ГДЗ по геометрии 9 класс Мерзляк Задание 219

\[Схематический\ рисунок.\]

\[Дано:\]

\[ABC\ldots - правильный\]

\[восьмиугольник;\]

\[O - центр\ опис.\ окружности;\]

\[AB = a.\]

\[Найти:\]

\[AC;\ AD;\ AE.\]

\[Решение:\]

\[1)\ 8 \bullet \angle B = 6 \bullet 180{^\circ}\]

\[8 \bullet \angle B = 1080{^\circ}\ \ \ \]

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

\[\angle D = \angle C = \angle B = 135{^\circ}.\]

\[2)\ AE - диаметр;\]

\[\angle ACE = \angle ADE = 90{^\circ}.\]

\[3)\ В\ \mathrm{\Delta}ABC:\]

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

\[= a^{2} + a^{2} - 2a \bullet a \bullet \cos{135{^\circ}} =\]

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

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

\[4)\ В\ \mathrm{\Delta}ACE:\]

\[CE = AC = a\sqrt{2 + \sqrt{2}};\]

\[AE = \sqrt{AC^{2} + CE^{2}} =\]

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

\[= a\sqrt{4 + 2\sqrt{2}}.\]

\[5)\ В\ \mathrm{\Delta}ADE:\]

\[AD = \sqrt{AE^{2} - DE^{2}} =\]

\[= \sqrt{a^{2}\left( 4 + 2\sqrt{2} \right) - a^{2}} =\]

\[= \sqrt{a^{2}\left( 3 + 2\sqrt{2} \right)} =\]

\[= a\sqrt{2 + 2\sqrt{2} + 1} =\]

\[= a\sqrt{\left( \sqrt{2} + 1 \right)^{2}} = a\left( \sqrt{2} + 1 \right).\]

\[Ответ:\ \ a\sqrt{2 + \sqrt{2}};\ a\left( \sqrt{2} + 1 \right);\ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a}\sqrt{4 + 2\sqrt{2}}.\]