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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.

Answer:

**Solution:**

Given: Two tangents are drawn from an external point A to the circle with centre O. Tangent BC is drawn at a point R and radius of circle = 5 cm.

Required to find : Perimeter of ∆ABC.

Proof:

We know that,

∠OPA = 90°[Tangent at any point of a circle is perpendicular to the radius through the point of contact]

OA² = OP² + PA² [by Pythagoras Theorem]

(13)² = 5² + PA²

⇒ PA² = 144 = 12²

⇒ PA = 12 cm

Now, perimeter of ∆ABC = AB + BC + CA = (AB + BR) + (RC + CA)

= AB + BP + CQ + CA [BR = BP, RC = CQ tangents from internal point to a circle are equal]

= AP + AQ = 2AP = 2 x (12) = 24 cm[AP = AQ tangent from internal point to a circle are equal]

**Therefore, the perimeter of ∆ABC = 24 cm.**

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