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# In the figure above, ABCD is a square with sides equal to

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In the figure above, ABCD is a square with sides equal to [#permalink]  04 Aug 2017, 07:52
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#GRepracticequestion In the figure above, ABCD is a square with sides equal to 1,.jpg [ 26.23 KiB | Viewed 1795 times ]

In the figure above, ABCD is a square with sides equal to 1, AFC is an arc of a circle centered at B, and AEC is an arc of a circle centered at D. What is the area of rhombus AECF?

A. 2 - $$\sqrt{2}$$

B. $$\sqrt{2}$$ - 1

C. $$\sqrt{2}(2 - \sqrt{2})$$

D. $$\sqrt{2}$$

E. 1 + $$\sqrt{2}$$
[Reveal] Spoiler: OA

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Director
Joined: 03 Sep 2017
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Re: In the figure above, ABCD is a square with sides equal to [#permalink]  19 Sep 2017, 08:10
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If the side of the square is equal to 1 its diagonal is $$sqrt(2)$$, which coincides with the rhombus main diagonal.

To find the smaller one, I exploit the fact that DE = 1 since it is a radius of the semicircle so as AD, side of the square. Thus, the smaller diagonal is equal to the diagonal of the square minus twice DF, i.e. $$EF=sqrt(2)-2*(sqrt(2)-1)$$.

Given the two diagonals we can compute the area of the rhombus as $$\frac{sqrt(2)*[sqrt(2)-2*(sqrt(2)-1]}{2}=sqrt(2)-1$$. Thus, the answer is B!
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Re: In the figure above, ABCD is a square with sides equal to [#permalink]  30 Sep 2017, 22:34
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fraction[sqrt(2)*[sqrt(2)-2*(sqrt(2)-1]]/[2]=sqrt(2)-1
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Re: In the figure above, ABCD is a square with sides equal to [#permalink]  09 Jan 2019, 13:23
done thank you
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Re: In the figure above, ABCD is a square with sides equal to [#permalink]  07 Sep 2019, 17:30
Hi,

Can somebody please post a better solution to this?

Thanks!
Re: In the figure above, ABCD is a square with sides equal to   [#permalink] 07 Sep 2019, 17:30
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