3. In a cyclic quadrilateral $ABCD$, $AB=3$, $BC=4$, $CD=5$, $DA=6$. Then the area of quadrilateral $ABCD$ is $\qquad$ .
$3.6 \sqrt{10}$.
From the cosine rule, we have
$$
\begin{array}{l}
A C^{2}=A B^{2}+B C^{2}-2 A B \cdot B C \cos B \\
=C D^{2}+D A^{2}-2 C D \cdot D A \cos D .
\end{array}
$$

Since points $A, B, C, D$ are concyclic,
$$
\angle B+\angle D=\pi \text {. }
$$

Then $\cos B=\frac{3^{2}+4^{2}-5^{2}-6^{2}}{2(3 \times 4+5 \times 6)}=-\frac{3}{7}$.
Thus, the area of quadrilateral $ABCD$ is
$$
\begin{array}{l}
=\frac{1}{2} A B \cdot B C \sin B+\frac{1}{2} C D \cdot D A \sin D \\
=\frac{1}{2}(3 \times 4+5 \times 6) \frac{\sqrt{40}}{7}=6 \sqrt{10} .
\end{array}
$$