ГДЗ по геометрии 9 класс Мерзляк Задание 226

\[Схематический\ рисунок.\]

\[Дано:\]

\[\mathrm{\Delta}ABC - равнобедренный;\]

\[AC = 24\ см;\]

\[AB = BC;\]

\[O - центр\ опис.\ окружности;\]

\[AO = OC = 13\ см.\]

\[Найти:\]

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

\[Решение.\]

\[\sin{\angle B} = \frac{\text{AC}}{2R} = \frac{24}{2 \bullet 13} = \frac{12}{13};\]

\[\cos{\angle B} = - \sqrt{1 - \sin^{2}{\angle B}} =\]

\[= - \sqrt{1 - \frac{144}{169}} = - \frac{5}{13}.\]

\[AC^{2} = AB^{2} + BC^{2} - 2AB \bullet BC\cos{\angle B}\]

\[AB^{2} + AB^{2} - 2AB \bullet AB \bullet \left( - \frac{5}{13} \right) = 576\]

\[2AB^{2} + \frac{10}{13}AB^{2} = 576\]

\[\frac{36}{13}AB^{2} = 576\ \ \ \]

\[AB^{2} = 208.\]

\[S_{\text{ABC}} = \frac{1}{2}AB \bullet BC\sin{\angle B} =\]

\[= \frac{1}{2}AB^{2} \bullet \frac{12}{13} = \frac{6}{13} \bullet 208 = 96\ см^{2}.\]

\[Ответ:\ \ 96\ см^{2}.\]