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:
Answer & explanation
Correct answer: option 2
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