O is the center of the circle above, OB=2, and angle AOB

Question

https://gmatclub.com/forum/download/file2.php?mode=view&id=20793 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 Screen Shot 2013-08-17 at 2.24.00 PM.png

blueseas · Accepted Answer

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

Chiranjeevee · Answer

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.

nsaruna · Answer

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

Konstantin1983 · Answer

Please move this question to PS section!

Bunuel · Answer

Done. Thank you for noticing.

keats · Answer

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)

T700ISB · Answer

Area = 1/2 AB Sin(included angle)
 = 1/2 x 4 x sin(180-60) 
 =1/2 x 4 x sin 60
 = Rt 3.

nightblade354 · Answer

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?

BrentGMATPrepNow · Answer

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. 
 https://s30.postimg.cc/xanlfb1v1/AOB_1.jpg

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!
 https://s30.postimg.cc/9l2r4csvh/AOB_2.jpg

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. 
 https://s30.postimg.cc/xrdefhezx/AOB_3.jpg

At this point, we can see that our original triangle has a base with length  2√3  and height  1 
 https://s30.postimg.cc/6s9jkbsj1/AOB_4.jpg

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

Cheers,
Brent

Basshead · Answer

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.

bumpbot · Answer

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O is the center of the circle above, OB=2, and angle AOB

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GMAT Problem Solving (PS) Questions

O is the center of the circle above, OB=2, and angle AOB

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