10. As shown in Figure 6, in two concentric circles with center 0, the chord $AB$ of the larger circle is tangent to the smaller circle at point $C$, and $AB=8 \text{ cm}$. Then the area of the shaded part in the figure is $\qquad$
$10.16 \pi \mathrm{cm}^{2}$.
Connect $O A$ and $O C$. By the property of tangents, $O C \perp A B$. By the perpendicular diameter theorem, we get $A C=B C=4(\mathrm{~cm})$. In the right triangle $\triangle O A C$, $A C^{2}=O A^{2}-O C^{2}$.
Therefore, the area of the shaded part is
$$
\begin{array}{l}
S=\pi \cdot O A^{2}-\pi \cdot O C^{2}=\pi\left(O A^{2}-O C^{2}\right) \\
=\pi \cdot A C^{2}=16 \pi\left(\mathrm{cm}^{2}\right) .
\end{array}
$$