\[\boxed{\mathbf{1226.ОК\ ГДЗ - домашка\ н}а\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\textbf{а)}\ Дано:\]
\[шар\ (O;R);\]
\[R = 4\ см.\]
\[Найти:\]
\[S,V - ?\]
\[Решение.\]
\[1)\ S = 4\pi R^{2} = 4\pi \bullet 4^{2} =\]
\[= 64\pi\ см^{2};\]
\[2)\ V = \frac{4}{3}\pi R^{3} = \frac{4}{3}\pi \bullet 4^{3} =\]
\[= \frac{256}{3}\text{π\ }см^{2}.\]
\[\textbf{б)}\ Дано:\]
\[шар\ (O;R);\]
\[V = 113,04\ см^{3}.\]
\[Найти:\]
\[R,S - ?\]
\[Решение.\]
\[1)\ V = \frac{4}{3}\pi R^{3}\]
\[R = \sqrt[3]{\frac{3V}{4\pi}} = \sqrt[3]{\frac{3 \bullet 113,04}{4 \bullet 3,14}} =\]
\[= \sqrt[3]{27} = 3\ см.\]
\[2)\ S = 4\pi R^{2} = 4\pi \bullet R^{2} =\]
\[= 4\pi \bullet 3^{2} = 36\pi\ см^{2}.\]
\[\textbf{в)}\ Дано:\]
\[шар\ (O;R);\]
\[S = 64\pi\ см^{2}.\]
\[Найти:\]
\[R,V - ?\]
\[Решение.\]
\[1)\ S = 4\pi R^{2}\]
\[R = \sqrt{\frac{S}{4\pi}} = \sqrt{\frac{64\pi}{4\pi}} = \sqrt{16} = 4\ см.\]
\[2)\ V = \frac{4}{3}\pi R^{3} = \frac{4}{3}\pi \bullet 4^{3} =\]
\[= \frac{256}{3}\text{π\ }см^{3}.\]
\[Ответ:а)\ S = 64\pi\ см^{2};\]
\[V = \frac{256}{3}\text{π\ }см^{3};\]
\[\textbf{б)}\ R = 3\ см;S = 36\pi\ см^{2};\]
\[\textbf{в)}\ R = 4\ см;V = \frac{256}{3}\ см^{3}.\]