Question

ABCD trapez isoscel, m(B)=60 de grade, AB=BC=AD=4cm.Pabcd=?

Answer 1

fie AM⊥DC si BN⊥DC=>AM si BN inaltimi

AB=MN=4 cm

In ΔBNC , m(∡N)=90°=>sin B=NC/BC=>NC=BCsinB=4√3/2=2√2 cm

NC=DM=2√2 cm

DC=NC+DM+MN=2√2+2√+4=4+4√2=4(1+√2) cm

P=AB+BC+CD+DA=4+4+4+4+4√2=16+4√2=4(4+√2)cm