[tex]\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ V=457.17\\ h = 15 \end{cases}\implies 457.17=\cfrac{\pi r^2(15)}{3}\implies 457.17=5\pi r^2 \\\\\\ \cfrac{457.17}{5\pi }=r^2\implies \sqrt{\cfrac{457.17}{5\pi }}=r\implies 5.39 \approx r~\hfill \boxed{\stackrel{diameter = 2r}{2(5.39) = 10.78}}[/tex]
