In the given figure, `ABCD` is a quadrilateral such that `/_D=90^(@)`. A circle with centre `O` and radius `r`, touches the sides `AB`, `BC`, `CD` and `DA` at `P`, `Q`, `R` and `S` respectively. If `BC=40cm`, `CD=25cm` and `BP=28cm`, find `r`.

Explanation

It is given that `/_D=90^(@)`
Also, `/_ORD=/_OSD=90^(@)`. [`:'` tangent at a point is perpendicular to the radius through the point of contact]
`:./_ROS=180^(@)-/_D=90^(@)`. [`:'` angle between the tangents from an external point is supplementary to the angle subtended by the line segments joining the points of contact to the centre]
And, `OR=OS=r`.
`:.ROSD` is a square and so `OR=DR`, i.e., `r=DR` ......`(i)`
We know that the lengths of tangents drawn from an exterior point to a circle are equal.
`:.BP=BQ` and `CQ=CR`
Now, `CQ=BC-BQ=BC-BP=(40-28)cm=12cm`
And so, `DR=CD-CR=CD-CQ=(25-12)cm=13cm`
`:.r=DR=13cm` [using `(i)`]