from a point P 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

60 sq.cm Solution: PO=13 (hypotenuse) OQ=5 PQ=√PQ^2-OQ^2=12 Area of quadrilateral=Area of triangle POQ+POR (And POQ=POR. CONGRUENT) Area of POQ=1/2*5*12=30 (Area of right angle triangle=1/2*a*b) AREA OF QUADRILATERAL PQOR=2*POQ=2*30= 60