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МатематикаГДЗ по геометрии 7 класс Атанасян ФГОС Задание 830
Задание 830
\[\boxed{\mathbf{830.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[M \in AC;K \in BC;\]
\[O = AK \cap BM;\]
\[S_{\text{OMA}} = S_{1};\ \]
\[S_{\text{OAB}} = S_{2};\]
\[S_{\text{OBK}} = S_{3}.\]
\[\mathbf{Найти:}\]
\[S_{\text{CMK}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ Проведем\ несколько\ \]
\[перепендикуляров:\]
\[\text{OD}\bot AC;\ \ BE\bot AC;KF\bot AC;\]
\[OG\bot BC;\ \ AP\bot BC;MH\bot BC.\]
\[2)\ Площади\ треугольников:\]
\[\ S_{\text{MOK}} = S_{4};\ \ S_{\text{ABC}} = S.\ \]
\[3)\ S_{\text{CMK}} =\]
\[= S - \left( S_{1} + S_{2} + S_{3} + S_{4} \right).\]
\[Преобразуем:\]
\[\frac{S_{\text{AMK}}}{S_{\text{AMB}}} = \frac{S_{1} + S_{4}}{S_{1} + S_{2}} = \frac{\text{KF}}{\text{BE}} =\]
\[= \frac{\text{KF}}{\text{BE}} \bullet \frac{\text{MC}}{\text{MC}} = \frac{S_{\text{CMK}}}{S - \left( S_{1} + S_{2} \right)} =\]
\[= \frac{S_{\text{CMK}}}{S_{3} + S_{4} + S_{\text{CMK}}}.\]
\[\frac{S_{3}}{S_{2}} = \frac{\text{OK}}{\text{OA}} = \frac{S_{4}}{S_{1}};\ \ \ \]
\[S_{4} = \frac{S_{1}S_{3}}{S_{2}}.\]
\[Решим\ уравнение\ \]
\[относительно\ S_{\text{CMK}}:\]
\[\left( S_{1} + \frac{S_{1}S_{3}}{S_{2}} \right)\left( S_{3} + \frac{S_{1}S_{3}}{S_{2}} + S_{\text{CMK}} \right) =\]
\[= \left( S_{1} + S_{2} \right) \bullet S_{\text{CMK}}\]
\[\left( S_{1} + \frac{S_{1}S_{3}}{S_{2}} \right)\left( S_{3} + \frac{S_{1}S_{3}}{S_{2}} \right) =\]
\[= (S_{1} + S_{2} - S_{1}) \bullet \left( 1 + \frac{S_{3}}{S_{2}} \right)S_{\text{CMK}}\]
\[S_{\text{CMK}} = \frac{\left( S_{1} + \frac{S_{1}S_{3}}{S_{2}} \right)\left( S_{3} + \frac{S_{1}S_{3}}{S_{2}} \right)}{S_{1} + S_{2} - S_{1}\left( 1 + \frac{S_{3}}{S_{2}} \right)} =\]
\[= \frac{S_{1}S_{3}\frac{\left( S_{2} + S_{3} \right)}{S_{2}^{2}}}{\frac{S_{2}\left( S_{1} + S_{2} \right) - S_{1}\left( S_{2} + S_{1} \right)}{S_{2}}} =\]
\[= \frac{S_{1}S_{3}\left( S_{1} + S_{2} \right)\left( S_{2} + S_{3} \right)}{S_{2}\left( S_{1}S_{2} + S_{2}^{2} - S_{1}S_{2} - S_{1}S_{3} \right)} =\]
\[= \frac{S_{1}S_{3}\left( S_{1} + S_{2} \right)\left( S_{2} + S_{3} \right)}{S_{2}(S_{2}^{2} - S_{1}S_{3})}.\]
\[Ответ:\]
\[S_{\text{CMK}} = \frac{S_{1}S_{3}\left( S_{1} + S_{2} \right)\left( S_{2} + S_{3} \right)}{S_{2}(S_{2}^{2} - S_{1}S_{3})}.\]
\[\boxed{\mathbf{830.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[AA_{1};BB_{1};CC_{1} - медианы.\]
\[\mathbf{Доказать:}\]
\[\frac{P}{2} < AA_{1} + BB_{1} + CC_{1} < P.\]
\[\mathbf{Доказательство.}\]
\[1)\ ABCD - параллелограмм:\ \]
\[AA_{1} = A_{1}D = \frac{1}{2}\text{AD.}\]
\[2)\ Из\ неравенства\ \]
\[треугольника\ \]
\[(при\ условии,\ что\ BD = AC):\]
\[AD < AB + BD;\ \ \ \]
\[AA_{1} < \frac{AB + AC}{2}\text{.\ }\]
\[3)\ Рассуждая\ аналогично:\]
\[BB_{1} < \frac{AB + BC}{2};\ \ \ \ \]
\[CC_{1} < \frac{AC + BC}{2}.\]
\[4)\ Складываем\ неравенства:\]
\[AA_{1} + BB_{1} + CC_{1} <\]
\[< \frac{AB + AC}{2} + \frac{AB + BC}{2} + \frac{AC + BC}{2} =\]
\[= AB + BC + AC = P.\]
\[5)\ Из\ неравенства\ \]
\[треугольников\ (\text{AB}A_{1}\ и\ \text{AC}A_{1}):\ \]
\[AA_{1} + A_{1}B > AB;\ \ \]
\[AA_{1} + A_{1}C > AC.\]
\[Следовательно:\]
\[2AA_{1} + A_{1}B + A_{1}C > AB + AC\]
\[AA_{1} > \frac{AB + BC - BC}{2}.\]
\[6)\ Аналогичным\ образом:\]
\[BB_{1} > \frac{AB + BC - AC}{2};\ \ \ \]
\[CC_{1} > \frac{AC + BC - AC}{2}.\]
\[7)\ Складываем\ три\ последних\ \]
\[неравенства:\]
\[AA_{1} + BB_{1} + CC_{1} <\]
\[8)\ Получаем:\]
\[\frac{P}{2} < AA_{1} + BB_{1} + CC_{1} < P.\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]
