Решебник по геометрии 9 класс Мерзляк Задание 125

\[Схематический\ рисунок.\]

\[Дано:\]

\[AB = BC = 20\ см;\]

\[CM - медиана;\]

\[AD - биссектриса\ \angle A;\]

\[\angle A = 70{^\circ}.\]

\[Найти:\]

\[1)\ AC;\ \]

\[2)\ CM;\]

\[3)\ AD;\ \]

\[4)\ R_{опис.\ окр.}.\]

\[Решение.\]

\[1)\ В\ \mathrm{\Delta}ABC:\]

\[AB = BC;\ \ \ \]

\[\angle C = \angle A = 70{^\circ};\]

\[\angle B = 180{^\circ} - 70{^\circ} - 70{^\circ} = 40{^\circ};\]

\[= 400 + 400 - 2 \bullet 20 \bullet 20 \bullet \cos{40{^\circ}} =\]

\[= 800 - 800 \bullet 0,7660 = 187,2;\]

\[AC = \sqrt{187,2} \approx 13,7\ см.\]

\[2)\ В\ \mathrm{\Delta}AMC:\]

\[AM = \frac{1}{2}AB = \frac{1}{2} \bullet 20 = 10\ см;\]

\[= 287,2 - 274 \bullet 0,3420 = 193,5;\]

\[CM = \sqrt{193,5} \approx 13,9\ см.\]

\[3)\ В\ \mathrm{\Delta}ADC:\]

\[\angle CAD = \frac{1}{2}\angle A = \frac{1}{2} \bullet 70{^\circ} = 35{^\circ};\]

\[\angle ADC = 180{^\circ} - 35{^\circ} - 70{^\circ} = 75{^\circ};\]

\[\frac{\text{AD}}{\sin{\angle C}} = \frac{\text{AC}}{\sin{\angle ADC}}\text{\ \ \ }\]

\[AD = \frac{AC \bullet \sin{\angle C}}{\sin{\angle ADC}}\]

\[AD = \frac{13,7 \bullet \sin{70{^\circ}}}{\sin{75{^\circ}}} \approx 13,3\ см.\]

\[4)\ В\ \mathrm{\Delta}ABC:\]

\[R = \frac{\text{AB}}{2\sin{\angle C}} = \frac{20}{2\sin{70{^\circ}}} \approx 10,6\ см.\]

\[Ответ:\ \ 1)\ 13,7\ см;\ 2)\ 13,9\ см;\ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3)\ 13,3\ см;\ 4)\ 10,6\ см.\]