Решебник по геометрии 9 класс Атанасян ФГОС Задание 1215

 

\[\boxed{\mathbf{1215.ОК\ ГДЗ - домашка\ на\ 5}}\]

\[\mathbf{Да}но:\]

\[в\ цилиндр\ вписана\ правильная\ \]

\[n - угольная\ призма.\]

\[Найти:\]

\[отношение\ объемов\ призмы\ и\ \]

\[цилиндра.\]

\[Решение.\]

\[\textbf{а)}\ n = 3:\]

\[V_{п} = \frac{3hr^{2}}{2} \bullet \sin{120{^\circ}} = \frac{3\sqrt{3} \bullet hr^{2}}{4};\]

\[\frac{V_{п}}{V_{ц}} = \frac{3\sqrt{3} \bullet hr^{2}}{4 \bullet \pi r^{2}h} = \frac{3\sqrt{3}}{4\pi}.\]

\[\textbf{б)}\ n = 4:\]

\[V_{п} = \frac{4hr^{2}}{2} \bullet \sin{90{^\circ}} = 2hr^{2};\]

\[\frac{V_{п}}{V_{ц}} = \frac{2hr^{2}}{\pi r^{2}h} = \frac{2}{\pi}.\]

\[\textbf{в)}\ n = 6:\]

\[V_{п} = \frac{6hr^{2}}{2} \bullet \sin{60{^\circ}} = \frac{3\sqrt{3} \bullet hr^{2}}{2};\]

\[\frac{V_{п}}{V_{ц}} = \frac{3\sqrt{3} \bullet hr^{2}}{2 \bullet \pi r^{2}h} = \frac{3\sqrt{3}}{2\pi}.\]

\[\textbf{г)}\ n = 8:\]

\[V_{п} = \frac{8hr^{2}}{2} \bullet \sin{45{^\circ}} = 2\sqrt{2} \bullet hr^{2};\]

\[\frac{V_{п}}{V_{ц}} = \frac{2\sqrt{2} \bullet hr^{2}}{\pi r^{2}h} = \frac{2\sqrt{2}}{\pi}.\]

\[\textbf{д)}\ \frac{V_{п}}{V_{ц}} = \frac{\text{nh}r^{2}}{2\pi r^{2}h} \bullet \sin\frac{360{^\circ}}{n} =\]

\[= \frac{n}{2\pi} \bullet \sin\frac{360{^\circ}}{n}.\]

\[Ответ:а)\ \frac{3\sqrt{3}}{4\pi};б)\ \frac{2}{\pi};в)\ \frac{3\sqrt{3}}{2\pi};\]

\[\textbf{г)}\ \frac{2\sqrt{2}}{\pi};д)\ \frac{n}{2\pi} \bullet \sin\frac{360{^\circ}}{n}.\]

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