In triunghiul ABD dreptunghic in D, AB este ipotenuza si m(<ABD)=30°, deci din teorema unghiului de 30°, AD = AB/2
[tex]AB=2AD=2\cdot3\sqrt{3}=6\sqrt{3}[/tex]
[tex]AD^2+BD^2=AB^2[/tex]
[tex](3\sqrt{3})^2+BD^2=(6\sqrt{3})^2[/tex]
[tex]27+BD^2=108[/tex]
[tex]BD^2=81[/tex]
[tex]BD=9~cm[/tex]
Din teorema inaltimii
[tex]AD=\sqrt{BD\cdot CD}[/tex]
[tex]3\sqrt{3}=\sqrt{9CD}[/tex]
[tex]9CD=27[/tex]
[tex]CD=3~cm[/tex]
[tex]BC=9+3=12~cm[/tex]
In triunghiul ACD dreptunghic in D, cu AC ipotenuza, m(<CAD)=30°, deci CD=AC/2
[tex]AC=2CD=2\cdot3=6~cm[/tex]
[tex]a)~P_{\triangle ABC}=AB+BC+AC=6\sqrt{3}+12+6=6\sqrt{3}+18[/tex]
b) Stim ca AC = 6 cm si AM = 6 cm
CM = MB = BC/2 = 12/2 = 6 cm
[tex]P_{\triangle ACM}=3\cdot6=18~cm[/tex]
