O is the center of the circle above, OB=2, and angle AOB : GMAT Problem Solving (PS)
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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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17 Aug 2013, 10:32
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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 [ 11.85 KiB | Viewed 5797 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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17 Aug 2013, 10:44
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DelSingh wrote:

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

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31 Aug 2014, 17:19
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: O is the center of the circle above, OB=2, and angle AOB [#permalink]

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17 Jun 2015, 05:22
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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16 Jul 2016, 01:15
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: O is the center of the circle above, OB=2, and angle AOB [#permalink]

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16 Jul 2016, 07:26
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Please move this question to PS section!
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Re: O is the center of the circle above, OB=2, and angle AOB [#permalink]

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16 Jul 2016, 07:33
Konstantin1983 wrote:
Please move this question to PS section!

Done. Thank you for noticing.
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Re: O is the center of the circle above, OB=2, and angle AOB [#permalink]

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20 Aug 2016, 14:10
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)
Re: O is the center of the circle above, OB=2, and angle AOB   [#permalink] 20 Aug 2016, 14:10
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