\[\boxed{\mathbf{514.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\text{ABCD} - ромб;\]
\[S_{\text{ABCD}} = 540\ см^{2};\]
\[\text{BD} = 4,5\ дм;\]
\[\text{BD}\bot\text{AC}.\]
\[\mathbf{Найти:}\]
\[\text{OH} - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{ABCD}} = \frac{1}{2}\text{BD} \bullet \text{AC}\]
\[540 = \frac{1}{2} \bullet 45 \bullet \text{AC}\]
\[\text{AC} = \frac{540 \bullet 2}{45} = 12 \bullet 2 = 24\ см.\]
\[2)\ \text{ABCD} - ромб:\]
\[\text{AB} = \text{BC} = \text{CD} = \text{AD};\]
\[\text{AO} = \text{OC};\]
\[\text{BO} = \text{OD}\ (по\ свойству\ ромба).\]
\[3)\ \text{AO} = \text{OC} = 24\ :2 = 12\ см.\]
\[\text{BO} = \text{OD} = 45\ :2 = 22,5\ см.\]
\[4)\ \ ⊿\text{DOC} - прямоугольный:\]
\[DC^{2} = OD^{2} + OC^{2}\]
\[DC^{2} = 506,25 \bullet 144 = 650,25\]
\[\text{DC} = 25,5\ см.\]
\[5)\ S_{\text{DOC}} = \frac{1}{2}\text{OH} \bullet \text{DC} =\]
\[= \frac{1}{2}\text{OC} \bullet \text{OD} = \frac{1}{2} \bullet 12 \bullet 22,5 =\]
\[= 135\ см^{2}.\]
\[6)\ \frac{1}{2} \bullet \text{OH} \bullet 25,5 = 135\]
\[\text{OH} \bullet 25,5 = 270\]
\[\text{OH} = \frac{270 \bullet 10}{255} = \frac{180}{17} =\]
\[= 10\frac{10}{17}\ см.\]
\[Ответ:10\frac{10}{17}\ см.\]