\[\boxed{\mathbf{234.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[\angle ACB = 90{^\circ};\]
\[AK = KC;\]
\[\text{KD}D_{1}K_{1}\bot AC;\]
\[AB = 20\ см;\]
\[BC = 21\ см;\]
\[AA_{1} = 42\ см.\]
\[Найти:\]
\[S_{\text{KD}D_{1}K_{1}}.\]
\[Решение.\]
\[1)\ AC = 25\ см;\ \ AK = 14,5\ см.\]
\[2)\ ⊿ABC\ подобен\ ⊿DKC -\]
\[по\ двум\ углам:\]
\[\frac{\text{AB}}{\text{KD}} = \frac{\text{BC}}{\text{KC}}\]
\[KD = \frac{20 \cdot 14,5}{21} = \frac{290}{21}\ см.\]
\[3)\ S_{\text{KD}D_{1}K_{1}} = \frac{290}{21} \cdot 42 = 580\ см^{2}.\]
\[Ответ:580\ см^{2}.\]