Решебник по геометрии 9 класс Мерзляк Задание 220

\[a_{12} = a;\ \ \ r_{12} = R_{6}.\]

\[Решение.\]

\[1)\ tg\ 15{^\circ} = \sqrt{\frac{1 - \cos{30{^\circ}}}{1 + \cos{30{^\circ}}}} =\]

\[= \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{1 + \frac{\sqrt{3}}{2}}} = \sqrt{\frac{2 - \sqrt{3}}{2 + \sqrt{3}}} =\]

\[= \sqrt{\frac{\left( 2 - \sqrt{3} \right)^{2}}{4 - 3}} = 2 - \sqrt{3}.\]

\[2)\ Для\ вписанной\ окружности:\]

\[r_{12} = \frac{a}{2\ tg\ \frac{180{^\circ}}{12}} = \frac{a}{2\ tg\ 15{^\circ}} =\]

\[= \frac{a}{2\left( 2 - \sqrt{3} \right)} = \frac{a\left( 2 + \sqrt{3} \right)}{2}.\]

\[3)\ Для\ описанной\ окружности:\]

\[a_{6} = 2R_{6}\sin\frac{180{^\circ}}{6} = 2R_{6}\sin{30{^\circ}} =\]

\[= 2 \bullet \frac{a\left( 2 + \sqrt{3} \right)}{2} \bullet \frac{1}{2} = \frac{a\left( 2 + \sqrt{3} \right)}{2}.\]

\[Ответ:\ \ \frac{a\left( 2 + \sqrt{3} \right)}{2}.\]