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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]

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Question Stats:

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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. \(\frac{\sqrt{3}}{2}\)
D. \(\sqrt{3}\)
E. \(2\sqrt{3}\)

Source: GMAT Prep Question Pack 1
Difficulty: Hard

[Reveal] Spoiler:
Attachment:
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 7664 times ]
[Reveal] Spoiler: OA

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Re: O is the center of the circle above, OB=2, and angle AOB [#permalink]

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New post 17 Aug 2013, 11:44
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DelSingh wrote:
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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 of\(1: \sqrt{3}: 2\)
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
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Re: O is the center of the circle above, OB=2, and angle AOB [#permalink]

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New post 18 Aug 2013, 09:32
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.

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Re: O is the center of the circle above, OB=2, and angle AOB [#permalink]

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New post 31 Aug 2014, 18:19
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Re: O is the center of the circle above, OB=2, and angle AOB [#permalink]

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New post 17 Jun 2015, 06:22
Extend OA to a point on the circle as point D. hence DAB is a right 30-60-90 triangle. Sin 60 will give AB to be 2sqrt3. Hence Area will be 1/2 * ((2sqrt3)/2) * ( sqrt(4-3)) = sqrt3

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Re: O is the center of the circle above, OB=2, and angle AOB [#permalink]

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New post 16 Jul 2016, 02:15
Hello from the GMAT Club BumpBot!

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Re: O is the center of the circle above, OB=2, and angle AOB [#permalink]

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New post 16 Jul 2016, 08:26
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Re: O is the center of the circle above, OB=2, and angle AOB [#permalink]

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Re: O is the center of the circle above, OB=2, and angle AOB [#permalink]

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New post 20 Aug 2016, 15:10
Area of triangle = 0.5*a*b*sin(theta)
= 0.5*2*2*sin(120)
= 2*sin(90+30)
= 2*cos 30
= 2*(1.732/2)
= root(3)

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Re: O is the center of the circle above, OB=2, and angle AOB [#permalink]

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New post 16 Sep 2017, 07:10
Area = 1/2 AB Sin(included angle)
= 1/2 x 4 x sin(180-60)
=1/2 x 4 x sin 60
= Rt 3.

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Re: O is the center of the circle above, OB=2, and angle AOB [#permalink]

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New post 19 Sep 2017, 09:53
I have a question about this problem. Because AOB is 120 degrees, doesn't that mean that the side across AOB (the base) is the largest side? In the problem, angle A should be smaller than angle O. A = 2, so O should be greater than 2. Given this, if we take 1/2 base * height, we get 2 * (at least 2)*1/2, which gives us an area of at least 2. Yet, the answer of root 3 is 1.7. How is this possible?
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Re: O is the center of the circle above, OB=2, and angle AOB   [#permalink] 19 Sep 2017, 09:53
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