In the given figure, $△ A B C$ is right-angles at $B$ , in which $A B = 15 c m$ and $B C = 8 c m$ . A circle with centre $O$ has been inscribed in $△ A B C$ . The value of $x$ , the radius of the inscribed circle is

6cm

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4cm

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3cm

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9cm

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Solution

The correct option is C 3cm
$G i v e n - A B C i s a r i g h t Δ w i t h ∠ A B C = 90^{o} . A c i r c l e w i t h c e n t r e O h a s b e e n i n s c r i b e d w i t h i n Δ A B C . T h e c i r c l e t o u c h e s A B, B C & A C a t P, Q & R r e s p e c t i v e l y . A B = 15 c m a n d B C = 8 c m . T o f i n d o u t - T h e r a d i u s o f t h e g i v e n c i r c l e = x = ? S o l u t i o n - A B C i s a r i g h t Δ w i t h A C a s h y p o t e n u s e . S o, a p p l y i n g P y t h a g o r a s t h e o r e m, w e g e t A C = \sqrt{{A B}^{2} + {B C}^{2}} = \sqrt{15^{2} + 8^{2}} c m = 17 c m N o w w e k n o w t h a t t h e r a d i u s t h r o u g h t h e p o i n t o f c o n t a c t o f a t a n g e n t t o a c i r c l e i s p e r p e n d i c u l a r t o t h e t a n g e n t . ∴ ∠ O Q B = 90^{o} = ∠ O P B . A l s o O P = O Q = x r a d i i o f t h e s a m e c i r c l e) . ∴ O P B Q i s a s q u a r e . i . e P B = B Q = x . N o w A P = A R a n d C Q = C R s i n c e t a n g e t s t o a c i r c l e f r o m a p o i n t a r e e q u a l . B u t A P = A B - P B = 15 - x = A R a n d C Q = B C - B Q = 8 - x = C R . ∴ A R + C R = A C = (15 - x) + (8 - x) = A C = 17 ⟹ 23 - 2 x = 17 ⟹ x = 3 c m . S o t h e r a d i u s o f t h e i n s c r i b e d c i r c l e i s 3 c m . A n s - O p t i o n C .$

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