The area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is $60^{0}$ , is

\frac{308}{3} - 98 \sqrt{3} c m^{2}

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\frac{308}{3} + 49 \sqrt{3} c m^{2}

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\frac{308}{3} + 98 \sqrt{3} c m^{2}

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\frac{308}{3} - 49 \sqrt{3} c m^{2}

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Solution

The correct option is C $\frac{308}{3} - 49 \sqrt{3} c m^{2}$
$G i v e n t h a t - t h e c e n t r a l a n g l e ∠ A O B o f a s e c t o r = 60^{o} a n d t h e r a d i u s = 14 c m . T o f i n d o u t - t h e a r e a o f t h e c o r r e s p o n d i n g m i n o r s e g m e n t . S o l u t i o n - I n Δ A O B w e h a v e O A = O B . ∴ ∠ O A B = ∠ O B A ⟹ ∠ O A B + ∠ O B A = 2 ∠ O A B o r 2 ∠ O B A . N o w ∠ A O B = 60^{o} . ∴ 2 ∠ O A B = 180^{o} - 60^{o} ⟹ ∠ O A B = 60^{o} = ∠ O B A . (t h e a n g l e s u m p r o p e r t y o f t r i a n g l e s) . S o Δ A O B i s e q u i l a t e r a l o n e w i t h s i d e a = 14 c m . ∴ a r . Δ A O B = \frac{\sqrt{3}}{4} \times a^{2} = \frac{\sqrt{3}}{4} \times 14^{2} {c m}^{2} = 49 \sqrt{3} {c m}^{2} . A l s o a r . s e c t o r = \frac{θ}{360^{o}} \times π \times r^{2} w h e n θ i s t h e c e n t r a l a n g l e a n d r = r a d i u s o f t h e s e c t o r = \frac{60^{o}}{360^{o}} \times \frac{22}{7} \times 14^{2} {c m}^{2} = \frac{308}{3} {c m}^{2} . N o w a r . m i n o r s e g m e n t = a r . s e c t o r - a r . Δ = \frac{308}{3} {c m}^{2} - 49 \sqrt{3} {c m}^{2} = (\frac{308}{3} - 49 \sqrt{3}) {c m}^{2} . A n s - O p t i o n D .$

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The area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 600, is

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