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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 1036 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 + \(\sqrt{2}\)
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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





