From a point which is at a distance of 13 cm from centre O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is

$$\because$$ PQ is a tangent to circle at Q,
$$\therefore \angle{OQP}=90$$
In $$\triangle{POQ}$$,
$$PO^{2}=OQ^{2}+PQ^{2}$$
$$13^{2}=5^{2}+PQ^{2}$$
$$PQ=12 cm$$
Similarly, $$PR=12 cm$$
$$\because \triangle{POQ}$$ is a right-angled triangle,
$$\therefore$$ Area of $$\triangle{POQ}$$=$$\frac{1}{2}\times{PQ}\times{OQ}$$
$$=\frac{1}{2}\times5\times12$$
$$=30$$
Area of quadrilateral POQR=$$2\times$$ Area of $$\triangle{POQ}$$
$$=2\times30$$
$$=60 cm^{2}$$
Hence, Option B is correct.