The radius of a circle is 40 cm and the length of perpendicular drawn from its centre to chord is 24 cm. The length of chord is

32cm

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64cm

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8cm

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16cm

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Solution

The correct option is B 64cm
$G i v e n - O i s t h e c e n t r e o f a c i r c l e o f r a d i u s 40 c m . A B i s a c h o r d o f t h e g i v e n c i r c l e a n d A B i s a t a d i s t a n c e o f 24 c m f r o m O . T o f i n d o u t - T h e l e n g t h o f A B = ? . S o l u t i o n - W e j o i n O A a n d d r o p a p e r p e n d i c u l a r O N t o A B . O N m e e t s A B a t N . s o O A i s t h e r a d i u s o f t h e c i r c l e a n d O A = 40 c m . A g a i n O N i s t h e p e r p e n d i c u l a r d r o p p e d f r o m O t o A B a t N . ∴ O N i s t h e p e r p e n d i c u l a r d i s t a n c e o f A B f r o m O . S o O N = 24 c m & ∠ O N A = 90^{o} . ∴ Δ O A N i s a r i g h t t r i a n g l e w i t h O A a s h y p o t e n u s e . S o, a p p l y i n g P y t h a g o r a s t h e o r e m, w e g e t A N = \sqrt{{O A}^{2} - {O N}^{2}} = \sqrt{40^{2} - 24^{2}} c m = 32 c m . N o w A B = 2 A N s i n c e t h e p e r p e n d i c u l a r, d r o p p e d f r o m t h e c e n t r e o f a c i r c l e t o i t s a n y c h o r d b i s e c t s t h e l a t t e r . ∴ A B = 2 \times 32 c m = 64 c m . A n s - O p t i o n B .$

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