\[\boxed{\mathbf{517.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ \ задачи:\]
\[\mathbf{Дано:}\]
\[\text{ABCD} - четырехугольник;\]
\[\text{AB} = 5\ см;\]
\[\text{BC} = 13\ см;\]
\[\text{CD} = 9\ см;\]
\[\text{DA} = 15\ см;\]
\[\text{AC} = 12\ см.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABCD}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{ABCD}} = S_{\text{ABC}} + S_{\text{ACD}}.\]
\[2)\ Рассмотрим\ ⊿\text{ABC}:\]
\[AC^{2} + AB^{2} = BC^{2}\]
\[144 + 25 = 169\]
\[верно.\]
\[⊿\text{ABC} - прямоугольный\ с\ \]
\[\angle\text{BAC} = 90{^\circ};\]
\[S_{\text{ABC}} = \frac{1}{2} \bullet \text{AB} \bullet \text{AC} = \frac{1}{2} \bullet 12 \bullet 5 =\]
\[= 30\ см^{2}.\]
\[3)\ Рассмотрим\ ⊿\text{ACD}:\]
\[AC^{2} + CD^{2} = AD^{2}\]
\[144 + 81 = 225\]
\[верно.\]
\[⊿\text{ACD} - прямоугольный\ с\ \]
\[\angle\text{ACD} = 90{^\circ};\]
\[S_{\text{ACD}} = \frac{1}{2} \bullet \text{AC} \bullet \text{CD} = \frac{1}{2} \bullet 12 \bullet 9 =\]
\[= 54\ см^{2}.\]
\[4)\ S_{\text{ABCD}} = 30 + 54 = 84\ см^{2}.\]
\[Ответ:84\ см^{2}.\]