a)
Construiesti inaltimea trapezului CM
DC//AM
DC=12 cm⇒AM=12cm
MB= AM-AM
MB=20-12=8cm
In ΔCMB, m<M=90
m<B=45
MB=8cm⇒ΔCMB- dreptunghic isocel⇒CM=8cm
In ΔΔCMB, m<M=90
MB=8cm
CM=8cm⇒Th Pitagora CB²=CM²+MB²⇒CB=8√2
CM//DA
CM=8cm⇒DA=8cm
P ABCD= 12+8√2+20+8=40+8√2=8(5+√2)
A ABCD=[tex] \frac{(AB+DC)*CM}{2} = \frac{(20+12)*8}{2} =128[/tex]