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In the figure above, if the area of the inscribed rectangula

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In the figure above, if the area of the inscribed rectangula [#permalink] New post 17 Dec 2016, 07:52
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In the figure above, if the area of the inscribed rectangular region is 32, then the circumference of the circle is

(A) 20pi
(B) 4pi\(\sqrt{5}\)
(C) 4pi\(\sqrt{3}\)
(D) 2pi\(\sqrt{5}\)
(E) 2pi\(\sqrt{3}\)
[Reveal] Spoiler: OA

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Re: In the figure above, if the area of the inscribed rectangula [#permalink] New post 29 Sep 2017, 07:49
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The area of the rectangular is \(x*2x = 32\), from which we get x = 4 (we exclude x = -4 because a side of a rectangular cannot have a negative length.

Then, using Pitagora's theorem we can find the length of the hypotenuse of the triangle, which is half of the rectangular and we get it equal to \(\sqrt(80) = 4sqrt(5)\).

Finally, the circumference of the circle is given by \(2r*\pi\) where 2r is the diameter that in this case equals \(4sqrt(5)\).

Thus, the circumference is equal to \(4\pi\sqrt(5)\). Answer B!
Re: In the figure above, if the area of the inscribed rectangula   [#permalink] 29 Sep 2017, 07:49
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In the figure above, if the area of the inscribed rectangula

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