Предмет: Геометрия, автор: julia14789

В прямом параллелепипеде АВСДА1В1С1Д1 ОСНОВАНИЕМ СЛУЖИТ КВАДРАТ. Найдите объем параллелепипеда если площадь сечения, проходящего через диагонали основания АС и вершину В1, равна S и это сечение наклонено к плоскости основания под углом 60 градусов.

Ответы

Автор ответа: ffAnon

Sabc=S*cos60 = S/2

Sabcd = S/2+S/2 = S

тогда AB=BC = sqrt(S)

B1C = 2*BC, так как BC - катет лежащий против угла в 30 градусов а B1C - гипотенуза

B1B=sqrt(3*S)

V=S*sqrt(3*S)

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