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Square ABCD and a circle with center C intersect as shown. [#permalink]
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Question Stats: 90% (01:29) correct 10% (02:56) wrong based on 20 sessions

Attachment: #GREpracticequestion Square ABCD and a circle with center C intersect as shown..jpg [ 11.02 KiB | Viewed 732 times ]

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.
Edited by Carcass Director Joined: 03 Sep 2017
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Re: Square ABCD and a circle with center C intersect as shown. [#permalink]
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KUDOS
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. 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.  Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.