Решебник по геометрии 9 класс Мерзляк Задание 217

\[Схематический\ рисунок.\]

\[Дано:\]

\[ABCD - квадрат;\]

\[AD = 6\ см;\]

\[EF = FM = MN.\]

\[Найти:\]

\[\text{EF.}\]

\[Решение.\]

\[1)\ ABCD - квадрат:\]

\[AF = MD = AE = DN;\]

\[AD = AF + FM + MD;\]

\[AF + EF + AF = 6\]

\[EF = 6 - 2AF.\]

\[2)\ В\ \mathrm{\Delta}AEF:\]

\[\angle EAF = 90{^\circ};\ \ \ \]

\[EF = \sqrt{AE^{2} + AF^{2}} =\]

\[= \sqrt{AF^{2} + AF^{2}} = AF\sqrt{2}.\]

\[\text{AF}\sqrt{2} = 6 - 2AF\]

\[\text{AF}\left( 2 + \sqrt{2} \right) = 6\ \ \ \]

\[AF = \frac{6}{2 + \sqrt{2}}.\]

\[EF = \frac{6\sqrt{2}}{2 + \sqrt{2}} = \frac{12}{2\sqrt{2} + 2} =\]

\[= \frac{6}{\sqrt{2} + 1} = 6\left( \sqrt{2} - 1 \right).\]

\(Ответ:\ \ 6\left( \sqrt{2} - 1 \right)\ см.\)