Решебник по геометрии 9 класс Мерзляк Задание 41

\[Схематический\ рисунок.\]

\[Дано:\]

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

\[BD = BC;\ \angle C = 90{^\circ};\]

\[BD = AC = BC = a.\]

\[Найти:\]

\[\text{CD.}\]

\[1)\ В\ \mathrm{\Delta}ABC:\]

\[\angle C = 90{^\circ};\ \ \ \]

\[\angle B = \angle A = 45{^\circ};\]

\[\angle CBD = 180{^\circ} - \angle B = 135{^\circ}.\]

\[2)\ В\ \mathrm{\Delta}CBD:\]

\[CD^{2} =\]

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

\[= a^{2} + a^{2} - 2 \bullet a \bullet a \bullet \left( - \frac{\sqrt{2}}{2} \right) =\]

\[= 2a^{2} + \sqrt{2}a^{2} = a^{2}\left( 2 + \sqrt{2} \right);\]

\[CD = a\sqrt{2 + \sqrt{2}}.\]

\[Ответ:\ \ a\sqrt{2 + \sqrt{2}}.\]