\[\boxed{\mathbf{513.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ \mathbf{задачи:}\]
\[\mathbf{Дано:}\]
\[\text{ABCD} - ромб;\]
\[\text{AC} \cap \text{BD} = O;\]
\[\text{BD} = 18\ м;\]
\[\text{AC} = 24\ м.\]
\[\mathbf{Найти:}\]
\[P_{\text{ABCD}} - ?;h - ?\]
\[\mathbf{Решение.}\]
\[1)\ \text{ABCD} - ромб:\]
\[\text{AB} = \text{BC} = \text{CD} = \text{AD};\text{AO} = \text{OC};\]
\[\text{BO} = \text{OD}(по\ свойству\ ромба).\]
\[2)\ \text{AO} = \text{OC} = 24\ :2 = 12\ м;\]
\[\text{BO} = \text{OD} = 18\ :2 = 9\ м.\]
\[3)\ \mathrm{\Delta}\text{ABO} - прямоугольный\ \]
\[\left( так\ как\ \text{BD}\bot\text{AC} \right):\]
\[AB^{2} = AO^{2} + OB^{2};\]
\[AB^{2} = 144 + 81 = 225\]
\[\text{AB} = 15\ м.\]
\[4)\ P_{\text{ABCD}} = 4 \bullet \text{AB} = 4 \bullet 15 =\]
\[= 60\ м.\]
\[5)\ S_{\text{ABCD}} = \frac{1}{2}\text{AC} \bullet \text{BD} =\]
\[= \frac{1}{2} \bullet 18 \bullet 24 = 216\ м^{2}.\]
\[6)\ S_{\text{ABCD}} = \text{AD} \bullet h = 216\ м^{2}.\]
\[h = \frac{216}{\text{AD}} = \frac{216}{15} = 14,4\ м.\]
\[Ответ:\ P_{\text{ABCD}} = 60\ м;h = 14,4\ м.\]