Решебник по геометрии 10 класс Атанасян ФГОС 803

\[\boxed{\mathbf{803.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\ \ \]

\[\text{ABC}D_{1} - тетраэдр;\ \ \]

\[AB = a;\ \ \]

\[CD_{1} = b;\ \ \]

\[p(AB,CD_{1}\ ) = c;\ \]

\[\angle\left( AB,CD_{1} \right) = \varphi.\]

\[Доказать:\ \text{\ \ }\]

\[V_{тетр} = \frac{1}{6}abc \bullet \sin\varphi.\]

\[Доказательство.\]

\[1)\ Достроим\ тетраэдр\ до\ \]

\[параллелепипеда\ \]

\[\text{ABCD}A_{1}B_{1}C_{1}D_{1}:\]

\[\text{AB}B_{1}A_{1} \parallel CD_{1};\ \ \]

\[\text{BC}C_{1}B_{1} \parallel AD_{1};\ \ \]

\[\text{CD}D_{1}C_{1} \parallel AB;\ \ \]

\[\text{AD}D_{1}A_{1} \parallel BC;\ \ \]

\[A_{1}B_{1}C_{1}D_{1} \parallel ABCD\ \ \]

\[(по\ свойству\ параллелепипеда).\]

\[2)\ Таким\ образом:\ \ \]

\[\angle A_{1}BA = \angle\left( AB,CD_{1} \right) = \varphi;\]

\[AD = p\left( AB,CD_{1}\ \right) = c.\]

\[3)\ A_{1}B - диагональ\ \]

\[прямоугольника\ AA_{1}BB_{1}:\]

\[S_{\text{AB}B_{1}A_{1}} = 2S_{\text{AB}A_{1}} =\]

\[= 2 \bullet \left( \frac{1}{2} \bullet AB \bullet BA_{1} \bullet \sin{\angle ABA_{1}} \right) =\]

\[= ab \bullet \sin\varphi;\]

\[V_{парал} = S_{\text{AB}B_{1}A_{1}} \bullet AD =\]

\[= abc \bullet \sin\varphi.\]

\[4)\ \frac{V_{парал}}{V_{тетр}} = \frac{S_{\text{ABCD}} \bullet DD_{1}}{\frac{1}{3} \bullet S_{\text{ABC}} \bullet DD_{1}} =\]

\[= \frac{2 \bullet S_{\text{ABC}}}{\frac{1}{3} \bullet S_{\text{ABC}}} = 6:\]

\[V_{тетр} = \frac{1}{6}V_{парал} = \frac{1}{6}abc \bullet \sin\varphi.\]

\[Что\ и\ требовалось\ доказать.\]