Решебник по геометрии 10 класс Атанасян ФГОС 787

\[\boxed{\mathbf{787.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[\mathrm{\Delta}ABC - правильный;\]

\[AS\bot ABC;\]

\[AS = AB = a.\]

\[Найти:\]

\[p(AB,SC);\]

\[\angle(AB,SC).\]

\[Решение.\]

\[1)\ Достроим\ \mathrm{\Delta}\text{ABC\ }до\ \]

\[параллелограмма\ ABCD:\ \]

\[DC \parallel AB;\ \]

\[\angle SCD - угол\ между\ \text{AB\ }и\ \text{SC.}\]

\[2)\ Построим\ AH\bot SCD:\ \]

\[p(AB,SCD) = AH.\]

\[3)\ AB \parallel DC\ \ и\ \ DC \in SCD:\]

\[CD \parallel AB.\]

\[4)\ AS = AB = BC = AD = a:\ \]

\[SB = SC = SD = a\sqrt{2}.\]

\[5)\ Построим\ SM\bot DC:\]

\[DM = MC = \frac{a}{2}.\]

\[Отсюда:\]

\[\cos\text{SCA} = \frac{\text{MC}}{\text{SC}} = \frac{a}{2}\ :a\sqrt{2} = \frac{\sqrt{2}}{4};\ \]

\[\angle(AB,SC) = \angle SCA = \arccos\frac{\sqrt{2}}{4}.\]

\[6)\ SM = \sqrt{SC^{2} - CM^{2}} =\]

\[= \sqrt{2a^{2} - \frac{a^{2}}{4}} = \frac{a\sqrt{7}}{2};\]

\[S_{\text{CSD}} = \frac{a}{2} \bullet \frac{a\sqrt{7}}{2} = \frac{a^{2}\sqrt{7}}{4};\ \]

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

\[7)\ AH - высота\ тетраэдра\ \]

\[ASCD;\]

\[AS - высота\ ASCD:\]

\[V_{\text{ASCD}} = \frac{1}{3}S_{\text{CSD}} \bullet AH = \frac{1}{3} \bullet S_{\text{ACD}} \bullet a\]

\[\frac{a^{2}\sqrt{7}}{4}AH = \frac{a^{2}\sqrt{3}}{4}a\]

\[p(AB,SC) = AH = \sqrt{\frac{3}{7}} \bullet a.\]

\[\mathbf{Ответ}\mathbf{:\ \ }\angle(AB,SC) = \arccos\frac{\sqrt{2}}{4};\ \ \]

\[p(AB,SC) = \sqrt{\frac{3}{7}} \bullet a.\]