\[\boxed{\mathbf{1289.ОК\ ГДЗ - домашка\ н}а\ 5}\]
\[\mathbf{Дано:}\]
\[EF\bot AB;\]
\[KC = LD;\ \]
\[\angle O = 90{^\circ};\]
\[F - фигура.\]
\[\mathbf{Доказать:}\]
\[S_{F} = \frac{\text{πE}F^{2}}{4}.\]
\[\mathbf{Доказательство.}\]
\[1)\ Пусть\ OA = R;\ OC = r:\ \]
\[AC = DB = \frac{R - r}{2};\ \]
\[EF = R + r.\]
\[2)\ S_{F} =\]
\[= \frac{1}{2}\pi R^{2} + \frac{1}{2}\pi r^{2} - \pi\left( \frac{R - r}{2} \right)^{2} =\]
\[= \frac{\pi}{4}(2\left( R^{2} + r^{2} \right) - (R - r)^{2}) =\]
\[= \frac{\pi}{4}\ \left( 2R^{2} + 2r^{2} - R^{2} + 2Rr - r^{2} \right) =\]
\[= \frac{\pi}{4}\left( R^{2} + 2Rr + r^{2} \right) =\]
\[= \frac{\pi}{4}(R + r)^{2} = \frac{\pi}{4} \cdot EF^{2}.\]
\[Что\ и\ требовалось\ доказать.\]