30. In $\triangle A B C \angle B A C=45^{\circ} . D$ is a point on $B C$ such that $A D$ is perpendicular to $B C$. If $B D=3$ $\mathrm{cm}$ and $D C=2 \mathrm{~cm}$, and the area of the $\triangle A B C$ is $x \mathrm{~cm}^{2}$. Find the value of $x$.
30. Ans: 15 .
Construct $C E$ which is perpendicular to $A B$ where $E$ is on $A B$. Since $\angle B A C=45^{\circ}$ (given), $A E=C E$. Thus the right $\triangle A E H$ is congruent to right $\triangle C E B$ So $A H=C B=5$.
Next $\triangle A D B$ is similar to $\triangle C D H$, thus
$$
\frac{B D}{A D}=\frac{H D}{C D}=\frac{A D-A H}{C D} \text { which implies } \frac{3}{A D}=\frac{A D-5}{2} .
$$
we get $A D^{2}-5 A D-6=0$ Solving $A D=6$ only as $A D>0$. The result follows.
Note, if one is familiar with trigonometry, one may have an alternative solution as follows: Let $\alpha$ and $\beta$ denote $\angle B A D$ and $\angle D A C$ respectively. Using $\tan (\alpha+\beta)=\tan 45^{\circ}=1$, one get
$$
\frac{\frac{3}{A D}+2 A D}{1-\frac{3}{A D} \frac{2}{A D}}=1,
$$
consequently $A D=6$.