Inscribed in a circle

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alex.gellatly: Master | Next Rank: 500 Posts; Posts: 435; Joined: Wed Nov 16, 2011 7:27 amFollowed by: 16 members; Thanked: 48 times

by alex.gellatly » Thu Jun 28, 2012 5:24 pm

Note: For this problem I will use ~ as a square root sign, sorry for the inconvenience.

An equilateral triangle that has an area of 9~3 is inscribe in a circle. What is the area of the circle?

A. 6pi
B. 9pi
C. 12pi
D. 9pi~3
E. 18pi~3

Thanks

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Source: Beat The GMAT — Problem Solving | Thu Jun 28, 2012 5:24 pm

eagleeye: Legendary Member; Posts: 520; Joined: Sat Apr 28, 2012 9:12 pmFollowed by: 49 members; Thanked: 339 times; GMAT Score:770

by eagleeye » Thu Jun 28, 2012 6:22 pm

alex.gellatly wrote:Note: For this problem I will use ~ as a square root sign, sorry for the inconvenience.

An equilateral triangle that has an area of 9~3 is inscribe in a circle. What is the area of the circle?

A. 6pi
B. 9pi
C. 12pi
D. 9pi~3
E. 18pi~3

Thanks

Nice sign convention. I will just use sqrt for clarity

We know that if side of an equilateral triangle is a, then.

Height = sqrt(3)*a/2.
Area = sqrt(3)*a^2/4.

Also, When an equilateral triangle is inscribed in a circle, the median (which is the same as the height) is divided into 1:2 ratio. so for the circle r = 2/3*sqrt(3)/2 = a/sqrt(3)
So for the circle we have:
r= a/sqrt(3)
Area of circle = pi*r^2 = pi* a^2/3.

Now we are given that for the triangle area = 9*sqrt(3).
So we have sqrt(3)*a^2/4 = 9* sqrt(3) => a^2 = 4*9.

So finally, Area of circle = pi* a^2/3 = pi*4*9/3 = 12pi.

Hence C is the right answer.

Let me know if this helps

Last edited by eagleeye on Thu Jun 28, 2012 6:41 pm, edited 1 time in total.

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Thu Jun 28, 2012 6:32 pm

alex.gellatly wrote:An equilateral triangle that has an area of 9âˆš3 is inscribe in a circle. What is the area of the circle?

Refer to the following figure,

If length of each side of the triangle is a, then area of the triangle is given by (âˆš3/4)aÂ²

Hence, (âˆš3/4)aÂ² = 9âˆš3 ---> a = âˆš(9*4) = 6

Therefore, AB = BC = AC = 6
Hence, AD = Height of the triangle = (2*Area)/(Base) = 2*9âˆš3/6 = 3âˆš3

Now, O is the centroid of the triangle, i.e. where the medians of the triangles intersect each other at 2:1 ratio.

Hence, radius of the circle = OA = 2/3 of AD = (2/3)*3âˆš3 = 2âˆš3

Hence, area of the circle = Ï€(2âˆš3)Â² = 12Ï€

The correct answer is C.

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Jeevanantham: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Sun Apr 08, 2012 12:44 am; Thanked: 1 times

by Jeevanantham » Fri Jun 29, 2012 2:29 am

I have one doubt here. Here Equilateral triangle is considered as 30 : 60 :90 . And taken as x:xsqrt3 :2x .. Why is it so?? Equilateral triangle will be 60 : 60 : 60 . right??

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Fri Jun 29, 2012 2:33 am

Jeevanantham wrote:I have one doubt here. Here Equilateral triangle is considered as 30 : 60 :90 . And taken as x:xsqrt3 :2x .. Why is it so?? Equilateral triangle will be 60 : 60 : 60 . right??

ABC is a equilateral triangle with all angles equal to 60Â°.
Whereas ABD and ACD are right angle triangle with angles 30Â°, 60Â° and 90Â°

Hope that helps.

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