A complete and practical solution book for the common school teacher . ft., required length. PROBLEM 379. A circular lot 15 rd. in diameter is to have three circular grass bedsjust touching each other and the larger boundary: what must be thedistance between their centers, and how much ground is left in the tri-angular space about the center? (Rays Higher Arith.) Solution (1) OD = 7^rd. = R, the radius of the lot. (2) AO = (3) AB—AOV3= rd. CD=22| rd. (4) Area of ABC=^(ABx CD) =+ sq. rd. (5) The area of the three tri- angles AOB, COB andCOA=J- of
A complete and practical solution book for the common school teacher . ft., required length. PROBLEM 379. A circular lot 15 rd. in diameter is to have three circular grass bedsjust touching each other and the larger boundary: what must be thedistance between their centers, and how much ground is left in the tri-angular space about the center? (Rays Higher Arith.) Solution (1) OD = 7^rd. = R, the radius of the lot. (2) AO = (3) AB—AOV3= rd. CD=22| rd. (4) Area of ABC=^(ABx CD) =+ sq. rd. (5) The area of the three tri- angles AOB, COB andCOA=J- of sq. rd. (6) Let SD, or SL=r, the radius of the small circle. (7) r=(AOBx2)-^AB+OB +AO, or twice the area FIG66, of AOB divided by the perimeter= (8) .\ SF= rd. The area of sector PSL small circle, because the angle PSL=60°. (9) ^27r=3 times ^27r=area of the three sectors in the tri- angle KSF. (10) r2V3—\r*ir=r2(y/3—iir) = , the area of thespace enclosed. .*. SF = rd., and sq. of the curvi-linear the MENSURATION. 186 PROBLEM 380. If 2 ft. be the radius of a circle, how far from the center must achord be drawn parallel! to divide the semicircle into two equal parts? Solution. (1) Let ADB be the semicircle, AB the diameter, GH the line thatdivides the semicircle into twoequal parts. (2) On OD, perpendicular to AB, let OF=;t, the distance required,and 0=angle HOB. r=OB, orradius. (3) Then *=rsin0, FH=rcos0. (4) Area of sector HOB=^-20. (5) Area of sector DOH=|r2(^r—0). (6) Area of triangle FOH=^r2sin0cos0. (7) We then have r20-r-r2sin0cos0=r2(^r— 0) — r2sin0cos0.
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