Răspuns:
Explicație pas cu pas:
Fie AE⊥BC
d(M,BC)=ME
[tex]AE=\frac{l\sqrt{3} }{2} =3\sqrt{3}[/tex]
In ΔMAE aplicam pitagora
ME²=MA²+AE²
ME²=9+27=36
Me=6cm
Fie AF⊥ME
d(A,(MBC))=AF
In ΔMAE, AF inaltime⇒ [tex]AF=\frac{MA\cdot AE}{ME} =\frac{3 \cdot 3\sqrt{3} }{6} =\frac{3\sqrt{3} }{2}[/tex]
