Two chords PQ and MN of length 11 cm and 5 cm respectively of a circle are parallel to each other and are one the opposite sides of its centre. If the distance between chord MN and chord PQ is 6 cm find the radius of the circle

\frac{10 \sqrt{5}}{2} c m

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\frac{5 \sqrt{5}}{4} c m

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\frac{5 \sqrt{5}}{2} c m

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\frac{5 \sqrt{10}}{2} c m

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Solution

The correct option is C $\frac{5 \sqrt{5}}{2} c m$
$G i v e n - O i s t h e c e n t r e o f a c i r c l e w i t h r a d i u s = r . P Q & M N a r e t w o p a r a l l e l c h o r d s w h e n P Q = 11 c m a n d M N = 5 c m . T h e d i s t a n c e A B b e t w e e n P Q & M N, t h r o u g h t h e c e n t r e i s 6 c m . T o f i n d o u t - r = ? S o l u t i o n - T h e l i n e A B i s t h e p e r p e n d i c u a r d i s t a n c e b t w e e n P Q & M N . A t t h e s a m e t i m e O A & O B a r e t h e d i s t a n c e s o f P Q & M N f r o m t h e c e n t e r O . ∴ ∠ O A Q = ∠ O B N = 90^{o} s i n c e t h e p e r p e d i c u l a r d i s t a n c e o f a c h o r d f r o m t h e c e n t r e b i s e c t s t h e c h o r d a t r i g h t a n g l e . i . e Δ O A Q & Δ O B N a r e r i g h t t r i a n g l e s w i t h h y p o t e n u s e s a s O Q = r a n d O N = r r e s p e c t i v e l y . A l s o O A = \frac{11}{2} c m a n d O B = \frac{5}{2} c m . L e t u s t a k e O A = x . T h e n O B = A B - O A = 6 - x . T h e n, a p p l y i n g P y t h a g o r a s t h e o r e m, w e h a v e {O A}^{2} + {A Q}^{2} = r^{2} a n d {O B}^{2} + {B N}^{2} = r^{2} . ∴ {O A}^{2} + {A Q}^{2} = {O B}^{2} + {B N}^{2} ⟹ x^{2} + {(\frac{11}{2})}^{2} = {(6 - x)}^{2} + {(\frac{5}{2})}^{2} ⟹ x = 1. ∴ r = \sqrt{{O A}^{2} + {A Q}^{2}} = \sqrt{x^{2} + {(\frac{11}{2})}^{2}} = \sqrt{1^{2} + {(\frac{11}{2})}^{2}} c m = \frac{5 \sqrt{5}}{2} c m . A n s - O p t i o n C .$

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