2 To fec.ang. A 28° 57' 10,05799 To area 290,4738 = Which is 29 ac. or. 7 falls, 21 ells, as before. 10,00000 9,68489 2,77815 2.46304 From these four different varieties, it appears, that the logarithmic operation is the easiest. It were to be wished, that all land furveyors would take the trouble of computing their measurements by logarithms; then would they agree in their calculations, and depend lefs upon the accuracy of their fcales. Ex. 2. Required the area of a triangle, whose three fides are 500, 300, and 400 links. Anf. 2 roads, 16 falls. Ex. 3. Required the area of a triangle, whofe fides are 80, 60, 100, feet. Anf. 2400 feet. Ex 4. How many fquare yards are in a triangular court, whofe three fides are 36, 24, and 30 feet? Ex. 5. Anf. 39 yards, 6.17 feet. How many square yards are in a triangle, whose three fides are 63, 123,5 and 148 yards? Anf. 4168 yards. Ex. 6. How many fquare yards are in a triangle, whofe fides are 39, 42, and 45 feet? Anf. 84 yards. Ex 7. Required the area of a triangle, whofe fides are 90,84. and 78 yards. Anf. 3024 yards. PRO PROBLEM VII. Plate 6. fig. 75. Two fides of a right-angled triangle being given, to find the other fide. RULE. To find the hypothenuse, add the square of both the legs, and the square root of the fum is the hypothenuse. To find one of the legs, fubtract the fquare of the given leg from the square of the hypothenufe, and the fquare root of the remainder is the leg required. EXAMPLE I. The hypothenufe is 60, and the bafe AC 45; required the perpendicular. Ex. 2. Required the length of a ladder, to reach the top of a tower 56 feet high, the foot of the ladder being 48 feet from Anf. 73 feet 9.072 inches. the wall. Ex. 3. The hypothenufe is 600, and one of the legs 360: Required the other leg. Anf. 480. Ex. 4. The legs of a right-angled triangle are 64, and 48: Required the hypothenuse. Ex. 5. The hypothenuse of a right-angled triangle one of the legs 80: Required the other leg. PROBLEM VIII. Plate 6. fig. 76. To find the area of a trapezoid. RULE. Anf. 80. is 100, and Anf. 60. Multiply one half of the sum of the parallel fides by the perpendicular distance between them, and the product will be the area. EXAMPLE I. Required the area of a trapezoid, whofe parallel fides are 15, 19 chains, and their perpendicular distance 14 chains. BC 15 2)34.5 17.25 14 6900 1725 10)241.50 24.15 Anf. 24 ac. o r. 24 falls. Ex. 2. Required the area of trapezoid, whofe fides are 12,18+ feet, and the perpendicular distance between 7 feet. Anf. 106 fquare feet 2 inches. Ex. 3. Required the area of a trapezoid, the parallel fides being 180 and 200 yards, and their perpendicular distance 100 yards. Anf. 19000 fquare yards. Ex. 4. How many fquare yards are in a trapezoid, whose parallel fides are 9 and 100 feet, and breadth 50 feet? Anf. 527 yards 7 feet. Ex. 5. Required the area of a trapezoid, whofe parallel fides are 3, 4 feet, and perpendicular breadth 3 feet. Anf. 10 feet. Ex. 6. How many fquare feet are in a plank, 13 inches broad at one end, and 15 at the other, the length being 16 feet 5 inches? Anf. 19 feet 1 inch 10 parts. Ex. 2. Required the expence of caufewaying a bridge 150 feet long and 30 broad, at 1s. 6d. per square yard. Anf. 37h 105. PROBLEM IX. Plate 6. fig. 77. To find the area of a trapezium. RULE. Refolve the trapezium into triangles; compute the area of each of the triangles feparately, and the fum will be the area of the trapezium. EXAMPLE I. Required the area of a trapezium ABCD, the diagonal AC 60, BF 50, and DE 40 feet. |