Решебник по геометрии 9 класс Атанасян ФГОС Задание 1200

\[\boxed{\mathbf{1200.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[\mathbf{Дано:}\]

\[n - угольная\ правильная\ \]

\[призма;\]

\[\textbf{а)}\ n = 3;\]

\[\textbf{б)}\ n = 4;\]

\[\textbf{в)}\ n = 6;\]

\[\textbf{г)}\ n = 8.\]

\[\mathbf{Найти:}\]

\[C - ?\]

\[\mathbf{Решение.}\]

\[\textbf{а)}\ V = S_{3} \bullet a =\]

\[= \frac{1}{2} \bullet a \bullet a^{2} \bullet \sin{60{^\circ}} = \frac{a^{3}\sqrt{3}}{4}.\]

\[\textbf{б)}\ V = S_{2} \bullet a = a^{2} \bullet a = a^{3}.\]

\[\textbf{в)}\ V = S_{6} \bullet a =\]

\[= a\left( \frac{1}{2} \bullet 6a^{2} \bullet \sin{60{^\circ}} \right) = \frac{3a^{3}\sqrt{3}}{2}.\]

\[\textbf{г)}\ V = S_{8} \bullet a =\]

\[= 8 \bullet \frac{\sqrt{2}}{4\left( 2 - \sqrt{2} \right)}a^{3} = \frac{2\sqrt{2}}{2 - \sqrt{2}}a^{3} =\]

\[= \frac{2}{\sqrt{2 - 1}}a^{3} =\]

\[= \frac{2\left( \sqrt{2} + 1 \right)}{\left( \sqrt{2} - 1 \right)\left( \sqrt{2} + 1 \right)}a^{3} =\]

\[= 2\left( \sqrt{2} + 1 \right)a^{3}.\]

\[Ответ:а)\frac{a^{3}\sqrt{3}}{4};б)\ a^{3};\]

\[\textbf{в)}\ \frac{3a^{3}\sqrt{3}}{2};г)\ 2\left( \sqrt{2} + 1 \right)a^{3}.\]