CD/4=AB/6=k
CD=4k
AB=6k
CE⊥AB ⇒ AECB este dreptunghi
DC=AE⊥AC=CE=12√2
EB=AB-DC=6k-4k=2k
ΔACB⇒teorema inaltimii CE²+AE²=EB
(12√2)²*4k*2k
k²=288/8 ⇒K=36⇒K=6
CD=4*6
CD=24
AB=6*6
AB=36
Aria trapezului= (B+b)*h/2=360√2cm²
d(D,AC)=36/6=6 cm
ai desenul mai jos. ai datele problemei. succes!