Question

$ABCD$ is a quadrilateral such that $\angle D={90}^o$. A circle $C(O,r)$ touch the sides $AB$, $BC$, $CD$ and $DA$ at $P,Q,R$ and $S$ respectively. If $BC=38cm$, $CD=25cm$ and $BP=27cm$, find $r$.

Answer 1

Since tangents to a circle is perpendicular to the radius through the point.
$\therefore \angle ORD=\angle OSD={90}^o$

It is given that $\angle D={90}^o$. Also, $OR=OS$. Therefore, $ORDS$ is a square.

Since tangents from an exterior point to a circle are equal in length.

$\therefore BP=BQ$ $CQ=CR$ and, $DR=DS$.

Now,
$BP=BQ$

$\Rightarrow$ $BQ=27$ [$\because BP=27cm$ $\left(Given\right)$]

$\Rightarrow BC-CQ=27$

$\Rightarrow 38-CQ=27$

$\because BC=38cm$ $\Rightarrow CQ=11cm$

$\Rightarrow CR=11$ $[\because CR=CQ]$

$\Rightarrow CD-DR=11$

$\Rightarrow 25-DR=11$ $[\because CD=25cm]$

$\Rightarrow DR=14cm$

But, $ORDS$ is a square. $\Rightarrow$ $OR=DR=14cm$.

$\Rightarrow$ $r=14cm$.