2A. Let $A B C D$ be a cyclic quadrilateral such that: $\overline{A B}=\overline{B C}=\sqrt{3}$, $\overline{C D}=2$ and $\measuredangle B A D=30^{\circ}$. Calculate the area of the quadrilateral $A B C D$.
Solution. Let $\overline{A B}=\overline{B C}=b=\sqrt{3}, \overline{C D}=c=2$ and $\measuredangle B A D=30^{\circ}$. Quadrilateral $A B C D$ is cyclic, so $\measuredangle B C D=180^{\circ}-30=150^{\circ}$. We will use the cosine theorem, applied to $\triangle B A D$ and $\triangle B C D$. From $\triangle B A D$ we have $\overline{B D}^{2}=b^{2}+\overline{A D}^{2}-2 b \overline{A D} \cos 30^{\circ}$, and from $\triangle B C D$ we have

$$
\overline{B D}^{2}=b^{2}+c^{2}-2 b c \cos 150^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_47715ec139c950209997g-08.jpg?height=635&width=642&top_left_y=1598&top_left_x=823)

By equating the two equations, we get

$$
b^{2}+\overline{A D}^{2}-2 b \overline{A D} \cos 30^{\circ}=b^{2}+c^{2}-2 b c \cos 150^{\circ}
$$

which simplifies to $\overline{A D}^{2}-3 \overline{A D}-10=0$. The solution for $\overline{A D}$ is the positive root, $\overline{A D}=5$. Now for the area of quadrilateral $A B C D$ we get

$$
\begin{aligned}
P_{\square A B C D} & =P_{\triangle A B D}+P_{\triangle B C D} \\
& =\frac{\overline{A B} \cdot \overline{A D}}{2} \sin 30^{\circ}+\frac{\overline{B C} \cdot \overline{D C}}{2} \sin 150^{\circ} \\
& =\frac{5 \sqrt{3}}{2} \cdot \frac{1}{2}+\frac{2 \sqrt{3}}{2} \cdot \frac{1}{2}=\frac{7 \sqrt{3}}{4} .
\end{aligned}
$$