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## Problem solving

tagged by: truongvi1982

truongvi1982 Newbie | Next Rank: 10 Posts
Joined
19 Nov 2017
Posted:
1 messages
Test Date:
02 Mar 2018
Target GMAT Score:
750

#### Problem solving

Wed Jan 03, 2018 3:41 am
If the circle with center O has area 9\piÏ€, 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!

_________________
Vi

### GMAT/MBA Expert

ceilidh.erickson GMAT Instructor
Joined
04 Dec 2012
Posted:
1805 messages
Followed by:
230 members
1443
Mon Feb 05, 2018 8:07 am
truongvi1982 wrote:
If the circle with center O has area 9\piÏ€, 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))
(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:

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DrMaths Senior | Next Rank: 100 Posts
Joined
15 Jan 2018
Posted:
82 messages
Thu Feb 01, 2018 9:57 am
truongvi1982 wrote:
If the circle with center O has area 9\piÏ€, 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!
36sqrt{3} = 12*3/sqrt[3] = 12*sqrt[3]

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