The two circles pictured have the same center $C$. Chord $\overline{AD}$ is tangent to the inner circle at $B$, $AC$ is $10$, and chord $\overline{AD}$ has length $16$. What is the area between the two circles?


$\textbf{(A)}\ 36 \pi \qquad\textbf{(B)}\ 49 \pi\qquad\textbf{(C)}\ 64 \pi\qquad\textbf{(D)}\ 81 \pi\qquad\textbf{(E)}\ 100 \pi$
Since $\triangle ACD$ is isosceles, $CB$ bisects $AD$. Thus $AB=BD=8$. From the Pythagorean Theorem, $CB=6$. Thus the area between the two circles is
$100\pi - 36\pi=64\pi$ $\boxed{\textbf{(C)}\ 64\pi}$
Note: The length $AC$ is necessary information, as this tells us the radius of the larger circle.  The area of the annulus is $\pi(AC^2-BC^2)=\pi AB^2=64\pi$.