side measuring 2.5 in., and the length of the prism is 10 in.; find its volume. Area of base =2·5 × 2·5×·433=2·70625 sq. in. (Art. 94). Ex. 3. Find the volume of an hexagonal prism whose length is 12 ft., each side of the base being 15 ft. Area of base 585 sq. ft. (Art. 94, worked example). Ex. LXIII. (1) Required the solidity of a triangular prism whose length is 18 ft., and each of the equal sides of the base 10 ft. (2) The height of a prism is 12 ft.; the base is a triangle whose sides are 1 ft. 8 in., 5 ft. 5 in., and 6 ft. 3 in.; calculate the solidity. (3) The base of a prism is an isosceles triangle, each of whose equal sides is 5 in., and the third side 6 in.; its length is 12 ft.; find its solid content. [Find the area of the base by Art. 69.] (4) Find the volume of a prism whose base is a regular hexagon, each of the equal sides being 5 ft., and the length of the prism 8 ft. (5) Find the volume of an hexagonal prism whose length is 10 in., the side of the hexagon being 6 in. (6) A wall is a mile and a half long, 8 ft. high, 2 ft. 3 in. wide at the top, and 4 ft. 6 in. at the bottom; how many cubic yards of material does it contain? Its [Each end of the wall forms a trapezoid whose parallel sides are 2 ft. 3 in. and 4 ft. 6 in. respectively, and its perpendicular height 8 ft. area may therefore be found by Art. 74.] (7) A canal, 4 miles long, is 30 ft. wide at the top, 18 ft. wide at the bottom, and 2 yds. deep. How many solid yards of earth were removed in digging it? (8) A railway cutting is 28 ft. wide at the bottom, 80 ft. at the top, and 36 ft. deep. The slopes are uniform and the dimensions the same throughout its entire length. How many cubic yards of earth were removed in every 100 yards? 151. To find the length of a right prism when the volume and area of the base are given. RULE. Divide the volume by the area of the base, and the quotient will be the length. Ex. 1. The volume of a right prism is 126 9 c. ft., and the area of its base 23.5 sq. ft.; find the height. Area of base x height = volume (Art. 150). height=volume÷area of base Ex. 2. In the walls of an Eastern town is a stone in the form of a square prism, whose solid content is 9072 c. ft.; each side of the ends measures 12 ft.; what is its length? Area of end-12 x 12=144 sq. ft. length=9072÷144-63 ft. Ex. LXIV. Find the heights of the prisms which have the following volumes and bases: (1) Volume, 68 c. ft.; area of base, 16 sq. ft. Volume, 20 c. ft. 1440 c. in.; area of base, 1000 sq. in. (4) Volume, 15 c. ft.; area of base, 20 sq. ft. (5) In a square prism each side of the base measures 13 in., and the volume is 1 c. ft. 1483 c. in.; find the height. (6) Find, in feet, the length of a prism whose ends are squares, each side measuring 18 in., and whose solid content is 31 c. ft. 864 c. in. (7) The volume of a prism whose ends are equilateral triangles, each side measuring 12 ft., is 498-816 c. ft.; find the length. (8) The volume of a triangular prism is 13860 c. ft.; the sides of the base measure 35 ft., 53 ft., and 66 ft. respectively; what is its length? (9) The solid content of an hexagonal prism is 208 c. in., the side of the hexagon being 4 in.; required the length of the prism. E A D E A 152. The whole surface of a right prism consists of the areas of the two ends, and the areas of the several side faces. Suppose ABCD EF to be a hollow triangular prism of A paper or cardboard. Let it be cut down one of the edges, A B, and then laid out flat. It will be seen to form the rectangle AB B'A', whose length B C B F B' I BB' is equal to the perimeter of the triangular base of the prism, and breadth A B cqual to the height of the prism. Now the area of the rectangle is found by multiplying the length by the breadth (Art. 50), hence the area of the side faces of a prism is found by multiplying the perimeter of the base by the height or length. To find the whole surface the areas of the two ends must be added. 