Question 19In 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 $2\sqrt3-3$$1-\frac{\sqrt3}{3}$$\frac{\sqrt3}{4}$$\frac{\sqrt2}{3}$$4-2\sqrt3$
