smc

The area of the square is $s^2$. The diagonal of a square with side $s$ bisects the square into two $45-45-90$ right triangles, so the diagonal has length $s\sqrt{2}$. The area of the triangle is $\frac{1}{2}bh$. The base $b$ of the triangle is the diagonal of the square, which is $b=s\sqrt{2}$. If the area of the triangle is equal to the area of the square, we have: $s^2=\frac{1}{2}bh$ $s^2=\frac{1}{2}s\sqrt{2}\cdot h$ $s=\frac{\sqrt{2}}{2}h$ $h=\frac{2}{\sqrt{2}}s$ $h=s\sqrt{2}$ This is option $\boxed{s\sqrt{2}}$