Предмет: Алгебра, автор: aoprya15

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В треугольнике а BC угол а равен 45°, BC равен 10 см, высота BD делит сторону а цены отрезке AD = 6 см, BC = 8 см. Найдите площадь треугольника и высоту, проведённая к стороне BA

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Автор ответа: gels5344Inetka

＜BDC＝90°
т.піфагора
ВС2＝ВD2+DC2
BD2＝100-64＝36
BD＝6
S＝1/2aha
S＝1/2*14*6＝42cm2

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