Task B-1.3. Master Zvonko cut a needle of length $30 \mathrm{~cm}$ into three parts to make a right-angled triangle from those parts. Zvonko first cut the needle into two parts, one of which was $6 \mathrm{~cm}$ shorter, and he took it as one of the legs. Determine the lengths of all three parts into which Zvonko cut the needle.
## Solution.

Let's denote the parts of the arrow with $a, b, c$. Let $a$ be the leg that Zvonko obtained after the first cut. Then we have

$$
\begin{gathered}
a+b+c=30 \\
b+c=a+6 \\
2 a+6=30, \\
a=12 \\
b+c=18 \\
a^{2}+b^{2}=c^{2} \\
12^{2}+(18-c)^{2}=c^{2} \\
c=13 \\
\quad b=5
\end{gathered}
$$

The parts of the arrow are $5 \mathrm{~cm}, 12 \mathrm{~cm}, 13 \mathrm{~cm}$.