In baza notatiilor din figura , avem:
m(< ABC)=30^o => m(< ACB)\60^o ;
AD_|_BC si CD=6cm => m(< CAD)=30^0
conform T.<.30^0 , obtinem: AC=2*6=12cm;
conf. T. catetei in triunghiul dr. ABC , AC^2=6*BC => BC=24cm;
inseamna ca AD^2=BD*DC=(24-6)*6-18*6=36*3 deci: AD=6rad.3cm ;
[tex]A_{AMC}=\frac{MC*AD}{2}=\frac{12*6\sqrt3}{2}=36\sqrt3cm^2[/tex]