| Lorenzo Fairbanks - 1875 - 472 pages
...perpendicular 18.25 chains ? PROBLEM III. 710. To find the area of a triangle when three sides are given. RULE. — From half the sum of the three sides subtract each side separately. Then multiply the half sum and the three remainders continually together, and the square root of the... | |
| Malcolm MacVicar - Arithmetic - 1876 - 412 pages
...the rafters ? 801. PROB. III. — When the three sides of a triangle are given, to find the area : From half the sum of the three sides subtract each...remainders together ; the square root of the product is the area. 1. Find the area of a triangle whose sides are 15, 20, 25 feet. 2. What is the area of... | |
| Daniel W. Fish - Arithmetic - 1876 - 296 pages
...an isosceles triangle whose base is 20 ft., and each of its equal sides 15 feet ? RULE.—From huff the sum of the three sides, subtract each side separately; multiply the half-surn and the three remainders together; the square root of the product is the area. 3. Find the... | |
| John Barter (of the science and art coll, Plymouth.) - 1877 - 328 pages
...the journey is completed ? EXERCISE CCXV. Having the three sides of any triangle given, to find its area. Rule. — From half the sum of the three sides...multiply the half sum and the three remainders together, and the square root of the last product will be the area of the triangle. Example. — Let AB = 30,... | |
| Samuel Mecutchen, George Mornton Sayre - Arithmetic - 1877 - 200 pages
...preceding right triangle, AB is the hypotenuse, and AC, the perpendicular. To find the area of a triangle. RULE. From half the sum of the three sides, subtract...multiply the half sum and the three remainders together, and the square root of the product will be the area required. Note. — When the base and altitude... | |
| Popular encyclopedia - 1877 - 526 pages
...half the perpendicular from the vertex. Area of a triangle when the lengths of the sides are known; from half the sum of the three sides subtract each...multiply the half sum and the three remainders together, and extract the square-root of the product. Area of any parallelogram = any side multiplied by the... | |
| William James Milne - Arithmetic - 1877 - 402 pages
...the product of the base by the altitude. When the three sides are given, the following is the rule: RULE. — From half the sum of the three sides subtract each side separately. Multiply together the half sum and the three remainders, and extract the square root of the product. The result... | |
| Stoddard A. Felter, Samuel Ashbel Farrand - Arithmetic - 1877 - 496 pages
...= 9, 2d remainder. 27 — 24 = 3, 3d remainder. 27 X 15 X 9 X 3 = 10935. y 10935 = 104.57 sq. rds. area. RULE. — From half the sum of the three sides subtract each side separately ; then multiply the continued product of these remainders by half the sum of the sides, and extract... | |
| William James Milne - Arithmetic - 1877 - 418 pages
...of the base by the attitude. When the three sides are given, the following is the rule: RULE.—From half the sum of the three sides subtract each side separately. Multiply together the half sum and tfie three remainders, and extract the square root of the product. The result... | |
| Alfred Hiley - 1879 - 228 pages
...base by the height, and divide the product by 2. (2) To find the area, when the three sides are given. From half the sum of the three sides, subtract each...Multiply the half sum and the three remainders together, and the square root of the product is the area. Note 1. — To find one dimension, either the base... | |
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