AHSME 1974 Problem 19

In the adjoining figure $ABCD$ is a square and $CMN$ is an equilateral triangle. If the area of $ABCD$ is one square inch, then the area of $CMN$ in square inches is

$\mathrm{(A)\ } 2\sqrt{3}-3 \mathrm{(B) \ }1-\frac{\sqrt{3}}{3} \mathrm{(C) \ } \frac{\sqrt{3}}{4} \mathrm{(D) \ } \frac{\sqrt{2}}{3} \mathrm{(E) \ }4-2\sqrt{3}$