2. What is the content of a field in the form of a rhombus, whose length is 7.6 chains, and perpendicular height 5.7 chains ? Ans. 43.32 chains. 3. What is the area of a rhombus, whose side is 7 feet 6 inches, and perpendicular height 3 feet 4 inches? Ans. 25 feet. 4. What is the area of a rhombus whose length is 3 yards, and perpendicular height 2 feet 3 inches? Ans. 20 feet 3 inches. PROBLEM IV. To find the area of a Triangle. RULE. Multiply the base by the perpendicular height, and divide the product by two for the area.† 1. The base of a triangle is 76.5 feet, and perpendicular 99.2 feet; what is its area? Ans. 76.5x92.2÷2=3526·65 square feet, the area. * See Appendix, Demonstration 9. 2. The base of a triangle is 727 yards, and the perpendicular height 36.5 yards, what is its area? Ans. 1326-775 yards. 3. The base of a triangular field is 1276 links, and perpendicular 976 links; how many acres in it? Ans. 6 acres, 36.3008 perches. 4. The base of a triangle measures 15 feet 6 inches, and the perpendicular 12 feet 7 inches; what is its area? Ans. 97 feet 6 inches. PROBLEM V. Having the three sides of any Triangle given, to find its area. RULE I. From half the sum of the three sides subtract each side separately, then multiply the half sum and the three remainders together, and the square root of the last product will be the area of the triangle.* RULE II. Divide the difference between the squares of two sides of the triangle by the third side; to half this third side add half the quotient, and deduct the square of this sum from the square of the greater side, the remainder will be the square of the perpendicular, the square root of which, multiplied by half the base, will give the area of the triangle.† See Appendix, Demonstration 11. + See Appendix, Demonstration 12. = 1. Given the side A B 9-2, BC= 7·5, and A C = 5.5. Required the area of the triangle? 9.2 7.5 5.5 Sum 22.2 Sum 11.1-9.2-1.9): then ✔ (11·1 x 1.9 x 3.6 x 5.6) 11.1-7.5 3.6 × 425.174420-619 the area 11·1-5.5=56) by RULE I. Again, 9-22-7.52 — 84·64 — 56.25 28.39; then 28.39 5.55.161818, quotient. Now (5-161818 ÷ 2) + (5·5 ÷ 2) = 2·580909 + 2·75 = 5.3309 half quot. plus half third side: then 84.64 28-41849481 = 56·22150519, and ✔ 56·22150519 = 7·498 = perpendicular; then 7-498 × 2·75 = 20-619 the area as before. 2. What is the area of a triangle whose sides are 50, 40, and 30? Ans. 600. 3. The sides of a triangular field are 4900, 5025, and 2569 links; how many acres does it contain? Ans. 61 acres 1 rood 39-68 perches. 4. What is the area of an isosceles triangle, whose base is 20, and each of its equal sides 15 ? Ans. 111.803. 5. How many acres are there in a triangle, whose three sides are 380, 420, and 765 yards ? Ans. 9 acres, 38 poles. 6. How many square yards in a triangle, whose three sides are 13, 14, and 15 feet? Ans. 9 square yards. 1 7. How many acres, &c. in a triangle, whose three sides are 49, 50-25, and 25.69 chains? Ans. 61 acres 1 rood 39-68 perches. PROBLEM VI. To find the area of an equilateral triangle. RULE. Square the side, and from this square deduct its fourth part; then multiply the remainder by the fourth part of the square of the side, and the square root of the product will give the area.* Or multiply by 3 for the area.t AB2 4 1. Each side of a triangular field, ABC, measures 4 perches, what is its area? 4216, then 164 = 4 and 16 4 12 then 12 × 16 = 12 X 4 = 48, and 48 6.928, the area. 2. How many acres in a field of a triangular form, each of whose sides measures 70 perches? Ans. 13 acres 1 rood 1 perch. 3. The perimeter of an equilateral triangle is 27 yards, what is its area? Ans. 35:074. Note. When the triangle is isosceles, the perpendicular is equal to the square root of the difference between the squares of either of the equal sides, and half the base. PROBLEM VII. Given the area and altitude of a triangle, to find the base· RULE. Divide the area by the altitude or perpendicular, and double the quotient will give the base.‡ 1. Given the area of a triangle 12 yards, and altitude = 4; what is its base ? Ans. 124=3; then 3 ×2=6 yards, the base AB. A C 2. A surveyor having lost his field book, and requiring * See Appendix, Demonstration 13. the base of a triangular field, whose content he knew from recollection was 14 acres, and altitude 7 yards, how much is the base ? Ans. 19360 yards. PROBLEM VIII. Given the area of a triangle, and its base, to find its altitude. RULE. Divide the area by the given base, and double the quotient will give the perpendicular. The reason of this rule is manifest, from the last. 1. Given the area of a triangle = 12, and its base = 6, what is its perpendicular height? Ans. 126=2; then 2 × 24 the altitude. PROBLEM IX. Given any two sides of a right angled triangle, to find the third side, and thence its area. RULE. I. To the square of the perpendicular add the square of the base, and the square root of the sum will give the hypothenuse. II. The square root of the difference of the squares of the hypothenuse, and either side will give the other. III. Or multiply the sum of the hypothenuse and either side, by their difference; and the square root of the product will give the other.* 1. Given the base A C 3, the perpendicular C B 4; required the hypothenuse A B? 324225; then / 25 = 5, the hypothenuse A B. 2. Given A B 5, A C3; required 8 x 2 = 16; then ✔ 164, as before. 3. Given A B 5, B C 4; 52 429; then A required A C. 9 B C 3 the side A C; or (5 + 4) × (5 — 4) = 9 × 1 = 9; then √ 9 = 3, as before. 3 × 426 the area of the triangle. * See Appendix, Demonstration 16. And |