A complete and practical solution book for the common school teacher . (1)(2)(3)(4)(5) (5) Then we have x* + 1600 = 4s- (6) {x+y)* + (40 — 29)* = y* _|_ 292 (7) j2 + 29* = ^2 + 462 . . (8) Reducing (2), x* + 2xy = 720 (9) From (3), z* = j>* — 1275 . (10) Substituting the value of z in (1), 2y = V (x* — 6700). (11) Substituting this value of 2y in (4), transposing, squar- ing and reducing, 407* = 25920; also x = * ft., v = ft., and 2z = ft. PROBLEM 311. A tree 125 ft. high stands on the bank of a river 105ft. wide; wheremust the tree break off so that it may remain connected
A complete and practical solution book for the common school teacher . (1)(2)(3)(4)(5) (5) Then we have x* + 1600 = 4s- (6) {x+y)* + (40 — 29)* = y* _|_ 292 (7) j2 + 29* = ^2 + 462 . . (8) Reducing (2), x* + 2xy = 720 (9) From (3), z* = j>* — 1275 . (10) Substituting the value of z in (1), 2y = V (x* — 6700). (11) Substituting this value of 2y in (4), transposing, squar- ing and reducing, 407* = 25920; also x = * ft., v = ft., and 2z = ft. PROBLEM 311. A tree 125 ft. high stands on the bank of a river 105ft. wide; wheremust the tree break off so that it may remain connected at the point ofbreaking- and its top just reach the opposite shore? First Solution. (1) Let AB = the height or sum of the altitudeand hypothenuse. (2) AG = the hypothe- nuse, and GB = thealtitude. (3) Complete the square ABFL; its area willbe 125* ft., or 15625sq. ft. (4) Let BC = GD; draw GK parallel to BFand NC parallel toAB. FIGp n>. 142 FAIRCHILDS SOLUTION BOOK. (5) GC = the square of the altitude, and DL = the square of the hypothenuse. (6) Take LH equal to the base of the triangle formed by the broken tree, and complete the square SL. (7) Then, SL is the square of the base, which we take from the square of DL; the difference is the gnomon. (8) hh == the square of the altitude GC. (9) But if from ABFL, or 15625 sq. ft., we takeSL, or 11025 sq. ft., the remainder is GC + DF and AD -f AM (AM= hh) = 4600 sq. ft. (10) Then we find \ of 4600 sq. ft., 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 + C
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