Предмет: Геометрия, автор: benlinus130

На рисунке изображена прямая шестиугольная призма, полученная из прямоугольного параллелепипеда выпиливанием меньшего прямоугольного параллелепипеда с указанием ее размеров...

Приложения:

siestarjoki: 35

siestarjoki: (10 -5/4)*4 =35

benlinus130: 5/4 это откуда?

Ответы

Автор ответа: siestarjoki

Рассмотрим параллелепипед ABCGA1B1C1G1.

Диагонали параллелепипеда пересекаются в одной точке и делятся это точкой пополам.

=> искомое сечение лежит в плоскости BB1G1G.

△ABG: BG=√(AB^2+AG^2) =10

Пусть BG пересекает прямую FE в точке K.

Искомое сечение - прямоугольник BB1K1K.

△FKG~△ABG, KG/BG =FG/AG =1/8

=> BK =7/8 BG =35/4

S(BB1K1K) =BK*BB1 =35

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