ГДЗ по геометрии 8 класс Атанасян Задание 1020

\[\boxed{\mathbf{1020.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условия\ задачи:\]

\[\mathbf{Дано:}\]

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

\[\textbf{а)}\ AB = 6\sqrt{8}\ см;\]

\[AC = 4\ см;\]

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

\[\textbf{б)}\ BC = 3\ см;\]

\[AB = 18\sqrt{2}\ см;\]

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

\[\textbf{в)}\ AC = 14\ см;\]

\[CB = 7\ см;\]

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

\[\mathbf{Найти:}\]

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

\[\mathbf{Решение.}\]

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

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

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

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

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

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

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

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

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

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

\[\textbf{в)}\ 36,41\ см^{2}.\]