A complete and practical solution book for the common school teacher . FIG (5) (6) The area ofZABK=(102xl7) 867 sq. sq. in. (7) Then, the area of DABC=867—147 (8) Area of FECD = 720+-2 = 360 sq. in. (9) The area of FEK = 360+147=507 sq. in. (10) .-. 867 : 507 : : 1022 : KE2. EK=78 in. (11) 102 in.—78 in. .*. EB = 2 ft., the distance the board must be cut from thelarge end. 160 FAIRCHILDS SOLUTION BOOK. PROBLEM 340. The sides of a triangular field are 39, 40 and 50 rd.: from whatpoint in one of the shorter sides must a line be drawn parallel with theother shorter side, so as to d
A complete and practical solution book for the common school teacher . FIG (5) (6) The area ofZABK=(102xl7) 867 sq. sq. in. (7) Then, the area of DABC=867—147 (8) Area of FECD = 720+-2 = 360 sq. in. (9) The area of FEK = 360+147=507 sq. in. (10) .-. 867 : 507 : : 1022 : KE2. EK=78 in. (11) 102 in.—78 in. .*. EB = 2 ft., the distance the board must be cut from thelarge end. 160 FAIRCHILDS SOLUTION BOOK. PROBLEM 340. The sides of a triangular field are 39, 40 and 50 rd.: from whatpoint in one of the shorter sides must a line be drawn parallel with theother shorter side, so as to divide the field into two equal parts? (i) (2) (3) (4)(5) Solution Let ABC be the triangle,and draw EK parallel toAB, cutting off the halfof AB or [(40X30)-r-2]^-2 = 300 sq. rd. Then we have the propor-tion, 600 : 300 :: 402 :(EK, or rd.). FIG. 34. EK= rd., length of the line that divides the tri-angle into equal parts; and it is parallel to AB, and AB=40 rd. Also, draw TP parallel to CB, whose side is 30 rd. .\ 600 : 300 :: 302 : CB2, from which CB= rd. .-. TP= rd., dividing the field into equal parts par-allel to CB. PROBLEM 341. Two trees stand on opposite sides of a stream 40 ft. wide; theheight of one tree is to the width of the stream as 8 is to 4, and thewidth of the stream is to the height of the other as 4 is to 5: what isthe distance between their tops? Solution. (1) From Fig. 9, AB=40 ft., the width of the stream. CB and FA = the height of the trees. (2) Then by the conditions of the problem, one tree is twice the width of the river. (3) .-. CB= FE=40ft„ FA=50 ft., the height of the 2d tree. (4) CE = 30 ft.; FC, the distance between their tops= V^FE2+CE2)= The required distance is 50 ft. PROBLEM 342. The lengths of the lines that bisect the acute angles of a rig4tt-angled tria
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