\[\boxed{\mathbf{628.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD,\ AEFK - квадраты;\]
\[AB = AE = a.\]
\[\mathbf{Найти:}\]
\[S_{\text{AEQD}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{AEQD}} = S_{\text{ACD}} - S_{\text{ECQ}}.\]
\[2)\ S_{\text{ACD}} = \frac{1}{2}AD \bullet DC = \frac{1}{2}a \bullet a =\]
\[= \frac{a^{2}}{2}.\]
\[3)\ AC^{2} = AD^{2} + DC^{2} =\]
\[= a^{2} + a^{2} = 2a^{2}\]
\[AC = a\sqrt{2}.\]
\[4)\ EC = AC - AE = a\sqrt{2} - a =\]
\[= a\left( \sqrt{2} - 1 \right).\]
\[5)\ ⊿ECQ - прямоугольный\ \]
\[(так\ как\ AC\bot EF):\]
\[\angle ECQ = 90{^\circ}:2 = 45{^\circ};\]
\[отсюда:\]
\[⊿ECQ - равнобедренный;\]
\[EQ = EC.\]
\[6)\ S_{\text{ECQ}} = \frac{1}{2}QE \bullet EC = \frac{1}{2} \bullet EC^{2}\]
\[S_{\text{ECQ}} = \frac{1}{2}a^{2}\left( \sqrt{2} - 1 \right)^{2} =\]
\[= \frac{1}{2}a^{2}\left( 2 - 2\sqrt{2} + 1 \right) =\]
\[= \frac{1}{2}a^{2}\left( 3 - 2\sqrt{2} \right).\]
\[7)\ S_{\text{AEQD}} = \frac{a^{2}}{2} - \frac{a^{2}}{2}\left( 3 - 2\sqrt{2} \right) =\]
\[= \frac{a^{2}}{2}\left( 1 - 3 + 2\sqrt{2} \right) =\]
\[= \frac{a^{2}}{2}\left( 2\sqrt{2} - 2 \right) = a^{2}\left( \sqrt{2} - 1 \right).\]
\[Ответ:a^{2}\left( \sqrt{2} - 1 \right).\]