Решебник по геометрии 11 класс. Атанасян ФГОС 234

\[\boxed{\mathbf{234.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

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

\[AK = KC;\]

\[\text{KD}D_{1}K_{1}\bot AC;\]

\[AB = 20\ см;\]

\[BC = 21\ см;\]

\[AA_{1} = 42\ см.\]

\[Найти:\]

\[S_{\text{KD}D_{1}K_{1}}.\]

\[Решение.\]

\[1)\ AC = 25\ см;\ \ AK = 14,5\ см.\]

\[2)\ ⊿ABC\ подобен\ ⊿DKC -\]

\[по\ двум\ углам:\]

\[\frac{\text{AB}}{\text{KD}} = \frac{\text{BC}}{\text{KC}}\]

\[KD = \frac{20 \cdot 14,5}{21} = \frac{290}{21}\ см.\]

\[3)\ S_{\text{KD}D_{1}K_{1}} = \frac{290}{21} \cdot 42 = 580\ см^{2}.\]

\[Ответ:580\ см^{2}.\]