3. As shown in Figure 1, it is easy to know that $O$ is the intersection of $A C$ and $B D$.
$$
\begin{array}{l}
\text { Then } S_{\triangle A O B} \\
=\frac{1}{2} \times 10 \times 10 \\
=50, \\
S_{\text { sector } A O B}=\frac{\pi}{4} \cdot 100 \\
=25 \pi .
\end{array}
$$

By the Pythagorean theorem, we know
$$
A B^{2}=A O^{2}+O B^{2}=200 \text {. }
$$

Then the area of the semicircle on $A B$ is $\frac{\pi}{2}\left(\frac{A B}{2}\right)^{2}=25 \pi$. Therefore, the required area is $25 \pi-(25 \pi-50)=50$.
3. Given that the radius of the circumcircle $\odot O$ of square $ABCD$ is 10. Draw a semicircle outside the square with $AB$ as the diameter. Find the area of the region of the semicircle that lies outside $\odot O$.