Решебник по геометрии 10 класс Атанасян ФГОС 350

\[\boxed{\mathbf{350.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[SB = SC = 2r.\]

\[Найти:\]

\[S_{\text{BSC}}.\]

\[Решение.\]

\[S_{\text{BSC}} = \frac{1}{2}SB \cdot SC \cdot \sin{\angle BSC} =\]

\[= \frac{1}{2} \cdot 2r \cdot 2r \cdot \sin{\angle BSC} =\]

\[= 2r^{2} \cdot \sin{\angle BSC}.\]

\[\textbf{а)}\ \angle BSC = 30{^\circ}:\]

\[S_{\text{BSC}} = 2r^{2} \cdot \sin{30{^\circ}} = 2r^{2} \cdot \frac{1}{2} =\]

\[= r^{2}.\]

\[\textbf{б)}\ \angle BSC = 45{^\circ}:\]

\[S_{\text{BSC}} = 2r^{2} \cdot \sin{45{^\circ}} = 2r^{2} \cdot \frac{\sqrt{2}}{2} =\]

\[= \sqrt{2}r^{2}.\]

\[\textbf{в)}\ \angle BSC = 60{^\circ}:\]

\[S_{\text{BSC}} = 2r^{2} \cdot \sin{60{^\circ}} = 2r^{2} \cdot \frac{\sqrt{3}}{2} =\]

\[= \sqrt{3}r^{2}.\]