bai 18
H hinh chieu S tren BC
D hinh chieu H tren AC
∆SBH=>BH =3a; SH=a√3; 
=> HC=a
∆ABC=>S=1/2.3a.4a=6a^2; AC=5a
∆ABC~HDC
HD/AB=HC/AC=1/5
HD=3a/5
SD=√(SH^2+HD^2)=a.√(9/25+3)=2a√(21)/5
d(B,(sac)=h
h/SH=S∆abc/S∆sac=12.a^2/[5a.2a√(21)/5]
h=12.a√3./[2√(21)=6a/√7=6a√7/7
(A)