juanierik wrote:

O is the center of the circle above, OB=2, and angle AOB measures 120 degrees. What is the area of triangular region AOB?

A) 1
B) 2
C) (√3)/2
D) √3
E) 2√3

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