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
