6. Given a cyclic quadrilateral $ABCD$ with side lengths $AB=1, BC=3, CD=DA=2$. Then the area of quadrilateral $ABCD$ is $\qquad$ .
6. $2 \sqrt{3}$.

Connect $B D$. Then
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D}=S_{\triangle A B D}+S_{\triangle C B D} \\
=\frac{1}{2} A B \cdot A D \sin A+\frac{1}{2} C B \cdot C D \sin C \\
=\sin A+3 \sin C=4 \sin A .
\end{array}
$$

By the cosine rule, we get
$$
\begin{array}{l}
B D^{2}=A B^{2}+A D^{2}-2 A B \cdot A D \cos A \\
=5-4 \cos A, \\
B D^{2}=C B^{2}+C D^{2}-2 C B \cdot C D \cos C \\
=13-12 \cos C=13+12 \cos A .
\end{array}
$$

Thus, $5-4 \cos A=13+12 \cos A$
$$
\Rightarrow \cos A=-\frac{1}{2} \text {. }
$$

Since $0<\angle A<\pi$, therefore, $\angle A=\frac{2 \pi}{3}$.
Hence, $S_{\text {quadrilateral } A B C D}=4 \sin A=2 \sqrt{3}$.