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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] New post 04 Aug 2017, 07:52
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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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Re: In the figure above, ABCD is a square with sides equal to [#permalink] New post 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] New post 30 Sep 2017, 22:34
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fraction[sqrt(2)*[sqrt(2)-2*(sqrt(2)-1]]/[2]=sqrt(2)-1
Re: In the figure above, ABCD is a square with sides equal to   [#permalink] 30 Sep 2017, 22:34
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In the figure above, ABCD is a square with sides equal to

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