a) m(BAC) =m(DAC) =30° m(BAD) = m(BAC) +m(DAC) =30°+30°=60° b) In ∆ AOB dr => sin BAO = BO/AB 1/2=3/AB =>AB= 6 cm P = 4×l=4×6=24 cm c) In ∆ AOB dr => tg 30°= BO/AO 1/√3= 3/AO => AO=3√3 BD=BO+OD =3+3=6 cm AC= AO+OC=3√3+3√3=6√3 cm A = (d1×d2/2)=( 6×6√3)/2= 18√3 cm²
