Решебник по геометрии 11 класс. Атанасян ФГОС 286

\[\boxed{\mathbf{286.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[SO = h - высота.\]

\[Решение.\]

\[В\ треугольнике\ ABC:\]

\[AO = R;\]

\[AL = \frac{m\sqrt{3}}{2}.\]

\[По\ теореме\ синусов:\]

\[2R = \frac{\text{BC}}{\sin{60{^\circ}}}\]

\[AO = R = \frac{m}{2 \cdot \frac{\sqrt{3}}{2}} = \frac{m\sqrt{3}}{3}.\]

\[\textbf{а)}\ По\ теореме\ Пифагора:\]

\[AS^{2} = SO^{2} + OA^{2}\]

\[m^{2} = \left( \frac{m\sqrt{3}}{3} \right)^{2} + h^{2}\]

\[h^{2} = m^{2} - \frac{3m^{2}}{9} = m^{2} - \frac{m^{2}}{3} =\]

\[= \frac{2m^{2}}{3}\]

\[h = m\sqrt{\frac{2}{3}}\]

\[m = \frac{h\sqrt{6}}{2}.\]

\[\textbf{б)}\ OL = LK:\]

\[OK \parallel AS.\]

\[⊿LOK\ подобен\ ⊿LAS\ \]

\[(\angle L - общий):\]

\[\frac{\text{OL}}{\text{AS}} = \frac{\text{OL}}{\text{LA}}\]

\[OL = AL - OA = \frac{m\sqrt{3}}{2} - \frac{m\sqrt{3}}{3} =\]

\[= \frac{3m\sqrt{3} - 2m\sqrt{3}}{6} = \frac{m\sqrt{3}}{6};\]

\[\frac{n}{m} = \frac{m\sqrt{3}}{6}\ :\frac{m\sqrt{3}}{2} = \frac{2}{6} = \frac{1}{3};\]

\[n = \frac{m}{3}.\]

\[Ответ:а)\ \frac{h\sqrt{6}}{2};\ \ б)\ \frac{m}{3}.\]