Решебник по геометрии 9 класс Мерзляк Задание 160

\[Схематический\ рисунок.\]

\[Дано:\]

\[BD - высота;\]

\[CE - высота;\]

\[\angle A = \alpha;\]

\[BD = h_{1};\]

\[CE = h_{2}.\]

\[Найти:\]

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

\[Решение.\]

\[S = \frac{1}{2}AC \bullet BD = \frac{1}{2}AC \bullet h_{1}\text{\ \ \ }\]

\[AC = \frac{2S}{h_{1}}.\]

\[S = \frac{1}{2}AB \bullet CE = \frac{1}{2}AB \bullet h_{2}\text{\ \ \ }\]

\[AB = \frac{2S}{h_{2}}.\]

\[S = \frac{1}{2}AB \bullet AC \bullet \sin{\angle A}.\]

\[\frac{2S^{2}}{h_{1}h_{2}} \bullet \sin\alpha = S\ \ \ \]

\[S = \frac{h_{1}h_{2}}{2\sin\alpha}.\]

\[Ответ:\ \ \frac{h_{1}h_{2}}{2\sin\alpha}.\]