ГДЗ по геометрии 7 класс Атанасян Задание 1098

\[\boxed{\mathbf{1098.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[\mathrm{\Delta}ABC - правильный;\]

\[окружность\ (O;r);\]

\[окружность\ (O;R).\]

\[\mathbf{Найти:}\]

\[a_{3};P\ и\ S:\]

\[\textbf{а)}\ через\ r;\]

\[\textbf{б)}\ через\ R.\]

\[\mathbf{Решение.}\]

\[\textbf{а)}1)\ r = R \bullet \cos\frac{180{^\circ}}{3}\]

\[r = R \bullet \cos{60{^\circ}} = R \bullet \frac{1}{2}\]

\[R = 2r.\]

\[2)\ a_{3} = 2R \bullet \sin\frac{180{^\circ}}{3} =\]

\[= 2 \bullet 2r \bullet \sin{60{^\circ}} = 4r \bullet \frac{\sqrt{3}}{2} =\]

\[= 2\sqrt{3}\text{r.}\]

\[3)\ P = 3 \bullet 2\sqrt{r} = 6\sqrt{3}\text{r.}\]

\[4)\ S = \frac{1}{2}P \bullet r = \frac{1}{2} \bullet 6\sqrt{3} \bullet r \bullet r =\]

\[= 3\sqrt{3}r^{2}.\]

\[\textbf{б)}\ 1)\ r = R \bullet \cos\frac{180{^\circ}}{3} =\]

\[= R \bullet \cos{60{^\circ}} = \frac{R}{2}.\]

\[2)\ a_{3} = 2R \bullet \sin\frac{180{^\circ}}{3} =\]

\[= 2R \bullet \sin{60{^\circ}} = 2R\frac{\sqrt{3}}{2} = \sqrt{3}\text{R.}\]

\[3)\ P = 3 \bullet \sqrt{3}R = 3\sqrt{3}\text{R.}\]

\[4)\ S = \frac{1}{2}P \bullet r = \frac{1}{2} \bullet 3\sqrt{3}R \bullet \frac{R}{2} =\]

\[= \frac{3\sqrt{3}}{4}R^{2}.\]

\[\mathbf{Ответ:}\mathbf{а})\ a_{3} = 2\sqrt{3}r;P = 6\sqrt{3}r;\]

\[S = 3\sqrt{3}r^{2};\]

\[\ б)\ a_{3} = \sqrt{3}R;P = 3\sqrt{3}R;\ \]

\[S = \frac{3\sqrt{3}}{4}R^{2}.\]