Macedonian Mathematical Competitions

From the condition in the problem we have that $m = \frac{2}{7}L$ where $L$ is the perimeter of the trapeze and $m$ the length of its middle line. We have $m = \frac{a+b}{2} = \frac{18+10}{2} = 14\,\text{cm}$ so $L = m \cdot \frac{7}{2} = 14 \cdot \frac{7}{2} = 49\,\text{cm}$. Because the trapeze is isosceles $L = a+b+2c$ and we have that $c = \frac{L-a-b}{2} = \frac{49-18-10}{2} = 10.5\,\text{cm}$.