Gọi H là tâm đáy \(\Rightarrow SH\perp\left(ABC\right)\)
Ta có: \(AH=\dfrac{2}{3}.\dfrac{a\sqrt{3}}{2}=\dfrac{a\sqrt{3}}{3}\)
Áp dụng định lý Pitago:
\(SH=\sqrt{SA^2-AH^2}=\dfrac{a\sqrt{33}}{3}\)
\(V=\dfrac{1}{3}SH.S_{ABC}=\dfrac{1}{3}.\dfrac{a\sqrt{33}}{3}.\dfrac{a^2\sqrt{3}}{4}=\dfrac{a^3\sqrt{11}}{12}\)