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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, 20:54
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An equilateral triangle of side 12 is inscribed in a circle, what is the area of the circle?
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Re: Gmat Geometry 2 [#permalink] New post   29 Feb 2012, 13:24
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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] New post   23 Sep 2009, 01:32
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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, 06:18
Was racking my brain on this one. Nice solution arkadiyua. :)
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Re: Gmat Geometry 2 [#permalink] New post   23 Sep 2009, 11:29
Soln. I too go with the Ans C - 48pi
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Re: Gmat Geometry 2 [#permalink] New post   30 Sep 2009, 14:11
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, 16:28
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, 16:48
Thank you! I got it now.
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Re: An equilateral triangle of side 12 is inscribed in a circle [#permalink] New post   27 Mar 2017, 12:58
C - 48P

The height of the triangle equals 63√63 since the sides are 6 and 12 (1:2: \sqrt{3} ratio). So the area of the triangle is (63√∗12)/2=363√(63∗12)/2=363

Since the triangle is equilateral, we can get 3 equal triangles with area of each 363√/2=123√363/2=123. From this we can get the heights of the smaller triangles: 123√∗2/12=23√123∗2/12=23. Thus, the radius is 63√−23√=43√63−23=43. The area of the cirle is (23√)2∗P=48P
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Re: An equilateral triangle of side 12 is inscribed in a circle [#permalink] New post   02 Apr 2021, 15:03
arkadiyua wrote:
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 \((2\sqrt{3})^2*P=48P\)



Hello,
if the area is equal for 3 triangles, shouldn't it be divided by 3 instead of 2?
also can you explain how did you get the value of height if smaller triangles if you knew the area 12root3?
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Re: An equilateral triangle of side 12 is inscribed in a circle [#permalink]
02 Apr 2021, 15:03
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