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(C) Given: Radius of the circle $OQ = OR = 5 \, cm$. Distance from centre $O$ to point $P$ is $OP = 13 \, cm$.Since $PQ$ and $PR$ are tangents,the rad

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$60$

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(C) Given: Radius of the circle $OQ = OR = 5 \, cm$. Distance from centre $O$ to point $P$ is $OP = 13 \, cm$.Since $PQ$ and $PR$ are tangents,the radius is perpendicular to the tangent at the point of contact. Therefore,$\angle OQP = 90^\circ$ and $\angle ORP = 90^\circ$.In the right-angled $	riangle OQP$,by the Pythagorean theorem:$OP^2 = OQ^2 + PQ^2$$13^2 = 5^2 + PQ^2$$169 = 25 + PQ^2$$PQ^2 = 169 - 25 = 144$$PQ = 12 \, cm$.Now,the area of $	riangle OQP = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes OQ 	imes PQ = \frac{1}{2} 	imes 5 	imes 12 = 30 \, cm^2$.The quadrilateral $PQOR$ is composed of two congruent right-angled triangles,$	riangle OQP$ and $	riangle ORP$.Therefore,the area of quadrilateral $PQOR = 2 	imes 	ext{Area}(	riangle OQP) = 2 	imes 30 = 60 \, cm^2$.

From a point $P$ which is at a distance of $13 \, cm$ from the centre $O$ of a circle of radius $5 \, cm$,the pair of tangents $PQ$ and $PR$ to the circle are drawn. Then the area of the quadrilateral $PQOR$ is (in $cm^2$)

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