ГДЗ по геометрии 9 класс Мерзляк Задание 223

\[Шестиугольник\ ABCDEF.\]

\[Схематический\ рисунок.\]

\[Решение.\]

\[1)\ В\ ABCDEF:\]

\[\angle ACD = \frac{1}{2} \cup AD = \frac{1}{2} \bullet 180{^\circ} = 90{^\circ};\]

\[\angle D = \frac{(6 - 2) \bullet 180{^\circ}}{6} = 120{^\circ};\]

\[\angle D = \angle E = \angle B = 120{^\circ}.\]

\[2)\ В\ \mathrm{\Delta}KDE - равносторонний:\]

\[\angle D = \angle E = 180{^\circ} - 120{^\circ} = 60{^\circ}.\]

\[Отсюда:\]

\[DK = KE = DE = 1.\]

\[3)\ В\ \mathrm{\Delta}ABC:\]

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

\[= 1 + 1 - 2 \bullet 1 \bullet 1 \bullet \cos{120{^\circ}} =\]

\[= 2 + 2 \bullet \frac{1}{2} = 2 + 1 = 3;\ \ \ \]

\[AC = \sqrt{3}.\]

\[4)\ В\ \mathrm{\Delta}ACK:\]

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

\[CK = CD + DK = 2;\]

\[AK = \sqrt{CK^{2} + AC^{2}} = \sqrt{4 + 3} = \sqrt{7}.\]

\[Ответ:\ \ AK - искомый\ отрезок.\]