In the given figure, a circle touches the side $B C$ of $△ A B C$ at $P$ and touches $A B$ and $A C$ produced at $Q$ and $R$ respectively. If $A Q = 5 c m$ , find the perimeter of $△ A B C$ .

6 c m

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10 c m

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12 c m

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15 c m

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Solution

The correct option is B $10 c m$
Given, $A Q = 5 c m$
$A R = A Q$ (tangent of point A)
also $C R = C P (1) B Q = B P (2)$
Adding $(1)$ and $(2)$
$C P + B P = C R + B Q B C = C R + B Q (3)$
Perimeter of $△ A B C$ $= A C + B C + A B = A C + C R + A B + B Q f r o m (3) = A R + A Q = 10 c m$

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