Question

The length of radius of a circle with centre $O$ is $5cm$. $P$ is a point at a distance of 13 from the point $O.PQ$ and $PR$ are two tangents from the point $P$ to the circles; Find the area of the quadrilaterals PQOR.

Answer 1

As we know that, Tangents from a point to a circle are equal.
$\Rightarrow PR=PQ$
and Angles made by the tangent and the radius at the point of contact is ${90}^{\circ }$ $\Rightarrow OQ\perp PQ\&OR\perp PR$
In right angled triangle $POQ$
$\begin{array}{cc}& PQ=\sqrt{{OP}^2-{OQ}^2}\left(\text{Using Pythagorean theorem}\right)\\ & \Rightarrow PQ=\sqrt{{13}^2-5^2}\\ & \Rightarrow PQ=\sqrt{169-25}\\ & \Rightarrow PQ=\sqrt{144}\\ & \Rightarrow PQ=12cm\\ & \because PR=PQ\\ & PR=12cm\end{array}$
Here, in $△POR$ and $△POQ$ we have,
(i) $PR=PQ$
(ii) $OQ=OR(\because$ radius of circle $)$
(iii) $OP=OP$ (Common)
$\therefore$ By SSS test $△POQ$ is congruent to $△POR$
ar $△POQ=$ ar $△POR$
$\therefore$ ar $△POQ=\frac{1}{2}\times PQ\times OQ$
$\Rightarrow$ ar $△POQ=\frac{1}{2}\times 5\times 12$
$\Rightarrow$ ar $△POQ=30{cm}^2$
$\therefore$ ar $△POR=30{cm}^2$
$\mathrm{arPQOR}=\mathrm{ar}△POQ+\mathrm{ar}△POR$
$\Rightarrow \mathrm{arPQOR}=30+30$
$\Rightarrow \mathrm{arPQOR}=60{cm}^2$