The area of the shaded region given in Fig. 11.20 is

180 - 4 π c m^{2}

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196 - 8 π c m^{2}

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196 - 4 π c m^{2}

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180 - 8 π c m^{2}

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Solution

The correct option is D $180 - 8 π c m^{2}$
$G i v e n - A s q u a r e o f s i d e = a = 14 c m h a s a n i n n e r r e g i o n a s s h o w n i n t h e f i g u r e .$
Where, b: side of inner square.
$T h e r e a r p o i n t s o f t h e i n n e r r e g i o n a r e a t a d i s t a n c e o f 3 c m f r o m t h e s i d e s . T o f i n d o u t - t h e a r e a o f t h e s h a d e d r e g i o n . S o l u t i o n - W e r e d r a w a n d n a m e t h e f i g u r e a s s h o w n . T h e i n n e r r e g i o n c o n s i s t s o f a s q u a r e M X Y N w i t h f o u r e q u a l s e m i c i r c l e s e a c h o f w h i c h i s a t t a c h e d t o t h e s i d e s o f t h e i n n e r s q u a r e . L e t t h e s i d e o f t h e o u t e r s q u a r e = a = 14 c m . ∴ a r . o u t e r s q u a r e = a^{2} = 14^{2} {c m}^{2} = 196 c m^{2} . N o w t h e r a d i u s o f e a c h s e m i c i r c l e = \frac{1}{2} b . s o a r . (4 s e m i c i r c l e s) = a r . (2 c i r c l e s o f r a d i u s = r) = \frac{1}{2} b = 2 \times π {(\frac{1}{2} b)}^{2} = π \times \frac{1}{2} b^{2} = 0.5 π b^{2} a n d a r M X Y N = b^{2} a r . i n n e r r e g i o n = 0.5 π b^{2} + b^{2} = (0.5 π + 1) b^{2} . ∴ A l s o t h e p o i n t s P & S a r e 3 c m w i t h i n t h e s u q a r e . ∴ P S = (14 - 2 \times 3) c m = 8 c m . s i m i l a r l y A D = 8 c m . B u t P S = P Q + Q R + R S = \frac{1}{2} b + b + \frac{1}{2} b = 2 b = 8 c m ⟹ b = 4 c m . ∴ a r . i n n e r r e g i o n = (0.5 π + 1) b^{2} = (0.5 π + 1) 4^{2} {c m}^{2} = (8 π + 16) {c m}^{2} . ∴ a r . s h a d e d r e g i o n = a r . o u t e r s q u a r e - a r . i n n e r r e g i o n = 196 {c m}^{2} - (8 π + 16) {c m}^{2} = (180 - 8 π) {c m}^{2} . A n s - O p t i o n D .$

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Similar questions

4 π c m^{2}

θ^{\circ}

\frac{r^{2}}{2} (t a n θ - \frac{π θ}{180})

A B = 20

B C = 14

180^{\circ}

Figure shows a sector of a circle, centre O, containing an angle $θ^{\circ}$ . Prove that:

(i) Perimeter of the shaded region is $r (t a n θ + s e c θ + \frac{π θ}{180} - 1)$

(ii) Area of the shaded region is $\frac{r^{2}}{2} (t a n θ - \frac{π θ}{180})$

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