18.85 As shown in the figure, $A B C D$ is a square, and $C M N$ is an equilateral triangle. If the area of $A B C D$ is one square inch, then the number of square inches in the area of $C M N$ is 
(A) $2 \sqrt{3}-3$.
(B) $1-\sqrt{3} / 3$.
(C) $\sqrt{3} / 4$.
(D) $\sqrt{2} / 3$.
(E) $4-2 \sqrt{3}$.
［Solution］Let $D M=N B=x$; then $A M=A N=1-x$. If we use $y$ to denote the side length of the equilateral $\triangle C M N$, by the Pythagorean theorem we have
$$
x^{2}+1^{2}=y^{2} \text { and }(1-x)^{2}+(1-x)^{2}=y^{2} \text {. }
$$

Substituting the first equation into the second, we get
$$
2(1-x)^{2}=x^{2}+1 \text {, i.e., } x^{2}-4 x+1=0 \text {. }
$$

Solving this, we get $x=2 \pm \sqrt{3}$. But since $2+\sqrt{3}>1$, we can only choose
$$
\begin{aligned}
& x=2-\sqrt{3} . \\
S_{\triangle C M N}= & S_{\square A B C D}-S_{\triangle A N M}-S_{\triangle N B C}-S_{\triangle C D M} \\
= & 1-\frac{1}{2}(1-x)^{2}-\frac{x}{2}-\frac{x}{2} \\
= & \frac{1}{2}\left(1-x^{2}\right) .
\end{aligned}
$$

Substituting $x=2-\sqrt{3}$, we get $S_{\triangle C M N}=2 \sqrt{3}-3$. Therefore, the answer is $(A)$.