## Task 1 - 060621

A distance of $20 \mathrm{~m}$ is divided into three sections. The first section is twice as long as the second, and the length of the third section is three times the length of the first section. Calculate the lengths of the individual sections!
If the length of the second segment is denoted by $a$, then the first segment has a length of $2a$ and the third segment has a length of $6a$. The total length of the entire segment is therefore

$$
a+2a+6a=9a
$$

According to the problem, this is $20 \mathrm{~m}$. Therefore, the length of the second segment is $20:9=\frac{20}{9}$, the length of the first segment is $\frac{40}{9}$, and the length of the third segment is $\frac{40}{3}$.

The lengths of the individual segments are thus $2 \frac{2}{9} \mathrm{~m}, 4 \frac{4}{9} \mathrm{~m}$, and $13 \frac{1}{3} \mathrm{~m}$.