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 

$\mathrm{(A)\ } 2\sqrt{3}-3 \qquad \mathrm{(B) \ }1-\frac{\sqrt{3}}{3} \qquad \mathrm{(C) \  } \frac{\sqrt{3}}{4} \qquad \mathrm{(D) \  } \frac{\sqrt{2}}{3} \qquad \mathrm{(E) \  }4-2\sqrt{3}$
Let $BN=x$ so that $AN=1-x$. From the Pythagorean Theorem on $\triangle NBC$, we get $CN=\sqrt{x^2+1}$, and from the Pythagorean Theorem on $\triangle AMN$, we get $MN=(1-x)\sqrt{2}$. Since $\triangle CMN$ is equilateral, we must have $\sqrt{x^2+1}=(1-x)\sqrt{2}\implies x^2+1=2x^2-4x+2\implies x^2-4x+1=0$. From the Pythagorean Theorem, we get $x=2-\sqrt{3}$, since we want the root that's less than $1$.
Therefore, $CN^2=x^2+1=(2-\sqrt{3})^2+1=8-4\sqrt{3}$. The area of an equilateral triangle with side length $x$ is equal to $\frac{x^2\sqrt{3}}{4}$, so the area of $\triangle CMN$ is $\frac{(8-4\sqrt{3})(\sqrt{3})}{4}=2\sqrt{3}-3, \boxed{\text{A}}$.