The perimeter of a certain right-angled triangle is $30 \mathrm{~cm}$, the height corresponding to the hypotenuse is $m_{c}=6 \mathrm{~cm}$. What are the lengths of the sides of the triangle?
Let the legs be $a$ and $b$, and the hypotenuse $c$. According to the problem,

\[
\begin{gathered}
a+b+c=30 \\
a^{2}+b^{2}=c^{2}
\end{gathered}
\]

and the double area of the triangle is

\[
a b=m_{c} \cdot c=6 c
\]

From (1), (2), and (3),

\[
30-c=a+b=\sqrt{a^{2}+b^{2}+2 a b}=\sqrt{c^{2}+12 c}
\]

thus,

\[
\begin{gathered}
900-60 c+c^{2}=c^{2}+12 c \\
72 c=900
\end{gathered}
\]

from which

\[
c=\frac{900}{72}=\frac{25}{2}=12.5
\]

(We verify that the value $c=12.5$ also satisfies the square root equation: $\sqrt{156.25+150}=\sqrt{306.25}=17.5=30-12.5$)

From (1) and (3),

\[
\begin{aligned}
a+b & =17.5 \\
a b & =75
\end{aligned}
\]

Subtracting 4 times (5) from the square of (4):

\[
(a-b)^{2}=306.25-300=6.25
\]

from which

\[
a-b= \pm 2.5
\]

From (4) and (6),

\[
a_{1}=b_{2}=10 \text { and } a_{2}=b_{1}=7.5
\]

Thus, the sides of the triangle are $10 \text{ cm}, 7.5 \text{ cm}$, and $12.5 \text{ cm}$.

László Kovács (Debrecen, Ref. koll. g. II. o. t.)