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Answer :
[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]
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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