A complete and practical solution book for the common school teacher . quare. (2) EF = 1 rd. and the triangle EFO = 16 A., or 2560 sq. rd. (3) -OF= (2560x2)^-1=5120 rd. (4) CB = 5120 rd. X2 = 10240 rd. (5) The perimeter of the field = 10240 rd. (6) Now as 16 rails fence a rod, and every rail will fence a rod; there will be as many acres in the field ABCD as theproduct of 40960x16=655360 A. Note.—The above problem has bothered many an applicant at thecounty examination. Now, my friend, do not miss it again. MENSURA TION. 213 PROBLEM 422. You have a square farm of 40 acres: find the
A complete and practical solution book for the common school teacher . quare. (2) EF = 1 rd. and the triangle EFO = 16 A., or 2560 sq. rd. (3) -OF= (2560x2)^-1=5120 rd. (4) CB = 5120 rd. X2 = 10240 rd. (5) The perimeter of the field = 10240 rd. (6) Now as 16 rails fence a rod, and every rail will fence a rod; there will be as many acres in the field ABCD as theproduct of 40960x16=655360 A. Note.—The above problem has bothered many an applicant at thecounty examination. Now, my friend, do not miss it again. MENSURA TION. 213 PROBLEM 422. You have a square farm of 40 acres: find the side of that farmwithout using square root. Solution. (1) In the diagram, let ABCD = 1 sq. mi. =640 A. (2) Then EFGD=TV sq. mi. = 40 A., the required farm. (3) But ED= PROBLEM 423. At the northwest corner of a rectangular field two men start towalk at the same rate, one east on the short boundary line and the otheron the diagonal: where will they meet if the one turns at the south-east corner to meet the other, the field being 96 rd. long and 28 Solution. (i) (2) (3)(4) (5) From Fig 94, let ABCD be the rectangle, AC the diago-nal, and P the point of meeting. VADa+DU2=100 rd. = AC. AD+DC=124 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::
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