153. To find the area of the whole surface of a right prism. RULE. Multiply the perimeter of the base by the length, and add to this the areas of the two ends. Ex. Each side of an hexagonal prism measures 5 ft., and the length is 10 ft.; find the area of the whole surface. Perimeter of base =6x5=30 ft. Areas of side faces=30x10=300 sq. ft. Areas of two ends = 2 × 5 × 5 × 2.6=130 sq. ft. (Art. 94). Area of whole surface=300+130=430 sq. ft. Ex. LXV. (1) The base of a prism is an equilateral triangle; each edge of the prism measures 2 ft.; find the area of the whole surface. (2) The height of an octagonal granite pillar is 11 ft.; each edge of the base measures 2 ft. 6 in.; find the cost of polishing the sides at 1s. 8d. per sq. ft. (3) Find the whole surface of an hexagonal prism 2 ft. 6 in. long, each side of the base measuring 1 ft. 3 in. (4) The height of a triangular prism is 6 ft., and the three sides of the base are respectively 1 ft. 4 in., 1 ft. 7 in., and 2 ft. 6 in.; find the whole surface. (5) The gallery of a church is supported by 18 octagonal stone pillars, each of which measures 3 ft. 6 in. in circumference, and 12 ft. in height; find the cost of painting and decorating them at 1s. 6d. per sq. yd. (6) The perpendicular height of a triangular stone pillar is 8 ft., and the edges of the base are 3, 4, and 5 ft. respectively; how many sheets of paper, each 1 ft. 6 in. by 1 ft. 4 in. will it take to cover the whole surface ?.. THE RIGHT CYLINDER. 154. A Cylinder is a solid whose two ends are equal and parallel circles. The axis of a cylinder is the straight line drawn through the centres of the two ends. When the axis is perpendicular to the ends of a cylinder, it is said to be a right cylinder. 155. To find the volume of a right cylinder. RULE. Multiply the area of the base by the height, and the product will be the volume. [The reason for the rule has been given in Art. 149.] A B C D when the diameter of the base A B D Solidity of cylinder =346·5 × 20=6930 c in. Ex. LXVI. Find the volumes of the right cylinders whose dimensions are as follows: A (1) Area of base, 35 sq. ft.; height, 2 ft. B (5) Circumference of base, 3 ft. 8 in.; height, 5 ft. 2 in. (6) How many cubic yards must be dug out to make a round well 3 ft. 6 in. in diameter and 30 ft. deep? (7) The diameter of a well is 4 ft. 8 in., and its depth 27 ft.; find the cost of excavating it, at 7s. 6d. per c. yd. (8) A cylindrical shaft is sunk to the depth of 120 ft.; it measures 14 ft. 8 in. round at the top; how many c. feet of earth are taken out? (9) The diameter of a well is 8.4 ft., and its depth 25 ft.; what did it cost sinking, at 4s. 6d. per c. yd.? (10) The diameter of a well is 2 ft. 11 in., and its depth 45 ft.; what did it cost sinking, at 1s. 4d. per c. ft.? (11) The gallery of a church is supported by 16 circular pillars of stone, each of which is 7 ft. 4 in. in circumference, and 12 ft. high; required the cost of the whole, at 74d. per c. ft. (12) The top of a circular table is 7 ft. in diameter and 1 in. thick; find its cost, at 1s. per c. ft. (13) The expense of excavating a round well, of which the depth is 24 ft., and the diameter 3 ft. 6 in., was £1 18s. 6d. ; what was charged per c. yd.? (14) How many pieces of money, 7 in. in diameter and in. thick, must be melted down in order to form a cube whose edge is 7 in. long? (15) A circular well 5 ft. 3 in. in diameter is 46 ft. 2 in. deep; how much water will it hold? [1 gallon of water contains 277 c. in.] (16) A cylindrical boiler 11 ft. in circumference is 11 ft. 6 in. long; how many gallons of water will fill it? 156. To find the height of a cylinder when its volume and the area of its base are given. RULE. Divide the volume by the area of its base, and the quotient will give the height. Ex. 1. The volume of a cylinder is 150 c. ft., and the area of the base is 25 sq. ft.; find the height. Area of base x height = solidity (Art. 155.) height=solidity area of base Ex. 2. The volume of a cylinder is 1232 c. in., and the radius of its base is 7 in.; find the height. Diameter of base = 14 in. area of base=×14×14=154 sq. in. (Art. 102.) height=1232÷154=8 ft. Ex. LXVII. Find the heights of the cylinders which have the following volumes and bases: (1) Volume, 1032 c. ft.; area of base, 129 sq. ft. (2) Volume, 38500 c. ft.; area of base, 962.5 sq. ft. (5) Volume, 42 c. ft. 1344 c. in.; radius of base, 2 ft. 4 in. Volume, 369.6 c. ft.; circumference of base, 17.6 ft. |