 2 ft. 1 in. (5) Radius of base, 10 ft. 6 in.; slant height, 17 ft. 6 in. (6) Circumference of base, 15-4 ft.; slant height, 8.75 ft. (7) In a cone the diameter of the base is 3.5 ft., and the perpendicular height is 4.5 ft.; find the solidity. (8) The slant height of a right cone being 50 in., and the radius of its base 14 in., find its solid content. (9) Find the volume of a right cone the height of which is 15 ft. and the circumference of the base 44 ft. (10) The circumference of the base of a cone is 12 ft. 10 in., and its perpendicular height is 10 ft. 6 in.; find its volume. (11) A conical wine-glass is 2 in. wide at the top, and 3 in. deep; how many c. in. of wine will it hold? (12) How many conical wine-glasses 2 in. wide at the top, and 3 in. deep, may be filled from a cylindrical vessel which is 3 ft. 6 in. deep, and has a diameter of 1 ft. 9 in.? (13) From a cubical block of granite whose side measures 5 ft. 10 in., the largest possible right cone is cut find the number of cubic inches of granite cut away. [Here, the diameter of the base and the perpendicular height of the cone will each measure 5 ft. 10 in. From the solid content of the cube subtract that of the cone, and the remainder will be the content of the material cut away.] 168. The whole surface of a right cone consists of a circular base and the curved or convex surface. Suppose A B יפ C' to be a hollow cone of paper or cardboard. Let it be cut down perpendicularly from the vertex to the base, as in A D, E and then laid out flat. It will be seen to form a sector of a circle A DD' (Art. 116), the circumference of the base of the cone being the arc of the sector, and the slant height of the cone the radius of the sector. Now the area of a sector is found by multiplying half the arc by the radius (Art. 119); hence the area of the curved surface of a cone is equal to half the product of the circumference of the base and the slant height. D If the perpendicular height be given, the slant height may be found by Art. 71. 169. To find the whole surface of a right cone. RULE. Multiply the circumference of the base by half the slant height, and to the product add the area of the base. Ex. The circumference of the base of a cone is 4 ft. 6 in., and the slant height 8 yds.; find the convex surface. Here, 4 ft. 6 in. =4 ft.; 8 yds. = 24 ft. Ex. 2. Find the curved surface of a cone whose diameter is 42, and perpendicular height 28. Circumference of base Square of radius of base Hence, convex surface =22×42=132, (Art. 99.) =×42=21. = 21 × 21=441. =441+784=1225.(Art. 71.) Ex. LXXIII. Find, in square feet and inches, the area of the curved surface of right cones having the following dimensions:(1) Circumference of base, 14.5 ft.; slant height, 12.8 ft. Circumference of base, 5.25 ft.; slant height, 8 ft. (3) Circumference of base, 3 ft.; slant height, 18 in. (4) Diameter of base, 5 ft. 3 in.; slant height, 2 ft. 8 in. (5) Radius of base, 2 yds. 1 ft.; slant height, 5 ft. (6) Radius of base, 2 ft. 4 in.; slant height, 6 ft. (7) Circumference of base, 7 ft. 4 in.; perpendicular height, 4 ft. (8) Diameter of base, 14 ft.; perpendicular height, 5 ft. 3 in. (9) Radius of base, 1 ft. 9 in.; perpendicular height, 2 ft. 4 in. Find the whole surface of right cones having the following dimensions: : (10) Circumference of base, 8.8 ft.; slant height, 6.25 ft. (13) Circumference of base, 18 ft. 4 in.; perpendicular height, 7 ft. (14) Diameter of base, 4 ft. 8 in.; perpendicular height, 16 ft. 3 in. (15) Radius of base, 5 ft. 3 in.; perpendicular height, 23 ft. 4 in. (16) What will be the cost of painting a conical spire whose slant height is 120 ft., and whose circumference at the base is 63 ft., at 2d. per sq. yd.? (17) The slant height of a marble cone is 6 yds., and the diameter of the base 23 yds.; find the cost of polishing the curved surface, at 1s. 3d. per sq. ft. (18) The radius of the base of a cone is 24 ft., and the perpendicular height 84 ft.; find the area of the convex surface. (19) The conical cap of a tower is 10 yds. 1 ft. 6 in. diameter at base, and 7 yds. high; find the expense of covering it with lead, at 12s. 6d. per sq. yd. (20) A conical spire is 66 ft. in circumference at the base, and 14 ft. high; find the cost of covering it with lead, at 1s. 6d. per sq. ft. (21) How many yards of canvas, 1 yd. wide, will be required to make a conical tent 30 ft. high, and 17 ft. 6 in. in diameter ? (22) What length of canvas, 33 yds. wide, is required to make a conical tent whose slant height is 27 ft., and diameter 49 ft. ? [See Art. 56.] (23) How many yards of canvas, yd. wide, will be required to make a round tent 14 ft. in diameter, whose side is perpen dicular to the height of 7 ft., the upper part forming a right cone whose slant height is 10 ft.? [Here we are to find the curved surface of a cylinder whose base is 14 ft. in diameter and height 7 ft. (Art. 159), and also the curved surface of a right cone whose base is 14 ft. in diameter, and slant height 10 ft. (Art. 169).] (24) A circular pillar of granite whose length is 30 ft. and perimeter 66 ft. is cut away at one end so as to form a cone, whose height is 14 ft. Find the cost of polishing the curved surfaces, at 3s. per sq. yd. [In this case also we are to find the curved surfaces of a cylinder and a cone. The height of the cylinder will be 30-14=16 ft.] THE SPHERE. 170. A Sphere or Globe is a solid bounded by a curved surface, every part of which is equally distant from a certain point within the figure, called the centre. The radius of a sphere is the straight line drawn from the centre to the surface. The diameter of a sphere is the straight line drawn through the centre and terminated at both ends by the surface. Suppose A B C D, a cylindrical vessel whose height E F is equal to the diameter C D of the base, to be filled with water, and to contain 3 pints. If now a sphere of the same diameter and height be carefully dropped into it, two pints, i.e. two-thirds of the contents of the cylinder, will be found to overflow, thus showing that the volume A E D B of a sphere is equal to two-thirds of the volume of a cylinder whose height and diameter are the same. Now the volume of a cylinder is equal to x diam. × diam. x height (Art. 155), and since the height of the cylinder and the diameter of the base are equal, this may be written, 11 × diam. × diam. × diam., or K |