1.4. In a right-angled triangle, the bisector of an acute angle divides the opposite leg into segments of 4 and 5 cm. Determine the area of the triangle.
1.4. Let $B C=x$ (Fig. 1.3). Then $A B=\sqrt{81+x^{2}}$ and $\frac{x}{4}=\frac{\sqrt{81+x^{2}}}{5}$ (since $B D$ is the bisector). From this, we have $25 x^{2}=$ $=16\left(81+x^{2}\right)$, i.e., $x=12$ (cm). Therefore,

$S_{\triangle A B C}=\frac{1}{2} B C \cdot A C=\frac{1}{2} \cdot 12 \cdot 9=54\left(\mathrm{~cm}^{2}\right)$.

Answer: $54 \mathrm{~cm}^{2}$.