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Square ABCD and a circle with center C intersect as shown.
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Updated on: 27 Dec 2018, 04:36
Updated on: 27 Dec 2018, 04:36
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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) \(2k^2\)
Re: Square ABCD and a circle with center C intersect as shown.
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24 Sep 2017, 08:58
24 Sep 2017, 08:58
2
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!
Answer E!
Re: Square ABCD and a circle with center C intersect as shown.
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27 Dec 2021, 03:00
27 Dec 2021, 03:00
Expert Reply
