a)Notam cu MN = d(M,AC) = 8 cm
MB ⊥ (ABC)
BN ⊂ (ABC)
=> MB ⊥ BN
In triunghiul MBN dreptunghic aplicam teorema lui Pitagora:
MN² = MB²+BN² => 64 = 16 + BN²=> BN²=48=> BN = 4√3 cm
Avand MB ⊥ (ABC) si MN⊥AC => BN⊥AC(Prima reciproca a T.3⊥)
[BN] este inaltime in triunghiul ABC echilateral => BN = [tex] \frac{l\sqrt{3}}{2} [/tex]
=> l√3/2=4√3=>l√3=8√3 => AB = 8 cm
b)Construim BP⊥MN.
BP⊥MN
MN⊥AC
MN,AC⊂(MAC)
BN⊥AC
=> BP⊥(MAC)(a 2-a reciproca a T.3⊥)
[tex]pr_{(MAC)}BM=MP =\ \textgreater \ MP = BM * cos(\ \textless \ BMN) [/tex]
In triunghiul BMN dreptunghic: cos(BMN)=BM/MN => cos(BMN)=4/8=>
=> cos(BMN)=1/2
=> MP = 4 * 1/2 => MP = 2 cm
