In the given figure $D C = C B = 4 c m$ and $A E = 2 c m .$ $D C ∥ A B$ . Then area of shaded portion is $7.44 C m^{2}$
If true then enter $1$ and if false then enter $0$

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Solution

$I t i s g i v e n t h a t i n A B C D D C ∥ A B . i . e i t i s a t r a p e z i u m a n d D C = 4 c m, A B = A E + E B = 2 c m + 4 c m = 6 c m . A l s o ∠ C B E = 90^{o} . i . e t h e h e i g h t h i s 4 c m . S o a r A B C D = \frac{1}{2} \times (D C + A B) \times h = \frac{1}{2} \times (4 + 6) \times 4 {c m}^{2} = 20 {c m}^{2} . N o w B C = 4 c m = B E & ∠ C B E = 90^{o} . ∴ B E F C i s a s e c t o r o f a n g l e θ = 90^{o} & r a d i u s r = 4 c m . ∴ a r B E F C = \frac{θ}{360^{O}} \times {π r}^{2} = \frac{90^{o}}{360^{o}} \times 3.14 \times 4^{2} {c m}^{2} = 12.56 {c m}^{2} . ∴ a r . s h a d e d p o r t i o n = a r . A B C D - a r . s e c t o r B E F C = (20 - 12.56) {c m}^{2} = 7.44 {c m}^{2} A n s - a r . s h a d e d p o r t i o n = 7.44 {c m}^{2} .$

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