Răspuns:
a) tr ABC :isoscel => AB=AC
P-perimetrul
BC=AB+8 cm
P ABC=AB+AC+BC
P ABC=2AB+AB+8
128=3AB+8
3AB=120cm
AB=40 cm=> AC=40 cm
BC=40+8=48 cm
b)Fie AD⊥BC ,D€( BC)
tr ABC -isoscel => AD -mediana => BD=DC=BC/2=48/2=24 cm
tr ABD :m(<D)=90° == T.P.==> AD²=AB²-BD²
AD²=40²-24²=1600-576=1024=> AD=32cm
A-Aria
A ABC=BC•AD/2=48•32/2=24•32=768 cm²
c) Fie BB'⊥AC, B'€(AC)=> d(B,AC)=BB'
Stiind ca Aria triunghiului este baza•inaltimea/2 ,alegem baza [AC] si inaltimea BB'
A ABC=AC•BB'/2
768 =40 •BB'/2
768=20• BB'
BB'=768/20
BB'=38,4 cm
Sper ca e bine...Se mai pot strecura greseli:))
