 | George Bruce Halsted - 1885 - 389 ページ
...vertex, to find the area of a triangle. 319 808. Given, the three sides, to find the area of a triangle. RULE. From half the sum of the three sides subtract each side separately ; multiply together the half -sum and the three remainders. The square root of this product... | |
 | George Bruce Halsted - 1886 - 394 ページ
...vertex, to find the area of a triangle. 319 808. Given, the three sides, to find the area of a triangle. RULE. From half the sum of the three sides subtract each side separately ; mult1ply together the half-sum and the three remainders. The square root of tltis product... | |
 | Christian Brothers - 1888 - 484 ページ
...triangle whose hase is 18 ft. 4 in., and height 11 ft. 10 in. Ans. 108 sq. ft. 68 sq. in. CASE II. 247. To find the area of a triangle when the three sides are given. 1. The sides of a triangle are, respectively, 25, 39, and 40 yd. ; what is the area ? OPERATION. 25... | |
 | James William Nicholson - 1889 - 408 ページ
...15. 15 — 5 = 10, 15 — 12 = 3, 15 — 13 = 2. 15 X 10 X 3 X 2 = 900; v'itOO = 30, Ans. in sq. in. RULE. — From half the sum of the three sides subtract each side; then multiply the half sum and the three remainders together , and extract the square root of the product.... | |
 | John Groesbeck - 1891 - 426 ページ
...at one end, and tapers to a point. Find the value at 4^ cents per square foot. Ans. 63 cents. 447. To find the Area of a Triangle when the Three Sides...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... | |
 | Thomas Baker - 1891 - 262 ページ
...adjacent angles 60° 14' and 59° 46'; -required the area by logarithms. Am. 17 a. Ir. I2p PROBLEM III, To find the area of a triangle when the three sides are given. Let a, b and c be the sides, and * their half sum, then A = •/ * (* — o) (* — b) (s — c). By... | |
 | Horatio Nelson Robinson - 1892 - 428 ページ
...+ 50) -4- 2 = 60 ; 60 - 30 = 30 ; 60 - 40 = 20 ; 60 - 50 = 10. V60x30x20xlo' = 600 sq. ft. , area. 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... | |
 | Horatio Nelson Robinson - 1892 - 428 ページ
...50) -=- 2 = 60 ; 60 - 30 = 30 ; 60 - 40 = 20; 60 - 50 = 10. \X60~x30 x 20 x 10 = 600 sq. ft., area. 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... | |
 | Massachusetts - 1893 - 988 ページ
...given. Rule. — Take one-half the product of the side and perpendicular, and divide by 160. (ft) When three sides are given. 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... | |
 | William Shaffer Hall - 1893 - 88 ページ
...and applying logarithms, the equation becomes, log 2 8 = log b -f log « -\- log sin A — 10. § 21. To find the area of a triangle when the three sides are given. [13] RULE : Find half the sum of the three sides, and subtract from it each side separately. Then find... | |
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To find the Area of a Triangle when the Three Sides are given. Rule. — From half the sum of the three sides subtract each side separately. 
