\[\boxed{\mathbf{349.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[\angle ASB = 90{^\circ};\]
\[AO = OB = 5\ см.\]
\[Найти:\]
\[S_{\text{ASB}}.\]
\[Решение.\]
\[S_{\text{ASB}} = \frac{1}{2}AS \cdot SB = \frac{1}{2}AS^{2}.\]
\[По\ теореме\ Пифагора\ \]
\[(из\ ⊿ASB):\]
\[AB^{2} = AS^{2} + SB^{2} = 2AS^{2};\ \ \ \]
\[AB = 10\ см.\]
\[AS^{2} = \frac{1}{2}AB^{2} = \frac{1}{2} \cdot 100 = 50\ \]
\[AS = \sqrt{50}\ см.\]
\[S_{\text{ASB}} = \frac{1}{2}AS^{2} = \frac{1}{2} \cdot 50 = 25\ см^{2}.\]
\[Ответ:25\ см^{2}.\]