Geometry

[juanierik]

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

I assumed that OB and AO are radii of the circle and thus have the same length. With this assumption I came up with area 2 (answer B). This is not correct.

Why can't I assume that OB and AO are radii?

OA: D

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Hi juanierik,

I'm going to give you a "nudge" so that you can retry this question.

You ARE correct that OA and OB are both radii, so they both have a length of 2.

1) Draw the triangle on your pad. Notice that it's an Isosceles triangle and fill in all 3 angles.

2) Cut the triangle in half, right down the middle. You'll now have two identical triangles with right angles. What else do you know about those triangles? Can you calculate all 3 sides? (Think about the angles; what kind of triangles are they?).

3) Now calculate the area of one of the smaller triangles....

GMAT assassins aren't born, they're made,
Rich

[Brent@GMATPrepNow]

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