Respuesta :

[tex]\bf \begin{array}{llll} \textit{surface area of a sphere}\\\\ A=4\pi r^2 \\\\\\ \textit{a hemisphere is half that}\\\\ A=\cfrac{4\pi r^2}{2}\implies A=2\pi r^2 \end{array}\qquad r=radius \\\\\\ \textit{now, we know the area is }256\pi \implies 256\pi =2\pi r^2 \\\\\\ \cfrac{256\pi }{2\pi }=r^2\implies \sqrt{128}=r\implies \boxed{8\sqrt{2}=r} \\\\ -----------------------------\\\\[/tex]

[tex]\bf \textit{volume of a sphere}\\\\ V=\cfrac{4}{3}\pi r^3\qquad r=radius \\\\\\ \textit{a hemisphere is half that}\\\\ V=\cfrac{\frac{4}{3}\pi r^3}{2}\implies V=\cfrac{\frac{4\pi r^3}{3}}{\frac{2}{1}}\implies V=\cfrac{4\pi r^3}{3}\cdot \cfrac{1}{2} \\\\\\ V=\cfrac{2\pi r^3}{3} \\\\\\ \textit{now, we know the radius is }8\sqrt{2}\implies V=\cfrac{2\pi (8\sqrt{2})^3}{3} \\\\\\ V=\cfrac{2\pi (8^3\sqrt{2^3})}{3}\implies V=\cfrac{2\pi \cdot 512\cdot 2\sqrt{2}}{3} \\\\\\ \boxed{V=\cfrac{2048\pi \sqrt{2}}{3}}[/tex]