6. Given a cyclic quadrilateral $ABCD$ with side lengths $AB=1, BC=3, CD=DA=2$, then the area of quadrilateral $ABCD$ is . $\qquad$
Let $\angle ABC = \alpha$, then $AC^2 = 1 + 9 - 6 \cos \alpha = 4 + 4 + 8 \cos \alpha \Rightarrow \cos \alpha = \frac{1}{7}$, so $S_{ABCD} = \frac{1}{2} \cdot (1 \cdot 3 + 2 \cdot 2) \cdot \frac{4 \sqrt{3}}{7} = 2 \sqrt{3}$.