A complete and practical solution book for the common school teacher . D—BE)+EB=40. (9) BC=AB—EB = 30. .-. ABC is a 30, 40 and 50 t/iangle. PROBLEM 326. The area of a right triangle is 924 sq. ft., the sum of the base andperpendicular is 89 ft.: find the sides of the triangle. Solution. (1) (2) From Fig. 24, ABC^the +BC=89, MN2=7921 sq. ft. (3) KLa = DN-4AABC = V4225, or 65 (4) GE = KC—4AABC = 529 sq. ft., EB= V529 = 23 ft. (5) BC=i(FN—GB)=33ft. (6) AB=AE, or BC+BE, or 56 ft. .-. ABC is a 33, 56 and 65 327. Show that the difference between the hypothenuse
A complete and practical solution book for the common school teacher . D—BE)+EB=40. (9) BC=AB—EB = 30. .-. ABC is a 30, 40 and 50 t/iangle. PROBLEM 326. The area of a right triangle is 924 sq. ft., the sum of the base andperpendicular is 89 ft.: find the sides of the triangle. Solution. (1) (2) From Fig. 24, ABC^the +BC=89, MN2=7921 sq. ft. (3) KLa = DN-4AABC = V4225, or 65 (4) GE = KC—4AABC = 529 sq. ft., EB= V529 = 23 ft. (5) BC=i(FN—GB)=33ft. (6) AB=AE, or BC+BE, or 56 ft. .-. ABC is a 33, 56 and 65 327. Show that the difference between the hypothenuse and the sum ofthe two legs of a right-angled triangle is equal to the diameter of theinscribed circle. MENSURA TION. 153 Solution. (1) Let ABC be the triangle, and O the center of theinscribed circle. BE,BF, AE, AK, KC andFC are tangents to thesame circle; hence EB= FB, AE = AK andKC=FC, also AE=OE=OK. (2) EB+KC=FB+CF. (3) EB+KC + AE + AK— (FB+CF) = AE+AK. .. AF+AK = FO + OE, the diameter of the —Solved bv the author for the Teachers FIG. 25. PROBLEM 328, From a right-angled triangular piece of tin whose sides measurea = 13, b = 12 and c = 5 inches, cut out. first, the greatest possible cir-cle; and after that, the greatest possible rectangle: find the length andbreadth of the rectangle. Solution.
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