Une solution complète et pratique pour le livre de l'école enseignant commun . Solution. (1) Let AC  = 20 pi., la base du triangle ABC et BC = 15 ft., l'altitude. (2) AB, l'hypothenuse  = V(AC2 -hBC2), ou 25 ft. (3) Ensemble FDE  = le cercle inscrit ; OD, D, OE  = les rayons fig. Est. de le cercle inscrit. (4) ^(AC X BC)  = 150 m², la zone de l'AACB. (5) le rayon d'un cercle inscrit est trouvé en divisant deux fois l'aire du triangle par son périmètre, ou (150 X 2) -^ 60  = 5 pi., le rayon Œ ou OD. (6) Par conséquent, 52 X n  = 78,53 m². .*. 78,53 m² est la surface requise. Problème 313. A l

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A complete and practical solution book for the common school teacher . Solution. (1) Let AC = 20 ft., the base of the triangle ABC, and BC= 15 ft., the altitude. (2) AB, the hypothenuse = V(AC2 -h~BC2), or 25 ft. (3) FDE = the inscribed circle; OD, OF, OE = the radii fig. is. of the inscribed circle. (4) ^(AC X BC) = 150 sq. ft., the area of the AACB. (5) The radius of an inscribed circle is found by dividing twice the area of the triangle by its perimeter, or (150X 2) -^ 60 = 5 ft., the radius OE or OD. (6) Hence, 52 X n = 78.53 sq. ft. .*. 78.53 sq. ft. is the required area. PROBLEM 313. A ladder stands against the side of a house; if it is pulled out at thebottom 16 ft., it will not reach the top by 4 ft., and it is long enoughto reach the top: find the length of the ladder. Solution. (1) Let DC = theheight of thehouse, and AB= the ladderdrawn out 16 ft.on CB. (2) DC = AB. (3) Construct t h esquare DF up-on DC = thesquare on AB. (4) The square SF= the square onAC, the perpen-dicular of theA CBA. (5). FIG. 14. Let b = AD, or 4 ft. = the distancedown when it was pulled out 16 ft. (6) aa = the area of the rectangles ES and SC the area of the square DS. (7) .-. FO — FG2 = aab or 16* = 256 sq. ft. the ladder slippedand b2 — 144 FAIRCHILDS SOLUTION BOOK. (8) aab1 — b2 = 240 sq. ft., the area of the two rectangles ES and SC. (9) .-. The area of SC = | of 240 sq. ft. = 120 sq. ft. (10) SA = AD, or 4 ft.; then AC = of 120 sq. ft., or 30 ft. (11) AC -4- AD = 34 *ft. = DC, or AB. .*. 34 ft. is the required length of the ladder. PROBLEM 314. Two lines are drawn from the acute angles of a right triangle tothe middle points of the opposite sides, and measure respectively, Vf3and V-^: find the base and perpendicular. Solution. (1) Let ABC =the triangle. (2) EC = V52 and AT = V73. (3) Let 0 repres e n t the square on the half ofBC and ED half thesquare on AB. (4) Also the squares AX, PX and XK are eachequal to ED. (5) Then, if we subtract 0 from either of thesquares ED

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