In the given figure, $△ A B C$ is circumscribed. The circle touches the sides $A B, B C$ and $C A$ at $P, Q, R$ respectively. If $A P = 5 c m, B P = 7 c m, A C = 14 c m$ and $B C = x c m$ . The value of $x$ is

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Solution

$G i v e n - A B C i s a t r i a n g l e, i n s c i r c l e o f w h i c h t o u c h e s A B a t P, A C a t Q a n d B C a t R . A P = 5 c m, B P = 7 c m, A C = 14 c m a n d B C = x c m . T o f i n d o u t - x = ? S o l u t i o n - W e k n o w t h a t t h e l e n g t h s o f t h e t a n g e n t s, f r o m a p o i n t t o a c i r c l e, a r e e q u a l . ∴ A P = A Q = 5 c m, B P = B R = 7 c m a n d C Q = C R . N o w C Q = A C - A Q = (14 - 5) c m = 9 c m = C R . S o B C = x = C R + B R = (9 + 7) c m = 16 c m . A n s - O p t i o n B .$

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