1. Given that there are 10 squares with a side length of 1 inscribed in square $A B C D$ (as shown in Figure 1). Then the area of square $A B C D$ is $\qquad$
$-1 . \frac{1521}{58}$.
As shown in Figure 3, let $P B$
$$
=a, Q B=b, A B=
$$
c. Then
$$
\begin{array}{l}
P E=4 b, \\
E F=4 a \text {, } \\
H F=2 b \text {, } \\
H I=2 a \text {, } \\
J I=2 b \text {, } \\
S J=2 a, D R=a, D S=b . \\
\text { Therefore }\left\{\begin{array}{l}
b+2 a+2 b+4 a=c \\
2 b-2 a+4 b+a=c
\end{array}\right. \text {, } \\
\Rightarrow\left\{\begin{array}{l}
3 b+6 a=c, \\
6 b-a=c
\end{array}\right. \\
\Rightarrow c=13 a, b=\frac{7 a}{3} \text {. } \\
\end{array}
$$

From $P B^{2}+Q B^{2}=P Q^{2}$, we get
$$
\left(\frac{7 a}{3}\right)^{2}+a^{2}=1 \Rightarrow a^{2}=\frac{9}{58} \text {. }
$$

Therefore, $S_{\text {square } A B C D}=c^{2}=\frac{1521}{58}$.