ГДЗ по геометрии 8 класс Атанасян Задание 1304

\[\boxed{\mathbf{1304.ОК\ ГДЗ - домашка\ н}а\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[OABC - тетраэдр;\]

\[все\ углы\ при\ O\ равны\ 90{^\circ}.\]

\[\mathbf{Доказать:}\]

\[S_{\text{ABC}}^{2} = S_{\text{AOB}}^{2} + S_{\text{BOC}}^{2} + S_{\text{AOC}}^{2}.\]

\[\mathbf{Доказательство.}\]

\[1)\ Пусть\ \text{OA} = a;\ \text{OB} = b;\ \]

\[\text{OC} = c:\]

\[S_{\text{AOB}} = \frac{1}{2}ab;\ \ \]

\[S_{\text{BOC}} = \frac{1}{2}bc;\ \ \]

\[S_{\text{AOC}} = \frac{1}{2}\text{ac.}\]

\[2)\ S_{\text{AOB}}^{2} + S_{\text{BOC}}^{2} + S_{\text{AOC}}^{2} =\]

\[= \frac{1}{4}\left( a^{2}b^{2} + b^{2}c^{2} + a^{2}c^{2} \right)\]

\[AB^{2} = a^{2} + b^{2};\ \]

\[BC^{2} = b^{2} + c^{2};\]

\[AC^{2} = a^{2} + c^{2}.\]

\[3)\ cosA = \frac{AB^{2} + AC^{2} - BC^{2}}{2 \cdot AB \cdot AC} =\]

\[= \frac{a^{2} + b^{2} + a^{2} + c^{2} - b^{2} - c^{2}}{2\sqrt{(a^{2} + b^{2})(a^{2} + c^{2})}} =\]

\[= \frac{a^{2}}{\sqrt{(a^{2} + b^{2})(a^{2}c^{2})}}\text{.\ }\]

\[= \frac{a^{4} + a^{2}b^{2} + b^{2}c^{2} + a^{2}c^{2} - a^{4}}{\left( a^{2} + b^{2} \right)\left( a^{2} + c^{2} \right)} =\]

\[= \frac{a^{2}b^{2} + b^{2}c^{2} + a^{2}c^{2}}{(a^{2} + b^{2})(a^{2} + c^{2})}.\]

\[5)\ S_{\text{ABC}}^{2} = \left( \frac{1}{2}AB \cdot AC \cdot sinA \right)^{2} =\]

\[= \frac{1}{4}\left( a^{2}b^{2} + b^{2}c^{2} + a^{2}c^{2} \right) =\]

\[= S_{\text{AOB}}^{2} + S_{\text{BOC}}^{2} + S_{\text{AOC}}^{2}.\]

\[Что\ и\ требовалось\ доказать.\]