fie tr dreptunghic ADB cu ungh B =30 => conform th ungh de 30 grade AB=2AD=> AB=40. aplicam th pitagora => BD²=AB²-AD² =1600-400=1200 => BD=20√3;;fie tr ADC - dreptunghic si ungh C =180-105-30 => ungh C=45 => ungh DAC=45 => tr ADC- dreptunghic si isoscel cu AD=DC=20 => BC=20√3+20. Aria=b*h/2=(20√3+20)*20/2 => Aria=200√3+200=200(√3+1)