In the given figure, a circle is inscribed in quad. $A B C D$ . If $B C = 38 c m, B Q = 27 c m, D C = 25 c m$ and $A D ⊥ D C$ . The radius of the circle is

28cm

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21cm

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14cm

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42cm

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Solution

The correct option is C 14cm
$G i v e n - O i s t h e c e n t r e o f a c i r c l e, i n s c r i b e d i n a q u a d r i l a t e r a l A B C D . T h e r a d i u s o f t h e c i r c l e i s 10 c m . D C, B C, A B & A D t o u c h t h e i n s c r i b e d c i r c l e a t P, Q, R & S r e s p e c t i v e l y . B Q = 27 c m, B C = 38 c m & D C = 25 c m . A D ⊥ D C . T o f i n d o u t - T h e r a d i u s o f t h e c i r c l e = ? S o l u t i o n - W e j o i n O S & O P . T h e n O S & O P a r e r a d i i o f t h e i n s c r i b e d c i r c l e t h r o u g h t h e p o i n t s o f c o n t a c t o f t h e t a n g e n t s A S & A P r e s p e c t i v e l y . ∴ O S = O P . A l s o ∠ A D C = 90^{o} s i n c e A D ⊥ D C . N o w i n O S D P, ∠ P A S = 90^{o} a n d O S = O P, D P = D S s i n c e t h e l e n g t h s o f t h e t a n g e n t s, f r o m a p o i n t t o a c i r c l e, a r e e q u a . ∴ O S D P i s a s q u a r e o f s i d e . S o D P = O S . A g a i n B Q = B R = 27 c m a n d C Q = C P s i n c e t h e l e n g t h s o f t h e t a n g e n t s, f r o m a p o i n t t o a c i r c l e, a r e e q u a l . ∴ C Q = B C - B Q = (38 - 27) c m = 11 c m . A l s o C P = C Q = 11 c m . ∴ D P = O S = D C - C P = (25 - 11) c m = 14 c m . S o t h e r a d i u s o f t h e c i r c l e i s 14 c m . A n s - O p t i o n C .$

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