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Let ABCD be a square. An arc of a circle with A as centre and AB as radius is drawn inside the square joining the points B and D. Points P on AB, S on AD,Q and R on are taken such that PQRS is a square. Further suppose that PQ and RS are parallel to AC. "Then"("areaPQRS")/("areaABCD") is |
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Answer» `1/8` ![]() Let `A(0,0), B(1,0), C(1,1), D(0,1)rArr" Area ABCD = 1"` Again let `Q (cosalpha,sinalpha)" & "R(cosbeta,sinbeta)` `rArr " COORDINATE of " P (cosalpha,sinalpha)" & "S(cosbeta,sinbeta)` PQRS is a square `rArrPQ_|_QRrArr" slope of "QR=-1=" slope of SP"` `rArr(sinbeta-sinalpha)/(cosbeta-cosalpha)=-1=(sinbeta-cosbeta)/(sinalpha-cosalpha)` `rArrsinbeta-sinalpha=-sinalpha+cosalpha` `rArrsinbeta-cosbeta=sinalpha+cosalpha.........(i)` and `sinalpha+sinbeta=cosalpha+cosbeta...........(ii)` `rArrcosalpha=sinbeta` `rArrcosalpha=cos(90-beta)` `rArralpha+beta=90^(@)` Also PQ QR `rArrtanalpha=1/2` Area of `PQRS=2sin^(2)alpha=2(1/5)` `("Area of PQRS")/("Area of ABCD")=("2/5")/1=2/5` |
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