A quadrilateral ABCD is drawn so that ∠D = 90º, BC = 38 cm and CD = 25 cm. A circle is inscribed in the quadrilateral and it touches the sides AB, BC, CD and DA at P, Q, R and S respectively. If BP = 27 cm, find the radius of the inscribed circle.
Let O be the centre and r be the radius of the inscribed circle. Join OR and OS.
Here, ∠D = 90º
∠ORD = ∠OSD = 90º
OR = OS.
Therefore, ORDS is a square.
⇒RD = SD = r..... (i)
BP and BQ are tangents to the circle from point P.