 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 252by William Seneca Sutton - 1896Full view - About this book
 | 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... | |
 | Harry Anson Finney, Joseph Clifton Brown - Business mathematics - 1916 - 506 pages
...= 2. 24 x 12 x 10 x 2 = 5760. V5760 = 75.88, the number of square feet in the area. Rule. From one half the sum of the three sides subtract each side separately. Find the product of the half sum and of the three remainders. Extract the square root of this product. The result... | |
 | 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.... | |
 | Hugh Findlay - Agriculture - 1920 - 584 pages
...take one-half the product of the side and perpendicular, and divide by one hundred and sixty. When 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 divided by... | |
 | Hugh Findlay - Agriculture - 1920 - 586 pages
...take one-half the product of the side and perpendicular, and divide by one hundred and sixty. When 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 divided by... | |
 | Joseph Gregory Horner - 1920 - 418 pages
...dimensions are given in terms of the length of the three sides, then the area is 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... | |
 | Adolf Hopp, William Neubecker - Sheet-metal work - 1921 - 436 pages
...the dimensions of the three sides are known or can be obtained. The rule in this case is as follows : From half the sum of the three sides subtract each side separately; find the continued product of the half sum of the sides and the three remainders, and the square root of this... | |
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