3. Given a cyclic quadrilateral $ABCD$ with side lengths $AB=2, BC=6, CD=DA=4$, find the area of quadrilateral $ABCD$.


3. $S_{\triangle B C D}=S_{\triangle A B D}+S_{\triangle C B D}=\frac{1}{2} A B \cdot A D \sin A+\frac{1}{2} B C \cdot C D \sin C$

Since $A+C=180^{\circ}$, we have $\sin A=\sin C$.
$$
S_{A B C D}=\frac{1}{2}(A B \cdot A D+B C \cdot C D) \sin A=16 \sin A
$$

In $\triangle A B D$ and $\triangle C B D$, by the cosine rule we get: $B D^{2}=2^{2}+4^{2}-2 \cdot 2 \cdot 4 \cos A=20-16 \cos A$
Similarly, $B D^{2}=52-48 \cos C=52+48 \cos A$
Thus, $\cos A=-\frac{1}{2}$, so $A=120^{\circ}, S=16 \sin A=8 \sqrt{3}$.