Respuesta :

that is the circumference of its circular base.

[tex]\textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ h=9\\ V=27\pi \end{cases}\implies 27\pi =\cfrac{\pi r^2 (9)}{3}\implies 27\pi =3\pi r^2 \\\\\\ \cfrac{27\pi }{3\pi }=r^2\implies 9=r^2\implies \sqrt{9}=r\implies \underline{3=r} \\\\[-0.35em] ~\dotfill\\\\ \textit{circumference of a circle}\\\\ C=2\pi r\qquad \qquad \qquad C=2\pi (\underline{3})\implies C=6\pi[/tex]

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