The pair of tangents  $PQ$ and $PR$ to the circle are drawn from a point $P$ that is $13 CM$ away from the centre $O$ of a circle of radius  $5 cm$. The area of the quadrilateral  $PQOR$ is then calculated.

Concept: If the diagonal and the length of the perpendiculars drawn from the vertices are known, the area of the quadrilateral is calculated as follows: Area of quadrilateral = (1/2) diagonal length sum of the length of the perpendiculars drawn from the remaining two vertices 

In right angled triangle  $OPQ$, we have

$$\begin{gathered} PQ=\sqrt{{\left(OP\right)}^2-{\left(OQ\right)}^2} \\ =\sqrt{169-25 }= \sqrt{144 }=12 cm \\ \Rightarrow PR =12 cm \end{gathered}$$

(Two tangents from the same external point to a circle are equal) 

$=(2\times \frac{1}{2}\times 12\times 5)c{m}^2=60 c{m}^2$

Hence, area of the required quadrilateral is  $60 c{m}^2$