\[\boxed{\mathbf{618.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ \mathbf{задачи:}\]
\[\mathbf{Дано:}\]
\[ABCD - ромб;\]
\[AC \cap BD = O;\]
\[BD = 18\ м;\]
\[AC = 24\ м.\]
\[\mathbf{Найти:}\]
\[P_{\text{ABCD}} - ?;h - ?\]
\[\mathbf{Решение.}\]
\[1)\ ABCD - ромб:\]
\[AB = BC = CD = AD;AO = OC;\]
\[BO = OD(по\ свойству\ ромба).\]
\[2)\ AO = OC = 24\ :2 = 12\ м;\]
\[BO = OD = 18\ :2 = 9\ м.\]
\[3)\ \mathrm{\Delta}ABO - прямоугольный\ \]
\[(так\ как\ BD\bot AC):\]
\[AB^{2} = AO^{2} + OB^{2};\]
\[AB^{2} = 144 + 81 = 225\]
\[AB = 15\ м.\]
\[4)\ P_{\text{ABCD}} = 4 \bullet AB = 4 \bullet 15 =\]
\[= 60\ м.\]
\[5)\ S_{\text{ABCD}} = \frac{1}{2}AC \bullet BD =\]
\[= \frac{1}{2} \bullet 18 \bullet 24 = 216\ м^{2}.\]
\[6)\ S_{\text{ABCD}} = AD \bullet h = 216\ м^{2}.\]
\[h = \frac{216}{\text{AD}} = \frac{216}{15} = 14,4\ м.\]
\[Ответ:\ P_{\text{ABCD}} = 60\ м;h = 14,4\ м.\]