Решебник по геометрии 10 класс Атанасян ФГОС 353

\[\boxed{\mathbf{353.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[Найти:\]

\[S_{сеч}.\]

\[Решение.\]

\[\textbf{а)}\ BC - хорда.\]

\[OK\bot BC;\ \ SK\bot BC:\]

\[\angle COB = 60{^\circ};\ \ \]

\[⊿COK - прямоугольный:\]

\[\angle COK = 30{^\circ};\]

\[CK = \frac{\text{OC}}{2} = \frac{r}{2};\]

\[S_{сеч} = S_{\text{BSC}} = \frac{1}{2}CB \cdot SK.\]

\[По\ теореме\ Пифагора\ \]

\[(из\ ⊿SOK):\]

\[SK^{2} = SO^{2} + OK^{2}.\]

\[По\ теореме\ Пифагора\ \]

\[(из\ ⊿CSK):\]

\[SK = \sqrt{CS^{2} - CK^{2}} =\]

\[= \sqrt{l^{2} - \left( \frac{r}{2} \right)^{2}} = \frac{\sqrt{4l^{2} - r^{2}}}{2}.\]

\[S_{сеч} = \frac{1}{2}r \cdot \frac{\sqrt{4l^{2} - r^{2}}}{2} =\]

\[= \frac{r}{4}\sqrt{4l^{2} - r^{2}}.\]

\[\textbf{б)}\ \angle COB = 90{^\circ};\ \ \angle COK = 45{^\circ}.\]

\[По\ теореме\ Пифагора:\]

\[OC^{2} = 2CK^{2}\]

\[CK = \frac{\text{OC}}{\sqrt{2}} = \frac{r}{\sqrt{2}}.\]

\[CB = 2CK = \frac{2r}{\sqrt{2}} = \sqrt{2}\text{r.}\]

\[По\ теореме\ Пифагора:\]

\[SK = \sqrt{l^{2} - \left( \frac{r}{\sqrt{2}} \right)^{2}} = \sqrt{\frac{2l^{2} - r^{2}}{2}}.\]

\[S_{сеч} = \frac{1}{2} \cdot \sqrt{2}r \cdot \frac{\sqrt{2l^{2} - r^{2}}}{\sqrt{2}} =\]

\[= \frac{r}{2}\sqrt{2l^{2} - r^{2}}.\]