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 160
FIG. 15. 3. AC = VBC2 + AB2. ABC is a 6, 8 and 10 triangle. V* + S = 4, 10. or PROBLEM 315. There are two ladders standing-]in the middle of a street; one is80 ft., and the other 90 ft., both lean ag-ainst a building- and the 90 ft.ladder reaches the top 20 ft. hig-her than the 80 ft. ladder: find… … FKr. V, Solution. (1) Let AC and DC be the position of the lad-ders. (2) BD* = the square LG, and BO — the squareBM. (3) Subtract the square BM from the square AG= the two rectanglesFL and LD plus thesquare AL = thesquare BM, = AC 2 —DC2, or 1700 sq. ft. (4) AD2 = 400 sq. ft., the square AL. (5)…
Arithmetic · Geometry
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