7. As shown in Figure 2, a circle and two semicircles are pairwise tangent, and their radii are all 1. Then the area of square $A B C D$ is $\qquad$
7. $3+\sqrt{2}+\sqrt{3}+\sqrt{6}$.

As shown in Figure 6, let $P$, $Q$, and $R$ be the centers of the circle and the two semicircles, respectively. This figure is symmetric about the diagonal $AC$ of the square, and $AP=\sqrt{2}$.
$$
\begin{array}{l}
\text { It is easy to see that } P S=\sqrt{3}, \\
S C=S R=1, \\
A C=1+\sqrt{2}+\sqrt{3} . \\
\text { Therefore, } S_{\text {square } A B C D}=\frac{1}{2} A C^{2}=\frac{1}{2}(1+\sqrt{2}+\sqrt{3})^{2} \\
=3+\sqrt{2}+\sqrt{3}+\sqrt{6} .
\end{array}
$$