ГДЗ по геометрии 11 класс. Атанасян ФГОС 103

\[\boxed{\mathbf{103.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[DABC - тетраэдр;\]

\[DM\ :\ MA = DN\ :\ NB =\]

\[= DP\ :\ PC;\]

\[S_{\text{ABC}} = 10\ см^{2};\]

\[DM\ :\ MA = 2\ :1.\]

\[Доказать:\]

\[MNP \parallel ABC.\]

\[Найти:\]

\[S_{\text{MNP}}.\]

\[Решение.\]

\[1)\frac{\text{DM}}{\text{MA}} = \frac{\text{DP}}{\text{PC}}:\ \]

\[\frac{DA = MA + MO}{DC = DP + PC\ \ };\]

\[\frac{\text{DM}}{AD - MP} = \frac{\text{DP}}{DC - DP}\ \]

\[или\]

\[\frac{AD - MD}{\text{MD}} = \frac{DC - DP}{\text{DP}}.\]

\[Отсюда:\ \]

\[\frac{\text{AD}}{\text{DM}} = \frac{\text{DC}}{\text{DP}}.\]

\[2)\ \mathrm{\Delta}CDA\sim\mathrm{\Delta}DPM:\]

\[\angle D - общий;\]

\[\frac{\text{AD}}{\text{DM}} = \frac{\text{DC}}{\text{DP}}.\]

\[Отсюда:\]

\[\angle DMP = \angle DAC;\ \]

\[\angle DPM = \angle DCA;\]

\[MP \parallel CA.\]

\[3)\ В\ треугольнике\ DCB:\ \]

\[NP \parallel CB.\]

\[4)\ MP \parallel CA\ и\ NP \parallel CN:\ \ \]

\[MNP \parallel ABC\ \]

\[(по\ двум\ пересекающимся\ прямым).\]

\[Что\ и\ требовалось\ доказать.\]

\[5)\ \mathrm{\Delta}MNP\sim\mathrm{\Delta}BCA - по\ двум\ \]

\[углам:\]

\[\frac{\text{MD}}{\text{AM}} = \frac{2}{1};\]

\[\frac{\text{MD}}{AD - MD} = \frac{2}{1};\]

\[\frac{AD - MD}{\text{MD}} = \frac{1}{2};\]

\[\frac{\text{AD}}{\text{DM}} - 1 = \frac{1}{2};\]

\[\frac{\text{AD}}{\text{DM}} = \frac{3}{2}.\]

\[6)\ \mathrm{\Delta}DAC\sim\mathrm{\Delta}MDP:\]

\[\frac{\text{AD}}{\text{MD}} = \frac{\text{AC}}{\text{MP}};\ \ \]

\[\frac{\text{AC}}{\text{MP}} = \frac{3}{2};\]

\[Площади\ подобных\ фигур\ \]

\[соотносятся,\ как\ квадрат\ \]

\[линейных\ размеров:\]

\[\frac{S_{\text{ABC}}}{S_{\text{MNP}}} = \left( \frac{\text{AC}}{\text{MP}} \right)^{2}.\]

\[7)\frac{10}{S_{\text{MNP}}} = \frac{9}{4}\]

\[S_{\text{MNP}} = 4\frac{4}{9}\ см^{2}.\]

\[Ответ:\ \ 4\frac{4}{9}\ см^{2}.\]