In the figure, ODCE is a square then the area of shaded region is

52.5 c m^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

24.5 c m^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

49 c m^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $52.5 c m^{2}$
$G i v e n - O D C E i s a s q u a r e o f s i d e = 7 c m . A C B i s a n a r c w i t h O a s c e n t r e a n d t h e p o i n t s A & B l i e o n O D e x t e n d e d a n d O E e x t e n d e d . T o f i n d o u t - T h e a r e a o f t h e s h a d e d r e g i o n A C B E D . S o l u t i o n - T h e a r c A C B p a s s e s t h r o u g h C a n d t h e c e n t r e i s a t O . ∴ I t s r a d i u s = r = d i a g o n a l O C o f t h e s q u a r e . N o w O D C E i s a s q u a r e o f s i d e = 7 c m . . ∴ T h e d i a g o n a l O C = s i d e \times \sqrt{2} = 7 \sqrt{2} c m . ∴ r = 7 \sqrt{2} c m . A l s o ∠ A O B = θ = 90^{o} . ∴ a r . s e c t o r A O B = \frac{θ}{360^{o}} \times π \times r^{2} = \frac{90^{o}}{360^{o}} \times \frac{22}{7} \times {(7 \sqrt{2})}^{2} {c m}^{2} = \frac{1}{4} \times \frac{22}{7} \times 98 {c m}^{2} = 77 {c m}^{2} . A g a i n a r . Δ E O D = \frac{1}{2} \times b a s e \times h e i g h t = \frac{1}{2} \times O D \times O E = \frac{1}{2} \times 7 \times 7 {c m}^{2} = \frac{49}{2} {c m}^{2} = 24.5 {c m}^{2} . N o w a r . A C B D = a r . s e c t o r A O B - a r . Δ E O D = (77 - 24.5) {c m}^{2} = 52.5 {c m}^{2} . S o a r . s h a d e d r e g i o n = 52.5 {c m}^{2} . A n s - O p t i o n A .$

Suggest Corrections

Similar questions

24.5 {c m}^{2}

A D ∥ B C, ∠ D A B = 90^{0}, A D = 10 c m, B C = 4 c m

A B E

π

\frac{22}{7}

Join BYJU'S Learning Program