 | George Edward Davis - Chemical engineering - 1904 - 560 pages
...be measured. (a) Multiply the base by the perpendicular height, and divide the product by two. (b) From half the sum of the three sides, subtract each side separately, then multiply the half sum by the three remainders consecutively. The square root of this product is the... | |
 | 1906 - 576 pages
...to the base.) The area may also be found if the lengths of the three sides of a triangle be given. 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... | |
 | International Correspondence Schools - Building - 1906 - 634 pages
...To find the area of a triangle from the lengths of its three sides, apply the following: 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.... | |
 | Joseph Gregory Horner - Engineering - 1906 - 572 pages
...to the base.) The area may also be found if the lengths of the three sides of a triangle be given. 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... | |
 | Stephen Christie - Steam-boilers - 1908 - 272 pages
...divide by given dimension. Rule to find area of triangle when dimensions of three sides are given : 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.... | |
 | 1908 - 560 pages
...Scale, 10 chains to 1 inch. xvy.-? The area of the resulting triangle may be found as follows : — 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 will be the... | |
 | Gustavus Sylvester Kimball - Business mathematics - 1911 - 444 pages
...Solution. (20+30+40) -5-2 =45; 45-20 = 25; 45-30 = 15; 45-40 = 5. ^45X25X15X5 = 290.4 + ft. 357. Rule. 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.... | |
 | George Morris Philips, Robert Franklin Anderson - Arithmetic - 1913 - 444 pages
..."" ~~2~• 405. To find the area of a triangle when its three sides are given, apply the following : From half the sum of the three sides, subtract each side separately; multiply together the number of units in the half sum and in each of the three remainders, and extract... | |
 | Joseph Gregory Horner - Iron-founding - 1914 - 460 pages
...breadth. 2. Triangle. Multiply the base by the perpendicular height, and take half the product. Or: 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 will be the... | |
 | United States - 1918 - 840 pages
..."perpendicular to it from the opposite vertex equals the area. Formula: — ^-£=area. (Fig. 6, page 51.) Or 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 the product is the area.... | |
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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. 
