20 PRELIMINARY OBSERVATIONS. Now the acre being equal to 10 square chains, and each chain containing 22 yards, or 4 poles, we have for the area, (1)2 chain x 10 10 square chains; or (66)2 feet×10=43560 square feet; or (100)2 links10=100,000 square links. Again, as there are 100 links and 66 feet in a chain, 1 chain 100 links 66 feet-792 inches; hence 1 link=7-92 or 8 inches nearly; and 1 link 11.88 inches=1 foot nearly; hence, with any quantity less than one chain, to bring feet into links, add more, to bring links into feet, take less. 18 Again, because the acre equals 160 square poles or perches, and there are 4 roods in an acre, there will be 40 square rods or perches in a rood. These are the aliquot parts of an acre. PROBLEM I. To bring acres therefore into roods and perches, multiply by 4 and by 40. EXAMPLE 1. In 8 acres, how many roods? Answer 32 roods, or 1280 perches. EXAMPLE 2. In 24 acres, how many perches? Answer 3920 perches. PROBLEM II. To bring square chains into acres, roods and perches. As almost all areas are obtained in square chains, by the multiplication of one side in chains, by another side in chains, this is perhaps the most practical case that can be given to a beginner. Rule. Divide the square chains by 10, to bring them to acres, and then multiply by 4 and 40 as before. EXAMPLE 1. Let the area of a field equal 8.2500 square chains (that is 8 square chains and 2500 square links). Required the number of acres. 10 | 8.2500 0.82500 4 3.30000 40 12.00000 A. R. P. Answer 0. 3. 12. EXAMPLE 2. How many acres, &c., will there be in 8250 chains; and in 82.50 chains. Answer 0. 0. 13. and 8. 1. 0. The difference depends upon the different positions of the decimal point. MENSURATION. PROBLEM I. To find the area of a triangle, when the base and perpendicular height are given. Rule. Multiply the base by the height. EXAMPLE 1. What is the area of a triangular field, whose base is 5 chains 50 links, and height 3 chains 20 links? EXAMPLE 2. Required the area of a triangle, whose base is 7 chains 25 links, and perpendicular height 90 links. A. R. P. Answer 0. 1. 12. EXAMPLE 3. How many square yards are contained in a triangular plot of ground, whose base and height measure respectively, 8 chains 50 links, and 5 chains 50 links? By preceding rule area 2.3375 acres. multiply by the sq. yards in 1 acre 4840 area=11313.5 sq. yards. PROBLEM II. To find the area of a triangle, when the three sides Rule. Take half the sum of the three sides, subtract each side severally from this sum; then multiply this and the three remainders together, and take the square root for the area. EXAMPLE 1. What is the area of a triangle, whose three sides are 30, 40, and 50 chains? Answer ✔ 360000 = 600 sq. chains 60 acres. EXAMPLE 2. Required the area of a triangular field whose three sides measure respectively 25 chains, 42 chains, and 56 chains? A. R. P. Answer 49. 0. 10. EXAMPLE 3. The three sides of a triangular plot of ground are respectively 20 chains 40 links, 25 chains 20 links, and 30 chains 50 links. What is the area? A. R. P. Answer 25. 2. 4. EXAMPLE. 4. Given the sides of a triangle 24 chains 72 links; 38 chains 75 links; and 44 chains 68 links; to ascertain the area in square yards. Answer 231384 square yards. PROBLEM III. To find the area of a quadrilateral right-angled figure. Rule. Multiply the length by the breadth, the product will be the area. EXAMPLE 1. What is the area of a rectangular field, whose length is 35 chains 40 links, and breadth 24 chains 36 links? chs. chs. sq. chs. 35.40 x 24.36862·3486.234 B acres. A 4 EXAMPLE 2. Required the area of a right-angled parallelogram, whose length is 56 chains 24 links, and breadth 35 chains 42 links. A. R. P. Answer 199. 0. 32. PROBLEM IV. To find the area of a trapezoid. Rule. Multiply half the sum of the two parallel sides by the distance between them. EXAMPLE 1. In a trapezoid, whose parallel sides are 7 chains 25 links and 8 chains 63 links, the distance being 11 chains 65 links, how many acres are there? |