Răspuns :
Ip: Concluzie:
D=_BC_ P ACD= ?
2
ΔABC= tr. drept.
BC= 50 mm
m(∡C) = 60°
Dem.
C
I Teorema ∡ de 30°: AC=_BC_
I 60° 2
I D AC=_50_
I 2
I AC= 25 mm
I__ 90° 30°
I__I___________________ CD= _BC_ ⇒ CD= 25 mm 2
A B
Proprietatea medianei!
AD=_ BC_
2
AD =_50_
2
AD= 25 mm
P ACD=AC+ AD+ DC
=25 mm+ 25mm+ 25mm
=75 mm
=7,5 cm
D=_BC_ P ACD= ?
2
ΔABC= tr. drept.
BC= 50 mm
m(∡C) = 60°
Dem.
C
I Teorema ∡ de 30°: AC=_BC_
I 60° 2
I D AC=_50_
I 2
I AC= 25 mm
I__ 90° 30°
I__I___________________ CD= _BC_ ⇒ CD= 25 mm 2
A B
Proprietatea medianei!
AD=_ BC_
2
AD =_50_
2
AD= 25 mm
P ACD=AC+ AD+ DC
=25 mm+ 25mm+ 25mm
=75 mm
=7,5 cm