\[\boxed{\mathbf{547.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}\text{ABC}\sim\mathrm{\Delta}A_{1}B_{1}C_{1}.\]
\[\mathbf{Доказать:}\]
\[\frac{P_{\text{ABC}}}{P_{A_{1}B_{1}C_{1}}} = k.\]
\[\mathbf{Доказательство.}\]
\[1)\ \mathrm{\Delta}\text{ABC}\sim\mathrm{\Delta}A_{1}B_{1}C_{1}\ (по\ условию):\]
\[\frac{\text{AB}}{A_{1}B_{1}} = \frac{\text{BC}}{B_{1}C_{1}} = \frac{\text{AC}}{A_{1}C_{1}} = k.\]
\[2)\ \text{AB} = k \bullet A_{1}B_{1};\]
\[\text{BC} = k \bullet B_{1}C_{1};\]
\[\text{AC} = k \bullet A_{1}C_{1}.\]
\[3)\ P_{\text{ABC}} =\]
\[= k \bullet A_{1}B_{1} + k \bullet B_{1}C_{1} + k \bullet A_{1}C_{1};\]
\[P_{\text{ABC}} = k\left( A_{1}B_{1} + B_{1}C_{1} + A_{1}C_{1} \right).\]
\[4)\ P_{A_{1}B_{1}C_{1}} = A_{1}B_{1} + B_{1}C_{1} + A_{1}C_{1}.\]
\[5)\frac{P_{\text{ABC}}}{P_{A_{1}B_{1}C_{1}}} =\]
\[= \frac{k\left( A_{1}B_{1} + B_{1}C_{1} + A_{1}C_{1} \right)}{A_{1}B_{1} + B_{1}C_{1} + A_{1}C_{1}} = k.\]
\[Что\ и\ требовалось\ доказать.\]
\[\boxed{\mathbf{547.еуроки - ответы\ на\ пятёрку}}\]
\[\textbf{а)}\ 1\ см^{2} = 100\ мм^{2}:\]
\[S = 24\ см^{2} = 24 \bullet 100 =\]
\[= 2400\ мм^{2}.\]
\[\textbf{б)}\ 1\ см^{2} = 0,01\ дм^{2}:\]
\[S = 24\ см^{2} = 24\ :100 =\]
\[= 0,24\ дм^{2}.\]