Ta có: \({S_{ABB'A'}} = AB\,.\,AA' = 2{a^2} \Rightarrow AA' = \frac{{2{a^2}}}{a} = 2a\).

\({S_{ABC}} = \frac{{A{B^2}\sqrt 3 }}{4} = \frac{{{a^2}\sqrt 3 }}{4}\).

Suy ra\(V = AA'\,.\,{S_{ABC}} = 2a\,.\,\frac{{{a^2}\sqrt 3 }}{4} = \frac{{{a^3}\sqrt 3 }}{2}\).