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МатематикаГДЗ по геометрии 7 класс Атанасян ФГОС Задание 831
Задание 831
\[\boxed{\mathbf{831.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[M \in AC;K \in BC;\]
\[P \in MK;\ \]
\[\frac{\text{AM}}{\text{MC}} = \frac{\text{CK}}{\text{KB}} = \frac{\text{MP}}{\text{PK}};\]
\[S_{\text{AMP}} = S_{1};\]
\[S_{\text{BKP}} = S_{2}.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABC}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ Допустим:\ \]
\[\ \frac{\text{AM}}{\text{MC}} = \frac{\text{CK}}{\text{KB}} = \frac{\text{MP}}{\text{PK}} = k.\ \]
\[Тогда:\]
\[\frac{S_{\text{AMP}}}{S_{\text{APK}}} = \frac{\text{MP}}{\text{PK}} = k = \frac{S_{1}}{S_{\text{APK}}}\]
\[S_{\text{APK}} = \frac{S_{1}}{k};\]
\[S_{\text{AMK}} = S_{\text{AMP}} + S_{\text{APK}} =\]
\[= \left( 1 + \frac{1}{k} \right) \bullet S_{1}.\]
\[\frac{S_{\text{KAM}}}{S_{\text{KMC}}} = \frac{\text{AM}}{\text{MC}} = k = \frac{\left( 1 + \frac{1}{k} \right) \bullet S_{1}}{S_{\text{KMC}}}\]
\[S_{\text{KMC}} = \frac{\left( 1 + \frac{1}{k} \right) \bullet S_{1}}{k} = \frac{(k + 1)}{k^{2}} \bullet S_{1}.\]
\[Получаем:\]
\[S_{\text{ACK}} = S_{\text{KAM}} + S_{\text{KMC}} =\]
\[= \frac{(k + 1)}{k^{2}} \bullet S_{1} + \frac{(k + 1)}{k^{2}} \bullet S_{1} =\]
\[= \frac{(k + 1)^{2}}{k} \bullet S_{1}.\]
\[Значит:\]
\[\frac{S_{\text{ACK}}}{S_{\text{AKB}}} = \frac{\text{CK}}{\text{KB}} = k = \frac{(k + 1)^{2} \bullet S_{1}}{S_{\text{AKB}}}\]
\[S_{\text{AKB}} = \frac{1}{k} \bullet \left( \frac{k + 1}{k} \right)^{2} \bullet S_{1}\]
\[S_{\text{ABC}} = S_{\text{ACK}} + S_{\text{AKB}} =\]
\[= \frac{(k + 1)^{3}}{k} \bullet S_{1}.\]
\[2)\ \frac{S_{\text{BMP}}}{S_{\text{BPK}}} = \frac{\text{MP}}{\text{PK}} = k = \frac{S_{\text{BMP}}}{S_{2}}\]
\[S_{\text{BMP}} = kS_{2}\]
\[S_{\text{BMK}} = S_{\text{BMP}} + S_{\text{BPK}} =\]
\[= (k + 1) \bullet S_{2}.\]
\[\frac{S_{\text{MCK}}}{S_{\text{MKB}}} = \frac{\text{CK}}{\text{KB}} = k = \frac{S_{\text{MCK}}}{(k + 1)S_{2}}\]
\[S_{\text{MCK}} = k(k + 1)S_{2}\]
\[S_{\text{MCB}} = S_{\text{MCK}} + S_{\text{MKB}} =\]
\[= (k + 1)^{2}S_{2}.\]
\[\frac{S_{\text{BAM}}}{S_{\text{BMC}}} = \frac{\text{AM}}{\text{MC}} = k = \frac{S_{\text{BAM}}}{(k + 1)^{2}S_{2}}\]
\[S_{\text{BAM}} = k \bullet (k + 1)^{2} \bullet S_{2}\]
\[S_{\text{ABC}} = S_{\text{BAM}} + S_{\text{BMC}} =\]
\[= (k + 1)^{3} \bullet S_{2}.\]
\[3)\ Получаем:\]
\[\frac{(k + 1)^{3}}{k} \bullet S_{1} = (k + 1)^{3} \bullet S_{2}\]
\[k^{3} = \frac{S_{1}}{S_{2}}\]
\[k = \sqrt[3]{\frac{S_{1}}{S_{2}}}.\]
\[4)\ S_{\text{ABC}} = (k + 1)^{3} \bullet S_{2} =\]
\[= \left( \sqrt[3]{\frac{S_{1}}{S_{2}}} + 1 \right)^{3} \bullet S_{2} =\]
\[= \frac{\left( \sqrt[3]{S_{1}} + \sqrt[3]{S_{2}} \right)^{3}}{S_{2}} \bullet S_{2} =\]
\[= \left( \sqrt[3]{S_{1}} + \sqrt[3]{S_{2}} \right)^{3}.\]
\[Ответ:S_{\text{ABC}} = \left( \sqrt[3]{S_{1}} + \sqrt[3]{S_{2}} \right)^{3}.\]
\[\boxed{\mathbf{831.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[ABCD - выпуклый\ \]
\[четырехугольник;\]
\[O = AC \cap BD.\]
\[P_{\text{AOB}} = P_{\text{BOC}} = P_{\text{COD}} = P_{\text{AOD}} = P.\]
\[\mathbf{Доказать:}\]
\[ABCD - ромб.\]
\[\mathbf{Доказательство.}\]
\[1)\ По\ условию;\]
\[OA + OB + AB =\]
\[= OB + OC + BC;\]
\[OC + OD + CD =\]
\[= OD + OA + AD;\]
\[тогда:\]
\[AC + BD + AB + CD =\]
\[= AC + BD + BC + AD.\]
\[Значит:\]
\[AB + CD = BC + AD.\]
\[2)\ Суммы\ противоположных\ \]
\[сторон\ равны,\ как\ и\ разности\ \]
\[смежных\ сторон:\]
\[AB - BC = AD - CD.\]
\[3)\ Пусть\ AB > BC \Longrightarrow\]
\[\Longrightarrow AD > CD;\ \ AO > OC:\]
\[Неравенство\ для\ сумм\ \]
\[периметров\]
\[P_{\text{AOB}} + P_{\text{AOD}} > P_{\text{BOC}} + P_{\text{COD}}\text{\ \ }\]
\[противоречит\ условию\ задачи.\]
\[4)\ Пусть\ AB < BC:\]
\[неравенство\ для\ сумм\ \]
\[периметров\ \ противоречит\ \]
\[условию;\]
\[P_{\text{AOB}} + P_{\text{AOD}} < P_{\text{BOC}} + P_{\text{COD}}.\]
\[Следовательно:\]
\[AB = BC;\ \ AD = CD.\]
\[5)\ Аналогично\ предыдущему,\ \]
\[из\ равенства\ AB - AD =\]
\[= BC - CD:\ \]
\[AB = AD;\ \ BC = CD.\]
\[Значит:\ \]
\[AB = BC = CD = AD.\]
\[6)\ \ ABCD - параллелограмм\]
\[\ (по\ третьему\ признаку):\]
\[AB = CD;BC = AD.\]
\[Если\ все\ стороны\ \]
\[параллелограмма\ равны,\ то\ \]
\[он\ ромб:\]
\[ABCD - ромб.\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]
