\[\boxed{\mathbf{531.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\text{ABCD} - прямоугольник;\]
\[\text{AB} = 6\ см;\]
\[\text{BC} = 8\ см;\]
\[a\bot\text{BD};C \in a;\]
\[a \cap \text{AD} = M;\]
\[\text{BD} \cap a = K.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABKM}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{ABD}} = S_{\text{BCD}}\ \]
\[\left( так\ как\ \text{BD} - диагональ \right).\]
\[2)\ BD^{2} = AB^{2} + AD^{2} =\]
\[= 36 + 64 = 100\]
\[\text{BD} = 10\ см.\]
\[3)\ S_{\text{ABD}} = S_{\text{BCD}} = \frac{1}{2}\text{BA} \bullet \text{AD} =\]
\[= \frac{1}{2} \bullet 6 \bullet 8 = 24\ см^{2}.\]
\[4)\ S_{\text{BCD}} = \frac{1}{2} \bullet \text{BD} \bullet \text{CK}\]
\[24 = \frac{1}{2} \bullet 10 \bullet \text{CK}\]
\[24 = 5\text{CK}\]
\[\text{CK} = 4,8\ см.\]
\[5)\ \mathrm{\Delta}\text{CKD} - прямоугольный:\]
\[KD^{2} = CD^{2} - CK^{2} =\]
\[= 36 - 23,04 = 12,96\]
\[\text{KD} = 3,6\ см.\]
\[6)\ S_{\text{CDM}} = \frac{1}{2}\text{CD} \bullet \text{DM} =\]
\[= \frac{1}{2} \bullet 6 \bullet \text{DM} = 3\text{DM};\]
\[S_{\text{CDM}} = \frac{1}{2} \bullet \text{CM} \bullet \text{KD} =\]
\[= \frac{1}{2} \bullet 3,6 \bullet \text{CM} = 1,8\ \text{CM}.\]
\[7)\ \text{CM} = \sqrt{CD^{2} + MD^{2}} =\]
\[= \sqrt{36 + MD^{2}}.\]
\[8)\ 3\text{MD} = 1,8\sqrt{36 + MD^{2}}\]
\[\text{MD} = 0,6\sqrt{36 + MD^{2}}\]
\[MD^{2} = 0,36\left( 36 + MD^{2} \right)\]
\[MD^{2} = 12,96 + 0,36MD^{2}\]
\[0,64MD^{2} = 12,96\]
\[MD^{2} = 20,25\]
\[\text{MD} = 4,5\ см.\]
\[9)\ KM^{2} = MD^{2} - KD^{2} =\]
\[= 20,25 - 12,96 = 7,29\]
\[\text{KM} = 2,7\ см.\]
\[10)\ S_{\text{DKM}} = \frac{1}{2}\text{KD} \bullet \text{KM} =\]
\[= \frac{1}{2} \bullet 3,6 \bullet 2,7 = 4,86\ см^{2}.\]
\[11)\ S_{\text{ABKM}} = S_{\text{ABD}} - S_{\text{DKM}} =\]
\[= 24 - 4,86 = 19,14\ см^{2}.\]
\[\mathbf{Ответ:}19,14\ см^{2}.\]
\[\boxed{\mathbf{531.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[D \in AC;\]
\[K \in BD;\]
\[BK = KD.\]
\[\mathbf{Доказать:}\]
\[AM = MB;\]
\[BN = NC.\]
\[\mathbf{Доказательство.}\]
\[1)\ Пусть\ точка\ K -\]
\[середина\ \text{BD.\ }\]
\[Проведем\ через\ точку\ \text{K\ }\]
\[прямую,\ параллельную\ AC:\]
\[h \cap AB = M\ и\ h \cap BC = N.\]
\[Следовательно:\ \ \ NM \parallel AC.\]
\[2)\ BK = KD;\ \ \ MN \parallel AC:\ \]
\[MB = MA\ (по\ теореме\ Фалеса).\]
\[Значит:\]
\[M - середина\ \text{AB.}\]
\[3)\ BK = KD;\ \ MN \parallel AC:\ \]
\[BN = NC\ (по\ теореме\ Фалеса)\text{.\ }\]
\[Значит:\]
\[N - середина\ \text{BC.}\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]