\[Схематический\ рисунок.\]
\[Дано:\]
\[\mathrm{\Delta}ABC - равнобедренный;\]
\[AC = 24\ см;\]
\[AB = BC;\]
\[O - центр\ опис.\ окружности;\]
\[AO = OC = 13\ см.\]
\[Найти:\]
\[S_{\text{ABC}}.\]
\[Решение.\]
\[\sin{\angle B} = \frac{\text{AC}}{2R} = \frac{24}{2 \bullet 13} = \frac{12}{13};\]
\[\cos{\angle B} = - \sqrt{1 - \sin^{2}{\angle B}} =\]
\[= - \sqrt{1 - \frac{144}{169}} = - \frac{5}{13}.\]
\[AC^{2} = AB^{2} + BC^{2} - 2AB \bullet BC\cos{\angle B}\]
\[AB^{2} + AB^{2} - 2AB \bullet AB \bullet \left( - \frac{5}{13} \right) = 576\]
\[2AB^{2} + \frac{10}{13}AB^{2} = 576\]
\[\frac{36}{13}AB^{2} = 576\ \ \ \]
\[AB^{2} = 208.\]
\[S_{\text{ABC}} = \frac{1}{2}AB \bullet BC\sin{\angle B} =\]
\[= \frac{1}{2}AB^{2} \bullet \frac{12}{13} = \frac{6}{13} \bullet 208 = 96\ см^{2}.\]
\[Ответ:\ \ 96\ см^{2}.\]