$A C$ and $B D$ are two perpendicular diameters of a circle $A B C D$ . It is given that the area of the shaded portion is $308 c m^{2}$ . The circumference of the circle is

44 cm

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66cm

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88cm

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77cm

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Solution

The correct option is C 88cm
$G i v e n - A B & C D a r e t w o p r p e n d i c u l a r l y b i s e c t i n g d i a m e t e r s o f a c i r c l e w i t h c e n t e r O . a r . A O B + a r . C O D = 308 {c m}^{2} .$

$T o f i n d o u t - T h e c i r c u m f e r e n c e = C = ?$

$S o l u t i o n - L e t t h e r a d i u s o f t h e c i r c l e = r . A B & C D a r e t w o p r p e n d i c u l a r l y b i s e c t i n g d i a m e t e r s o f t h e c i r c l e .$

$S o ∠ A O B = 90^{o} = ∠ C O D a n d O A = O B = O C = O D (r a d i i o f t h e s a m e c i r c l e) . ∴ A O B & C O D a r e t w o e q u a l s e c t o r s w i t h c e n t r a l$

$a n g l e θ = 90^{o} a n d r a d i u s = r .$

$S o a r . s e c t o r A O B = a r . s e c t o r C O D$

$i . e a r . s e c t o r A O B + a r . s e c t o r C O D = 2 \times a r . s e c t o r A O B .$

$N o w a r s e c t o r = \frac{θ}{360^{o}} \times π \times r^{2} w h e n θ i s t h e c e n t r a l a n g l e$

$a n d r i s t h e r a d i u s o f t h e s e c t o r .$

$H e r e θ = 90^{o} .$

$∴ b y t h e g i v e n c o n d i t i o n,$

$2 \times a r . s e c t o r A O B = 308$

$\Rightarrow 2 \times \frac{90^{o}}{360^{o}} \times \frac{22}{7} \times r^{2} = 308$

$\Rightarrow r^{2} = \frac{308 \times 2 \times 7}{22} {c m}^{2} \Rightarrow r = 14 c m$

$∴ C = 2 π r = 2 \times \frac{22}{7} \times 14 c m = 88 c m .$

$A n s - O p t i o n C .$

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