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