ГДЗ по геометрии 9 класс Атанасян ФГОС Задание 636

\[\boxed{\mathbf{636.еуроки - ответы\ на\ пятёрку}}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[ABCD - прямоугольник;\]

\[AB = 6\ см;\]

\[BC = 8\ см;\]

\[a\bot BD;C \in a;\]

\[a \cap AD = M;\]

\[BD \cap a = K.\]

\[\mathbf{Найти:}\]

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

\[\mathbf{Решение.}\]

\[1)\ S_{\text{ABD}} = S_{\text{BCD}}\ \]

\[(так\ как\ BD - диагональ).\]

\[2)\ BD^{2} = AB^{2} + AD^{2} =\]

\[= 36 + 64 = 100\]

\[BD = 10\ см.\]

\[3)\ S_{\text{ABD}} = S_{\text{BCD}} = \frac{1}{2}BA \bullet AD =\]

\[= \frac{1}{2} \bullet 6 \bullet 8 = 24\ см^{2}.\]

\[4)\ S_{\text{BCD}} = \frac{1}{2} \bullet BD \bullet CK\]

\[24 = \frac{1}{2} \bullet 10 \bullet CK\]

\[24 = 5CK\]

\[CK = 4,8\ см.\]

\[5)\ \mathrm{\Delta}CKD - прямоугольный:\]

\[KD^{2} = CD^{2} - CK^{2} =\]

\[= 36 - 23,04 = 12,96\]

\[KD = 3,6\ см.\]

\[6)\ S_{\text{CDM}} = \frac{1}{2}CD \bullet DM =\]

\[= \frac{1}{2} \bullet 6 \bullet DM = 3DM;\]

\[S_{\text{CDM}} = \frac{1}{2} \bullet CM \bullet KD =\]

\[= \frac{1}{2} \bullet 3,6 \bullet CM = 1,8\ CM.\]

\[7)\ CM = \sqrt{CD^{2} + MD^{2}} =\]

\[= \sqrt{36 + MD^{2}}.\]

\[8)\ 3MD = 1,8\sqrt{36 + MD^{2}}\]

\[MD = 0,6\sqrt{36 + MD^{2}}\]

\[MD^{2} = 0,36\left( 36 + MD^{2} \right)\]

\[MD^{2} = 12,96 + 0,36MD^{2}\]

\[0,64MD^{2} = 12,96\]

\[MD^{2} = 20,25\]

\[MD = 4,5\ см.\]

\[9)\ KM^{2} = MD^{2} - KD^{2} =\]

\[= 20,25 - 12,96 = 7,29\]

\[KM = 2,7\ см.\]

\[10)\ S_{\text{DKM}} = \frac{1}{2}KD \bullet KM =\]

\[= \frac{1}{2} \bullet 3,6 \bullet 2,7 = 4,86\ см^{2}.\]

\[11)\ S_{\text{ABKM}} = S_{\text{ABD}} - S_{\text{DKM}} =\]

\[= 24 - 4,86 = 19,14\ см^{2}.\]

\[\mathbf{Ответ:}19,14\ см^{2}.\]