\[\boxed{\mathbf{555.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[куб,\ шар,\ цилиндр,\ конус;\]
\[S - равные;\]
\[h = d_{осн}.\]
\[Найти:\]
\[\text{V.}\]
\[Решение.\]
\[1)\ Пусть\ a - сторона\ куба;\ \]
\[R - радиус\ шара,\]
\[r_{1} - радиус\ основания\ \]
\[цилиндра\ и\ \ r_{2} - конуса;\ \]
\[h_{1} - высота\ цилиндра\ и\ \ \]
\[h_{2} - конуса\ (h = 2r).\]
\[2)\ Найдем\ сторону\ куба:\]
\[S_{куба} = S_{шара}\]
\[4a^{2} = 4\pi R^{2}\ \]
\[a = R\sqrt{\pi}.\]
\[3)\ Радиус\ основания\ \]
\[цилиндра:\]
\[S_{цилиндра} = S_{шара}\]
\[\ 2\pi r_{1}\left( h_{1} + r_{1} \right) = 4\pi R^{2}\]
\[r_{1}\left( 2r_{1} + r_{1} \right) = 2R^{2}\]
\[r_{1} = R\sqrt{\frac{2}{3}}.\]
\[4)\ Радиус\ основания\ конуса:\]
\[S_{конуса} = S_{шара}\ \]
\[\pi r_{2}\left( r_{2} + \sqrt{r_{2}^{2} + h_{2}^{2}} \right) = 4\pi R^{2}\]
\[r_{2}\left( r_{2} + \sqrt{r_{2}^{2} + 4r_{2}^{2}} \right) = 4R^{2}\]
\[6r_{2}^{2} = 4R^{2}\]
\[r_{2} = R\sqrt{\frac{2}{3}}\ \]
\[r_{1} = r_{2}.\]
\[5)\ Отношение\ объемов\ тел\ к\ \]
\[площадям\ их\ поверхностей:\]
\[\frac{V_{куба}}{S_{куба}} = \frac{a^{3}}{6\ a^{2}} = \frac{R^{3}\sqrt{\pi^{3}}}{6 \bullet R^{2}\pi} = \frac{R\sqrt{\pi}}{6};\]
\[\frac{V_{шара}}{S_{шара}} = \frac{\frac{4}{3}\pi R^{3}}{4\pi R^{2}} = \frac{4R}{3 \bullet 4} = \frac{R}{3};\]
\[\frac{V_{цил}}{S_{цил}} = \frac{\pi r_{1}^{2} \bullet h_{1}}{2\pi r_{1}\left( h_{1} + r_{1} \right)} =\]
\[= \frac{r_{1}^{2} \bullet 2r_{1}}{2r_{1} \bullet \left( 2r_{1} + r_{1} \right)} = \frac{r_{1}^{3}}{3r_{1}^{2}} = \frac{r_{1}}{3} =\]
\[= \frac{R\sqrt{\frac{2}{3}}}{3};\]
\[\frac{V_{конуса}}{S_{конуса}} = \frac{\frac{1}{3}\pi r_{2}^{2}h_{2}}{\pi r_{2}\left( r_{2} + \sqrt{r_{2}^{2} + h_{2}^{2}} \right)} =\]
\[= \frac{\frac{1}{3} \bullet r_{2}^{2} \bullet 2r_{2}}{r_{2}\left( r_{2} + \sqrt{r_{2}^{2} + 4r_{2}^{2}} \right)} = \frac{\frac{2}{3} \bullet r_{2}^{3}}{6 \bullet r_{2}^{2}} =\]
\[= \frac{2}{18} \bullet r_{2} = \frac{2}{18} \bullet R\sqrt{\frac{2}{3}} = \frac{R\sqrt{\frac{2}{3}}}{9}.\]
\[7)\ Объемы\ тел:\]
\[\frac{R}{3} > \frac{R\sqrt{\frac{2}{3}}}{3} > \frac{R\sqrt{\pi}}{6} > \frac{R\sqrt{\frac{2}{3}}}{9};\]
\[V_{шара} > V_{цил} > V_{куба} > V_{конуса}.\]
\[\mathbf{Отв}ет:\ \]
\[\ V_{шара} > V_{цил} > V_{куба} > V_{конуса}.\]