m(∡DAB)=90-60=30°
In Δ DAB-dreptunghic
m(DAB)=30⇒(TEOREMA ∡30):DB=AB/2=6√3
cos de 30=AD/AB⇒√3/2=AD/12√3⇒AD=18
In ΔABC⇒(teorema catetei): AB^2=BD×BC
(12√3)^2=6√3×BC
432=6√3×BC
BC=24√3
⇒(teorema lui Pitagora):AC^2=BC^2-AB^2
AC^2=1728-432
AC=36
P ABC=36+24√3+12√3
=36+36√3
=36(1+√3)
Sper ca te-am ajutat! ^..^
