In the given figure, a circle touches the side BC of $△$ ABC at P and AB and AC produced at Q and R respectively. If $A Q = 15 c m,$ then the perimeter of $△$ ABC is ___ $c m$

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Solution

We know that

$A Q = A R$ [Tangents from an external point on the circle are equal in length]

$\Rightarrow A Q = A C + C R$

But, $C R = C P$ [Tangents from an external point on the circle are equal in length]

Therefore, $A Q = A C + C P$ …..(i)

Also, $A Q = A B + B Q$

But, $B Q = B P$ [Tangents from an external point on the circle are equal in length]

$∴ A Q = A B + B P$ .….(ii)

Adding equations (i) and (ii)

$2 A Q = A C + C P + B P + A B$

$\Rightarrow 2 A Q =$ Perimeter of $△$ ABC

$\Rightarrow$ Perimeter of $△$ ABC $= 2 \times 15 = 30 c m$

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In the given figure, a circle touches the side BC of △ABC at P and AB and AC produced at Q and R respectively. If AQ=15 cm, then the perimeter of △ABC is ___ cm

In the given figure, a circle touches the side BC of $△$ ABC at P and AB and AC produced at Q and R respectively. If $A Q = 15 c m,$ then the perimeter of $△$ ABC is ___ $c m$