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AB=100 cm
AC=60 cm
a. AE=12 cm
EF=16 cm
AF=20 cm
[tex]P_{BEFC}=BE+EF+FC+BC[/tex]
BE=AB-AE=100-12=88 cm
FC=AC-AF=60-20=40 cm
[tex]\frac{AE}{AC} =\frac{AF}{AB} \\\\\frac{12}{60} =\frac{20}{100}[/tex]⇒ΔAEF~ ΔACB⇒
[tex]\frac{AE}{AC} =\frac{EF}{BC} \\\\\frac{1}{5} =\frac{16}{BC}[/tex]
BC=80 cm
[tex]P_{BEFC}=BE+EF+FC+BC=88+16+40+80=224 cm[/tex]
Ca o mica paranteza, ΔABC in acest caz este dreptunghic in C, pentru ca AB²=BC²+AC²
b. AE=15 cm
EF=20 cm
AF=25 cm
AB=100 cm
AC=60 cm
EB=AB-AE=100-15=85 cm
FC=AC-AF=60-25=35 cm
[tex]\frac{AE}{AC} =\frac{AF}{AB} \\\\\frac{15}{60} =\frac{25}{100}[/tex]
ΔAEF~ ΔACB⇒
[tex]\frac{AE}{AC} =\frac{EF}{BC} \\\\\frac{1}{4} =\frac{20}{BC}[/tex]
BC=80 cm
[tex]P_{BEFC}=BE+EF+FC+BC=85+20+35+80= 220cm[/tex]
Ca o mica paranteza, ΔABC in acest caz este dreptunghic in C, pentru ca AB²=BC²+AC²
