A complete and practical solution book for the common school teacher . ., or 2300 sq. ft. = GC + CK or AD + AM. (11) But BC + CF = 125 ft. (12) Then 2300 sq. ft. ^- 125 = 18| ft., the altitude, or BC. .. The tree must break off 18$ ft. from the ground. Note.—This solution was prepared by the author for the SchoolVisitor. Second Solution. (1) Let AB = b, the width of the stream, and draw BD perpendicular to it =the height of the tree. (2) Join AD, and bisect it in E by the per- pendicular EC; join AC. (3) It is evident that AC = CD and AC + CB = BD = a, or the sum of the hy-pothenuse and perpen
A complete and practical solution book for the common school teacher . ., or 2300 sq. ft. = GC + CK or AD + AM. (11) But BC + CF = 125 ft. (12) Then 2300 sq. ft. ^- 125 = 18| ft., the altitude, or BC. .. The tree must break off 18$ ft. from the ground. Note.—This solution was prepared by the author for the SchoolVisitor. Second Solution. (1) Let AB = b, the width of the stream, and draw BD perpendicular to it =the height of the tree. (2) Join AD, and bisect it in E by the per- pendicular EC; join AC. (3) It is evident that AC = CD and AC + CB = BD = a, or the sum of the hy-pothenuse and perpendicular. (4) If x — the perpendicular, a — x will be the hypothenuse AC. (5) (a — x) 2 = b2 + x\ or a* + 2ax + x2 ^Q n = b2 — x2. a2 fr2 (6) Whence x = -—^ . Now substitute the values of a2, b2 and a, and x == 18| ft. Utile.—From the square of the height subtract the square of thebase, and divide the difference by twice the height. PROBLEM 312. Find the area of a circle inscribed in a right-angled triangle whoselegs measure 15 and 20 ft., MENSURA riON. 143
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