A complete and practical solution book for the common school teacher . From AD to P=112 rd. PC=[(96+28)—100] = 12 rd., the point of meeting northof the southeast coiner. PROBLEM 424. Schwarts has a garden in the form of a square; from the corners toa spring within the garden, the distances are a=40, £=50, r=80 ft.:find the side of the garden. Solution. (1) Let ABCD be the square and F the spring. (2) DF = 50 ft., CF=40 ft. and FB = 80 ft. (3) Draw DE=50, EC = 40, CT=40, and TB=80 ft. (4) Suppose S is a spring whose distances are AS=40, DS = 50, and SB = 80 ft. (5) VSB~*+BT* = V802+80*=
A complete and practical solution book for the common school teacher . From AD to P=112 rd. PC=[(96+28)—100] = 12 rd., the point of meeting northof the southeast coiner. PROBLEM 424. Schwarts has a garden in the form of a square; from the corners toa spring within the garden, the distances are a=40, £=50, r=80 ft.:find the side of the garden. Solution. (1) Let ABCD be the square and F the spring. (2) DF = 50 ft., CF=40 ft. and FB = 80 ft. (3) Draw DE=50, EC = 40, CT=40, and TB=80 ft. (4) Suppose S is a spring whose distances are AS=40, DS = 50, and SB = 80 ft. (5) VSB~*+BT* = V802+80*= (6) SE=VSDa+DE2, orV502+502= ft. 214 FAIRCHILDS SOLUTION BOOK. (7) (8) (9) (10) (ID(12)(13) (14) In the scalene tri-angle STE, EK isits perpendicular,and now we willfind SK and KT bya well known theo-rem in geometry. TS : TE + ES:: TE—ES : KT—SK, :: : Half the differenceof the segments ad-ded to half theirsum gives the great-er segment, and subtracted gives the lesser segment. Therefore. KT is ft., and SK= FIG. 99. KE = VKT2—ET2= ft. Area of STE^jSTx EK = sq. ft. The area of SBT = 3200 sq. ft, and area of EDS = 12500 sq. , the area of DSBTE = area of ABCD = sq. *. AB, the side of the square garden = 85-f- ft. PROBLEM square farm contains 40 acres. It is required to lay off anotherpark containing- the same area, enclosed by an iron fence forming- cir-cular arcs only, and to find the cost of such a fence at $2 per rd. Solution (1) Let ABCD represent the square farm, whosesides AB, BC=V40X160=80 rd. (2) With a radius equal to that of the circum-scribing circle drawthe arcs APB andBRC. (3) The area of the pelicoid APRCD is equal tothat of the squarefarm, and equal to thearea of the park,bounded by circulararcs only.
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Keywords: ., bookcentury1800, bookdecade1890, booksubject, booksubjectgeometry
