G reprezinta punctul de intersectie al medianelor si se numeste centrul de greutate. Acesta se afla la o treime de baza si doua treimi de varf
a) GD=2 cm
[tex]GD=\frac{1}{3} \times AD\\\\2=\frac{1}{3} \times AD\\\\AD=6\ cm[/tex]
Fiindca este triunghi echilateral, medianele sunt egale, deci GE=GD=2 cm
b) AD+BE=24 cm
AD si BE sunt egale, fiind un triunghi echilateral.
AD=BE
AD+AD=24 cm
AD=BE=12 cm
Notam d(G,AB)=GH
Aflam Aria triunghiului ABC
[tex]A_{ABC}=\frac{l^2\sqrt{3} }{4} =36\sqrt{3} \ cm^2[/tex]
Stim ca mediana imparte aria triunghiului in 2 arii egale
[tex]GD=\frac{1}{3} \times 12=4\ cm[/tex]
BD=12:2=6 cm
[tex]A_{GBD}=\frac{GD\times BD}{2} =\frac{4\times 6}{2} =12\ cm^2[/tex]
[tex]A_{ABD}=A_{ABC}:2=36\sqrt{3} :2=18\sqrt{3} \ cm^2[/tex]
[tex]A_{AGB}=18\sqrt{3} -12=\frac{GH\times AB}{2} \\\\18\sqrt{3} -12=\frac{GH\times 12}{2}\\\\18\sqrt{3} -12=6GH\\\\\\\GH=3\sqrt{3} -2[/tex]
