Two concentric circles are of radii $13 c m$ and $5 c m$ . The length of the chord of the outer circle which touches the inner circle is

12cm

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24cm

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18cm

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6cm

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Solution

The correct option is B 24cm
$G i v e n - O i s t h e c e n t r e o f t w o c o n c e n t r i c c i r c l e s . T h e r a d i u s o f t h e o u t e r c i r c l e = O Q = r_{1} = 13 c m a n d t h e r a d i u s o f t h e i n n e r c i r c l e = r_{2} = 5 c m . A c h o r d A B o f t h e o u t e r c i r c l e t o u c h e s t h e i n n e r o n e a t P . T o f i n d o u t - T h e l e n g t h o f A B = ? S o l u t i o n - W e j o i n O A & O P . ∴ O A = r_{2} & O P r_{1} . N o w O P ⊥ A B i . e O P ⊥ A P s i n c e t h e r a d i u s t h r o u g h t h e p o i n t o f c o n t a c t o f a t a n g e n t t o a c i r c l e i s p e r p e n d i c u l a r t o t h e t a n g e n t . ∴ Δ O A P i s a r i g h t o n e w i t h O A a s h y p o t e n u s e . S o, a p p l y i n g P y t h a g o r a s t h e o r e m, w e g e t A P == \sqrt{{O A}^{2} - {O P}^{2}} = \sqrt{13^{2} - 5^{2}} c m = 12 c m . A g a i n A B = 2 \times A P = 2 \times 12 c m = 24 c m s i n c e t h e p e r p e n d i c u l a r, f r o m t h e c e n t e r o f a c i r c l e t o a n y o f i t s c h o r d, b i s e c t s t h e l a t t e r . ∴ T h e l e n g t h o f A B = 24 c m . A n s - O p t i o n B .$

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