Centrul de greutate se afla la intersectia medianelor, la doua treimi de varf si o treime de baza
a)
AM=12 cm
[tex]AG=\frac{2}{3} \cdot AM=\frac{2}{3} \cdot 12\\\\AG=8\ cm[/tex]
b)
BG=10 cm
[tex]BG=\frac{2}{3} \cdot BN\\\\10=\frac{2}{3} \cdot BN\\\\BN=15\ cm[/tex]
BN=BG+GN
GN=15-10=5 cm
c)
PG=3 cm
[tex]PG=\frac{1}{3} \cdot PC\\\\3=\frac{1}{3} \cdot PC\\\\PC=9\ cm[/tex]
d)
AG=6 cm
[tex]AG=\frac{2}{3}\cdot AM\\\\ 6=\frac{2}{3}\cdot AM\\\\AM=9\ cm[/tex]
Un alt exercitiu de geometrie gasesti aici: https://brainly.ro/tema/2741822
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