Решебник по геометрии 9 класс Мерзляк Задание 126

\[Схематический\ рисунок.\]

\[Дано:\]

\[ABCD - трапеция;\]

\[AC = 8\ см;\]

\[AB = CD;\]

\[\angle CAD = 38{^\circ};\]

\[\angle BAD = 72{^\circ}.\]

\[Найти:\]

\[1)\ AB,\ BC,\ AD;\]

\[2)\ R_{опис.\ окр.}.\]

\[Решение.\]

\[1)\ Рассмотрим\ ABCD:\]

\[\angle B = 180{^\circ} - \angle A =\]

\[= 180{^\circ} - 72{^\circ} = 108{^\circ};\]

\[\angle BAC = \angle A - \angle CAD =\]

\[= 72{^\circ} - 38{^\circ} = 34{^\circ};\]

\[\angle BCA = 180{^\circ} - \angle B - \angle BAC = 38{^\circ};\]

\[\angle C = \angle B = 108{^\circ};\ \ \ \]

\[\angle D = \angle A = 72{^\circ}.\]

\[2)\ В\ \mathrm{\Delta}ABC:\]

\[AB = \frac{AC \bullet \sin{\angle BCA}}{\sin{\angle ABC}} =\]

\[= \frac{8 \bullet \sin{38{^\circ}}}{\sin{108{^\circ}}} \approx 5,2\ см;\]

\[BC = \frac{AC \bullet \sin{\angle BAC}}{\sin{\angle ABC}} =\]

\[= \frac{8 \bullet \sin{34{^\circ}}}{\sin{108{^\circ}}} \approx 4,7\ см.\]

\[3)\ В\ \mathrm{\Delta}ACD:\]

\[\angle ACD = \angle C - \angle BCA =\]

\[= 108{^\circ} - 38{^\circ} = 70{^\circ};\]

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

\[= \frac{8 \bullet \sin{70{^\circ}}}{\sin{72{^\circ}}} \approx 7,9\ см.\]

\[4)\ В\ \mathrm{\Delta}ABC:\]

\[R = \frac{\text{AC}}{2\sin{\angle ABC}} =\]

\[= \frac{8}{2 \bullet \sin{108{^\circ}}} = 4,2\ см.\]

\[Ответ:\ \ \]

\[1)\ 5,2\ см;\ 4,7\ см;\ 5,2\ см;\ 7,9\ см;\ \]

\[2)\ 4,2\ см.\]