If O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, a point P is taken. From this point, two tangents PQ and PR are drawn to the circle. Then , the area of quadrilateral PQOR is

Given : OQ = 5 cm and OP = 13 cm

To find : ar(PQOR) = ?

Solution :  The radius intersects the tangent at the circumference of the circle at right angle.

=> $$\angle OQP=90^\circ$$

In $$\triangle$$ PQO

=> $$(PQ)^2=(OP)^2-(OQ)^2$$

=> $$(PQ)^2=(13)^2-(5)^2$$

=> $$(PQ)^2=169-25=144$$

=> $$PQ=\sqrt{144}=12$$ cm

Similarly, PR = 12 cm

$$\therefore$$ Area of quad(PQOR) = $$ar(\triangle POQ)+ar(\triangle POR)$$

= $$PQ \times OQ$$

= $$12 \times 5=60$$ $$cm^2$$

=> Ans - (A)