Triangle inside a circle that is inside a square

Let the side length of the equilateral triangle be $a$ . Then its area is $\dfrac {a^2\sqrt 3}4 = \dfrac {75\sqrt 3}4\implies a = 5\sqrt 3$ . The length of its median is $\dfrac {\sqrt 3}2 a = \dfrac {15}2$ . We know the centroid of the equilateral triangle is the center of the circumcircle. Therefore the radius of the circle is $\dfrac 23$ the length of the medium or $r = \dfrac 23 \times \dfrac {15}2 = 5$ . Since the side length of the square is $2r$ , its perimeter is $8r = \boxed{40}$ .

If s is the side of the equilateral triangle, we see that (Area of triangle) = (sqrt(3)) (s^2) / 4. But, this area is given to be 75 (sqrt(3))/4. Setting these two equal yields s^2 = 75. If r is the radius of the circle, an application of the Law of Cosines shows that s^2 = 2 r^2 - 2 (r^2) cos(120 degrees). This simplifies to s^2 = 3 r^2. Therefore, 3 r^2 = 75, or r = 5. We then see that the side of the square is 2 r = 10. Therefore, the perimeter is 4*10 = 40