 | Daniel W. Fish - 1874 - 302 páginas
...What is the area of an isosceles triangle whose base is 20 ft., and each of its equal sides 15 feet 1 RULE. — From half the sum of the three sides, subtract each side teparately ; multiply the half -sum and the three remainders together; the square root of the product... | |
 | Henry Lewis (M.A.) - 1875 - 104 páginas
...perpendicular height. CASE II. — The area of a triangle may, however, be determined from its three sides by the following rule: — 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... | |
 | Horatio Nelson Robinson - 1875 - 468 páginas
...of an isosceles triangle whose base is 20 ft., each of its equal sides 15 ft. ? Ans. 111.85 sq. ft. RULE, from half the sum of the three sides subtract each side separately ; multiply the half-sum and the three remainders together ; the square root of the product is the area. 3. How... | |
 | Lorenzo Fairbanks - 1875 - 472 páginas
...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 - 1876 - 412 páginas
...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 side separately. Multiply the half sum and the three remainders together ; the square root of the product is the area. 1. Find... | |
 | William James Milne - 1877 - 418 páginas
...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 will be the area of the... | |
 | Samuel Mecutchen, George Mornton Sayre - 1877 - 200 páginas
...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 each side separately; multiply the half sum and the three remainders together, and the square root of the product will be the area... | |
 | John Barter (of the science and art coll, Plymouth.) - 1877 - 328 páginas
...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 subtract each side separately, multiply the half sum and the three remainders together, and the square root of the last product will be the... | |
 | Popular encyclopedia - 1877 - 526 páginas
...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 side separately; multiply the half sum and the three remainders together, and extract the square-root of the product. Area of... | |
 | Stoddard A. Felter, Samuel Ashbel Farrand - 1877 - 496 páginas
...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... | |
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RULE. — From half the sum of the three sides subtract each side separately ; multiply the half -sum and the three remainders together ; the square root of the product is the area. 
