A complete and practical solution book for the common school teacher . FIG. 93. Solution. (1) Let ABCD represent the lot; HGLE, the path. (2) EI, HS, GF and LK are each perpendicular toDB, and are each 18 in. (3) By similar triangles, DC : CB :: DS : HS,and DS = ft. (4) DC :CB::GF: FB, from which FB—1 ft. (5) The total area of the four triangles is sq. ft. (6) DC2 + BC* = DB2, from this we find DB=+ ft. (7) 593 23 ft.—3 25 ft. (DS+1) =+ ft., SFor IK. (8) ft. =+ sq. ft., area of the two rectangles. (9) sq. ft.+ sq. ft. = sq. ft., tot
A complete and practical solution book for the common school teacher . FIG. 93. Solution. (1) Let ABCD represent the lot; HGLE, the path. (2) EI, HS, GF and LK are each perpendicular toDB, and are each 18 in. (3) By similar triangles, DC : CB :: DS : HS,and DS = ft. (4) DC :CB::GF: FB, from which FB—1 ft. (5) The total area of the four triangles is sq. ft. (6) DC2 + BC* = DB2, from this we find DB=+ ft. (7) 593 23 ft.—3 25 ft. (DS+1) =+ ft., SFor IK. (8) ft. =+ sq. ft., area of the two rectangles. (9) sq. ft.+ sq. ft. = sq. ft., total area. PROBLEM 414. Find the diagonal of a rectangular field containing lHi A., is to its breadth as 12 to 5. fk;. *>4. (1) Let A BCD be the field. (2) Its area is13J- X 160 =2160 sq. rd. (3) AssumeABCD asa similar fieldwhose sidesare DA = 5, AB=lL\ (4) Then, similar figures are to each other as the squares oftheir like dimensions. (5) Area of ABCD=5X 12=60 sq. rd. (6) We have ABCD : ABCD : A^2 : AD2, or60 : 2160 ::52 : AD2. (7) From this we find AD—30 rd. (8) ABCD: ABCD:;DC2 : DC2, or 60 : 2160:: 12* : DC2. (9) From this DC=72 rd. (10) AC = V(AB* + BO)=:7tf The diagonal AC = 78 rd. 210 FAIR CHILDS SOLUTION BOOK. PROBLEM 415. The diagonal of a rectangle is 109 and its perimeter 302; required,the sides. Solution. (1) The diagonal of a rectangle divides it into two equal right- angled triangles. (2) Let x represent the perpendicular, y the base and h the hypothenuse of one of the triangles.{3) Then, we have x-\-y—lbl ... (1). (4) ^2+^2^1092 ... (2). (5) Subtracting (2) from the square of (1), we have 2xy= 10920 ... (3). (6) Subtracting (3) from (2), we have x2—2xy+y2 =961. (7) E
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