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

Options

A. `1/8`
B. `1/5`
C. `1/4`
D. `2/5`

Answer

D

Explanation

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`