AB=10 cm
AC=10 cm
BC=12 cm
a.
D mijlocul lui BC⇒ BD=DC=6 cm
M mijlocul lui DC⇒ DM=MC=3 cm
[tex]P_{ABD}=AB+BD+AD[/tex]
In Δ ABD dr in D aplicam Pitagora
AB²=AD²+BD²
100=AD²+36
AD=8 cm
[tex]P_{ABD}=AB+BD+AD=10+6+8=24\ cm[/tex]
P=24 cm
b.
d(M,AC)=MN
Fie DP⊥AC
DP inaltime in triunghi dreptunghic
[tex]DP=\frac{AD\times CD}{AC} =\frac{6\times8}{10} =4,8 \ cm[/tex]
DP=4,8 cm
DP║MN⇒
[tex]\frac{CM}{CD} =\frac{MN}{DP}\\\\ \frac{3}{6} =\frac{MN}{4,8}[/tex]
MN=2,4 cm
