Respuesta :

[tex]\bf \textit{area of a sector of a circle}\\\\ A=\cfrac{\theta\pi r^2}{360}\implies \sqrt{\cfrac{360A}{\theta\pi }}=r\quad \begin{cases} r=radius\\ \theta=\textit{angle in degrees}\\ ----------\\ \theta=60\\ A=25\pi \end{cases} \\\\\\ \sqrt{\cfrac{360\cdot 25\pi }{60\pi }}=r\implies \sqrt{150}=r\implies 5\sqrt{6}=\boxed{r}\\\\ -----------------------------\\\\ \textit{circumference of a circle}\\\\ C=2\pi r\implies C=2\pi \boxed{5\sqrt{6}}\implies C=10\pi \sqrt{6}[/tex]
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