Question 563793
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The larger circle circumscribes an equilateral triangle, which circumscribes a small circle. 
The area of the larger circle is 12pi. What is the triangle's perimeter. 
I not sure how to do this problem and how to work it ou
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In this problem, the smaller circle is irrelevant, so we can omit any mention about it.
Therefore, I will reformulate the problem by removing everything that does not matter:


<pre>
    The circle circumscribes an equilateral triangle. The area of the circle is 12pi. 
    What is the triangle's perimeter ?
</pre>

Below is the solution for this modified formulation.



<pre>
For any triangle with the side lengths 'a', 'b' and 'c', the radius of the circumscribed circle is

    R = {{{(a*b*c)/(4*A)}}},     (1)


where A is the area of the triangle.  In our case, all three sides are congruent a = b = c,
so the radius of the circumscribed circle is 

    R = {{{a^3/(4A)}}}.     (2)


The area of the circumscribed circle is  A = {{{a^2*(sqrt(3)/4)}}} square units.

Substituting it into the formula (1), we get

    R = {{{a^3/(4*a^2*(sqrt(3)/4))}}} = {{{a/sqrt(3)}}}.    (3)
 

We are given that  the area of the circle is {{{12*pi}}}, so we can write this equation

    {{{pi*R^2}}} = {{{12*pi}}}.


Cancel  {{{pi}}}  in both sides to get

    {{{R^2}}} = 12.


Substitute here R from (3).  You will get

    {{{a^2/3}}} = 12.


Simplify and find 'a'

    {{{a^2}}} = 3*12 = 36,   a = {{{sqrt(36)}}} = 6 units.


Hence, the perimeter of the triangle is 3*6 = 18 units.


<U>ANSWER</U>.  The perimeter of the triangle is 18 units.
</pre>

Thus we produced short, straightforward, compact and elegant Geometry solution.