Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. The length of common chord is

6 cm

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4 cm

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12 cm

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8 cm

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Solution

The correct option is B 6 cm
$G i v e n - A & B a r e t w o i n t e r s e c t i n g c i r c l e s, t h e c o m o n c h o r d o f w h i c h i s C D . C D i n t e r s e c t s A B o r e x t e n d e d A B a t E . A B = 4 c m, r a d i u s o f t h e c i r c l e w i t h c e n t r e A i s A C = 5 c m a n d r a d i u s o f t h e c i r c l e w i t h c e n t r e B i s B C = 3 c m . T o f i n d o u t - T h e l e n g t h o f t h e c o m m o n c h o r d C D . S o l u t i o n - ∠ A E C = 90^{o} s i n c e t h e l i n e j o i n i n g t h e c e n t r e s o f o f t w o i n t e r s e c t i n g c i r c l e s i s p e r p e n d i c u l a r t o t h e i r c o m m o n c h o r d . S o w e h a v e t w o r i g h t t r i a n g l e s, Δ A C E & Δ B C E w i t h h y p o t e n u s e s A C & B C r e s p e c t i v e l y . L e t B E = x c m . T h e n A E = (4 + x) c m a n d B E = x c m . ∴ B y P y t h a g o r a s t h e o r e m, w e g e t C E = \sqrt{A E^{2} - {A C}^{2}} = \sqrt{{B C}^{2} - {B E}^{2}} ⟹ \sqrt{5^{2} - {(4 + x)}^{2}} = \sqrt{3^{2} - x^{2}} ⟹ x = 0. i . e B & E a r e s a m e p o i n t . ∴ A B = A E = 4 c m . S o C E = \sqrt{A E^{2} - {A C}^{2}} = \sqrt{5^{2} - {(4 + 0)}^{2}} c m = 3 c m . N o w C D = 2 C E s i n c e t h e l i n e j o i n i n g t h e c e n t r e s o f o f t w o i n t e r s e c t i n g c i r c l e s b i s e c t s t h e c o m m o n c h o r d . ∴ T h e l e n g t h o f t h e c o m m o n c h o r d C D = 2 \times 3 c m = 6 c m . A n s - O p t i o n A .$

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Question 1
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