\[\boxed{\mathbf{476.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\text{ABCD} - ромб;\]
\[\textbf{а)}\ \text{AC} = 14\ см;\]
\[\text{BD} = 3,2\ дм;\]
\[\textbf{б)}\ \text{AC} = 2\ дм;\]
\[\text{BD} = 4,6\ дм.\]
\[\mathbf{Доказать:}\]
\[S_{\text{ABCD}} = \frac{\text{BD} \bullet \text{AC}}{2}.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABCD}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ По\ свойству\ диагоналей\ \]
\[ромба:\]
\[\text{AO} = \text{OC};\]
\[\text{BO} = \text{OD};\]
\[\text{BD}\bot\text{AC}.\]
\[Треугольники\ \text{AOD},\text{DOC},\text{AOB},\]
\[\text{BOC} - прямоугольные.\]
\[2)\ ⊿\text{AOD} = ⊿\text{DOC} = ⊿\text{AOB} =\]
\[= ⊿\text{BOC}\ (по\ двум\ катетам).\]
\[3)\ S_{\text{AOD}} = \frac{1}{2} \bullet \frac{\text{BD}}{2} \bullet \frac{\text{AC}}{2} = \frac{\text{BD} \bullet \text{AC}}{8}.\]
\[4)\ S_{\text{ABCD}} =\]
\[= S_{\text{AOD}} + S_{\text{DOC}} + S_{\text{AOB}} + S_{\text{BOC}} =\]
\[= 4 \bullet S_{\text{AOD}} = 4 \bullet \frac{\text{BD} \bullet \text{AC}}{8} =\]
\[= \frac{\text{BD} \bullet \text{AC}}{2}.\]
\[Что\ и\ требовалось\ доказать.\]
\[\textbf{а)}\ S_{\text{ABCD}} = \frac{14 \bullet 32}{2} = 224\ см^{2}.\]
\[\textbf{б)}\ S_{\text{ABCD}} = \frac{2 \bullet 4,6}{2} = 4,6\ дм^{2}.\]