7. In $\triangle A B C, \tan \angle C A B=\frac{22}{7}$ and the altitude from $A$ to $B C$ divides $B C$ into segments of lengths 3 and 17. Find the area of $\triangle A B C$.
(1 mark)
在 $\triangle A B C$ 中, $\tan \angle C A B=\frac{22}{7}$, 且從 $A$ 到 $B C$ 的高把 $B C$ 分成長度 3 和 17 的兩段。求 $\triangle A B C$ 的面積。
（１分）
7. 110
7. Let $D$ be the foot of perpendicular from $A$ to $B C$ and let $A D=x$. Using the compound angle formula, we have
$$
\begin{aligned}
\frac{22}{7} & =\tan \angle C A B \\
& =\frac{\tan \angle B A D+\tan \angle C A D}{1-\tan \angle B A D \tan \angle C A D} \\
& =\frac{\frac{3}{x}+\frac{17}{x}}{1-\frac{3}{x} \times \frac{17}{x}}=\frac{20 x}{x^{2}-51}
\end{aligned}
$$

After simplification, we get $11 x^{2}-70 x-561=0$, i.e. $(x-11)(11 x+51)=0$. The only positive solution is $x=11$. Thus the area of $\triangle A B C$ is $\frac{(3+17) \times 11}{2}=110$.