Consider a right-angled triangle with an area of $t=84 \mathrm{~cm}^{2}$ and a perimeter of $k=56 \mathrm{~cm}$. Calculate the lengths of the sides without using the Pythagorean theorem.
I. Solution: Let the legs be $a, b$, and the hypotenuse $c$. It is known that

$$
\operatorname{tg} \frac{\gamma}{2}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}=\sqrt{\frac{s(s-a)(s-b)(s-c)}{s^{2}(s-c)^{2}}}=\frac{t}{s(s-c)}
$$

Since in this case $\gamma=90^{\circ}, t=84, s=28$, then

$$
28(28-c)=84, \quad \text { or } \quad 28-c=3
$$

from which

$$
c=25 \text{ cm}
$$

and thus

$$
a+b=56-c=31
$$

Considering that according to the problem $a b=168$, then $a$ and $b$ are the roots of the following quadratic equation:

$$
x^{2}-31 x+168=0
$$

from which

$$
a_{1}=b_{2}=7, \quad a_{2}=b_{1}=24
$$

Gémesi Gabriella (Bp. VIII., Ságvári lg. II. o. t.)

II. Solution: Let the radius of the inscribed circle of the right triangle be $\varrho$, then from the known $t=\varrho s$ relationship

$$
\varrho=\frac{t}{s}=\frac{84}{28}=3
$$

Furthermore

$$
c=a-\varrho+b-\varrho=2 s-c-2 \varrho
$$

from which

$$
c=s-\varrho=28-3=25
$$

The continuation of the solution follows the I. solution.

Rátkay Zsolt (Bp. VI., Kölcsey g. II. o. t,)