\[\boxed{\mathbf{564.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[AB = 8\ см;\]
\[AC = 7\ см;\]
\[A_{1} \in BC;\]
\[B_{1} \in AC;\]
\[C_{1} \in AB;\]
\[A_{1},B_{1},C_{1} - середина.\]
\[\mathbf{Найти:}\]
\[P_{A_{1}B_{1}C_{1}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ A_{1},B_{1},C_{1} - середины\ сторон\ \]
\[треугольника:\]
\[A_{1}C_{1},\ \text{\ B}_{1}C_{1},\text{\ A}_{1}B_{1} - среднии\ \]
\[линии\ \mathrm{\Delta}\text{ABC.}\]
\[A_{1}C_{1} = \frac{\text{AC}}{2} = \frac{7}{2} = 3,5\ см.\]
\[B_{1}C_{1} = \frac{\text{BC}}{2} = \frac{5}{2} = 2,5\ см.\]
\[A_{1}B_{1} = \frac{\text{AB}}{2} = \frac{8}{2} = 4\ см;\]
\[2)\ P_{A_{1}B_{1}C_{1}} =\]
\[= A_{1}C_{1} + B_{1}C_{1} + A_{1}B_{1} =\]
\[= 3,5 + 2,5 + 4 = 10\ см.\]
\[Ответ:P_{A_{1}B_{1}C_{1}} = 10\ см.\]
\[\boxed{\mathbf{564.еуроки - ответы\ на\ пятёрку}}\]
\[\textbf{а)}\ a = 7\ см;\ \ h = 11\ см:\]
\[S = \frac{1}{2} \bullet ah = \frac{1}{2} \bullet 7 \bullet 11 =\]
\[= 38,5\ см^{2}.\]
\[\textbf{б)}\ a = 2\sqrt{3};\ \ h = 5\ см:\]
\[S = \frac{1}{2} \bullet ah = \frac{1}{2} \bullet 2\sqrt{3} \bullet 5 =\]
\[= 5\sqrt{3}\ см^{2}.\]
\[\textbf{в)}\ S = 37,8\ см^{2};\ a = 14\ см:\]
\[h = \frac{2S}{a} = \frac{2 \bullet 37,8}{14} = 5,4\ см.\]
\[\textbf{г)}\ S = 12\ см^{2};\ \ \ h = 3\sqrt{2}\ см:\]
\[a = \frac{2S}{h} = \frac{2 \bullet 12}{3\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}.\]