[tex]\text{Punctul de intersectie al medianelor este centrul de} \\
\text{greutate al triunghiului, care se afla la o treime de baza} \\
\text{si doua treimi de varf } \\ \displaystyle
\text{Asta inseamna ca: } \\ \\
PM = \frac{1}{3} \times AM; ~~AP = \frac{2}{3} \times AM;~~PN = \frac{1}{3} \times BM; ~~BP = \frac{2}{3} \times BM \\ \\
Rezulta: \\ \\
PM = \frac{AP}{2};~~~AP=2PM;~~~ PN = \frac{BP}{2};~~~BP=2PN
[/tex]
[tex]Rezolvare: \\ \displaystyle \\
a) \\
PM = \frac{AP}{2}=\frac{24}{2}=\boxed{12\;cm} \\ \\
PN = \frac{BP}{2}=\frac{30}{2}=\boxed{15\;cm} \\ \\
b) \\
AP=2PM = 2 \times \sqrt{6} = \boxed{2\sqrt{6}\;cm} \\
BP=2PN = 2 \times \sqrt{7} = \boxed{2\sqrt{7}\;cm}[/tex]
