Решебник по геометрии 9 класс Мерзляк Задание 221

\[a_{8} = a;\ \ \ r_{8} = R_{4}.\]

\[1)\ tg\ 22,5{^\circ} = \sqrt{\frac{1 - \cos{45{^\circ}}}{1 + \cos{45{^\circ}}}} =\]

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

\[= \sqrt{\frac{\left( \sqrt{2} - 1 \right)^{2}}{2 - 1}} = \sqrt{2} - 1.\]

\[2)\ Для\ вписанной\ окружности:\]

\[r_{8} = \frac{a}{2\ tg\ \frac{180{^\circ}}{8}} = \frac{a}{2\ tg\ 22,5{^\circ}} =\]

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

\[3)\ Для\ описанной\ окружности:\]

\[a_{4} = 2R_{4}\sin\frac{180{^\circ}}{4} = 2R_{4}\sin{45{^\circ}} =\]

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

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