Решебник по геометрии 7 класс Атанасян ФГОС Задание 844

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Год:2020-2021-2022
Тип:учебник

Задание 844

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Геометрия 7 класс Атанасян ФГОС Просвещение
 
фгос Геометрия 7 класс Атанасян ФГОС, Бутузов Просвещение
Издание 1
Геометрия 7 класс Атанасян ФГОС Просвещение

\[\boxed{\mathbf{844.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[ABCD - прямоугольник;\]

\[MB = a;\ \ MC = b;\]

\[MD = c.\]

\[\mathbf{Найти:}\]

\[MA - ?\]

\[\mathbf{Решение.}\]

\[1)\ Проведем\ через\ ( \bullet )\text{M\ }\]

\[прямые\ EF \parallel AD;\ \ GH \parallel AB.\]

\[2)\ Пусть\ MA = x;\ \ \]

\[AE = HM = FD = m;\ \]

\[EB = MG = FC = n;\ \ \]

\[BG = EM = AH = p;\]

\[GC = MF = HD = q.\]

\[3)\ По\ теореме\ Пифагора:\]

\[a^{2} = n^{2} + p^{2};\]

\[b^{2} = n^{2} + q^{2}.\]

\[Получаем:\]

\[a^{2} + c^{2} = n^{2} + p^{2} + m^{2} + q^{2} =\]

\[= x^{2} + b^{2};\]

\[c^{2} = m^{2} + q^{2};\]

\[x^{2} = m^{2} + p^{2}.\]

\[4)\ x^{2} = a^{2} - b^{2} + c^{2}\]

\[x = \sqrt{a^{2} - b^{2} + c^{2}}.\]

\[Ответ:\ \sqrt{a^{2} - b^{2} + c^{2}}.\]

Издание 2
фгос Геометрия 7 класс Атанасян ФГОС, Бутузов Просвещение

\[\boxed{\mathbf{844.еуроки - ответы\ на\ пятёрку}}\]

\[\mathbf{Дано:}\]

\[\mathrm{\Delta}ABC;\]

\[M \in AC;K \in BC;\]

\[P \in MK;\ \]

\[\frac{\text{AM}}{\text{MC}} = \frac{\text{CK}}{\text{KB}} = \frac{\text{MP}}{\text{PK}};\]

\[S_{\text{AMP}} = S_{1};\]

\[S_{\text{BKP}} = S_{2}.\]

\[\mathbf{Найти:}\]

\[S_{\text{ABC}} - ?\]

\[\mathbf{Решение.}\]

\[1)\ Допустим:\ \ \frac{\text{AM}}{\text{MC}} = \frac{\text{CK}}{\text{KB}} =\]

\[= \frac{\text{MP}}{\text{PK}} = k.\ \]

\[Тогда:\]

\[\frac{S_{\text{AMP}}}{S_{\text{APK}}} = \frac{\text{MP}}{\text{PK}} = k = \frac{S_{1}}{S_{\text{APK}}}\]

\[S_{\text{APK}} = \frac{S_{1}}{k};\]

\[S_{\text{AMK}} = S_{\text{AMP}} + S_{\text{APK}} =\]

\[= \left( 1 + \frac{1}{k} \right) \bullet S_{1}.\]

\[\frac{S_{\text{KAM}}}{S_{\text{KMC}}} = \frac{\text{AM}}{\text{MC}} = k = \frac{\left( 1 + \frac{1}{k} \right) \bullet S_{1}}{S_{\text{KMC}}}\]

\[S_{\text{KMC}} = \frac{\left( 1 + \frac{1}{k} \right) \bullet S_{1}}{k} =\]

\[= \frac{(k + 1)}{k^{2}} \bullet S_{1}.\]

\[Получаем:\]

\[S_{\text{ACK}} = S_{\text{KAM}} + S_{\text{KMC}} =\]

\[= \frac{(k + 1)}{k^{2}} \bullet S_{1} + \frac{(k + 1)}{k^{2}} \bullet S_{1} =\]

\[= \frac{(k + 1)^{2}}{k} \bullet S_{1}.\]

\[Значит:\]

\[\frac{S_{\text{ACK}}}{S_{\text{AKB}}} = \frac{\text{CK}}{\text{KB}} = k = \frac{(k + 1)^{2} \bullet S_{1}}{S_{\text{AKB}}}\]

\[S_{\text{AKB}} = \frac{1}{k} \bullet \left( \frac{k + 1}{k} \right)^{2} \bullet S_{1}\]

\[S_{\text{ABC}} = S_{\text{ACK}} + S_{\text{AKB}} =\]

\[= \frac{(k + 1)^{3}}{k} \bullet S_{1}.\]

\[2)\ \frac{S_{\text{BMP}}}{S_{\text{BPK}}} = \frac{\text{MP}}{\text{PK}} = k = \frac{S_{\text{BMP}}}{S_{2}}\]

\[S_{\text{BMP}} = kS_{2}\]

\[S_{\text{BMK}} = S_{\text{BMP}} + S_{\text{BPK}} =\]

\[= (k + 1) \bullet S_{2}.\]

\[\frac{S_{\text{MCK}}}{S_{\text{MKB}}} = \frac{\text{CK}}{\text{KB}} = k = \frac{S_{\text{MCK}}}{(k + 1)S_{2}}\]

\[S_{\text{MCK}} = k(k + 1)S_{2}\]

\[S_{\text{MCB}} = S_{\text{MCK}} + S_{\text{MKB}} =\]

\[= (k + 1)^{2}S_{2}.\]

\[\frac{S_{\text{BAM}}}{S_{\text{BMC}}} = \frac{\text{AM}}{\text{MC}} = k = \frac{S_{\text{BAM}}}{(k + 1)^{2}S_{2}}\]

\[S_{\text{BAM}} = k \bullet (k + 1)^{2} \bullet S_{2}\]

\[S_{\text{ABC}} = S_{\text{BAM}} + S_{\text{BMC}} =\]

\[= (k + 1)^{3} \bullet S_{2}.\]

\[3)\ Получаем:\]

\[\frac{(k + 1)^{3}}{k} \bullet S_{1} = (k + 1)^{3} \bullet S_{2}\]

\[k^{3} = \frac{S_{1}}{S_{2}}\]

\[k = \sqrt[3]{\frac{S_{1}}{S_{2}}}.\]

\[4)\ S_{\text{ABC}} = (k + 1)^{3} \bullet S_{2} =\]

\[= \left( \sqrt[3]{\frac{S_{1}}{S_{2}}} + 1 \right)^{3} \bullet S_{2} =\]

\[= \frac{\left( \sqrt[3]{S_{1}} + \sqrt[3]{S_{2}} \right)^{3}}{S_{2}} \bullet S_{2} =\]

\[= \left( \sqrt[3]{S_{1}} + \sqrt[3]{S_{2}} \right)^{3}.\]

\[Ответ:S_{\text{ABC}} = \left( \sqrt[3]{S_{1}} + \sqrt[3]{S_{2}} \right)^{3}.\]

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