ГДЗ по геометрии 9 класс Мерзляк Задание 163

\[Схематический\ рисунок.\]

\[Дано:\]

\[AD - биссектриса\ \angle A;\]

\[AB = 6\ см;\]

\[AC = 8\ см;\]

\[\angle BAC = 120{^\circ}.\]

\[Найти:\]

\[\text{AD.}\]

\[Решение.\]

\[1)\ В\ \mathrm{\Delta}ABC:\]

\[\angle BAD = \angle CAD = \frac{1}{2}\angle A = 60{^\circ}.\]

\[S_{\text{ABC}} = \frac{1}{2}AB \bullet AC \bullet \sin{\angle A} =\]

\[= \frac{1}{2} \bullet 6 \bullet 8 \bullet \sin{120{^\circ}} = 12\sqrt{3}\ см^{2}.\]

\[2)\ В\ \mathrm{\Delta}BAD:\]

\[S_{\text{BAD}} = \frac{1}{2}AB \bullet AD \bullet \sin{\angle BAD} =\]

\[= \frac{1}{2} \bullet 6AD \bullet \sin{60{^\circ}} = 1,5AD\sqrt{3}.\]

\[3)\ В\ \mathrm{\Delta}CAD:\]

\[S_{\text{CAD}} = \frac{1}{2}AC \bullet AD \bullet \sin{\angle CAD} =\]

\[= \frac{1}{2} \bullet 8AD \bullet \sin{60{^\circ}} = 2AD\sqrt{3}.\]

\[S_{\text{ABC}} = S_{\text{BAD}} + S_{\text{CAD}}:\]

\[1,5AD\sqrt{3} + 2AD\sqrt{3} = 12\sqrt{3}\]

\[3,5AD\sqrt{3} = 12\sqrt{3}\]

\[7AD\sqrt{3} = 24\sqrt{3}\]

\[AD = \frac{24}{7}\ см.\]

\[Ответ:\ \ \frac{24}{7}\ см.\]