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\( A \) is a point at a distance \( 13 \mathrm{~cm} \) from the centre \( O \) of a circle of radius \( 5 \mathrm{~cm} \). \( A P \) and \( A Q \) are the tangents to the circle at \( P \) and \( Q \). If a tangent \( B C \) is drawn at a point \( R \) lying on the minor arc \( P Q \) to intersect \( A P \) at \( B \) and \( A Q \) at \( C \), find the perimeter of the \( \triangle A B C \).
Given:
\( A \) is a point at a distance \( 13 \mathrm{~cm} \) from the centre \( O \) of a circle of radius \( 5 \mathrm{~cm} \). \( A P \) and \( A Q \) are the tangents to the circle at \( P \) and \( Q \).A tangent \( B C \) is drawn at a point \( R \) lying on the minor arc \( P Q \) to intersect \( A P \) at \( B \) and \( A Q \) at \( C \).
To do:
We have to find the perimeter of the \( \triangle A B C \).
Solution:
$\angle OPA = 90^o$ (Tangent at a point on a circle is perpendicular to the radius through the point of contact)
In $\triangle OPA$,
$OA^2 = OP^2 + PA^2$ (By Pythagoras theorem)
$(13)^2 = 5^2 + PA^2$
$PA^2 = 169-25$
$=144$
$=(12)^2$
$\Rightarrow PA = 12\ cm$
Perimeter of $\triangle ABC = AB + BC + CA$
$= (AB + BR) + (RC + CA)$
$= AB + BP + CQ + CA$ (since $BR = BP$ and $RC = CQ$, Tangents from internal point to a circle are equal)
$= AP + AQ$
$= 2AP$ (Tangents from internal point to a circle are equal)
$= 2 \times 12$
$= 24\ cm$
The perimeter of $\triangle ABC$ is $24\ cm$.