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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]  22 Sep 2009, 19:54
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67% (02:20) correct 33% (00:39) wrong based on 9 sessions
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]  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]  23 Sep 2009, 05:18
Was racking my brain on this one. Nice solution arkadiyua.
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Re: Gmat Geometry 2 [#permalink]  23 Sep 2009, 10:29
Soln. I too go with the Ans C - 48pi
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Re: Gmat Geometry 2 [#permalink]  30 Sep 2009, 13:11

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]  30 Sep 2009, 15:28
konayuki wrote:

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]  30 Sep 2009, 15:48
Thank you! I got it now.
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Re: Gmat Geometry 2 [#permalink]  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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