A complete and practical solution book for the common school teacher . and standing- 20 feet from the well, and the foot of the sweep to strike the ground 20 feet from the foot of the upright post? Solution. (1) AE = 20 ft.; FE, the post = 20 ft.; BE = 20 ft.; andCB will be the requiredlength of^the rope. (2) By the similar triangles ABC and AEF, we have AE :EF :: AB : CB, or 20 ft.: 20 ft. :: 40 ft. : (CB =40 ft.) Note.—This solution was prepared by the author for the AmericanMathematical Monthly. PROBLEM hypothenuse of a right-angled triangle is 97 ft., and the per-pendicular is 65 f
A complete and practical solution book for the common school teacher . and standing- 20 feet from the well, and the foot of the sweep to strike the ground 20 feet from the foot of the upright post? Solution. (1) AE = 20 ft.; FE, the post = 20 ft.; BE = 20 ft.; andCB will be the requiredlength of^the rope. (2) By the similar triangles ABC and AEF, we have AE :EF :: AB : CB, or 20 ft.: 20 ft. :: 40 ft. : (CB =40 ft.) Note.—This solution was prepared by the author for the AmericanMathematical Monthly. PROBLEM hypothenuse of a right-angled triangle is 97 ft., and the per-pendicular is 65 ft.: required the base, without squaring either of thegiven numbers. Solution.[ (97 + 65) (97 — 65) ] = 72 ft., the base. PROBLEM long a ladder will be required to reach a window 40 feet fromthe ground, if the distance of the foot of the ladder from the wall isone-fifth of the length of the ladder? Solution. (1) From Fig. 5, AB, or 5*, represents the length of the lad- der, and AC, or x, the distance the foot of the ladder isfrom the wall. (2) CB = 40 FIG. 7. MENSURA TWN. 139 (3) Then we have 25*2 + x* — 402. (4) Whence, 5* = ft., the required length. Note.—This solution was prepared by the author for the SchoolVisitor. PROBLEM 306. A right-angled triangle, altitude 7 ft., and hypothenuse 25 ft., hasthe same area as a rectangle whose sides are as 7 to 3: find dimensionsof the rectangle. Solution. 72) 24 ft. (1) The base of the triangle is V (252 (2) Its area is 7 X 24 -f- 2 = 84 sq. ft. (3) Assuming a rectangle 7 ft. long and 3 ft. wide which is sim- ilar to the given rectangle, we fino its area to be 21 sq. ft. (4) The areas of similar figures are to each other as the squares of their homologous sides; then 21 sq ft. : 84sq. ft :: 72 : (the required length = 14 ft.) (5) 21 sq. ft. : 84 sq. ft. :: 32 : (the required width = 6 ft.) .*. The dimensions of the rectangle are 14 ft. and 6 ft. PROBLEM 307. Two trees, the first of which is 60, and the seco
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