Решебник по геометрии 9 класс Мерзляк Задание 188

\[R_{1} = \frac{6\sqrt{3}}{2\sin{60{^\circ}}} = 3\sqrt{3}\ :\frac{\sqrt{3}}{2} = 6;\ \ \ \]

\[r_{1} = \frac{6\sqrt{3}}{2\ tg\ 60{^\circ}} = 3\sqrt{3}\ :\sqrt{3} = 3;\]

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

\[= 8\sqrt{3} \bullet \frac{\sqrt{3}}{2} = 12;\ \ \]

\[r_{2} = \frac{12}{2\ tg\ 60{^\circ}} = \frac{6}{\sqrt{3}} = 2\sqrt{3};\]

\[a_{3} = 2 \bullet 2\ tg\ 60{^\circ} = 4\sqrt{3};\ \ \]

\[R_{3} = \frac{4\sqrt{3}}{2\sin{60{^\circ}}} = 2\sqrt{3}\ :\frac{\sqrt{3}}{2} = 4.\]

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