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GMAT Set 7 Q8

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by Abhijit K » Thu Feb 19, 2015 1:49 am
An equilateral triangle that has an area of 9root3 is inscribed in a circle. What is the area
of the circle?
A. 6*3.14
B. 9*3.14
C. 12*3.14
D. 9*3.14*root3
E. 18*3.14*root3

Approach please....
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by GMATGuruNY » Thu Feb 19, 2015 7:16 am
An equilateral triangle can be divided into 30-60-90 triangles:
Image
In a 30-60-90 triangle, the sides are in the following ratio:
x : x√3 : 2x.
This ratio can also be represented as follows:
x/2 : (x/2)√3 : x.
An equilateral triangle that has an area of 9√3 is inscribed in a circle. What is the area of the circle?

A. 6Ï€
B. 9Ï€
C. 12Ï€
D. 9π√3
E. 18π√3
Image

In the figure above:
AC = 2(AD) = 2(r/2)√3 = r√3.
BD = r + r/2 = 3r/2.

Since the area of ∆ABC is 9√3, we get:
(1/2)(AC)(BD) = 9√3
(1/2)(r√3)(3r/2) = 9√3
3r²/4 = 9
3r² = 36
r² = 12.

Thus:
Area of circle = πr² = 12π.

The correct answer is C.
Last edited by GMATGuruNY on Thu Feb 19, 2015 7:31 am, edited 1 time in total.
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by regor60 » Thu Feb 19, 2015 7:16 am
Abhijit K wrote:An equilateral triangle that has an area of 9root3 is inscribed in a circle. What is the area
of the circle?
A. 6*3.14
B. 9*3.14
C. 12*3.14
D. 9*3.14*root3
E. 18*3.14*root3

Approach please....
Find length of one side via pythagoras = 6

since each of the angles are the same, they subtend 120 degrees of the circle

that is equivalent to a triangle centered at center of circle with an angle of 120

long leg = 6 as previously determined

two other legs = radius, so 30 degrees each other interior angle

sin 120/6 = sin 30/R,etc.

12pi
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