Ev. 2. What is the area of a rectangle ABCD, whose length AB Icing 12f. 6i., and the breadth BC 2f. gi. By duodecimals. By decimals. 12 6i 12:5 2-75 .,C 9 4 6 625 25 0 875 B f. 34 375 Ex. 3. What is the area of a rhombus ABCD, the length AB being 12f. 6i, and the height BE 6t. 3i. ? By duodecimals. By decimals. 12 6 12.5 D E 6 3 6 25 N 3 16 625 75 0 250 A B 750 Ans. 78 ] 6 f. f. 78:125 Er. 4. What is the area of a rhomboides ABCD, whose length AB is 16f. 3i. and the height DE 5f. 6i. ? By duodecimals. By decimals. 16 3 16:25 5 6 5 5 D 8 1 6 8125 8125 B 89 4 0 f. 89.375 f. i ii Ans. 89 ft. 4 in. 6 parts. Prob. 2. To find the area of a triangle. Multiply the base by the perpendicular height, and half the product will be the area. Er. What is the area of a triangle ABC, the base AB being 12f. 3i. and the height CD 8f. 6i. ? By duodecimals. By decima's. 12:25 85 Prob. 3. To find the area of a triangle, whose three sides only are given. From the half sum of the three sides, subtract each side severally; multiply the half sum and the three remainders together, and the square root of the product will be the area required. Ex. Requireth the area of a triangle ABC, whose three sides AB BC, and CA, are respectively 13, 14, and 15 feet. Prob. 4. Any two sides of a right-angled triangle being given, to find a third side. Case ).-When the two sides are given, to find the hypothenuse. Add the squares of the two legs together, and the square root of the sum will be the hypothenuse. Er. 1. Requireth the hypothenuse AC of the right angie ABC, the above AB being 4, and the perpendicular BC 3 4x4=16 the square of AB. 25(5 the answer. Ex. 2. There is a roof of which the span AB is 30 feet, and the height CD 12 feet; required the length of a rafter AC or BC? This is only two right-angled triangles of one-continued base, joined together at their perpendicular. Er. 3. The span AB of a roof is 48 feet, and the height EF 18 teet, standing upon a rectangle plan hipped at each end, I demand the length of the cominon rafters, and likewise the hip ratters. 18 2)48 19 AE or EB=24 144 24 A 18 E 96 924 48 +576 D +324 Geometrical construction. Draw AD and BC perpendicular to AB, 9:00(30 and make AD equal to AE or EB, 9 and draw DC parallel to AB : join ED and EC; produce CE to G, 50) 00 making EG equal to EF, and join 00 GD; then will GD be the length of the hip rafter. Therefore, the length of the common rafiers are 30 feet each. But because the hip rafters are the hypothenuses of right-angled triangles, having a common rafter for one of the perpendicular legs, and the other leg being equal to half the width of the roof. Hence 30 x 30=C00 And 24 x 245576 1976(38:41 feet the length of a bid rafter. 9 68) 576 544 764( 3200 3056 7681) 14400 7681 6719 Case 2.-The hypothenuse and one of the legs being given, to find the other leg. From the square of the hypothenuse, take the square of the given leg, and the square root of the remainder will be equal to the other leg. Er. 1. In the right-angled triangle ABC, the base AB being 40, and the hypothenuse AC 50, I demand the perpendicular BC. (50) = 2500 remainder=900(30=BC 9 60) 00 B 00 Ex. 2. The width AB of a roof being 36 feet, and the length of a raster AC or BC 25 feet, I demand the height of the roof. 25 2)36 25 18 the base 125 18 50 144 625 18 324 324 remainder=301(17.3 the height nearly. 1 с 27) 201 189 343) 1200 1029 А B 17100 Prob. 5. To find the area of a trapezium. Multiply the diagonal by half the sum of the two perpendiculars, falling upon it from the opposite angles, and the product will be the Er. What is the area of a trapezium ABCD, the diagonal AC being 36 feet, the perpendicular DE 16 feet, and BF 12 feet? 16 +12 area. |