18.94 As shown in the figure, the 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. Then, the side length of the square $O P Q R$ is
(A) $\sqrt{2}$.
(B) $\sqrt{3}$.
(C) 2 .
(D) 3 .
(China Junior High School Mathematics League, 1991)
［Solution］Let the side length of the square $O P Q R$ be $x$, that is, $O P$
$$
=P Q=Q R=O R=x \text {. }
$$

Draw the altitude $A D$ of $\triangle A B C$, intersecting $O R$ at $F$.
In $\triangle A O R$, given $S_{1}=1, O R=x$.
Thus, we have
$$
A F=\frac{2 S_{1}}{O R}=\frac{2}{x} .
$$

Similarly, we get $B P=\frac{6}{x}, Q C=\frac{2}{x}$.
Therefore, $S_{\triangle A B C}=\frac{1}{2} A D \cdot B C=\frac{1}{2}\left(\frac{2}{x}+x\right) \cdot\left(\frac{6}{x}+x+\frac{2}{x}\right)$
$$
=\frac{1}{2}\left(x^{2}+10+\frac{16}{x^{2}}\right) \text {, }
$$

Also, $S_{\triangle A B C}=S_{1}+S_{2}+S_{3}+S_{\text {OFQR }}=5+x^{2}$.
From $\frac{1}{2}\left(x^{2}+10+\frac{16}{x^{2}}\right)=5+x^{2}$,
we get $x^{4}=16$, then $x=2$.
Therefore, the answer is $(C)$.