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Re: O is the center of the circle above, OB=2, and angle AOB [#permalink]
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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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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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]
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]
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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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]
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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. \(\frac{\sqrt{3}}{2}\)
D. \(\sqrt{3}\)
E. \(2\sqrt{3}\)

Source: GMAT Prep Question Pack 1
Difficulty: Hard

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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Re: O is the center of the circle above, OB=2, and angle AOB [#permalink]
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. \(2\sqrt{3}\)



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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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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