\[\boxed{\mathbf{1304.ОК\ ГДЗ - домашка\ н}а\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[OABC - тетраэдр;\]
\[все\ углы\ при\ O\ равны\ 90{^\circ}.\]
\[\mathbf{Доказать:}\]
\[S_{\text{ABC}}^{2} = S_{\text{AOB}}^{2} + S_{\text{BOC}}^{2} + S_{\text{AOC}}^{2}.\]
\[\mathbf{Доказательство.}\]
\[1)\ Пусть\ \text{OA} = a;\ \text{OB} = b;\ \]
\[\text{OC} = c:\]
\[S_{\text{AOB}} = \frac{1}{2}ab;\ \ \]
\[S_{\text{BOC}} = \frac{1}{2}bc;\ \ \]
\[S_{\text{AOC}} = \frac{1}{2}\text{ac.}\]
\[2)\ S_{\text{AOB}}^{2} + S_{\text{BOC}}^{2} + S_{\text{AOC}}^{2} =\]
\[= \frac{1}{4}\left( a^{2}b^{2} + b^{2}c^{2} + a^{2}c^{2} \right)\]
\[AB^{2} = a^{2} + b^{2};\ \]
\[BC^{2} = b^{2} + c^{2};\]
\[AC^{2} = a^{2} + c^{2}.\]
\[3)\ cosA = \frac{AB^{2} + AC^{2} - BC^{2}}{2 \cdot AB \cdot AC} =\]
\[= \frac{a^{2} + b^{2} + a^{2} + c^{2} - b^{2} - c^{2}}{2\sqrt{(a^{2} + b^{2})(a^{2} + c^{2})}} =\]
\[= \frac{a^{2}}{\sqrt{(a^{2} + b^{2})(a^{2}c^{2})}}\text{.\ }\]
\[= \frac{a^{4} + a^{2}b^{2} + b^{2}c^{2} + a^{2}c^{2} - a^{4}}{\left( a^{2} + b^{2} \right)\left( a^{2} + c^{2} \right)} =\]
\[= \frac{a^{2}b^{2} + b^{2}c^{2} + a^{2}c^{2}}{(a^{2} + b^{2})(a^{2} + c^{2})}.\]
\[5)\ S_{\text{ABC}}^{2} = \left( \frac{1}{2}AB \cdot AC \cdot sinA \right)^{2} =\]
\[= \frac{1}{4}\left( a^{2}b^{2} + b^{2}c^{2} + a^{2}c^{2} \right) =\]
\[= S_{\text{AOB}}^{2} + S_{\text{BOC}}^{2} + S_{\text{AOC}}^{2}.\]
\[Что\ и\ требовалось\ доказать.\]
\[\boxed{\mathbf{1304.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC\sim\mathrm{\Delta}A_{1}B_{1}C_{1};\]
\[\frac{\text{BC}}{B_{1}C_{1}} = k;\]
\[BC = 1,4\ м;\]
\[B_{1}C_{1} = 56\ см.\]
\[\mathbf{Найти:}\]
\[\frac{P_{\text{ABC}}}{P_{A_{1}B_{1}C_{1}}} - ?\]
\[\mathbf{Решение.}\]
\[1)\frac{\text{BC}}{B_{1}C_{1}} = \frac{140}{56} = \frac{5}{2};\]
\[k = 2,5.\]
\[2)\ \mathrm{\Delta}ABC\sim\mathrm{\Delta}A_{1}B_{1}C_{1}\ (по\ условию):\]
\[\frac{P_{\text{ABC}}}{P_{A_{1}B_{1}C_{1}}} = k\]
\[\frac{P_{\text{ABC}}}{P_{A_{1}B_{1}C_{1}}} = \frac{5}{2}.\]
\[\mathbf{Ответ:}5\ :2\mathbf{.}\]