Решебник по геометрии 11 класс. Атанасян ФГОС 503

\[\boxed{\mathbf{503.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Площадь\ шара\ равна\ S = 4\pi R^{2}:\ \]

\[R = \sqrt{\frac{S}{4\pi}}.\]

\[Объем\ шара\ равен\ V = \frac{4}{3}\pi R^{3}:\ \]

\[R = \sqrt[3]{\frac{3V}{4\pi}}.\]

\[\textbf{а)}\ R = 4\ см:\]

\[S = 4\pi \bullet 4^{2} = 4 \bullet 16 \bullet \pi =\]

\[= 64\pi\ см^{2};\]

\[V = \frac{4}{3}\pi \bullet 4^{3} = \frac{4 \bullet 64}{3}\pi =\]

\[= \frac{256}{3}\text{π\ }см^{3}.\]

\[\textbf{б)}\ V = 113,04\ см^{3}:\]

\[R = \sqrt[3]{\frac{3 \bullet 113,04}{4 \bullet \pi}} \approx \sqrt[3]{\frac{339,12}{12,56}} =\]

\[= \sqrt[3]{27} = 3\ см;\]

\[S = 4\pi \bullet 3^{2} = 4 \bullet 9 \bullet \pi = 36\pi\ см^{2}.\]

\[\textbf{в)}\ S = 64\pi\ см^{2}:\]

\[R = \sqrt{\frac{64\pi}{4\pi}} = \sqrt{\frac{64}{4}} = \sqrt{16} = 4\ см;\]

\[V = \frac{4}{3}\pi \bullet 4^{3} = \frac{4 \bullet 64}{3}\pi =\]

\[= \frac{256}{3}\text{π\ }см^{3}.\]