17. $A B C D$ is a rectangle; $P$ and $Q$ are the mid-points of $A B$ and $B C$ respectively. $A Q$ and $C P$ meet at $R$. If $A C=6$ and $\angle A R C=150^{\circ}$, find the area of $A B C D$.
(2 marks)
$A B C D$ 是長方形, $P$ 和 $Q$ 分別是 $A B$ 和 $B C$ 的中點, 且 $A Q$ 交 $C P$ 於 $R \circ$ 若 $A C$ $=6$ 而 $\angle A R C=150^{\circ}$, 求 $A B C D$ 的面積。
17. $8 \sqrt{3}$
17. Let $S$ be the mid-point of $A C$. Then $R$ lies on $B S$ with $B R: R S=2: 1$ since $R$ is the centroid of $\triangle A B C$. We have $B S=3$ (as $S$ is the centre and $A B$ is a diameter of the circumcircle of $\triangle A B C$ ) and so $R S=1$. Let $A R=x$ and $C R=y$. Since the area of $\triangle A R C$ is $\frac{1}{2} x y \sin 150^{\circ}=\frac{1}{4} x y$, the area of $\triangle A B C$ is $\frac{3}{4} x y$ and hence the area of $A B C D$ is $\frac{3}{2} x y$.

To find $x y$, we first apply the cosine law in $\triangle A R C, \triangle A R S$ and $\triangle C R S$ to get
$$
\begin{aligned}
36 & =x^{2}+y^{2}-2 x y \cos 150^{\circ} \\
& =\left(3^{2}+1^{2}-2 \cdot 3 \cdot 1 \cdot \cos \angle R S A\right)+\left(3^{2}+1^{2}-2 \cdot 3 \cdot 1 \cdot \cos \angle R S C\right)+\sqrt{3} x y . \\
& =20+\sqrt{3} x y
\end{aligned}
$$
which gives (note that $\cos \angle R S A=-\cos \angle R S C$ ). It follows that $x y=\frac{16}{\sqrt{3}}$ and hence the answer is $\frac{3}{2}\left(\frac{16}{\sqrt{3}}\right)=8 \sqrt{3}$.