smc

Let the points be labeled as in the diagram, with $O$ being the center of the circle. Because we know that the small square has an area of $40$, it must have a side length of $\sqrt{40}=2\sqrt{10}$. It is simple to prove that $O$ is the midpoint of the bottom side of the small square, so $OB=\frac{2\sqrt{10}}{2}=\sqrt{10}$. By the Pythagorean Theorem, $AO=\sqrt{50}=5\sqrt{2}$, which is the radius of the circle. Thus, $OC=5\sqrt{2}$, and so the diagonal $\overline{CD}$ of the big square has length $10\sqrt{2}$. Thus, the big square has side length $\frac{10\sqrt{2}}{\sqrt{2}}=10$, and, consequently, it has an area of $\boxed{100}$.