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

Can somebody please post a better solution to this?

Thanks!
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Re: In the figure above, ABCD is a square with sides equal to [#permalink] New post 04 May 2020, 01:07
The key to this problem is finding the length of Rhombus diagonal EF, which is the base of both upper and lower triangles of the Rhombus.

The diagonal of the square is root(2). This length of root(2) is comprised of 2 segments of 1 (radius of the circle) "merged partially". So consider this as a Venn Diagram problem.

Full length = Sum of overlapping lengths - common length.
Diagonal = (Radius1 + Radius2) - EF.
root(2) = (1 + 1) - EF.
EF = 2 - root(2).

Now we have the base of the 2 triangles. The height of the triangles are half the length of the diagonal, which is root(2) / 2 = 1 / root(2).

Area of one triangle = 1/2 * base * height = 1/2 (EF) (half diagonal) = 1/2 (2 - root(2)) (1 / root(2)) = (2-root(2))/2root(2)

Area of rhombus = sum of areas of two triangles = 2 * Area of one triangle = 2 * (2 - root(2)) / 2root(2) = (1 - root(2)).

While this solution is very simple, it took me long enough to come up with it, sadly. The main point of this solution is using the Venn Diagram principle to find the length of the overlapping section of 2 line segments.
Re: In the figure above, ABCD is a square with sides equal to   [#permalink] 04 May 2020, 01:07
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In the figure above, ABCD is a square with sides equal to

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