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1. S = ½×(4+8)×5 = ½×6×5 = 3×5 = 15 см².
2. S=150, h=S:(½×(a+b)) = 150:(½×(9+11)) = 150:(½×20) = 150:10 = 15 см.
3. Пусть высота будет BH(нужно отметить Н на рисунке). Проведём высоту из точки С, будет она СЕ. Т.к. трапеция равнобедренная, то АН=DE. AH=BH=4 см, ведь угол А=45°, угол Н=90°, соответственно угол В=45° и треугольникк АВН равнобедренный. Из этого, AD=4+5+4 = 13 см.
Найдём площадь: S=½×(5+13)×4 = ½×18×4 = 9×4 = 36 см².
4. Пусть одна часть будет х, тогда BC=3x, AD=4x.
S=½×(3x+4x)×5 = ½×7x×5 = 3,5x×5 = 17,5x -> 17,5x = 35.
x=2 см.
AD=4x = 4×2 = 8 см.
ответ: V=6000см³; Sсеч=400см²; Sосн=300см²; Sбок.пов=1600см²;
Sпол=2200см², высота=20см
Объяснение: найдём площадь основания параллелепипеда. Диагональ ВД делит основание на 2 равных треугольника, в которых известны стороны. Рассмотрим полученный ∆АВД. В нём АВ=15см, ВД=20см, АД=25см. Найдём периметр этого треугольника:
Р=15+20+25=60см; р/2=60/2=30см.
Найдём площадь ∆АВД по формуле:
S∆АВД=√((р(р-АВ)(р-ВД)(р-АД))=
=√((30(30-15)(30-20)(30-25))=
=√(30×15×10×5)=√22500=150см²
S∆АВД=S∆ВСД=150см²
Зная площади 2-х треугольников, найдём площадь основания АВСД:
Sосн=150×2=300см².
Так как диагональное сечение параллелепипеда - квадрат, то
ВД=В1Д1=ВВ1=ДД1=20см
Площадь диагонального сечения:
Sсеч=20²=400см²
ВВ1, ДД1, АА1, СС1 также являются высотами параллелепипеда, поэтому мы може найти площадь каждой боковой грани по формуле прямоугольника:
Sбок.гр.АА1ДД1=АД×АА1=25×20=500см²
Sбок.гр.АА1ДД1=Sбок.грВВ1СС1=500см²
Sбок.гр.АА1В1В=Sбок.гр.ДД1С1С=
=15×20=300см²
Так как каждой грани по 2 найдём площадь боковой поверхности параллелепипеда:
Sбок.пов=500×2+300×2=1000+600=
=1600см²
Площадь полной поверхности параллелепипеда- это сумма всех площадей его граней:
Sпол=Sбок.пов.+ S2-х.осн=1600+300×2=
=1600+600=2200см²
Теперь найдём объем параллелепипеда по формуле: V=Sосн×АА1=300×20=6000см³