truongvi1982 wrote:If the circle with center O has area 9\piπ, what is the area of equilateral triangle ABC?


I got r = 3 --> d= 6 = AD
I tried to solve the problem by calculating CD based on the ratio of a 30-60-90 triangle
DC: AD:AC = 1:$$\sqrt{3}$$:2
AD = d = 6 --> DC = 6/$$\sqrt{3}$$
As 2DC = BC --> BC = 2 x 63 = 12$$\sqrt{3}$$
Area of ABC = (BC x AD) / 2 = (12$$\sqrt{3}$$ x 6) / 2 = 36$$\sqrt{3}$$

Sorry that I am totally newbie and don't know how to transfer the square root symbol correctly to my post. And moreover, I don't know why I got this wrong. Can someone help? Thank you very much!

Your logic is totally correct here! You correctly calculated that CD = 6/(sqrt(3)), so the base CB = 12/(sqrt(3)).

The area of the triangle = (1/2)bh --->
base = CB = 12/(sqrt(3))
height = AD = 6
(1/2)bh = (1/2)(12/(sqrt(3)))(6) = 36/(sqrt(3))

So far, you were totally correct. You didn't show us the answer options that went with this problem, so I'm guessing that you didn't see this answer there and assumed you were wrong. Not true! You just needed to find your answer in a different format.

The GMAT doesn't like square roots in DENOMINATORS. The way to get rid of the square root in the denominator is to multiply by (sqrt(3))/(sqrt(3)). In other words, we're multiplying by 1 (so we haven't changed the value), but it's the version of 1 that will get rid of the root in the denominator:


Now, simplify:


Same answer, different format.