PROBLEM I. Plate 5. fig. 70. To find the area of a square. RULE, Multiply the fide by itself, and the product will be the area. EXAMPLE I. How many square yards are in a square, whose fide is 15 feet? 3 inch. Ex. 2. Required the area of a square, whose fide is 12 feet? Anf. 144 Square feet. Ex. 3. How many square feet are in a square, whose side is 6 feet 3 inches ? Anf. 39 feet o in. 9 pts. Ex. 4. How many square yards are in a square court, whose fide is 804 feet? Anf: 715 yds. 5 feet o in. 9 pts. Ex. 5. How many square chains are in a field, whose fide is mile? Ans. 6400 sq. chains. Ex. 6. Required the area of a square, whose fide is 3 chains Anf. 9/a. cho PROBLEM II. Plate 5. fig. 71. To find the area of a rectangle. RULE. Multiply the length by the breadth, and the product is the area. EXAMPLE I. Required the area of a rectangle, whose height is 3000 línksj and breadth 1670 links of the English chain. Here, because the chain is divided into 100 links, and that 100 * 100 is 10000 (the number of square links in one square chain) and 10000 X 10 = 100000 (the number of square links in one acre) divide the product by 100000, the quot gives acres and decimals of an acre; and this decimal is reduced to value by multiplying by 4, by 4o, by 30. Or, instead of dividing the square links by 190000, cut off five decimal places towards the right hand, the integral part gives acres, and those cut off are decimals of an acre, and are reduced to value accordingly. Ex, 2. Required the area of a rectangular field, whose sides are 5.5 and 2.54 Scots chains. Ans. 1 acre I rood 23 falls 18.72 ells. Ex. 3. Required the area of a rectangle, whose length is 151 feet, and breadth 12 feet. Anf. 186 square feet. Ex: 4. Required the area of a rectangle, whose length is 12 inches, and breadth 6 inches. Ans. 60 inches. Ex. 5. Required the area of a rectangle, whose fides are 56 feet, and 18 feet 6 inches? Anf. 1036 square feet. Ex. 6. Required the area of a rectangle, whose length is 16, and breadth ro's yards Anf. 168.3 sds. PROBLEM JII. Plate 5. Fig. 72 To find the area of a rhombus or rhomboid. RULE. Multiply the length by the perpendicular breadth, and the product is the area. EXAMPLE I. Required the area of a rhombus, whose fide is 750 links, and one of its acute angles 60°. When one of the angles of a rhombus or rhomboid are gia ven, the area may be found by the following proportion. As rad : fine included ang. : : the prod. containing sides : area.. As rad. 90° 10.00000 4,87100 is to line 60° 9.93753 4 $o is prod. 562500 5.75012 3,48400 To the area = 487100 5.68765 40 19,36000 36 216000 108000 12,9600e Anf. 4 acres 3 roods 19 falls 13 ells nearly. 1 Ex. 2. Required the area of a rhombus, whose length is 15 feet, and perpendicular breadth 12 feet. Anf. 180 square feet. Ex. 3. Required the area of a rhomboid, its length being 24, and perpendicular 14 Scots chains. Anf. 33 acres 2 roods 16 falls. ells. Anf. 38 a. I r. 24 p. Ex. 6. Required the area of a rhomboid, when the length is 50 feet 6 inches, and perpendicular breadth 6 feet 6 inches. Anf-328 feet. Ex. 7. How many square yards are in a rhombus, whose lide is 151 feet, and perpendicular height 3 feet ? Anf 5 yards 5 feet 4 inches 6 parts. PROBLEM IV. To find the side, or the perpendicular breadth of any parallelogram, the other side and the area being given. RULE. Divide the area by the given fide, and the quotient will be the other side. EXAMPLE I. The area of a rectangle being 100000 square links, and one of its fides 1000 links, required the other fide. 1000)100000 100 links, or i chain length, 1000 |