<s>[INST] Finding the radius of the smallest circle that can circumscribe an equilateral triangle Q:A puzzle board is in the form of an equilateral triangle that has an area of $7\sqrt{3}$ if the board is placed on a circular table, what should be the min area of the table so that the whole board fits inside the table. [/INST] Consider this image:

The distance from the center of the triangle (a.k.a. apothem) to the midpoint of one of its side is $h/3$, where $h$ is the height of the triangle. So the radius of the circle r is the distance from the center of the triangle to a vertex, and this distance is $2h/3$.
Finding the side of the triangle (L)
We know that the area of the triangle is $7\sqrt{3}$, so we do:
$Area = \frac{1}{2}(base)(height) = \frac{1}{2}L\left(\frac{L\sqrt{3}}{2}\right) = 7\sqrt{3}$
Solving for L, we get $L = 2\sqrt{7}$.
Finding the radius of the circle (r)
As you already calculated, we can use the pythagorean theorem to find the height of the triangle:
$L^2 = \left(\frac{L}{2}\right)^2 + h^2$
$h = \sqrt{L^2 - \frac{L^2}{4}} = \frac{L\sqrt{3}}{2}$
The radius of the circle is then:
$r = \frac{2}{3}h = \frac{2}{3}\frac{L\sqrt{3}}{2} = \frac{L\sqrt{3}}{3} = \frac{2\sqrt{7}\sqrt{3}}{3} = \frac{2\sqrt{21}}{3}$
Finding the area of the circle
$Area_{circle} = \pi r^2 = \pi \left(\frac{2\sqrt{21}}{3}\right)^2 = \pi \frac{(4)(21)}{9} = \frac{28\pi}{3} \approx \frac{88}{3}$
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