fie trapezul dreptunghic ABCD, BC║AD, ∡C=∡D=90°, ∡BAD=45°
AD=8 cm
AB=4√2 cm
ducem BE⊥AD, E∈AD
tr. ABE e dreptunghic isoscel pt. ca are ∡ de 45°, rezulta:
BE=AE si cu pitagora
AB^2=2BE^2
4√2=BE√2
BE=AE=4 cm
BC=AD-AE=8-4=4 cm
perimetru
P=AD+DC+BC+AB=AD+BE+BC+AB=8+4+4+4√2=4(4+√2) cm