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Question
A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at 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, find the perimeter of the ∆ABC.
Solution
OP ⊥ AP
∴ ∠OPA = 90° ...[Tangent at any point of a circle is perpendicular to the radius through the point of contact]
In ∆OAP,
OA2 = OP2 + PA2
⇒ 132 = 52 + PA2
⇒ PA = 12 cm
Now, perimeter of ∆ABC = AB + BC + CA
= AB + BR + RC + CA
= (AB + BR) + (RC + CA)
= (AB + BP) + (CQ + CA) ...[∵ BR = BP, RC = CQ i.e., tangents from external point to a circle are equal]
= AP + AQ
= 2AP ...[∵ AP = AQ]
= 2 × 12
= 24 cm
Hence, the perimeter of ∆ABC = 24 cm.
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