What are the sides of a right-angled triangle, given that its area is $37.5 \mathrm{~m}^{2}$ and its sides form an arithmetic progression?
Let the sides of the triangle be $a-d, a, a+d$; then

$$
(a+d)^{2}=(a-d)^{2}+a^{2}
$$

from which

$$
d=\frac{a}{4}
$$

Thus, the area of the triangle is:

$$
t=\frac{3 a}{4} \cdot \frac{a}{2}
$$

from which

$$
a=\sqrt{\frac{8 t}{3}}=\sqrt{\frac{37.5 \cdot 8}{3}}=10
$$

Therefore, $d=2.5$ and the sides of the triangle are: $7.5 \text{ m}, 10 \text{ m}, 12.5 \text{ m}$.

(Lajos Pollák, Kaposvár.)

The problem was also solved by: Appel S., Barcsay Á., Bayer B., Bender E., Benedek Zs., Burján K., Csete F.A., Czank K., Fekete S., Fleischer F., Freibauer E., Grosz K., Kerekes T., Kiss A., Kornis F., Krausz B., Krisztián Gy., Kohn b., Lindtner M., Lukhaub Gy., Lupsa Gy., Obláth R., Ovenden S., Perl Gy., Sasvári G., Schuller I., Spitzer V., Stern D., Szabó J., Szentiványi L., Szibelth S., Szőllősy J., Vajda Ö., Weisz J.