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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, and the square root of the product will be the area required. "
Higher Book - Page 252
by William Seneca Sutton - 1896
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Commercial class book

Alfred Newsom Niblett - 1861 - 204 pages
...the base by the perpendicular, and half the product will be the area. CASE 2.—The three sides being given. From half the sum of the three sides subtract each side separately, multiply the half sum and the three remainders continually together, and the square root of the last...
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A handbook of practical gauging

Janes Boddely Keene - 1861 - 104 pages
...area in square yards ? Ans. 875 one-9th yds. When the Three Sides of a Triangle are given. 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...
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Adams's Improved Arithmetic: Arithmetic, in which are Combined the Analytic ...

Daniel Adams - Arithmetic - 1861 - 452 pages
...point at the other ? Ans. 131 feet. To find the area of a triangle when the length of each side is given, — From half the sum of the three sides subtract each side severally ; then multiply the half sum and the three remainders continually together, and the square...
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Elements of Plane and Spherical Trigonometry: With Practical Applications

Benjamin Greenleaf - Geometry - 1862 - 532 pages
...angle 40 rods ? Ans. 12 A. 20 P. PROBLEM IX. 622. To find the area of a TRIANGLE, the three sides being given. From half the sum of the three sides subtract each side ; multiply the half sum and the three remainders together, and the square root of the product will...
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The Practical Model Calculator, for the Engineer, Mechanic, Machinist ...

Oliver Byrne - Engineering - 1863 - 600 pages
...= 38 fe. 11 in. A " " 7J pa. = area required. To find the area of a triangle whose three sides only are given. — From half the sum of the three sides subtract each side severally. Multiply the half sum and the three remainders continually together, and the square root...
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The Analysis of Written Arithmetic: Book Second, Designed for Public and ...

Stoddard A. Felter - Arithmetic - 1864 - 412 pages
...8ft. X 2 = 16 sq.ft. PROB. II. — The sides of a triangle being given to find the 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 the...
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Elements of Geometry: With Practical Applications to Mensuration

Benjamin Greenleaf - Geometry - 1866 - 328 pages
...angle 40 rods ? Ans. 12 A. 20 P. PROBLEM IX. 622. To find the area of a TRIANGLE, the three sides being given. From half the sum of the three sides subtract each side ; multiply the half sum and the three remainders together, and the square root of the product will...
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Griswold's Railroad Engineers' Pocket Companion for the Field: Comprising ...

Whiting Griswold - Railroad engineering - 1866 - 144 pages
...angle, and one-half the product equals the area. To find the area of a triangle by its sides. RULE 8. 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...
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The advanced lesson book, by E.T. Stevens and C. Hole

Edward Thomas Stevens - 1866 - 434 pages
...hypoteneuse required. To Jind the area of any triaiu]k when the three sides only are gicen. KULE : — From half the sum of the three sides subtract each side separately, then multiply the half sum and the three remainders together. The square root of the product is the...
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The Crittenden Commercial Arithmetic and Business Manual: Designed for the ...

John Groesbeck - Arithmetic - 1867 - 226 pages
...length by the breadth. To find the area of a triangle. Multiply the base by one-half the altitude. Or, 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....
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