\[\boxed{\mathbf{552.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - трапеция;\]
\[BD \cap AC = O;\]
\[\textbf{а)}\ OB = 4\ см;\]
\[OD = 10\ см;\]
\[DC = 25\ см;\]
\[\textbf{б)}\ AB = a;DC = b;\]
\[\textbf{в)}\ AB = 9,6\ дм;\]
\[DC = 24\ см;\]
\[AC = 15\ см.\]
\[\mathbf{Найти:}\]
\[\textbf{а)}\ AB - ?\]
\[\textbf{б)}\frac{\text{AO}}{\text{OC}} - ?;\frac{\text{BO}}{\text{OD}} - ?\]
\[\textbf{в)}\ AO - ?\]
\[\mathbf{Решение.}\]
\[1)\ DC \parallel AB\ и\ CA - секущая:\]
\[\angle DCA = \angle CAB\ \]
\[(как\ накрестлежащие).\]
\[2)\ \mathrm{\Delta}DOC\sim\mathrm{\Delta}AOB\ \]
\[(по\ двум\ углам):\]
\[\angle DOC = \angle AOB\ \]
\[(как\ вертикальные);\ \]
\[\angle DCA = \angle CAB.\]
\[Отсюда:\]
\[\frac{\text{AB}}{\text{DC}} = \frac{\text{AO}}{\text{OC}} = \frac{\text{OB}}{\text{OD}} = k.\]
\[\textbf{а)}\ \frac{\text{AB}}{25} = \frac{\text{AO}}{\text{OC}} = \frac{4}{10} \Longrightarrow k = 0,4;\]
\[\frac{\text{AB}}{25} = 0,4 \Longrightarrow\]
\[\Longrightarrow AB = 0,4 \bullet 25 = 10\ см.\]
\[\textbf{б)}\ \frac{\text{AB}}{\text{DC}} = \frac{a}{b} \Longrightarrow k = \frac{a}{b};\]
\[\frac{\text{AO}}{\text{OC}} = \frac{\text{BO}}{\text{OD}} = \frac{a}{b}.\]
\[\textbf{в)}\ \frac{96}{24} = \frac{\text{AO}}{\text{OC}} = \frac{\text{OB}}{\text{OD}} \Longrightarrow k = 4;\]
\[\frac{\text{AO}}{AC - AO} = 4\]
\[AO = 4(15 - AO)\]
\[AO = 60 - 4AO\]
\[5AO = 60\]
\[AO = 12\ см.\]
\[\mathbf{Ответ:}а)\ AB = 10\ см;\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ \frac{\text{AO}}{\text{OC}} = \frac{\text{BO}}{\text{OD}} = \frac{a}{b};\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ в)\ AO = 12\ см\mathbf{.}\]
\[\boxed{\mathbf{552.еуроки - ответы\ на\ пятёрку}}\]
\[1)\ S_{плитки} = 15 \bullet 15 = 225\ см^{2}.\]
\[2)\ S_{стены} = 3 \bullet 2,7 = 8,1\ м^{2} =\]
\[= 81000\ см^{2}.\]
\[3)\ S_{стены}\ :S_{плитки} =\]
\[= 81000\ :225 = 360\ (штук) -\]
\[плитки.\]
\[Ответ:для\ облицовки\ стены\ \]
\[необходимо\ 360\ штук\ плитки.\]