To solve the exercise we need to know the perimeter of the circle
[tex]\begin{gathered} L_c=2\cdot\pi\cdot r \\ d=2\cdot r \\ L_c=\pi\cdot d \end{gathered}[/tex]
From the exercise we know that there are 2 lengths of 82m and the diameter of the semicircle is 50, so we generate our equation. Since we have 2 semicircles with the same diameter, then the corridor is as if it ran a complete circle.
[tex]\begin{gathered} L_f\to\text{Lenght of the training field} \\ d\to\text{diameter} \\ l\to\text{length of the rectangle} \\ L_c\to\text{perimeter of the circle} \end{gathered}[/tex][tex]L_f=2l+L_c[/tex][tex]\begin{gathered} l=82m \\ d=50m \\ \pi=3.14 \\ L_c=50\mleft(3.14\mright)m \\ L_c=157m \end{gathered}[/tex][tex]\begin{gathered} L_f=2(82)+157 \\ L_f=164+157 \\ L_f=321m \end{gathered}[/tex]
The runner will run 321.0 m in the training field
