 | Colin Arrott R. Browning - 1884 - 274 pages
...heightTT . , - 2 area He'Sht = T5T(15) When we know the length of each side, but not the perpendicular. Rule : — 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... | |
 | George Bruce Halsted - 1885 - 389 pages
...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 pages
...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 pages
...— 25 = 2? 52 — 39 = 13 52 — 40 = 12 52 x 27 x 13 x 12 = 219024 Area = -v/219024 = 468 sq. yd. 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... | |
 | James William Nicholson - 1889 - 408 pages
...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.... | |
 | Thomas Baker - 1891 - 262 pages
...has been made, and the work must be repeated. TO FIND THE AREA OF A TRIANGLE FROM THE THREE SIDES. RULE. From half the sum of the three sides subtract each side severally and reserve the three remainders ; multiply the half sum continually by the three remainders, and the... | |
 | Horatio Nelson Robinson - 1892 - 428 pages
...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... | |
 | Horatio Nelson Robinson - 1892 - 428 pages
...+ 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... | |
 | William Kent - 1895 - 1234 pages
...altitude. RULE a. Multiply half the product of two sides by the sine of the Included angle. Ri'LE 3. From half the sum of the three sides subtract each side severally; multiply together the half sum and the three remainders, and extract the square root of the product. The area... | |
 | Peder Lobben - 1899 - 460 pages
...circle of the same area. To Figure the Area of Any TriangIe when Only the Length of the Three Sides is Given. RULE. From half the sum of the three sides subtract each side separately ; multiply these three remainders with each other and the product by half the sum of the... | |
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From half the sum of the three sides, subtract each side severally; multiply the half sum, and the three remainders together, and the square root of the product will be the area required. Example. — Required the area of a triangle, whose sides are 50,... 
