A right-angled triangle has one leg of $15 \mathrm{~cm}$ and the height corresponding to the hypotenuse is $9 \mathrm{~cm}$; what is the area of the triangle?
Let $AC$ be the hypotenuse of the right triangle $ABC$, and $BD$ the altitude to the hypotenuse. If $BC=15 \text{~cm}$, then

$$
DC=\sqrt{\overline{BC}^{2}-\overline{BD}^{2}}=12 \text{~cm}
$$

Since

$$
AD=\frac{\overline{BC}^{2}}{DC}=6.75 \text{~cm}
$$

therefore $AC=18.75 \text{~cm},$ and thus the sought area is

$$
\frac{AC \times BD}{2}=84.375 \text{~cm}^{2}
$$

(Imre Kürti, Eger.)

Number of solutions: 25.