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An equilateral triangle of side 12 is inscribed in a circle

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An equilateral triangle of side 12 is inscribed in a circle [#permalink] New post 22 Sep 2009, 19:54
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An equilateral triangle of side 12 is inscribed in a circle, what is the area of the circle?
[Reveal] Spoiler: OA

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Re: Gmat Geometry 2 [#permalink] New post 23 Sep 2009, 00:32
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Thanks for the morning warmup! :)

C - 48P

The height of the triangle equals 6\sqrt{3} since the sides are 6 and 12 (1:2: \sqrt{3} ratio). So the area of the triangle is (6\sqrt{3}*12)/2=36\sqrt{3}

Since the triangle is equilateral, we can get 3 equal triangles with area of each 36\sqrt{3}/2=12\sqrt{3}. From this we can get the heights of the smaller triangles: 12\sqrt{3}*2/12=2\sqrt{3}. Thus, the radius is 6\sqrt{3}-2\sqrt{3}=4\sqrt{3}. The area of the cirle is (2\sqrt{3})^2*P=48P
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Re: Gmat Geometry 2 [#permalink] New post 23 Sep 2009, 05:18
Was racking my brain on this one. Nice solution arkadiyua. :)
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Re: Gmat Geometry 2 [#permalink] New post 23 Sep 2009, 10:29
Soln. I too go with the Ans C - 48pi
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Re: Gmat Geometry 2 [#permalink] New post 30 Sep 2009, 13:11
Hi arkadiyua.

I need your help. What do you mean by "we can get 3 equal triangles with area of each..."? I am lost in how to get this three triangles.

Since the triangle is equilateral, we can get 3 equal triangles with area of each 36\sqrt{3}/2=12\sqrt{3}.
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Re: Gmat Geometry 2 [#permalink] New post 30 Sep 2009, 15:28
konayuki wrote:
Hi arkadiyua.

I need your help. What do you mean by "we can get 3 equal triangles with area of each..."? I am lost in how to get this three triangles.

Since the triangle is equilateral, we can get 3 equal triangles with area of each 36\sqrt{3}/2=12\sqrt{3}.


If you draw a line from the centre of the circle to each of the triangle vertices you will see that the triangle is divided into 3 equal triangles. In fact if you just draw a triangle… and draw a line from the centre to each of the vertices that will have the same result.

Helps to draw and visualise it.
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Re: Gmat Geometry 2 [#permalink] New post 30 Sep 2009, 15:48
Thank you! I got it now.
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Re: Gmat Geometry 2 [#permalink] New post 29 Feb 2012, 12:24
remember these formulae for equilateral triangles and you'll save important seconds.
s = side

area of equilateral triangle = \frac{s^2}{4}\sqrt{3}, height = \frac{s}{2}\sqrt{3}

radius of circumscribe circle = \frac{s}{\sqrt{3}}, radius of circumscribe circle = \frac{s}{2\sqrt{3}}

radius of circle = \frac{12}{\sqrt{3}} = 4\sqrt{3} .... s=12

area of circle = 48\pi

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Re: Gmat Geometry 2   [#permalink] 29 Feb 2012, 12:24
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