【Solution】Solution: By rotating triangle $E B D$ to the position of triangle $A B C$, the shaded area can be combined into a segment of an annulus between two sectors.

As shown in the figure below:
The area of the shaded part AMNE is:
$S_{A M N E}=S_{\text {sector } A B E}-S_{\text {sector } M B N}=\frac{1}{3} \times \pi \times 10^{2}-\frac{1}{3} \times \pi \times 5^{2}=25 \pi$
Taking $\pi$ as 3, the area is:
$S_{A M N E}=25 \times 3=75$
Therefore, the answer is: 75.
8. (5 points) As shown in the figure, in the right triangle $\triangle ABC$, the hypotenuse $AB=10, BC=5, \angle ABC=60^{\circ}$. With point $B$ as the center, $\triangle ABC$ is rotated clockwise by $120^{\circ}$, and points $A$ and $C$ reach points $E$ and $D$, respectively. The area swept by side $AC$ (i.e., the shaded area in the figure) is $\qquad$. (Take $\pi$ as 3)