The radii of two concentric circles are 17 cm and 10 cm. A line segment PQRS cuts the larger circle at P and S and the smaller circle at Q and R. If QR= 12 cm. The length of PQ is

9 cm

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18cm

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15cm

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8cm

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Solution

The correct option is A 9 cm
$G i v e n - O i s t h e c e n t r e o f t w o c o n c e n t r i c c i r c l e s o f r a d i i 10 c m a n d 17 c m . T h e l i n e ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P Q R S i n t e r s e c t s t h e o u t e r c i r c l e a t P & S a n d t h e i n n e r c i r c l e a t Q & R . Q R = 12 c m T o f i n d o u t - P Q = ? S o l u t i o n - W e j o i n O P, O S, O R, O Q . ∴ O P = O S = 17 c m & O Q = O R = 10 c m . A l s o w e d r o p a p e r p e n d i c u l a r O N f r o m O t o ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P Q R S . O N q u a d m e e t s ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P Q R S a t N . N o w O N ⊥ Q R a t N . ∴ N i s t h e m i d p o i n t o f Q R s i n c e t h e p e r p e n d i c u l a r, d r o p p e d f r o m t h e c e n t e r o f a c i r c l e t o i t s a n y c h o r d b i s e c t s t h e l a t t e r . ∴ Q N = \frac{1}{2} Q R = \frac{1}{2} \times 12 c m = 6 c m . A l s o ∠ O N Q = 90^{o} s i n c e O N ⊥ Q R . S o Δ O Q N i s a r i g h t o n e w i t h O Q a s h y p o t e n u s e . ∴ B y P y t h a g o r a s t h e o r e m, w e g e t O N = \sqrt{{O Q}^{2} - {Q N}^{2}} = \sqrt{10^{2} - 6^{2}} c m = 8 c m . N o w O N ⊥ Q S . i . e N i s t h e m i d p o i n t o f P S s i n c e t h e p e r p e n d i c u l a r, d r o p p e d f r o m t h e c e n t e r o f a c i r c l e t o i t s a n y c h o r d b i s e c t s t h e l a t t e r . ∴ P N = \frac{1}{2} P S . A g a i n Δ O P N i s a r i g h t o n e w i t h O Q a s h y p o t e n u s e . (s i n c e O N ⊥ P S) ∴ B y P y t h a g o r a s t h e o r e m, w e g e t P N = \sqrt{O P^{2} - {O N}^{2}} = \sqrt{17^{2} - 8^{2}} c m = 15 c m . ∴ P Q = P N - Q N = (15 - 6) c m = 9 c m . A n s - O p t i o n A .$

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