Боковая сторона равнобокой трапеции равна 18 см а большее основание 32см.угол между ними равен 60 градусов найдите среднюю линию трапеции
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Ответы на вопрос:
Решение: проведем высоты bf и ck получили прямоугольный δabf: ∠a = 60°, ∠f =90°, ∠b = 180 - (60+90)=30° катет напротив угла 30° равен половине гипотенузы af = ab/2 af = 12/2 = 6 см kd = af = 6 см (т.к. трапеция равнобедренная) bc = ad - (af + kd) = 30 - (6 + 6) = 18 mn - средняя линия трапеции mn = (ad + bc) / 2 mn = (30 + 18)/2 mn = 24 см ответ: 24 см
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