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# In the figure above, the two axes divide the enclosed region into four

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Math Expert
Joined: 02 Sep 2009
Posts: 43349
In the figure above, the two axes divide the enclosed region into four [#permalink]

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13 Dec 2017, 19:57
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Difficulty:

35% (medium)

Question Stats:

67% (00:49) correct 33% (00:56) wrong based on 24 sessions

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In the figure above, the two axes divide the enclosed region into four regions that have the same size and shape. Of the following, which is closest to the area of the entire enclosed region?

(A) 7
(B) 10
(C) 19
(D) 22
(E) 29

[Reveal] Spoiler:
Attachment:

2017-12-12_2124_001.png [ 42.26 KiB | Viewed 278 times ]
[Reveal] Spoiler: OA

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Joined: 14 Oct 2015
Posts: 227
In the figure above, the two axes divide the enclosed region into four [#permalink]

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13 Dec 2017, 22:03
Bunuel wrote:

In the figure above, the two axes divide the enclosed region into four regions that have the same size and shape. Of the following, which is closest to the area of the entire enclosed region?

(A) 7
(B) 10
(C) 19
(D) 22
(E) 29

[Reveal] Spoiler:
Attachment:
2017-12-12_2124_001.png

It should be D

A quick glance reveals the square region is $$4x4$$ (area $$16$$) and is domed by four semicircles of radius $$1$$. These semicircles add up to make $$2$$ circles of radius $$1$$ with an area of $$(πr^2 + πr^2) = 2π= 6.28$$. Total area would be a little over $$16+2π$$ or just over $$22$$.
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In the figure above, the two axes divide the enclosed region into four   [#permalink] 13 Dec 2017, 22:03
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# In the figure above, the two axes divide the enclosed region into four

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