smc

Draw in the diameter through A perpendicular to chords BD and CE. Label the intersection points of the diameter with the chords F and G. Then, AF=4, AG=4, and AFB ,AFD, AEG. and AGC are right triangles. We're going to find the area inside sectors BAD and CAE but outside triangles BAD and CAE. By pythagorean theorem, BF=$4\sqrt{3}$, as are DF, EG, and GC. It then follows that the area of triangles BAD and CAE are $16\sqrt{3}$. Since AFB and AFD are 30-60-90 triangles, the area of sector BAD is $\frac{64\pi }{3}$, as is sector CAE. Thus, the area outside of the two chords is $\frac{128\pi }{3}-32\sqrt{3}$. Since we want the area inside the two chords, you can subtract the outside from the whole circle, which is $64\pi -(\frac{128\pi }{3}-32\sqrt{3})$ $\Rightarrow \frac{64\pi }{3}+32\sqrt{3}$, or $\boxed{32\sqrt{3}+21\frac{1}{3}\pi }$.