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

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Joined: 18 Apr 2015
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Kudos [?]: 1022 [0], given: 4624

In the figure above, if the area of the inscribed rectangula [#permalink]  17 Dec 2016, 07:52
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Question Stats:

100% (00:47) correct 0% (00:00) wrong based on 4 sessions

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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Joined: 03 Sep 2017
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Kudos [?]: 330 [1] , given: 66

Re: In the figure above, if the area of the inscribed rectangula [#permalink]  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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