Решебник по геометрии 11 класс. Атанасян ФГОС 537

\[\boxed{\mathbf{537.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[\text{ABC}A_{1}B_{1}C_{1} - усеченная\ \]

\[пирамида;\]

\[OH_{1} = OH;\]

\[MNK - сечение;\]

\[O \in MNK;\]

\[\frac{a}{b} = \frac{5}{2}.\]

\[Найти:\]

\[\frac{V_{1}}{V_{2}}.\]

\[Решение.\]

\[1)\ Пусть\ AC = a\ \ и\ \ A_{1}C_{1} = b:\]

\[\frac{\text{AC}}{A_{1}C_{1}} = \frac{a}{b} = \frac{5}{2}.\]

\[2)\ В\ трапеции\ A_{1}H_{1}HA:\ \ \]

\[MO\bot H_{1}H;\ \]

\[MO \parallel AH \parallel A_{1}H_{1}.\]

\[Отсюда:\ \]

\[OH_{1} = OH.\]

\[3)\ В\ трапеции\ AA_{1}CC_{1}:\ \ \]

\[MK - средняя\ линия\ \]

\[(MK \parallel AC \parallel A_{1}C_{1});\]

\[MK = \frac{AC + A_{1}C_{1}}{2} = \frac{a + b}{2} =\]

\[= \frac{a + \frac{2}{5}a}{2} = \frac{7}{10}\text{a.}\]

\[4)\ \mathrm{\Delta}ABC\sim\mathrm{\Delta}MKN\sim\mathrm{\Delta}A_{1}B_{1}C_{1}\ \]

\[(соответствующие\ стороны\ \]

\[лежат\ в\ параллельных\ \]

\[плоскостях).\]

\[5)\ \frac{S_{\text{ABC}}}{S_{A_{1}B_{1}C_{1}}} = \left( \frac{a}{b} \right)^{2} = \frac{25}{4}:\ \]

\[S_{A_{1}B_{1}C_{1}} = 0,16\ S_{\text{ABC}}.\]

\[\frac{S_{\text{ABC}}}{S_{\text{MNK}}} = \frac{AC^{2}}{MK^{2}} = a^{2} \bullet \frac{100}{49a^{2}} = \frac{100}{49}:\]

\[S_{\text{MKN}} = 0,49\ S_{\text{ABC}}.\]

\[\mathbf{Отв}ет:\ \ \frac{31}{73}.\]