7. As shown in the figure, square $O P Q R$ is inscribed in $\triangle A B C$. It is known that the areas of $\triangle A O R$, $\triangle B O P$, and $\triangle C R Q$ are $S_{1}=1$, $S_{2}=3$, and $S_{3}=1$, respectively. Therefore, the side length of square $O P Q R$ is
(A) $\sqrt{2}$;
(B) $\sqrt{3}$;
(C) 2;
(D) 3.
7. (C)

Solution: Let the side length of the square $O P Q R$ be $x$, i.e., $O P=P Q=Q R=O R=x$. Draw the altitude $A D$ of $\triangle A B C$, intersecting $O R$ at $F$. In $\triangle A O R$, $A F=\frac{2 S_{1}}{O R}=\frac{2}{x}$.
Similarly, we get $B P=\frac{6}{x}$,
$$
Q C=\frac{2}{x},
$$

Thus, $S_{\triangle A B C}=\frac{1}{2} A D \cdot B C$
$$
\begin{array}{l}
=\frac{1}{2}\left(\frac{2}{x}+x\right)\left(\frac{6}{x}+x+\frac{2}{x}\right) \\
=\frac{1}{2}\left(x^{2}+10+\frac{16}{x^{2}}\right) .
\end{array}
$$

On the other hand, $S_{\triangle A B C}=S_{1}+S_{2}+S_{3}+S_{O P Q R}=5+x^{2}$.
Therefore, $\frac{1}{2}\left(x^{2}+10+\frac{16}{x^{2}}\right)=5+x^{2}, x^{4}=16, x=2$.