12. $\triangle A B C$ is a triangle such that $\angle C=90^{\circ}$. Suppose $A C=156 \mathrm{~cm}, A B=169 \mathrm{~cm}$ and the perpendicular distance from $C$ to $A B$ is $x \mathrm{~cm}$. Find the value of $x$.
12. Ans: 60 By Pythagoras' Theorem, $B C=\sqrt{169^{2}-156^{2}}=65$. Let the perpendicular from $C$ to $A B$ meet $A B$ at $D$. Then $\triangle A B C \sim \triangle A C D$. Thus,
$$
\frac{C D}{A C}=\frac{B C}{A B} \Longrightarrow \frac{x}{156}=\frac{65}{169} \Longrightarrow x=60 .
$$