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Relationship between Equilateral Triangle and Circle

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Relationship between Equilateral Triangle and Circle

by cr1985 » Wed Aug 17, 2011 12:32 pm
If I was to place an equilateral triangle within a circle, what relationship exists? So far, I've been able to establish that 2/3's of the triangle's height might be the radius of the circle? Getting to my question, if I was given the perimeter of the triangle and asked to find the radius of the circle, how would I go about doing that?

Thanks.
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by rppala90 » Wed Aug 17, 2011 12:47 pm
Recently I came across the following formulas related to this:

Let's say "side" is the length of equilateral triangle's side.

Area of triangle = (sqrt(3) * square(side) ) /4.

Height of triangle = Height = (sqrt(3) * side ) / 2.

Radius of circle = ( 2 * Height ) / 3.

sqrt = square root.

Back to your question:

"side" here will be ( perimeter of triange /3 ).
Rest is a simple substitution of "side" in above formulas.
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by gmatboost » Wed Aug 17, 2011 2:39 pm
When you draw the picture, draw a line through the middle of the equilateral triangle to create two 30-60-90 triangles. Then, draw the 3 radii that go to the corners of the triangle. This will create two more, smaller 30-60-90 triangles.

Now, label the radii r. From here, use the side ratios of a 30-60-90, which are 1 : root(3) : 2, to fill in every side length in the picture. It's good practice, and it will show you all of the relationships mentioned by rppala90.
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by rppala90 » Wed Aug 17, 2011 2:52 pm
Thanks gmatboost.
Can you also please explain how radius of circle = (2 * Height) / 3.
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by gmatboost » Wed Aug 17, 2011 10:37 pm
Sure. Imagine (or draw) the equilateral triangle with one side along the bottom AB, and the other corner, C, at the top.

Now, let M be the midpoint of AB. The height is CM. Draw this line segment.
And let the center of the triangle/circle by O. So OA = OB = OC = r.

Now, OMB is a 30-60-90 triangle. Angle OBM is 30, because it is half of 60 degree angle ABC.
Similarly, angle BOM is 60, because it is half of 120 degree angle AOB.

In 30-60-90 triangle OMB, we know that OB is r. That means OM is r/2, since the side opposite the 30 degree angle is always half of the hypotenuse. That means the height CM is 1.5 r.

So, height = (3/2)r
Which is the same as r = (2/3) * height
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