ГДЗ по геометрии 8 класс Атанасян Задание 454

\[\boxed{\mathbf{454.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[{а)\ S}_{\text{ABCD}} = 250\ см^{2};\]

\[a = 2,5b.\]

\[{б)\ S}_{\text{ABCD}} = 9\ м^{2};\]

\[P_{\text{ABCD}} = 12\ м.\]

\[\mathbf{Найти:}\]

\[\mathbf{Решение.}\]

\[\textbf{а)}\ S_{\text{ABCD}} = 2,5b \bullet b = 2,5b^{2}\]

\[b^{2} = 100\ \]

\[b = 10\ см;\ \ \ \]

\[a = 2,5 \bullet 10 = 25\ см.\]

\[\textbf{б)}\ S_{\text{ABCD}} = \text{ab} = 9\ м^{2}.\]

\[P_{\text{ABCD}} = 2a + 2b = 12\ м:\]

\[a + b = 6\ \ \]

\[\ a = 6 - b.\]

\[S_{\text{ABCD}} = (6 - b)b = 9\]

\[6b - b^{2} = 9\]

\[b^{2} - 6b + 9 = 0\]

\[(b - 3)^{2} = 0\]

\[b = 3\ м;\ \ \]

\[\ a = 6 - 3 = 3\ м.\]

\[Ответ:а)\ a = 25\ см;b = 10\ см;\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ a = 3\ м;b = 3\ м.\]