SH=SO/sin60 ° =6/sqrt(3)/2=12/sqrt(3)=[b]4sqrt(3)[/b]
OH=(1/2)SH=2*sqrt(3)
OH=r
r=a*sqrt(3)/6 ⇒ 2*sqrt(3)=a*sqrt(3)/6 ⇒ a=12 - сторона основания
S_(пол)=S_(бок)+S_(осн)
S_(осн)=S_( Δ ABC)=a^2*sqrt(3)/4=12^2*sqrt(3)/4=36sqrt(3)
S_(бок)=3*S_( Δ SBC)=3*S_( Δ SBC)=3*(1/2)*12*4sqrt(3)
S_(пол)=S_(бок)+S_(осн)=