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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 \((2\sqrt{3})^2*P=48P\)

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.

Re: An equilateral triangle of side 12 is inscribed in a circle [#permalink]

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29 Apr 2016, 15:03

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Re: An equilateral triangle of side 12 is inscribed in a circle [#permalink]

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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

gmatclubot

Re: An equilateral triangle of side 12 is inscribed in a circle
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27 Mar 2017, 12:58

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