Question

In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides AB, BC, CD and AD at P, Q, R and S respectively. If the radius of the circle is $10cm,BC=38cm,$ $PB=27cm$ and $AD\perp CD$ then the length of $CD$ is

(a) $11cm$ (b) $15cm$ (c) $20cm$ (d) $21cm$

Answer 1

The correct option is (d): $21cm$

Given that ABCD is a quadrilateral such that $\angle D={90}^{\circ }.$

$BC=38cm,$ and $BP=27cm$ radius of the circle, $r=OS=10cm$

Join OR.

From the figure, we have

$BP=BQ=27cm$ [Tangents from an external point are equal ]

$BC=38$ [Given]

$\Rightarrow BQ+QC=38$

$\Rightarrow 27+QC=38$

$\Rightarrow QC=38-27$

$\Rightarrow QC=11cm$

$\therefore QC=CR=11cm$ [ Tangents from an external point are equal ]

OR and OS are radii of the circle.

$OR\perp CD,OS\perp AD.$

Since all the angles in the quadrilateral DSOR is right angles. So it is a rectangle.

$OR=DS=10cm$ [Opposite Sides of the rectangle are equal ]

$OS=DR=10cm$

So, $CD=DR+CR$

$=10+11cm=21cm$