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O is the center of the circle above, OB=2, and angle AOB

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O is the center of the circle above, OB=2, and angle AOB [#permalink] New post 17 Aug 2013, 10:32
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  35% (medium)

Question Stats:

70% (02:37) correct 30% (01:18) wrong based on 63 sessions
Image

O is the center of the circle above, OB=2, and angle AOB measures 120 degrees. What is the area of the triangular region AOB?

a) 1

b) 2

c) (sq root 3) /2

d) sq root 3

e) 2(sq root 3)




Source: GMAT Prep Question Pack 1
Difficulty: Hard
[Reveal] Spoiler: OA

Attachments

File comment: Circle
Screen Shot 2013-08-17 at 2.24.00 PM.png
Screen Shot 2013-08-17 at 2.24.00 PM.png [ 11.85 KiB | Viewed 1245 times ]


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Re: O is the center of the circle above, OB=2, and angle AOB [#permalink] New post 17 Aug 2013, 10:44
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DelSingh wrote:
Image

O is the center of the circle above, OB=2, and angle AOB measures 120 degrees. What is the area of the triangular region AOB?

a) 1

b) 2

c) (sq root 3) /2

d) sq root 3

e) 2(sq root 3)




Source: GMAT Prep Question Pack 1
Difficulty: Hard

SINCE ANGLE AT CENTRE = 120
when you draw perpendicular OC on AB
it divides in two equal 30 - 60 - 90 triangle see figure
sides are in ratio of1: \sqrt{3}: 2
therefore OC = 1 and CA = CB = \sqrt{3}
THEREFORE AB = 2\sqrt{3}
Hence AREA Of triangle = 1/2*base*height
= 1/2 * 2\sqrt{3}*1 = \sqrt{3}
hence D
Attachments

circle.png
circle.png [ 13.72 KiB | Viewed 954 times ]


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Re: O is the center of the circle above, OB=2, and angle AOB [#permalink] New post 18 Aug 2013, 08:32
Here radius is the side of the triangle. So, Area of the traingle = (1/2)*r*r*sin(angle between those two sides)
so area = (1/2)*2*2*sin120 = (1/2)*2*2*sin60 = sqrt3.
Re: O is the center of the circle above, OB=2, and angle AOB   [#permalink] 18 Aug 2013, 08:32
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O is the center of the circle above, OB=2, and angle AOB

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