 | John Farrar - 1822 - 270 páginas
...since any triangle whatever is equal to a rightangled triangle of the same base and altitude (Geoni. 170), we can make use of the following simple rule,...AB (fig. 65) = 12,38 ch., AC — 6,78 ch., and the angle Fig. 65. A = 46° 24' to find the area. 12,38 .... log 1,09272 6,78 .... log 0,83123 40° 24'... | |
 | John Farrar - 1822 - 244 páginas
...simple rule, where the known parts admit of it, as equivalent to the foregoing ; namely, the area nf a triangle is equal to the product of the base by half its altitude. Given AB (Jig. 65) = 12,38 ch., AC= 6,78 ch., and the angle Fig. 65. Jf = 46° 24' to find the area. 12,38 ....... | |
 | Peter Nicholson - 1823 - 210 páginas
...put A for their common altitude, and B, b, for their bases ; then BxA:fixA::B:5. THEOREM 53. 149. The area of a triangle is equal to the product of the base by half its altitude. For the triangle ABC is half the parallelogram ABCE, which has the same base, BC, and the same altitude... | |
 | John Farrar - 1833 - 274 páginas
...its contiguous side = 11,46 ch., the included angle being 47° 30', to find the area. Ans. 12"- 3r- 36P124. Since every parallelogram is divided by its...the area. 12,38 . . log. . . 1,09272 6,78 . . . log. . . . 0,83123 40° 24' . . log. sin . 9,85984 6,0,8986 1,78389 4 35944 40 14,37760 Area = 6a- 0r- 14P125.... | |
 | John Farrar - 1840 - 270 páginas
...and its contiguous side = 11,46 ch., the included angle being 47° 30', to find the area. Ans. 12' 3' 36P124. Since every parallelogram is divided by its...the base by half its altitude. Given AB (fig. 65) = 12, 38 ch., AC = 6,78 ch., and the Fig. 65. angle A = 46° 24' to find the area. 12,38 . . log. .... | |
 | Peter Nicholson - 1856 - 518 páginas
...their common altitude, and B, b, for their bases : then Bx A : их А :: B : b. THEOREM 44. 113. The area of a triangle is equal to the product of the base by half its altitude. For the triangle ABC is half the parallelogram ABCE, which has the same base, BC, and the same altitude... | |
 | William Schofield Binns - 1861 - 238 páginas
...applies to the triangles DEG, AEG; and, therefore, the triangle AFG is equal to AB c D E. Since the area of a triangle is equal to the product of the base multiplied by half its perpendicular height, (see Prob. 26 and Eue. I., 41), we can thus find the area... | |
 | Edward Brooks - 1895 - 424 páginas
...Altitude is a line perpendicular to the base drawn from the angle opposite; as, CD. Principle. — The area of a triangle is equal to the product of the base by one-half of the altitude. For it may readily be shown that the triangle ABC is J the parallelogram... | |
 | George William Myers, William Rockwell Wickes, Ernst Rudolph Breslich, Harris Franklin MacNeish, Ernest August Wreidt - 1907 - 208 páginas
...altitude; ie, A =ba. Find.4,if (1) 6 = 13' ft, and a = 24 ft. ; (2) 6 = 10.2 in., and a=3.5 inches. 3. The area of a triangle is equal to ^ the product of the base by the altitude; ie, A = ^ ba. Find A, if (1) b = 12 ft., and a = 16 ft.; (2) 6=8.2 rd., and a = 7.78... | |
 | George William Myers - 1909 - 394 páginas
...is, A =6a. Find A, if (1) 6 = 13 ft., and « = 24 feet; (2) 6 = 10.2 in., and a=3.5 inches. 3. The area of a triangle is equal to \ the product of the base by the altitude; that is, A = \ba. Find A, if (1) 6 = 12 ft., and a = 16 feet; (2) 6=8.2 rd., and a=7.78... | |
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... the same thing, the product of one side by half the other. Moreover, since any triangle whatever is equal to a right-angled triangle of the same base and altitude (Geom. 170), we can make use of the following simple rule, where the known parts admit... 
