coborim inaltimea BH si obtinem triunghiul dreptunghic BHC
BH=1/2BC
BH=6cm
HC=[tex] \sqrt{BC^2-BH^2}= \sqrt{144-36}=6 \sqrt{3} [/tex]
DC=DH+HC=4+6√3
[tex]A= \frac{AB+DC}{2}*BH= \frac{4+4+6 \sqrt{3} }{2}*6 = \frac{8+6 \sqrt{3} }{2}*6=3(8+6 \sqrt{3})cm^2= \\ =6(4+3 \sqrt{3})cm^2 [/tex]