A complete and practical solution book for the common school teacher (1899)
Fairchild, J[ohn] T[heodore] [from old catalog] · Crawfis College, Ohio, The author · leaf 152
AB+ED)or |(AB + AC)2; it is also equal to|(BCxCD) +twice the A ABC or |BC2 +(AB X AC). Therefore, KAB + AC)2 = 4rBC2 +whence, = |BC2 + 2 (ABX AC), or AB2 + AC2 = BC2. O. E. D. PROBLEM 300. Given the right-angled triangle ABC, the base ACtitude BC = <>: what is the hypothenuse? Solution. (1)… … PIG. . 41 = the hypothenuse. PROBLEM 301. The hypothenuse of a right-angled triangle is 97, and the altitude05: what is the base? Solution. (1) AB = 97, the hypothenuse. (2) BC = 65, the altitude. (3) 972 = 9409, the square of the hypothenuse AB. (4) 652 = 4225, the square of the altitude BC. (5)…
Arithmetic · Geometry
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