\[Схематический\ рисунок.\]
\[Дано:\]
\[\mathrm{\Delta}ABC - прямоугольный;\]
\[\angle C = 90{^\circ};\ \]
\[AB = 13\ см;\]
\[AC = 12\ см;\]
\[BD = 26\ см.\]
\[Найти:\]
\[\text{CD.}\]
\[Решение.\]
\[1)\ В\ \mathrm{\Delta}ABC:\]
\[\angle C = 90{^\circ};\ \ \ \]
\[BC = \sqrt{AB^{2} - AC^{2}};\]
\[BC = \sqrt{169 - 144} = \sqrt{25} = 5\ см.\]
\[\cos{\angle B} = \frac{\text{BC}}{\text{AB}} = \frac{5}{13};\]
\[\angle CBD = 180{^\circ} - \angle B;\]
\[\cos{\angle CBD} = - \cos{\angle B} = - \frac{5}{13}.\]
\[2)\ В\ \mathrm{\Delta}CBD:\]
\[CD^{2} =\]
\[= BC^{2} + BD^{2} - 2BC \bullet BD\cos{\angle B} =\]
\[= 25 + 676 - 2 \bullet 5 \bullet 26 \bullet \left( - \frac{5}{13} \right) =\]
\[= 701 + 100 = 801;\]
\[CD = \sqrt{801} = 3\sqrt{89}\ см.\]
\[Ответ:\ \ 3\sqrt{89}\ см.\]