Найдите периметр и площадь прямоугольной трапеции, основания которой составляют 8 см и 12 см, а углы - 135 градусов буду плакать

S=40см², Р=4√2+20=29,64см

Объяснение:

обозначим вершины трапеции А В С Д с основаниями ВС и АД и тупым углом В=135°. Так как сумма углов трапеции, прилегающих к одной боковой стороне составляет 180°, то <А=180–135=45°

Проведём к нижнему основанию АД высоту ВН. Она делит АД так что НД=ВС=8см, тогда АН=12–8=4см. Рассмотрим полученный ∆АВН. Он прямоугольный и АН и ВН катеты а АВ - гипотенуза. В прямоугольном треугольнике сумма острых углов составляет 90°, поэтому <АВН=90–45°=45°, значит <АВН=<А=45°, следовательно ∆АВН прямоугольный равнобедренный, поэтому АН=ВН=4см. Площадь трапеции вычисляется по формуле:

вставим в формулу наши данные:

Итак: S=40см²

Теперь в ∆АВН найдём гипотенузу АВ, которая является боковой стороной трапеции, по теореме Пифагора:

АВ²=АН²+ВН²=4²+4²=16+16=32; АВ=СД=√32=4√2см. Высота ВН=СД=4 см. Теперь найдём периметр трапеции зная её стороны: Р=4√2+8+12+4=4√2+24см

Если нужно вычислить полностью, то √2≈1,41, тогда:

Р=4×1,41+24=5,64+20=29,64см

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Дано: АВСВ - паралелограм, Р=48 см., СМ - бісектриса,  DM : MA = 5 : 2. Знайти АВ=СD,  AD=BC.

∠BCM=∠MCD за умовою

∠CMD=∠МСВ як внутрішні різносторонні при АD║ВС та січній  СМ, отже

∠CMD=∠MCD,  ΔMCD - рівнобедрений,  СD=MD.

Нехай АМ=2х см,  МD=5х см,  СD=5х см,  ВС=2х+5х=7х см,  АВ=5х см.

2(5х+7х)=48

12х=24;  х=2.

АВ=СD=5*2=10 см,  AD=BC=7*2=14 см.