A complete and practical solution book for the common school teacher . FIG. 4. (AB X AC)(AB + AC)5 Let ABC be the given right-angled tri-angle. Produce AC to E, making CE =AB. At E draw ED perpendicular to CEand make it equal to AC. Draw DC andBD; then CD = CB and A BCD = ^|AE(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) AC = 4
A complete and practical solution book for the common school teacher . FIG. 4. (AB X AC)(AB + AC)5 Let ABC be the given right-angled tri-angle. Produce AC to E, making CE =AB. At E draw ED perpendicular to CEand make it equal to AC. Draw DC andBD; then CD = CB and A BCD = ^|AE(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) AC = 40, the base. (2) BC = 9, the altitude. (3) 402 = 1600, the square of the base, AC. (4) 92 = 81, the square of the alti- tude, BC. (5) 1600+ 81 = 1681, AC2 + BC2. (6) V1681 = 41, AB. 40 and the al-. 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) 9409 — 4225 = 5184, the difference of the squares of the hypothenuse and altitude. (6) V5184 = 72 = 72 = the base. PROBLEM 302. What is the area of a right triang-le whose base is 24 feet and alti-tude 7 feet? 138 FAIRCHILDS SOLUTION BOOK.
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