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O is the center of the circle above, OB=2, and angle AOB
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17 Aug 2013, 11:32
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71% (01:14) correct 29% (01:10) wrong based on 395 sessions
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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 Attachment: File comment: Circle
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Re: O is the center of the circle above, OB=2, and angle AOB
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17 Aug 2013, 11:44
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
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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.



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Re: O is the center of the circle above, OB=2, and angle AOB
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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 306090 triangle. Sin 60 will give AB to be 2sqrt3. Hence Area will be 1/2 * ((2sqrt3)/2) * ( sqrt(43)) = sqrt3



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Re: O is the center of the circle above, OB=2, and angle AOB
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16 Jul 2016, 08:26
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
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16 Jul 2016, 08:33



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Re: O is the center of the circle above, OB=2, and angle AOB
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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
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16 Sep 2017, 07:10
Area = 1/2 AB Sin(included angle) = 1/2 x 4 x sin(18060) =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
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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
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19 Apr 2018, 13:56
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. \(\frac{\sqrt{3}}{2}\) D. \(\sqrt{3}\) E. \(2\sqrt{3}\) Source: GMAT Prep Question Pack 1 Difficulty: Hard Attachment: Screen Shot 20130817 at 2.24.00 PM.png First, since OA and OB are radii, they both have the same length (length 2) This means ∆AOB is an ISOSCELES triangle, which means the other 2 angles are 30º each. Once I see the 30º angles, I start thinking of the "special" 306090 right triangle, which we know a lot about. If we draw a line from point O so that it's PERPENDICULAR to AB, we can see that we have two 306090 right triangles hiding within this diagram. PERFECT! I've added (in purple) a 306090 right triangle with measurements. We can use this to determine the lengths of some important sides. At this point, we can see that our original triangle has a base with length 2√3 and height 1Area = (base)(height)/2 So, the area of ∆AOB = ( 2√3)( 1)/2 = √3 = D Cheers, Brent
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