Решебник по геометрии 11 класс. Атанасян ФГОС 530

\[\boxed{\mathbf{530.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[ABCDM - правильная\ \]

\[пирамида;\]

\[\angle MAC = \varphi;\]

\[MO - высота;\]

\[EK = m.\]

\[Найти:\]

\[\text{V.}\]

\[Решение.\]

\[1)\ ABCDM - правильная\ \]

\[пирамида:\]

\[\text{AB} = \text{BC} = \text{CD} = \text{DA} = a;\]

\[2)\ E - середина\ отрезка\ AM;\ \]

\[EK\bot AO\ и\ EK = m;\]

\[E \in AM\ и\ K \in AO:\ \]

\[EK \in AMO.\]

\[3)\ \mathrm{\Delta}\ AMO - прямоугольный:\]

\[MO\bot AO;\ \ \ EK\bot MO;\ AE = EM;\ \]

\[EK - средняя\ линия.\]

\[Отсюда:\ \]

\[MO = 2m;\]

\[AK = \frac{m}{\text{tg\ φ}};\ \ \ \]

\[AO = 2AK = \frac{2m}{\text{tg\ φ}};\ \ \ \ \]

\[AC = 2AO = \frac{4m}{\text{tg\ φ}}.\]

\[4)\ \mathrm{\Delta}ABC - прямоугольный\ \]

\[(\angle B = 90{^\circ}):\]

\[AC = a\sqrt{2} = \frac{4m}{\text{tg\ φ}}\]

\[a = \frac{2\sqrt{2}m}{\text{tg\ φ}}.\]

\[5)\ V = \frac{1}{3} \bullet S_{осн} \bullet \frac{1}{3} \bullet a^{2} \bullet MO =\]

\[= \frac{1}{3} \bullet \left( \frac{2\sqrt{2}m}{\text{tg\ φ}} \right)^{2} \bullet 2m =\]

\[= \frac{2m}{3} \bullet \frac{8m^{2}}{tg^{2}\text{\ φ}} = \frac{16m^{3}}{3 \bullet tg^{2}\ \varphi\ }.\]

\[\mathbf{Отв}ет:\ \ V = \frac{16m^{3}}{3 \bullet tg^{2}\ \varphi\ }.\]