Question

In the given figure, ABCD is a quadrilateral in which ∠D = 90°. A circle C(O, r) touches the sides AB, BC, CD and DA at P, Q, R, S, respectively. If BC = 38 cm, CD = 25 cm and BP = 27 cm, find the value of r.

Answer 1

$$\begin{gathered} \begin{array}{l}\text{Given,}ABCD\text{}\text{is a quadrilateral in which}\angle D={90}^0.\text{A circle}C(O,r)\text{touches the sides}\\ AB,BC,CD\text{and}DA\text{at}P,Q,R,S\text{respectively and}BC=38\text{cm,}CD=25\text{cm and}BP=27\text{cm}\text{.}\end{array} \\ \text{Now,}OR=OS\left(\text{Radii of the same circle}\right) \\ \text{Tangents drawn from an external point are perpendicular to the radius} \\ \text{at the point of contact}\text{.} \\ \therefore OR\perp DR\text{and}OS\perp DS. \\ \therefore ORDS\text{is a square}\text{.} \\ \text{Tangents drawn from an external point are equal.} \\ \therefore BP=BQ,CQ=CR\text{and}DR=DS \\ \therefore BQ=BP=27\text{cm} \\ \begin{array}{l}\\ \text{=>}BC-CQ=27\\ =>38-CQ=27\\ =>CQ=38-27\\ =>CQ=11\text{cm}\end{array} \\ \therefore CR=11\text{cm} \\ \text{=>}CD-DR=11\text{cm} \\ \text{=>25}-DR=11 \\ \text{=>}DR=25-11=14\text{cm} \\ \text{Now,}DR=DS=OR=OS(\because \text{Sides of a square are equal}) \\ \therefore OR=OS=14\text{cm} \\ \therefore r=14\text{cm} \\ \therefore \text{The radius is 14 cm}\text{.} \end{gathered}$$