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-3284 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 side. 1000) 100000 100 links, or i chain length, 1000 Ex. 2. A square field contains 120 acres, required the length of its fide. 120 acres = 12000000 square links. 3464 links. And 12000000 = Ex. 3. The area of a rhomboid is 63; square feet, and the length 10 feet. Required the perpendicular breadth. Anf. 6 feet 4.2 inches. Ex. 4. The perpendicular breadth of a rhombus is 4. feet, and area 30 sq. feet. Required the side. Anf: 7.0588 feet. Ex. 5. The area of a square is 6740 square yards. Required its side. Anf. 82.097 gards. Examples in this problem and in the three preceding problems prove each other. PROBLEM V. Plate 6. fig. 74. To find the area of a triangle, its perpendicular height and base be ing given. RULE. Multiply the base by half the perpendicular height, and the product will be the area. EXAMPLE. Required the area of a triangle, whose base is 64, and per. pendicular 80 feet. 64 base. 80 perp 2)5120 2560 area. Ex. 2 How many square yards are in a triangle, whose base is 99 feet, and perpendicular 85 feet 6 inches. Dee Ex. 3. Required the area of a triangle, whose base is 12 feet 3 inches, and perpendicular 8 feet 9 inches. Anf. 53 feet 7 in. 1 pt. 6". Ex. 4. How many square yards are contained in a triangular garden, the length of one of its sides being 80 yards, and the perpendicular distance between that side and the opposite angle 70 yards? Ans 2800 square yards. Ex. 5. What is the expence of paving a triangular court, at 4s. 6d. per square yard, one of its fides being 48 feet 6 inches, and perpendicular 304 feet? Anf. 181. 9. 9 d. PROBLEM VI. Plate 6. fig. 74. One of the angles of a triangle and the containing fides being given, to find the area. RULE I. As radius, is to the fine of the included angle, so is half the product of the containing sides, to the area. RULE 2. Find the perpendicular by trigonometry, and proceed as in the preceding problem. EXAMPLE I. Required the area of a triangle, whose included angle is 63° 30', and the containing fides 806 and 70c links of the English chain. By Rule I. =!0.00000 To area 2.52500 = 5.40219 Ang 2 acres 2 roods 4 perches. By Ex. 2. How many square yards are in a triangle, whose sides are 100.98, feet, and included angle 45° ? Ans: 384 59. yds 8. or 8,8 feet. Ex. 3. Required the area of a triangle, when the containing sides are 409 and 220 yards, and the included angle 30 Anf. 22495 sq. yds. Ex. 4 Required the area of a triangular field ABC, AB= 6000, AC 8000 links of the Scots chain, and angle A 39° 36'. Anf. 153 acres. Ex. 5. Required the area of a triangle, the containing sides being 21 and 25 yards, and the contained angle 50°. Anf. 2031 yards. PROBLEM VII. Plate 6. fig. 75. The three sides of any triengle being given, to find the area. RULE. Add the three given Gides, and from half their sum subtract the fides severally: Multiply the half sum and the three re mainders |