\[\boxed{\mathbf{477.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\text{ABCD} - ромб;\]
\[\text{AC}\ и\ \text{BD} - диагонали;\]
\[\text{BD} = \text{AC} \bullet 1,5;\]
\[S_{\text{ABCD}} = 27\ см^{2}.\]
\[\mathbf{Найти:}\]
\[\text{AC}\ и\ \text{BD}.\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{ABCD}} = \frac{\text{AC} \bullet \text{BD}}{2} =\]
\[= \frac{\text{AC} \bullet 1,5\text{AC}}{2} = 27\ см^{2}.\]
\[2)\ AC^{2} = \frac{2 \bullet 27}{1,5} = \frac{54}{1,5} = 36 \Longrightarrow\]
\[\Longrightarrow \text{AC} = 6\ см.\]
\[3)\ \text{BD} = 1,5 \bullet 6 = 9\ см.\]
\[\mathbf{Ответ:\ }\text{AC} = 6см;\text{BD} = 9\ см.\]