\[\boxed{\mathbf{998.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - четырехугольник;\]
\[A( - 2; - 3);B(1;4);\]
\[C(8;7);D(5;0).\]
\[\mathbf{Доказать:}\]
\[ABCD - ромб.\]
\[Найти:\]
\[S_{\text{ABCD}}.\]
\[\mathbf{Решение.}\]
\[1)\ AB =\]
\[= \sqrt{( - 2 - 1)^{2} + ( - 3 - 4)^{2}} =\]
\[= \sqrt{58};\]
\[BC = \sqrt{(1 - 8)^{2} + (4 - 7)^{2}} =\]
\[= \sqrt{58};\]
\[CD = \sqrt{(8 - 5)^{2} + (7 - 0)^{2}} =\]
\[= \sqrt{58};\]
\[AD = \sqrt{( - 2 - 5)^{2} + ( - 3 - 0)^{2}} =\]
\[= \sqrt{58}.\]
\[2)\ AB = BC = CD = AD \Longrightarrow\]
\[\Longrightarrow ABCD - ромб.\]
\[Что\ и\ требовалось\ доказать;\]
\[3)\ S_{\text{ABCD}} = \frac{1}{2}BD \bullet AC:\ \]
\[BD = \sqrt{(1 - 5)^{2} + (4 - 0)^{2}} =\]
\[= \sqrt{32} = 4\sqrt{2};\]
\[AC = \sqrt{( - 2 - 8)^{2} + ( - 3 - 7)^{2}} =\]
\[= \sqrt{200} = 10\sqrt{2}.\]
\[4)\ S_{\text{ABCD}} = \frac{1}{2} \bullet 4\sqrt{2} \bullet 10\sqrt{2} =\]
\[= 20 \bullet 2 = 40\ кв.ед.\]
\[Ответ:S_{\text{ABCD}} = 40\ кв.ед.\]