The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon? 

$\text{(A)}\ \frac{1}{2}\qquad\text{(B)}\ \frac{\sqrt 3}{3}\qquad\text{(C)}\ \frac{2}{3}\qquad\text{(D)}\ \frac{3}{4}\qquad\text{(E)}\ \frac{\sqrt 3}{2}$
$\triangle OCD$ is copied six times to form the hexagon, so if we find the ratio of the area of the kite inside $\triangle OCD$ to the the area of $\triangle OCD$ itself, it will be the same ratio.
Let $OC=CD=OD =2$ so that the area of the triangle is $\frac{\sqrt{3}s^2}{4} = \sqrt{3}$.
Notice that $\triangle OCD$ is made up of a kite and two $30-60-90$ triangles.  The two hypotenuses of these two triangles form $CD$, so the hypotenuse of each triangle must be $\frac{2}{2} = 1$.  Thus, the legs of each triangle are $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$, and the area of two of these triangles is $\frac{1}{2}\cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}$.
Subtracting the area of the two triangles from the area of the equilateral triangle, we find that the area of the kite is $\sqrt{3} - \frac{\sqrt{3}}{4} = \frac{3\sqrt{3}}{4}$.
Thus, the ratio of areas is $\frac{3}{4}$, which is option $\boxed{D}$.