a.
AB=15 cm
AD=10 cm
Aplicam Pitagora in ΔBAD
BD²=AD²+AB²
BD²=100+225=325
BD=5√13 cm≈18 cm
[tex]P_{ABD}=AB+AD+BD=10+15+18=43\ cm[/tex]
P=43 cm< 45 cm
b.
d(E,CD)=EF
AE bisectoarea ∡BAD ⇒
[tex]\frac{DE}{EB} =\frac{AD}{AB} \\\\\frac{DE}{EB} =\frac{10}{15}\\\\ \frac{DE}{EB} =\frac{2}{3}[/tex]
[tex]\frac{DE}{EB} =\frac{2}{3}\\\\\frac{DE}{DB} =\frac{2}{5}[/tex]
BC⊥DC si EF⊥DC⇒ EF║BC⇒
[tex]\frac{DE}{DB} =\frac{DF}{DC}=\frac{EF}{BC}[/tex]
[tex]\frac{2}{5} =\frac{DF}{15}=\frac{EF}{10}[/tex]
DF=30:5=6 cm
EF=20:5=4 cm
c.
DC=DF+FC
15=6+FC
FC=15-6=9 cm
InΔEFC aplicam Pitagora
EC²=EF²+FC²
EC²=16+81
EC=√97 cm
