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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, 11:44

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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, 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.

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

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31 Aug 2014, 18:19

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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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

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

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16 Jul 2016, 02:15

Hello from the GMAT Club BumpBot!

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