Предмет: Геометрия, автор: mea2508041

в основании прямой призмы лежит ромб с диагоналями равными 16 и 30 определите площадь полной поверхности если боковые 20 см​

Приложения:

Аноним: Теорема Пифагора √(8²+15²)=17. Р=17*4=68. Sбок=Росн*20=68*20=... Sосн=16*30/2=... Sпол=2*Sосн+Sбок=...

Ответы

Автор ответа: Аноним
1

Ответ:

1840см²

Объяснение:

Диагонали ромба пересекаются перпендикулярно и точкой пересечения делятся пополам.

АО=АС:2=30/2=15см

ВО=BD:2=16/2=8/см.

∆АВО- прямоугольный треугольник

По теореме Пифагора

АВ=√(АО²+ВО²)=√(15²+8²)=√(225+64)=

=√289=17см.

Р(ABCD)=4*AB=4*17=68см периметр основания.

Sбок=Р(ABCD)*AA1=68*20=1360см².

S(ABCD)=1/2*AC*BD=1/2*16*30=240см² площадь ромба

Sпол=2*S(ABCD)+Sбок=2*240+1360=

=480+1360=1840см²

Приложения:

bekk18929292: помогитее если не сложно пожалуйста
mea2508041: спасибо большое
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