\[\boxed{\mathbf{622.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - параллелограмм;\]
\[AK = \frac{1}{4}KD;\]
\[S_{\text{APK}} = 1\ см^{2}.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABCD}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ \mathrm{\Delta}APK\sim\mathrm{\Delta}CBP\ \]
\[(по\ двум\ углам):\]
\[\angle APK = \angle CPB\ \]
\[(как\ вертикальные);\ \]
\[\angle A = \angle C\ \]
\[(как\ накрестлежащие).\]
\[Отсюда:\]
\[\frac{\text{BP}}{\text{PK}} = \frac{\text{BC}}{\text{AK}}.\]
\[2)\ Так\ как\ AK = \frac{1}{4}KD:\ \]
\[AK = \frac{1}{5}AD = \frac{1}{5}\text{BC}\]
\[\frac{\text{BC}}{\text{AK}} = 5;\]
\[\frac{\text{BP}}{\text{PK}} = 5.\]
\[3)\ \frac{\mathrm{\Delta}APK}{\mathrm{\Delta}APB} = \frac{\text{BP}}{\text{PK}} = 5\ \]
\[(так\ как\ PK - общая\ высота):\ \]
\[S_{\text{APB}} = 5S_{\text{APK}} = 5\ см^{2}.\]
\[4)\ \mathrm{\Delta}APK\sim\mathrm{\Delta}CBP\ и\ k = \frac{\text{BC}}{\text{AK}} = 5:\]
\[\frac{S_{\text{CPB}}}{S_{\text{APK}}} = 25\]
\[S_{\text{CPB}} = 25\ см^{2}.\]
\[5)\ S_{\text{ABC}} = S_{\text{APB}} + S_{\text{CPB}} =\]
\[= 5 + 25 = 30\ см^{2}.\]
\[6)\ S_{\text{ABCD}} = 2 \bullet S_{\text{ABC}} = 60\ см^{2}.\]
\[\mathbf{Ответ:\ }\ S_{\text{ABCD}} = 60\ см^{2}.\]