60. To find the base of a rhombus or rhomboid when the area and perpendicular height are given. RULE. Divide the area by the perpendicular height, and the quotient will be the base. Ex. The area of a rhombus is 144 sq. ft., and the perpendicular height is 8 ft.; find the base. Base x height= =area (Art. 59). =144818 ft. 61. To find the perpendicular height when the area and base are given. RULE. Divide the area by the base, and the quotient will be the perpendicular height. Ex. The area of a rhomboid is 120 sq. ft., and the base is 15 ft. long; find the height. (1) The area of a field in the shape of a rhomboid is 4 ac. 1 ro. 34 po., and the base is 714 links, find the perpendicular breadth. (2) The area of a rhombus is 63 sq. yds. 5 sq. ft., and the perpendicular height is 7 yds. 1 ft., what is the length of the base? (3) The area of a rhomb-shaped field is 5 acres, and the length of the base 35 chains, find the perpendicular breadth. (4) The area of a rhomboid is 7481 sq. yds,, and the height 261 yds., find the base. (5) Each side of a rhombus is 20·5 yds., and the area contains 374-125 sq. yds., find the perpendicular height. (6) A rhomb-shaped field whose base is 12 ch. 85 lks. is let for £12 17s. at the rate of £4 per acre; find the perpendicular breadth. (7) The area of a rhombus is 17 sq. ft. 36 sq. in., and the length of the base 4 ft. 6 in.; find the height. (8) The area of a rhomboid is 32.625, and the perpendicular height 7.5; find the base. THE TRIANGLE. 62. A Triangle is a figure bounded by three straight lines. BC is the base and AD the perpendicular height, or altitude. B D 63. Triangles have three sides and three angles. 64. An Equilateral Triangle has its three sides equal. Equilateral. C 65. An Isosceles Triangle has two sides equal. Isosceles. 66. To find the area of a triangle when its base and perpendicular height are given. RULE. Multiply the base by the perpendicular height, and half the product will be the arca. Let ABC be a tri- E angle, and EBCF a For the triangle ABD is half the rectangle EBDA, and the triangle A D C is equal to half the rectangle ADCF. Therefore the whole triangle ABC is equal to half the rectangle EBCF. Hence it follows that the area of a triangle is equal to half the area of a rectangle having the same base and perpendicular height. = Ex. In the triangle ABC, the base B C 8 ft., and the perpendicular height A D=7 ft.; required the area. Area of rectangle E B C F=8x7=56 sq. ft., but the triangle A B C contains half this surface, hence area of triangle ABC =56÷2=28 sq. ft. Ex. XXIV. Find the areas of the triangles having the following dimensions: (1) Base, 24 ft.; height, 8 ft. (2) Base, 13 yds. 2 ft.; height, 6 yds. 1 ft. (3) Base, 3 yds. 2 ft. 6 in.; height, 2 yds. 1 ft. 9 in. (4) Base, 7 yds.; height, 4 yds. (5) Base, 15.625 ft.; height, 8.125 ft. Find the area in acres, roods, and poles, of triangles having the following dimensions : (6) Base, 750 lks.; height, 350 lks. (7) Base, 10 ch. 50 lks.; height, 4 ch. 75 lks. (8) Base, 18 ch. 20 lks.; height, 12 ch. 50 lks. (9) Find the area of a triangle whose base is 10 and height 4. (10) A triangular plot of ground whose base is 85 yds. 2 ft. and perpendicular height 72 yds. 1 ft. is to be paved at the rate of 1s. 6d. per sq. yd.; find the expense. (11) A triangular field whose base is 850 links and perpendicular height 275 links sells for £1870; find the price per sq. pole. (12) How much land is there in a triangular field whose base measures 4 ch. 25 lks. and height 2 ch. 50 lks., and what is the rental at 15s. per rood? 67. To find the base of a triangle when the area and perpendicular height are given. RULE. Divide double the area by the height, and the quotient will be the base. Ex. The area of a triangle is 643 sq. in. and the height is 27 in., find the base. Base x height = twice the area. .. base =twice the area÷height. =1296÷27=48 in. [Compare Art. 51.] 68. To find the perpendicular height of a triangle when the area and base are giren. RULE. Divide double the area by the base and the quotient will be the perpendicular height. Ex. The area of a triangle is 191-25 sq. ft. and the base i≤ 25-5 ft., required the perpendicular height. Base x height=twice the area. height=twice the area÷base. Ex. XXV. (1) A triangular field contains 31 acres, and its base is 1750 links, find the height. (2) An enclosure in the form of an equilateral triangle is paved at the rate of 9d. per sq. foot. Its perimeter is 63 yds., and the cost of paving the whole £35 18s. 9d. Find the height. (3) The perimeter of an equilateral triangle is 24 yds. and its area is 24 sq. yds., what is the height? (4) The base of a triangular field is 475 links; the field is let for £23 15s. a year, at the rate of £4 per acre. What is the height? (5) The area of a triangle is 1056 sq. ft. and its height 11 yds., find the length of the base. 69. To find the area of a triangle when the three sides are given. RULE. 1°. From half the sum of the three sides subtract each side separately. 2o. Multiply the half sum and the three remainders continually together. 3°. Take the square root of the last product. Ex. What is the area of a triangle whose sides are 13, 14, 15? Find the areas of the triangles having the following sides: (1) 50, 40, 30. (2) 13, 20, 21. (7) 71, 71, 9. (3) 21, 10, 17. (4) 35, 66, 53. (5) 11, 13, 20. (6) 10, 10, 12. (13) Find the area of a triangle whose sides are 3.27 ft., 4.36 ft., and 5.45 ft. (14) Find the area in acres, roods, and poles, of a triangular field, whose sides are 300 lks., 975 lks., and 1125 lks. (15) Find the area of a triangle whose sides are 319 ft., 444 ft., and 455 ft. (16) The sides of a triangular field are 5 ch., 16 ch. 25 lks., and 18 ch. 75 lks.; find the acreage. (17) Required the expense of reaping the corn in a triangular field whose sides are 150, 200, and 250 lks. long, at 13s. 4d. per acre. (18) A triangular field whose sides are 350, 440, and 750 yds. is let for £26 5s. a year. What is that per acre? (19) What rent should be paid for a triangular field whose sides are 175, 220, and 375 yds. at £6 12s. per acre? (20) A triangular piece of ground, whose sides are 800, 500, and 500 links, is let for £3; find the letting price per acre. (21) Find the area of an isosceles triangle whose base is 10 ft. and each of whose equal sides is 13 ft. (22) What is the rent of a triangular field whose sides are 550, 550, and 880 yds., at £2 15s. per acre? (23) Find the side of a square field whose area is half as much again as that of a triangular field whose sides are 20, 65, and 75 yards. 70. A right-angled triangle is one that has a right angle. AB is the hypotenuse, BO is the base, and A C the perpendicular. AC and CB, which include the right angle, are also called the sides of the right-angled triangle. B Hypotenuse. Base. Perpendicular. |