ГДЗ по геометрии 9 класс Атанасян Задание 514

\[\boxed{\mathbf{514.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[\text{ABCD} - ромб;\]

\[S_{\text{ABCD}} = 540\ см^{2};\]

\[\text{BD} = 4,5\ дм;\]

\[\text{BD}\bot\text{AC}.\]

\[\mathbf{Найти:}\]

\[\text{OH} - ?\]

\[\mathbf{Решение.}\]

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

\[540 = \frac{1}{2} \bullet 45 \bullet \text{AC}\]

\[\text{AC} = \frac{540 \bullet 2}{45} = 12 \bullet 2 = 24\ см.\]

\[2)\ \text{ABCD} - ромб:\]

\[\text{AB} = \text{BC} = \text{CD} = \text{AD};\]

\[\text{AO} = \text{OC};\]

\[\text{BO} = \text{OD}\ (по\ свойству\ ромба).\]

\[3)\ \text{AO} = \text{OC} = 24\ :2 = 12\ см.\]

\[\text{BO} = \text{OD} = 45\ :2 = 22,5\ см.\]

\[4)\ \ ⊿\text{DOC} - прямоугольный:\]

\[DC^{2} = OD^{2} + OC^{2}\]

\[DC^{2} = 506,25 \bullet 144 = 650,25\]

\[\text{DC} = 25,5\ см.\]

\[5)\ S_{\text{DOC}} = \frac{1}{2}\text{OH} \bullet \text{DC} =\]

\[= \frac{1}{2}\text{OC} \bullet \text{OD} = \frac{1}{2} \bullet 12 \bullet 22,5 =\]

\[= 135\ см^{2}.\]

\[6)\ \frac{1}{2} \bullet \text{OH} \bullet 25,5 = 135\]

\[\text{OH} \bullet 25,5 = 270\]

\[\text{OH} = \frac{270 \bullet 10}{255} = \frac{180}{17} =\]

\[= 10\frac{10}{17}\ см.\]

\[Ответ:10\frac{10}{17}\ см.\]