Question 1064039
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Construct a line segment from one of the vertices of the equilateral triangle through the center of the circle and continuing to the midpoint of the opposite side of the triangle.  Repeat for the other two vertices.  You will have divided the equilateral triangle into six congruent 30-60-90 right triangles.


The distance from any one of the vertices of the triangle to the center of the circle is equal to the radius of the circle.  Note that this is also the hypotenuse of two of the 30-60-90 triangles.  Also, the distance from the center of the circle to the midpoint of one of the sides of the triangle is the short leg of two of the 30-60-90 triangles, and since the short leg of a 30-60-90 triangle is 1/2 of the measure of the hypotenuse, the measure of this short segment must be 1/2 of the radius.


Since the area of a circle is given by *[tex \Large \pi{r^2}] and *[tex \Large 36\pi] is the area of the given circle, *[tex \Large r^2\ =\ 36] and therefore *[tex \Large r\ =\ 6].  Hence, the hypotenuse of any one of the 30-60-90 triangles is 6 and the short leg is 3.  Then, since the ratio of the sides of a 30-60-90 triangle is *[tex \Large 1\,:\,\frac{\sqrt{3}}{2}\,:\,\frac{1}{2}], the third side of any of the 30-60-90 triangles must be *[tex \Large 3\sqrt{3}].


Then the area of one of the six 30-60-90 triangles that comprise the larger equilateral triangle must be *[tex \Large \frac{1}{2}(3\sqrt{3})(3)\ =\ \frac{9\sqrt{3}}{2}].  Multiply by 6 and you have your answer.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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