fie ABCD -trapez dreptunghic cu m(<A)=m(<B)=90° si m(<D)=30°
BC=22cm si AD=28 cm
A(trapez)=[(BC+AD)/2]·h
din virful C coborim inaltimea CH⊥AD
AH=BC=22 cm ⇒HD=AD-AH=28-22=6 cm
in ΔCHD-dreptunghic, tg30°=CH/HD⇒√3/3=CH/6⇒CH=2√3
A=[(22+28)/2]·2√3=50√3 cm²
