Se da:
ΔABC echilateral cu AB = 18 m
D ∈ BC si este situat pe prelungirea laturii BC cu CD = 9 m.
E ∈ AD astfel incat <ACE congruent <DCE
Se cere:
a) Aria DABC = 81√3 m²
b) EC || AB
c) Perimetrul ΔEAC = 6(4+√7) m
Rezolvare:
[tex]\displaystyle\\
a)\\
\Delta ABC\text{este echilateral.}\\\\
A_{\Delta ABC}=\frac{l^2\sqrt{3}}{4}=\frac{AB^2\sqrt{3}}{4}=\frac{18^2\sqrt{3}}{4}=\frac{324\sqrt{3}}{4}=\boxed{\bf81\sqrt{3}~m^2}
[/tex]
[tex]\displaystyle\\
b)\\
\text{Unghiul ACD este un unghi exterior triunghiului ABC.}\\
\Longrightarrow~m(\angle ACD)=m(\angle CAB)+m(\angle ABC)=60^o+60^o=120^o\\
\ \textless \ ACE\text{ congruent cu }\ \textless \ DCE~~\text{(din enunt)}\\
\Longrightarrow~~CE\text{ este bisectoarea unghiului ACD.}\\
\Longrightarrow~~m(\angle ECD)=\frac{\angle ACD}{2}=\frac{120^o}{2}=60^o\\
m(\angle ABC)=60^o~~\text{triunghiul ABC fiind echilateral}\\
\Longrightarrow~~m(\angle ECD)=m(\angle ABC)=60^o\\
\Longrightarrow~~\boxed{\bf EC~||~AB}[/tex]
[tex]\displaystyle\\
c)\\
\text{In triunghiul DAB avem:}\\
CD=9~m\\
DB=BC+CD=18+9=27~m\\
EC~||~AB\\\\
\text{Aplicam T. Thales in triunghiul DAB pentru a calcula segmentul EC.}\\\\
\frac{DC}{DB}=\frac{EC}{AB}\\\\
\frac{9}{27}=\frac{EC}{18} \\\\
EC = \frac{18\times 9}{27}=\frac{18}{3}=6~m\\\\
\text{In triunghiul EAC avem: }\\
AC=18~m\\
EC=6~m\\
m(\angle ACE)= m(\angle DCE)=60^o\text{ CE fiind bisectoare.}[/tex]
[tex]\displaystyle\\
\text{Aplicam T. Pitagora generalizata pentru a calcula latura AE.}\\\\
AE^2 = AC^2+CE^2 - 2\times AC \times CE\times cos(\widehat{ACE})\\
AE^2 = 18^2+6^2 - 2\times 18\times 6\times cos(60^o)\\\\
AE^2 = 18^2+6^2 - 2\times 18\times 6\times \frac{1}{2} \\
AE^2 = 324+36 - 18\times 6 \\
AE^2 = 324+36 - 108 \\
AE^2 = 252 \\
AE = \sqrt{252}=\sqrt{36\times7}=6\sqrt{7} \\\\
P_{\Delta ACE}=AC + CE + AE \\
P_{\Delta ACE}=18+6+6\sqrt{7}=24+6\sqrt{7}=\boxed{6\Big(\bf 4+\sqrt{7}\Big)~m}
[/tex]
