Four circles of radius $1$ are each tangent to two sides of a square and externally tangent to a circle of radius $2$, as shown. What is the area of the square?





$\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}$
Draw a square connecting the centers of the four small circles of radius $1$. This square has a diagonal of length $6$, as it includes the diameter of the big circle of radius $2$ and two radii of the small circles of radius $1$. Therefore, the side length of this square is \[\frac{6}{\sqrt{2}} = 3\sqrt{2}.\] The side length of the larger square is $2$ units greater than the one found by connecting the midpoints, so its side length is \[2 + 3\sqrt{2}.\] The area of the larger square is $(2+3\sqrt{2})^2 = 22 + 12\sqrt{2}$ $(B).$