Решебник по геометрии 9 класс Мерзляк Задание 48

\[Дано:\]

\[\mathrm{\Delta}ABC;\]

\[\angle B = 120{^\circ};\]

\[BC\ :AB = 5\ :3;\]

\[P_{\text{ABC}} = 30\ см.\]

\[Решение:\]

\[P_{\text{ABC}} = AB + BC + AC = 30\]

\[AB + \frac{5}{3}AB + AC = 30\]

\[AC = 30 - \frac{8}{3}\text{AB.}\]

\[900 - 160AB + \frac{64}{9}AB^{2} = \frac{49}{9}AB^{2}\]

\[\frac{5}{3}AB^{2} - 160AB + 900 = 0\]

\[AB^{2} - 96AB + 540 = 0\]

\[D = 96^{2} - 4 \bullet 540 =\]

\[= 9216 - 2160 = 7056\]

\[AB_{1} = \frac{96 - 84}{2} = 6\ см;\]

\[AB_{2} = \frac{96 + 84}{2} = 90\ см;\]

\[BC_{1} = \frac{5}{3} \bullet 6 = 10\ см;\]

\[BC_{2} = \frac{5}{3} \bullet 90 = 150\ см;\]

\[AC_{1} = 30 - \frac{8}{3} \bullet 6 = 14\ см;\]

\[AC_{2} = 30 - \frac{8}{3} \bullet 90 = - 210\ см.\]

\(Ответ:\ \ 6\ см;\ 10\ см;\ 14\ см.\)