3. Let the sides of the cyclic quadrilateral $ABCD$ be $AB=3, BC=4, CD=5, DA=6$, then the area of quadrilateral $ABCD$ is $\qquad$ .
Given $\angle A D C=\alpha$, then $9+16+24 \cos \alpha$ $=25+36-60 \cos \alpha \Rightarrow \cos \alpha=\frac{3}{7} \Rightarrow \sin \alpha=\frac{2 \sqrt{10}}{7}$.
Therefore, $S_{A B C D}=\frac{1}{2}(12+30) \cdot \frac{2 \sqrt{10}}{7}=6 \sqrt{10}$.