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In the figure above, ABCD is a square with sides equal to [#permalink]
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Question Stats: 50% (03:21) correct 49% (03:03) wrong based on 59 sessions

Attachment: #GRepracticequestion In the figure above, ABCD is a square with sides equal to 1,.jpg [ 26.23 KiB | Viewed 2742 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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GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests. Director Joined: 03 Sep 2017
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Re: In the figure above, ABCD is a square with sides equal to [#permalink]
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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! Intern Joined: 21 Apr 2017
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Re: In the figure above, ABCD is a square with sides equal to [#permalink]
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fraction[sqrt(2)*[sqrt(2)-2*(sqrt(2)-1]]/=sqrt(2)-1
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Re: In the figure above, ABCD is a square with sides equal to [#permalink]
done thank you
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Re: In the figure above, ABCD is a square with sides equal to [#permalink]
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]
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.
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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