Triunghiul este isoscel.
Aplicam teorema sinusurilor:
[tex]\displaystyle\frac{a}{sin120}=\frac{b}{sin30} \\ \\ \implies a=\frac{bsin120}{sin30}= \\ \\ \frac{bsin(180-120)}{sin30}= \\ \\ \frac{bsin60}{sin30}= \\ \\ \frac{\frac{b\sqrt3}{2}}{\frac{1}{2}}=b\sqrt3 \\ \\ a+2b=12 \\ b\sqrt3+2b=12 \\ b(\sqrt3+2)=12 \\ \\ b=\frac{12}{2+\sqrt3}=12(2-\sqrt3) \\ \\ A=\frac{b^2sin120}{2}= \\ \\ \frac{b^2sin(180-120)}{2}= \\ \\ \frac{144(7-4\sqrt3)\frac{\sqrt3}{2}}{2}= \\ \\ \frac{1008\sqrt3-1728}{4} \\ \\ \approx 4,5[/tex]