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.
A: $\frac{88}{3}$
I get that the side of the triangle is $2\sqrt{7}$ and also that in an equilateral triangle the median, perpendicular bisector, altitude and angle bisector are the same. I'm however still stuck with how to get the radius without resorting to sin/cos etc.
Best Answer
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
ris 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
$Area_{circle} = \pi r^2 = \pi \left(\frac{2\sqrt{21}}{3}\right)^2 = \pi \frac{(4)(21)}{9} = \frac{28\pi}{3} \approx \frac{88}{3}$