A is point at a distance 26 cm from the centre O of a circle of radius 10 cm. AP and AQ are the tangents to the circle at the point of contacts P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, then the perimeter of ΔABC is:

48 cm

We know that,

(Hypotnuese)² = (perpendicular)² + (Base)²

We have,

Radius of circle = 10 cm 

 BP = BR and CR = CQ 

Perimeter of ΔABC = AB + BR + RC + CA

also, BP = BR and RC = CQ   [equal tangent theorem]

So, ⇒ AB + BP + QC + CA

= AP + QA = 2AP (AP = QA, From equal tangent theorem) 

= In ΔAPO, =26² = AP² + 10²

= AP² = 576

= AP = 24

Perimeter of ΔABC = 24 + 24 = 48 cm