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555-605 Level|   Geometry|      
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DelSingh
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O is the center of the circle above, OB=2, and angle AOB 17 Aug 2013, 11:32
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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25% (medium)

Question Stats:

74% (01:47) correct 26% (01:52) wrong based on 618 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

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Screen Shot 2013-08-17 at 2.24.00 PM.png [ 11.85 KiB | Viewed 26320 times ]
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D
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O is the center of the circle above, OB=2, and angle AOB 17 Aug 2013, 11:44
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DelSingh


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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Chiranjeevee
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O is the center of the circle above, OB=2, and angle AOB 18 Aug 2013, 09:32
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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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O is the center of the circle above, OB=2, and angle AOB 17 Jun 2015, 06:22
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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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O is the center of the circle above, OB=2, and angle AOB 16 Jul 2016, 08:26
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Please move this question to PS section!
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O is the center of the circle above, OB=2, and angle AOB 16 Jul 2016, 08:33
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Konstantin1983
Please move this question to PS section!

Done. Thank you for noticing.
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O is the center of the circle above, OB=2, and angle AOB 20 Aug 2016, 15:10
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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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O is the center of the circle above, OB=2, and angle AOB 16 Sep 2017, 07:10
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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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O is the center of the circle above, OB=2, and angle AOB 19 Sep 2017, 09:53
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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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O is the center of the circle above, OB=2, and angle AOB 19 Apr 2018, 13:56
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DelSingh


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

Show Spoiler
Attachment:
Screen Shot 2013-08-17 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" 30-60-90 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 30-60-90 right triangles hiding within this diagram. PERFECT!


I've added (in purple) a 30-60-90 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 1


Area = (base)(height)/2
So, the area of ∆AOB = (2√3)(1)/2
= √3
= D

Cheers,
Brent
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O is the center of the circle above, OB=2, and angle AOB 24 Nov 2020, 14:41
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DelSingh


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. \(2\sqrt{3}\)


Since AOB measures 120 degrees, we can split triangle AOB into two 30:60:90 right triangles.

The ratio of a \(30:60:90\) triangle is \(1:\sqrt{3}:2\)

Since we are told the hypotenuse is 2, we have the following ratios: \(1:\sqrt{3}:2\)

\(\frac{1}{2}b * h * 2 = \frac{\sqrt{3}}{2} * 2 = \sqrt{3}\)

Answer is D.
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O is the center of the circle above, OB=2, and angle AOB 21 May 2022, 14:50
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