A is a point at a distance 13 cm from the centre 0 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 ΔABC.
Given Two tangents are drawn from an external point A to the circle with centre 0,
$O A=13 \mathrm{~cm}$
Tangent $B C$ is drawn at a point $R$. radius of circle equals $5 \mathrm{~cm}$. To find perimeter of $\triangle A B C$.
Proof $\angle O P A=90^{\circ}$
[tangent at any point of a circle is perpendicular to the radius through the point of contact]
$\therefore$ $O A^{2}=O P^{2}+P A^{2}$ [by Pythagoras theorm]
$(13)^{2}=5^{2}+P A^{2}$
$\Rightarrow$ $P A^{2}=144=12^{2}$
$\Rightarrow \quad P A=12 \mathrm{~cm}$
Now, perimeter of $\triangle A B C=A B+B C+C A$
$=(A B+B R)+(R C+C A)$
$=A B+B P+C Q+C A$
$[\because B R=B P, A C=C Q$ tangents from internal point to a circle are equai]
$=A P+A Q$
$=2 A P$
$=2(12)$
$=24 \mathrm{~cm}$
$[A P=A Q$ tangent from internal point to a circle are equal]
Hence, the perimeter of $\triangle A B C=24 \mathrm{~cm}$.