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# A circle is inscribed in an equilateral triangle whose side

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A circle is inscribed in an equilateral triangle whose side [#permalink]

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01 Jun 2005, 12:10
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A circle is inscribed in an equilateral triangle whose side is 4 x square root (3) each. The center of the circle is how high from the base of the triangle ?

A. 2 x square root (3)
B. 2
C. 3 x square root (3)
D. 2 x square root (2)
E. 3 x square root (2)
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Re: PS: Circle inscribed in equilateral triangle [#permalink]

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01 Jun 2005, 14:32
In this case, the centroid is the same as the center of the circle.
The height of the triangle = sqrt[48 - 12] = 6.
Therefore, the required distance = 1/3 x 6 = 2.
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01 Jun 2005, 15:55
wow. quite a problem!!

So i started out drawing a circle within the triangle. I figured that since all 3 sides are equal the circle must sit perfectly in the center of the triangle meaning the center of the circle = center of the triangle.

Cant say that I have ever tried to find the center of a triangle but I figured it would have something to do with the height.

So height = 6

If in fact the center is found by dividing by 3 ??? then I guess the answer is 2.

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01 Jun 2005, 16:27
Kaps:
is there a rule for this?

I knew that the radius of the circle has to be less than 3, i was stuck between 2 and 2(sqrt(2))!
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02 Jun 2005, 06:56
fresinha12 wrote:
Kaps:
is there a rule for this?

I knew that the radius of the circle has to be less than 3, i was stuck between 2 and 2(sqrt(2))!

B.

The lenght of each median of an equilater triangle is (sqrt(3)*side)/2.
For this problem, the centre is the centroid which is always 1/3 the lenght of the median from the base.
So (1/3)(sqrt(3)*side)/2
= (1/3)(sqrt(3)*4*sqrt(3))/2
= 2.

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02 Jun 2005, 14:59
B)... i divided the triangle in 3 smaller ones. the base is 4sqrt3. the smaller legs that enclose the base are both 1/4sqrt3. so this triangle can again be divided into 2 smaller one with 30:60:90. so its 2sqrt3:2:1/4sqrt3. its 2...
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02 Jun 2005, 14:59
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# A circle is inscribed in an equilateral triangle whose side

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