Решебник по геометрии 8 класс Атанасян ФГОС Задание 1109

 

\[\boxed{\mathbf{1109.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Дано:\]

\[R = 6\ см\ \ и\ \ \text{α.}\]

\[Найти:\]

\[l - ?.\]

\[Решение.\]

\[l = \frac{\text{πR}}{180{^\circ}} \bullet \alpha.\]

\[\textbf{а)}\ \alpha = 30{^\circ}:\]

\[l = \frac{6\pi}{180{^\circ}} \bullet 30{^\circ} = \frac{6\pi}{6} = \pi\ см;\]

\[\textbf{б)}\ \alpha = 45{^\circ}:\]

\[l = \frac{6\pi}{180{^\circ}} \bullet 45{^\circ} = \frac{6\pi}{4} = \frac{3}{2}\text{π\ }см;\]

\[\textbf{в)}\ \alpha = 60{^\circ}:\]

\[l = \frac{6\pi}{180{^\circ}} \bullet 60{^\circ} = \frac{6\pi}{3} = 2\pi\ см;\]

\[\textbf{г)}\ \alpha = 90{^\circ}:\]

\[l = \frac{6\pi}{180{^\circ}} \bullet 90{^\circ} = \frac{6\pi}{2} = 3\pi\ см.\]

\[\boxed{\mathbf{1109.еуроки - ответы\ на\ пятёрку}}\]

\[Рисунок\ по\ условия\ задачи:\]

\[\mathbf{Дано:}\]

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

\[\textbf{а)}\ AB = 6\sqrt{8}\ см;\]

\[AC = 4\ см;\]

\[\angle A = 60{^\circ};\]

\[\textbf{б)}\ BC = 3\ см;\]

\[AB = 18\sqrt{2}\ см;\]

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

\[\textbf{в)}\ AC = 14\ см;\]

\[CB = 7\ см;\]

\[\angle C = 48{^\circ}.\]

\[\mathbf{Найти:}\]

\[S_{\text{ABC}} - ?\]

\[\mathbf{Решение.}\]

\[\textbf{а)}\ S_{\text{ABC}} = \frac{1}{2} \bullet AB \bullet AC \bullet \sin{\angle A};\]

\[S_{\text{ABC}} = \frac{1}{2} \bullet 6\sqrt{8} \bullet 4 \bullet \sin{60{^\circ}} =\]

\[= 12\sqrt{8} \bullet \frac{\sqrt{3}}{2} = 12\sqrt{6}\ см^{2}.\]

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

\[S_{\text{ABC}} = \frac{1}{2} \bullet 3 \bullet 18\sqrt{2} \bullet \sin{45{^\circ}} =\]

\[= 27\sqrt{2} \bullet \frac{\sqrt{2}}{2} = 27\ см^{2}.\]

\[\textbf{в)}\ S_{\text{ABC}} = \frac{1}{2} \bullet AC \bullet BC \bullet \sin{\angle C};\]

\[S_{\text{ABC}} = \frac{1}{2} \bullet 14 \bullet 7 \bullet \sin{48{^\circ}} =\]

\[= 49 \bullet 0,74 \approx 36,41\ см^{2}.\]

\[Ответ:а)12\sqrt{6}\ см^{2};\ б)\ 27\ см^{2};\ \]

\[\textbf{в)}\ 36,41\ см^{2}.\]