Răspuns :
AO = AC/2 = 3 cm
m(<ABO)=m(<BDC)=30° rezulta AB = 2*AO = 2*3 = 6 cm
P = 4*l = 4*6 = 24 cm
m(<ODC)=30° rezulta OC = CD/2 = 6/2 = 3 cm
m(<OCE)=60° rezulta m(<COE)=30° rezulta CE = OC/2 = 3/2 = 1,5 cm
m(<ABO)=m(<BDC)=30° rezulta AB = 2*AO = 2*3 = 6 cm
P = 4*l = 4*6 = 24 cm
m(<ODC)=30° rezulta OC = CD/2 = 6/2 = 3 cm
m(<OCE)=60° rezulta m(<COE)=30° rezulta CE = OC/2 = 3/2 = 1,5 cm

m∡BDC=30°⇒m∡ADC=60°(in romb diag.sunt si bisect.)
⇒ΔACDsiΔABC tr isoscele cu un unghide 60°,deci echilaterale⇒AC=6=AB
⇒Perim=6*4=24cm
m∡CIOE=m∡BDC (∡ cu laturi ⊥)⇒(teo ∡30° in tr.drOCE) CE=OC/2
dar OC=(teo∡30° in tr.dr.OCD)=CD/2
atunciCE=AB/4=6/4=1,5cm
⇒ΔACDsiΔABC tr isoscele cu un unghide 60°,deci echilaterale⇒AC=6=AB
⇒Perim=6*4=24cm
m∡CIOE=m∡BDC (∡ cu laturi ⊥)⇒(teo ∡30° in tr.drOCE) CE=OC/2
dar OC=(teo∡30° in tr.dr.OCD)=CD/2
atunciCE=AB/4=6/4=1,5cm

