in triunghiul ABC se aplica t lui Pitagora
BC²=AC²+AB²
BC²=36+64=100⇒BC=10
M mijlocul lui BC⇒CM≡MB⇒MB=BC/2=5
ΔABC asemenea cuΔMNB triunghiuri dreptunghice
[tex] \frac{AC}{MN} = \frac{BC}{NB} = \frac{AB}{MB} [/tex]
[tex] \frac{AC}{MN} = \frac{AB}{MB} [/tex]
MN=ACxMB/AB=15/4
Aria BMN=MNxMB/2=75/8
BC/NB=AB/MB
NB=MBxBC/AB⇒50/8
P=MN+NB+MB
P=15/4+50/8+5
