Remember that the formula for the lenght of an arc is:
[tex]L=\theta\cdot(\frac{\pi}{180})\cdot r[/tex]
Where theta is the central angle in degrees.
Solving this formula for the radius r :
[tex]\begin{gathered} L=\theta\cdot(\frac{\pi}{180})\cdot r\rightarrow\frac{L}{\theta}=\frac{\pi}{180}\cdot r\rightarrow\frac{L\cdot180}{\theta}=\pi\cdot r \\ \\ \Rightarrow r=\frac{L\cdot180}{\theta\cdot\pi} \end{gathered}[/tex]
Plugging in the data given,
[tex]\begin{gathered} r=\frac{L\cdot180}{\theta\cdot\pi}\rightarrow r=\frac{(7.34)\cdot180}{(81)\cdot\pi} \\ \\ \Rightarrow r=5.19 \end{gathered}[/tex]
Therefore, we can conclude that the arpoximate radus
