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In the figure above, point O is the center of the circle

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Intern
Intern
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Joined: 18 Mar 2012
Posts: 48
GMAT 1: 690 Q V
GPA: 3.7
Followers: 0

Kudos [?]: 41 [0], given: 117

In the figure above, point O is the center of the circle [#permalink] New post 05 Mar 2013, 10:38
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Difficulty:

  65% (hard)

Question Stats:

67% (03:30) correct 33% (04:09) wrong based on 52 sessions
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In the figure above, point O is the center of the circle and line segment BD is tangent to the to the circle at point C. If BC = 4, OB = 8, and OC = CD, then what is the area of the region whose perimeter is radius OA, arc ACE, and radius OE?

A. 8π
B. 10π
C. 12π
D. 16π
E. 20π
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Mar 2013, 01:25, edited 1 time in total.
Edited the question.
1 KUDOS received
Intern
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Joined: 18 Feb 2013
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GMAT 1: 710 Q49 V38
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Kudos [?]: 12 [1] , given: 11

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Re: In the figure above, point O is the center of the circle [#permalink] New post 05 Mar 2013, 13:13
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My question is how did you know where to put point A and point E? Was the image included in the problem or did you draw it?

Angle OCD is 90, and given OC=CD, we know angle COD and CDO =45 degrees.
OB=8, CB=4, and properities of the special 30, 60, 90 triangle. x, 2x, x\sqrt{3}, we find out the radius is 4\sqrt{3}.
We also know angle COB, which is opposite of x in the triangle = 30 degrees
Radius= 4\sqrt{3} Area=48pie
Angle AOE =45+30=75 degree

\frac{75}{360}*48pie=10pie
Intern
Intern
avatar
Joined: 18 Mar 2012
Posts: 48
GMAT 1: 690 Q V
GPA: 3.7
Followers: 0

Kudos [?]: 41 [0], given: 117

Re: In the figure above, point O is the center of the circle [#permalink] New post 05 Mar 2013, 13:14
DoItRight wrote:
My question is how did you know where to put point A and point E? Was the image included in the problem or did you draw it?

Angle OCD is 90, and given OC=CD, we know angle COD and CDO =45 degrees.
OB=8, CB=4, and properities of the special 30, 60, 90 triangle. x, 2x, x\sqrt{3}, we find out the radius is 4\sqrt{3}.
We also know angle COB, which is opposite of x in the triangle = 30 degrees
Radius= 4\sqrt{3} Area=48pie
Angle AOE =45+30=75 degree

\frac{75}{360}*48pie=10pie


It was part of the question I did not draw it
Re: In the figure above, point O is the center of the circle   [#permalink] 05 Mar 2013, 13:14
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In the figure above, point O is the center of the circle

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