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 153
he 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… … Solution. (1) We can see at once that ACB = AEB. (2) AC = 24fi., the base. (3) BC = 7 ft., the altitude. (4) 24 X 7 = 168 sq. ft., the product of the base andaltitude. (5) \ of 168 sq. ft. = 84 sq. ft., half the product of the base and altit .. 84 sq. ft. = the area of ABC. PROBLEM 303. What…
Arithmetic · Geometry
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