jmc

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