Question

In quadrilateral $ABCD$, angle $D={90}^o,BC=3cm$ and $DC=25cm$. A circle is inscribe in this quadrilateral which touches $AB$ at point $Q$ such that $QB=27cm$ and BC = $38$ cm. Find the radius of the circle.

Answer 1

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

$BC=38$cm, $CD=25$cm and $BP=27$cm

$\therefore$ From the figure,

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

Also, $BC=38$cm

$\Rightarrow BQ+QC=38$

$\Rightarrow 27+QC=38$

$\Rightarrow QC=38-27$

$\Rightarrow QC=11$cm

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

$CD=25$cm

$CR+RD=25$

$\Rightarrow 11+RD=25$

$\Rightarrow RD=14$cm

Also,

$RD=DS=14$cm [ Tangents from an external point are equal ]

$\therefore OR$ and $OS$ are radii of the circle.

From tangents $R$ and $S$, $\angle ORD=\angle OSD={90}^{\circ }$

Now, $ORDS$ is a square.

$\therefore OR=DS=14$ cm

Hence the radius of the circle , $r=14$cm