1. a) 3
BD/AB = 3x/x = 3
2. a) 70°
∠AOB + ∠BOC= 180° <=>
<=> ∠AOB= 180° - 40°=140°
OD bisectoare=>
<BOD=140° : 2 = 70°
3. c) 5cm
AM=AC/2=6/2=3 (M mijlocul lui AC)
Folosim Teorema lui Pitagora.
BM² = AB² + AM² <=>
<=> BM² = 4²+3² <=>
<=> BM²=16+9
<=> BM = √25 <=>
<=> BM = 5cm
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Verified answer
1. a) 3
BD/AB = 3x/x = 3
2. a) 70°
∠AOB + ∠BOC= 180° <=>
<=> ∠AOB= 180° - 40°=140°
OD bisectoare=>
<BOD=140° : 2 = 70°
3. c) 5cm
AM=AC/2=6/2=3 (M mijlocul lui AC)
Folosim Teorema lui Pitagora.
BM² = AB² + AM² <=>
<=> BM² = 4²+3² <=>
<=> BM²=16+9
<=> BM = √25 <=>
<=> BM = 5cm