Question

In $△PQR,PQ=24cm,QR=7cm$ and $\angle PQR={90}^{\circ }$. Find the radius (in $cm$) of the inscribed circle.

Answer 1

By Pythagoras' theorem,

$P{R}^2=P{Q}^2+Q{R}^2$

$P{R}^2={24}^2+7^2=576+49=625$

$\therefore PR=25cm$

Let the inradius of $△PQR$ be $xcm$.

$\square OAQC$ is a square. Hence $QA=xcm$ and $AR=(7-x)cm$

$RA$ and $RB$ act as tangents to the incircle from point $R$, hence their lengths are equal. $\therefore RB=AR=(7-x)cm$.

Similarly, $PB=PC=(24-x)cm$

$PR=PB+RB$

$\Rightarrow 25=(24-x)+(7-x)$

$\Rightarrow 25=31-2x$

$\Rightarrow x=3cm$