A complete and practical solution book for the common school teacher . MENSURA riON. 143. 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 = sq. ft. .*. sq. ft. is the required are
A complete and practical solution book for the common school teacher . MENSURA riON. 143. 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 = sq. ft. .*. 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 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)
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