IIT JEE Advanced 2007 Paper 2

In the quadrilateral ABCD, let CD = x and AD = 2r. Therefore, the area of the ABCD = 18. That is,

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(x + 2x) (2r) = 18
xr = 16

Now, AP = AT = DS = DT = r
BP = BR = 2x - r
CS = CR = x - r
In ΔBCQ, we get
CQ2+BQ2 = BC2
⇒ (2r)2+x2 = (3x−2r)2
⇒ 4r2+x2 = 9x2+4r2 - 12xr
⇒ 8x2 = 12xr ⇒ 8x2 = 72
⇒ x = 3 ⇒ r = 2