smc

To solve for the perimeter of the triangle we plug in the formula for the area of an equilateral triangle which is $\frac{x^2\sqrt{3}}{4}$. This has to be equal to $9\sqrt{3}$, which means that $x=6$, or the side length of the triangle is $6$. Thus, the triangle (and the square) have a perimeter of $18$. It follows that each side of the square is $\frac{18}{4}=4.5$. If we draw the diagonal, we create a 45-45-90 triangle, whose hypotenuse (also the diagonal of the square) is $\boxed{\frac{9\sqrt{2}}{2}}$.