A complete and practical solution book for the common school teacher . FIG. 84. PROBLEM 405. If the square ABCD has 40 acres in it, find the distance from C toa point & the distance from B to A, as F. 204 FAIRCHILDS SOLUTION BOOK. (1)(2)(3)(4) acres, or 6400 sq. = V6400=80 rd. FB iof 80rd. = 16rd. Now, as FCB is a right-angled tri angle, FC= VCB2+FB2 = V802+162=+ CF=81f5+ rd. PROBLEM FIG. 85. From a point in the side and 8 ch. from the corner of a square fieldcontaining- 40 A., a line is run, cutting- off 19| A.: how long is the line? [R. H. A.) Solution.
A complete and practical solution book for the common school teacher . FIG. 84. PROBLEM 405. If the square ABCD has 40 acres in it, find the distance from C toa point & the distance from B to A, as F. 204 FAIRCHILDS SOLUTION BOOK. (1)(2)(3)(4) acres, or 6400 sq. = V6400=80 rd. FB iof 80rd. = 16rd. Now, as FCB is a right-angled tri angle, FC= VCB2+FB2 = V802+162=+ CF=81f5+ rd. PROBLEM FIG. 85. From a point in the side and 8 ch. from the corner of a square fieldcontaining- 40 A., a line is run, cutting- off 19| A.: how long is the line? [R. H. A.) Solution. (1) Let ABCD be the squarefield, and P the point 8ch. from D. (2) (3) (4) Draw PE parallel to DC,PF cutting off 19i the line, or PFCD. DC, or DA = 20 ch.; then,the area of the rectangleDPEC=20x8 =160 , or 16 A. The triangle EPF con-tains 19|-16=3| A.(5) EF, the side of the triangle=3^X2
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Keywords: ., bookcentury1800, bookdecade1890, booksubject, booksubjectgeometry
