#### 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 $\bigtriangleup$ABC.

Solution
Let us make figure according to question

Given : OA = 13 cm, Radius = 5 cm
Here AP = AQ            (tangent from same point)
OP $\perp$ PA, OQ $\perp$ QA  (  AP, AQ are tangents)
In $\bigtriangleup$OPA using Pythagoras theorem
$\left ( H \right )^{2}= \left ( B \right )^{2}+\left ( P \right )^{2}$
$\left ( QP \right )^{2}= \left ( OP \right )^{2}+\left ( PA \right )^{2}$
$\left ( 13 \right )^{2}= \left ( 5 \right )^{2}+\left ( PA \right )^{2}$
$\left ( PA \right )^{2}= 169-25$
$PA= \sqrt{144}= 12\, cm$ ........(i)
Perimeter of $\bigtriangleup$ABC = AB + BC + CA
= AB + BR + RC + CA
= AB + BD + CQ + CA
[ BP and BR are tangents from point B and CP and CQ are tangents from point C]
= AP + AQ      [  AP = AB + BP, AQ = AC + CQ]
= AP + AP      [  AP = AQ]
= 2AP
= 2 × 12           [using (i)]
= 24 cm