## Task B-2.6.

Determine the lengths of the sides of a right-angled triangle with a perimeter of $24 \mathrm{~cm}$ and an inradius of $2 \mathrm{~cm}$.
## First Solution.

Let $a$ and $b$ be the legs of the right triangle, and $c$ the hypotenuse.

Then the radius of the inscribed circle is $r=\frac{a+b-c}{2}=2$.

From the system $\left\{\begin{array}{l}a+b+c=28 \\ a+b-c=4\end{array}\right.$ we get that $c=10 \text{ cm}$.

Applying the Pythagorean theorem, we have

$$
a^{2}+b^{2}=c^{2} \text { and } a+b-c=4, \text { i.e., } a+b=14
$$

so we need to solve the system

$$
\left\{\begin{array}{c}
a^{2}+b^{2}=100 \\
a+b=14
\end{array}\right.
$$

By substituting $b=14-a$ into the equation $a^{2}+b^{2}=100$, we reduce the system to the equation

$$
a^{2}+(14-a)^{2}=100
$$

This leads to $a^{2}-14 a+48=0$. The solutions to this equation are $a_{1}=6, a_{2}=8$.

Then $b_{1}=8, b_{2}=6$.

Thus, the lengths of the sides of the given triangle are $6 \text{ cm}, 8 \text{ cm}, 10 \text{ cm}$.