ГДЗ по геометрии 9 класс Мерзляк Задание 216

\[Схематический\ рисунок.\]

\[Дано:\]

\[ABCDEF - правильный\]

\[шестиугольник;\]

\[O - центр\ опис.\ окружности;\]

\[AB = a.\]

\[Найти:\]

\[S;\ CF;\ AC.\]

\[Решение.\]

\[1)\ R = \frac{a}{2\sin\frac{180{^\circ}}{6}} = a;\]

\[OC = OF = R = a;\]

\[AF = BC = AB = a.\]

\[6 \bullet \angle A = (6 - 2) \bullet 180{^\circ}\]

\[6\angle A = 720{^\circ}\ \ \ \]

\[\angle A = 120{^\circ}.\]

\[2)\ В\ \mathrm{\Delta}ABC:\]

\[S_{\text{ABC}} = \frac{1}{2}AB \bullet BC \bullet \sin{\angle B} =\]

\[= \frac{1}{2} \bullet a \bullet a \bullet \sin{120{^\circ}} =\]

\[= \frac{1}{2}a^{2} \bullet \frac{\sqrt{3}}{2} = \frac{a^{2}\sqrt{3}}{4}.\]

\[3)\ В\ \mathrm{\Delta}ACF:\]

\[CF = OC + OF = 2a;\]

\[AC = \sqrt{CF^{2} - AF^{2}} =\]

\[= \sqrt{4a^{2} - a^{2}} = a\sqrt{3}.\]

\[4)\ S = 6S_{\text{ABC}} = \frac{3a^{2}\sqrt{3}}{2}.\]

\[Ответ:\ \ a\sqrt{3};\ 2a;\ \frac{3a^{2}\sqrt{3}}{2}.\]