Решебник по геометрии 7 класс Атанасян ФГОС Задание 998

 

\[\boxed{\mathbf{998.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[ABCD - четырехугольник;\]

\[A( - 2; - 3);B(1;4);\]

\[C(8;7);D(5;0).\]

\[\mathbf{Доказать:}\]

\[ABCD - ромб.\]

\[Найти:\]

\[S_{\text{ABCD}}.\]

\[\mathbf{Решение.}\]

\[1)\ AB =\]

\[= \sqrt{( - 2 - 1)^{2} + ( - 3 - 4)^{2}} =\]

\[= \sqrt{58};\]

\[BC = \sqrt{(1 - 8)^{2} + (4 - 7)^{2}} =\]

\[= \sqrt{58};\]

\[CD = \sqrt{(8 - 5)^{2} + (7 - 0)^{2}} =\]

\[= \sqrt{58};\]

\[AD = \sqrt{( - 2 - 5)^{2} + ( - 3 - 0)^{2}} =\]

\[= \sqrt{58}.\]

\[2)\ AB = BC = CD = AD \Longrightarrow\]

\[\Longrightarrow ABCD - ромб.\]

\[Что\ и\ требовалось\ доказать;\]

\[3)\ S_{\text{ABCD}} = \frac{1}{2}BD \bullet AC:\ \]

\[BD = \sqrt{(1 - 5)^{2} + (4 - 0)^{2}} =\]

\[= \sqrt{32} = 4\sqrt{2};\]

\[AC = \sqrt{( - 2 - 8)^{2} + ( - 3 - 7)^{2}} =\]

\[= \sqrt{200} = 10\sqrt{2}.\]

\[4)\ S_{\text{ABCD}} = \frac{1}{2} \bullet 4\sqrt{2} \bullet 10\sqrt{2} =\]

\[= 20 \bullet 2 = 40\ кв.ед.\]

\[Ответ:S_{\text{ABCD}} = 40\ кв.ед.\]

\[\boxed{\mathbf{998.еуроки - ответы\ на\ пятёрку}}\]

\[\overrightarrow{h} = k\overrightarrow{h}\]

\[\textbf{а)}\ 2 = |k| \bullet 0,5 \Longrightarrow |k| = 4;\]

\[\overrightarrow{m}\ \uparrow \downarrow \ \overrightarrow{h} \Longrightarrow k < 0 \Longrightarrow k = - 4.\]

\[\textbf{б)}\ 240 = |k| \bullet 12 \Longrightarrow |k| = 20;\]

\[\overrightarrow{m}\ \uparrow \uparrow \ \overrightarrow{h} \Longrightarrow k > 0 \Longrightarrow k = 20.\]

\[\textbf{а)}\ 400 = |k| \bullet 400 \Longrightarrow |k| = 1;\]

\[\overrightarrow{m}\ \uparrow \downarrow \ \overrightarrow{h} \Longrightarrow k < 0 \Longrightarrow k = - 1.\]

\[\textbf{а)}\ \sqrt{50} = |k| \bullet \sqrt{2} \Longrightarrow\]

\[\Longrightarrow |k| = \sqrt{25} = 5;\]

\[\overrightarrow{m}\ \uparrow \uparrow \ \overrightarrow{h} \Longrightarrow k > 0 \Longrightarrow k = 5.\]