\[\boxed{\mathbf{555.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[M \in AB;N \in BC;\]
\[P \in AC;MN \parallel AC;\]
\[NP \parallel AB;\]
\[\textbf{а)}\ AB = 10\ см;\]
\[AC = 15\ см;\]
\[PN\ :MN = 2\ :3;\]
\[\textbf{б)}\ AM = AP;\]
\[AB = a;\]
\[AC = b.\]
\[\mathbf{Найти:}\]
\[MN - ?;AM - ?;\]
\[AP - ?;PN - ?\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\]
\[1)\ AM \parallel NP,\ MN \parallel AP:\]
\[AMNP - параллелограмм\ \]
\[(по\ определению);\]
\[MN = AP\ и\ \]
\[AM = NP\ (по\ свойству).\]
\[2)\ \mathrm{\Delta}ABC\sim\mathrm{\Delta}MBN\ \]
\[(по\ двум\ углам):\]
\[\angle B - общий;\]
\[\angle BMN = \angle BAC\ \]
\[(как\ соответственные).\]
\[3)\ Пусть\ PN = 2x;\ \ \ MN = 3x:\]
\[\frac{10}{10 - 2x} = \frac{15}{3x}\]
\[\frac{5}{5 - x} = \frac{5}{x}\]
\[5 - x = x\]
\[2x = 5.\]
\[x = 2,5\ (см)\text{.\ \ }\]
\(\ PN = 2 \bullet 2,5 = 5\ см.\)
\[MN = 3 \bullet 2,5 = 7,5\ см.\]
\[\textbf{б)}\ \]
\[1)\ AM \parallel NP,\ MN \parallel AP:\]
\[AMNP - параллелограмм\ \]
\[(по\ определению);\]
\[AM = AP.\]
\[Значит:\]
\[AMNP - ромб\ \]
\[(по\ определению).\]
\[2)\ Пусть\ AM = x:\]
\[\frac{\text{AB}}{\text{MB}} = \frac{\text{BC}}{\text{BN}} = \frac{\text{AC}}{\text{MN}} = k\]
\[\frac{a}{a - x} = \frac{b}{x}\]
\[ax = ab - bx\]
\[(a + b)x = ab\]
\[x = \frac{\text{ab}}{a + b}\]
\[MN = AM = \frac{\text{ab}}{a + b}.\]
\[\mathbf{Ответ:}а)\ MN = AP = 7,5\ см;\]
\[AM = NP = 5\ см;\]
\[\textbf{б)}\ MN = AP = AM = NP =\]
\[= \frac{\text{ab}}{a + b}.\]
\[\boxed{\mathbf{555.еуроки - ответы\ на\ пятёрку}}\]
\[\textbf{а)}\ a = 15\ см;h = 12\ см:\]
\[S = 15 \bullet 12 = 180\ см^{2}.\]
\[\textbf{б)}\ S = 34\ см^{2};h = 8,5\ см:\]
\[a = S\ :h = 34\ :8,5 = 4\ см.\]
\[\textbf{в)}\ S = 162\ см^{2};\ h = \frac{1}{2}a:\]
\[S = ah = a \bullet \frac{1}{2}a = \frac{1}{2}a^{2} =\]
\[= 162\ см^{2}\]
\[a^{2} = 324\]
\[\ a = 18\ см.\]
\[\textbf{г)}\ S = 27;h = 3a:\]
\[S = a \bullet h = 3a^{2} = 27\]
\[a^{2} = 9\]
\[a = 3.\]
\[h = 3a = 3 \cdot 3 = 9.\]