jmc

First, we label the diagram as shown below:



All four right triangles are 30-60-90 triangles. Therefore, the length of the shorter leg in each triangle is half the hypotenuse, and the length of the longer leg is $\sqrt{3}$ times the length of the shorter leg. We apply these facts to each triangle, starting with $△AOB$ and working clockwise.

From $△AOB$, we find $AB=AO/2=4$ and $BO=AB\sqrt{3}=4\sqrt{3}$.

From $△BOC$, we find $BC=BO/2=2\sqrt{3}$ and $CO=BC\sqrt{3}=2\sqrt{3}\cdot \sqrt{3}=6$.

From $△COD$, we find $CD=CO/2=3$ and $DO=CD\sqrt{3}=3\sqrt{3}$.

From $△DOE$, we find $DE=DO/2=3\sqrt{3}/2$ and $EO=DE\sqrt{3}=(3\sqrt{3}/2)\cdot \sqrt{3}=(3\sqrt{3}\cdot \sqrt{3})/2=\boxed{\frac{9}{2}}$.