Two equal parallel chords are drawn $8$ inches apart in a circle of radius $8$ inches. The area of that part of the circle that lies between the chords is:
$\textbf{(A)}\ 21\frac{1}{3}\pi-32\sqrt{3}\qquad \textbf{(B)}\ 32\sqrt{3}+21\frac{1}{3}\pi\qquad \textbf{(C)}\ 32\sqrt{3}+42\frac{2}{3}\pi \qquad\\ \textbf{(D)}\ 16\sqrt {3} + 42\frac {2}{3}\pi \qquad \textbf{(E)}\ 42\frac {2}{3}\pi$
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 
$\fbox{B}$.