SSC CHSL 19 Apr 2021 Shift 3 Paper

Given,Since, the tangent at any point of a circle is perpendicular to the radius through the point of contact.
$\therefore \angle \mathit{P}\mathit{Q}\mathit{O}=\angle \mathit{P}\mathit{R}\mathit{O}={90}^{\circ }$
Since, $△\mathit{P}\mathit{Q}\mathit{O}$ and $△\mathit{P}\mathit{R}\mathit{O}$ are right angle triangle
$\therefore \mathit{P}\mathit{Q}=\mathit{P}\mathit{R}=\sqrt{{\left(15\right)}^2-{\left(9\right)}^2}$
$=\sqrt{225-81}=12\mathit{c}\mathit{m}.$
$\therefore$ Area of the quadrilateral $\mathit{P}\mathit{Q}\mathit{O}\mathit{R}=$ Area of $△\mathit{P}\mathit{Q}\mathit{O}+$ Area of $△\mathit{P}\mathit{R}\mathit{O}$
$\therefore$ Area of the quadrilateral $\mathit{P}\mathit{Q}\mathit{O}\mathit{R}$
$=\frac{1}{2}\times 12\times 9+\frac{1}{2}\times 12\times 9$
$=108{\mathit{c}\mathit{m}}^2$