Question

ma ajutati putin si pe mine?problema de cl 7 .Triunghiul ABC are laturile AB=26cm,AC=28 cm si BC=30 cm.Fie[BD bisectoarea unghiului ,ABC cu D∈(AC) si DE||AB,E∈(BC).Calculati perimetrul trapezului ABED.

Answer 1

m(<ABD)=m(<DBC)   BD fiind bisectoare
DEIIAB
m(<ABD)=m(<BDE) rezulta triunghiul BDE isoscel BE=DE
ΔDCE si ΔABC asemenea  caz UUU
teorema lui Thales
DC/AC=EC/BC=DE/AB
DC/28=30-BE/30=DE/26    dar BE=DE
DC/28=30-DE/30=DE/26
30DE=780-26DE
DE=780/56=195/14
AC-AD/28=(195/14)/26
(28-AD)/28=(195/14)/26
28-AD= 28 *195/364 
28-AD= 195/13 
28-AD= 15 
AD=28-15=13 cm

P=26+2*195/14+13=39+195/7=468/7=66 intregi si 6/7

Answer 2

cf teoremei bisectoarei
AD/DC=AB/DC=26/30
AD/(AD+DC)=26/(30+26)
AD/28=26/56
AD=13
DE paralel AB
<EDB=<DBA=<DBE⇒DEB tr ISOSCEL⇒DE=EB
in tr ABC cf thales
CD/AD=CE/EB; (CD+AD)/AD=(CE+EB)/EB
28/13=30/EB
EB=195/14
P=AD+AB+BE+ED=13+26+195/14+195/14=39+195/7=$66\frac{6}{7}$