avem trapez isoscel, deci AD=CB=4 dm
in tr. dreptunghic ACB ∡CAB=90-60=30° ⇒ AB=2CB=8 dm (teorema ∡ 30°)
ducem
DE⊥AB, E∈AB si
CF⊥AB, F∈AB
∡CFB=∡ADE=90-60=30° (in trapez isoscel unghiurile de la baza sunt congruente)
rezulta ca FB=AE=AD/2=4/2=2 dm (vezi teorema ∡ 30°)
in consecinta
DC=AB-AE-FB=8-2-2=4 dm
perimetru P
P=DC+CB+AB+AD=4+4+8+4=20 dm
inaltimeatrapezului CF
CF=√(CB^2-FB^2)=√(16-4)
CF=2√3 dm
aria trapez A
A=(AB+DC) x CF/2=(8+4) x 2√3/2
A=12√3 dm2
aria tr.ADC=DC x DE/2 (formula clasica)
aria ADC =4 x 2√3/2 =4√3 dm2
