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Square ABCD and a circle with center C intersect as shown.

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Square ABCD and a circle with center C intersect as shown. [#permalink] New post 06 Jun 2017, 06:44
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90% (01:29) correct 10% (02:56) wrong based on 20 sessions


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Square ABCD and a circle with center C intersect as shown. If point E is at the center of ABCD and if the radius of circle C is k, then what is the area of ABCD, in terms of k ?

A) \(\frac{k^2}{2\pi}\)

B) \(\pi\) \(\frac{k^2}{2}\)

C) \(\pi\)\(k^2\)

D) \(k^2\)

E) \(2k^2\)
[Reveal] Spoiler: OA

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Last edited by Carcass on 27 Dec 2018, 04:36, edited 2 times in total.
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Re: Square ABCD and a circle with center C intersect as shown. [#permalink] New post 24 Sep 2017, 08:58
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The radius of the circle is also half of the square diagonal. Thus, the diagonal of the square is 2k. Then, we know that the diagonal is equal to sqrt(2)l where l is the side of the square. We can, therefore, derive the length of the side of the square rearranging sqrt(2)l = 2k as l = 2k/sqrt(2). Then, the area of the square is l^2 = 4*k^2/2 = 2*k^2.
Answer E!
Re: Square ABCD and a circle with center C intersect as shown.   [#permalink] 24 Sep 2017, 08:58
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Square ABCD and a circle with center C intersect as shown.

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