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The shaded region in the gure above represents a circular

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Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 533

Kudos [?]: 4209 [1], given: 217

Location: United Kingdom
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)

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31 Mar 2012, 06:52
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Question Stats:

80% (01:32) correct 20% (01:35) wrong based on 124 sessions

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The shaded region in the gure above represents a circular tire. If the distance from point O to point T is 18 inches and the area of the shaded region is equal to the area of the unshaded region, what is the radius of the unshaded region, in inches?

(A) $$\frac{9}{2}\sqrt{2}$$

(B) $$\frac{9}{2}\sqrt{2}$$

(C) 9

(D) $$9\sqrt{2}$$

(E) $$9\sqrt{3}$$

[Reveal] Spoiler:
Attachment:

Circle.GIF [ 5.99 KiB | Viewed 3331 times ]
[Reveal] Spoiler: OA

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MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Kudos [?]: 4209 [1], given: 217

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Re: The shaded region in the gure above represents a circular [#permalink]

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31 Mar 2012, 07:19
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enigma123 wrote:
The shaded region in the gure above represents a circular tire. If the distance from point O to point T is 18 inches and the area of the shaded region is equal to the area of the unshaded region, what is the radius of the unshaded region, in inches?

(A) $$\frac{9}{2}\sqrt{2}$$
(B) $$\frac{9}{2}\sqrt{2}$$
(C) 9
(D) 9$$\sqrt{2}$$
(E) 9$$\sqrt{3}$$

How to solve?

The area of the big circle is $$\pi{R^2}=324\pi$$;
The area of the small circle (the unshaded region) is $$\pi{r^2}$$;
The area of the shaded region is $$324\pi-\pi{r^2}$$;

We are told that the area of the shaded region is equal to the area of the unshaded region, so $$\pi{r^2}=324\pi-\pi{r^2}$$ --> $$2\pi{r^2}=324\pi$$ --> $$r^2=162$$ --> $$r=9\sqrt{2}$$.

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Re: The shaded region in the gure above represents a circular [#permalink]

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31 Mar 2012, 14:17
Outer ring's area = Bigger circle area - inner circle area = pi*R_out ^2 - pi*R_in ^2
it also equals to the inner circle area, by the question stem: pi*R_out ^2 - pi*R_in ^2 = pi*R_in ^2

So:
2*pi*R_in^2 = pi R_out^2
R_in^2 = (R_out^2)/2 => (18^2)/2
R_in = sqrt[ (18^2)/2 ] = 18/sqrt(2) = 9*sqrt(2) ===> answer D.

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Re: The shaded region in the gure above represents a circular [#permalink]

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06 Mar 2014, 02:33
Bumping for review and further discussion.

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Re: The shaded region in the gure above represents a circular [#permalink]

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28 Oct 2017, 20:34
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: The shaded region in the gure above represents a circular   [#permalink] 28 Oct 2017, 20:34
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