ГДЗ по геометрии 7 класс Атанасян Задание 1071

\[\boxed{\mathbf{1071.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

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

\[\angle A < 90{^\circ};\]

\[S_{\text{ABC}} = 3\sqrt{3};\]

\[AB = 4\sqrt{3};\]

\[AC = 3.\]

\[\mathbf{Найти:}\]

\[R - ?\]

\[\mathbf{Решение.}\]

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

\[3\sqrt{3} = \frac{1}{2} \bullet 4\sqrt{3} \bullet 3\sin{\angle A}\]

\[3\sqrt{3} = 6\sqrt{3}\sin{\angle A}\]

\[\sin{\angle A} = \frac{3\sqrt{3}}{6\sqrt{3}} = \frac{1}{2}\]

\[\angle A = 30{^\circ}.\]

\[2)\ По\ теореме\ косинусов:\]

\[BC^{2} =\]

\[= AC^{2} + AB^{2} - 2AC \bullet AB \bullet \cos{30{^\circ}}\]

\[BC^{2} = 9 + 48 - 24\sqrt{3} \bullet \frac{\sqrt{3}}{2} =\]

\[= 57 - 12 \bullet 3 = 57 - 36 = 21\]

\[BC = \sqrt{21}.\]

\[3)\ По\ теореме\ синусов:\]

\[\frac{\text{BC}}{\sin{\angle A}} = 2R\]

\[\frac{\sqrt{21}}{\sin{30{^\circ}}} = 2R\]

\[\sqrt{21}\ :\frac{1}{2} = 2R\]

\[R = \frac{2\sqrt{21}}{2} = \sqrt{21}.\]

\[Ответ:\ R = \sqrt{21}.\]