\(V=\dfrac{1}{3}S_{ABC}.SH\)
Kẻ đường cao \(SH\) của hình chóp => H là tâm đáy .
Có : \(S_{ABC}=\dfrac{1}{2}a^2.sin60^o=\dfrac{1}{2}a^2\cdot\dfrac{\sqrt{3}}{2}=\dfrac{a^2\sqrt{3}}{4}\)
Có : \(HM=\dfrac{1}{3}AM=\dfrac{1}{3}\cdot\dfrac{a\sqrt{3}}{2}=\dfrac{a\sqrt{3}}{6}\)
Có : \(tan60^o=\dfrac{SM}{BM}\rightarrow SM=BM\cdot tan60^o=\dfrac{a}{2}\cdot tan60^o=\dfrac{a\sqrt{3}}{2}\)
\(\Rightarrow SH=\sqrt{\dfrac{3a^2}{4}-\dfrac{3a^2}{36}}=\sqrt{\dfrac{24a^2}{36}}=\dfrac{a\sqrt{2}}{\sqrt{3}}\)
\(\Rightarrow V_{SABC}=\dfrac{1}{3}\cdot\dfrac{a^2\sqrt{3}}{4}\cdot\dfrac{a\sqrt{2}}{\sqrt{3}}=\dfrac{a^3\sqrt{6}}{12\sqrt{3}}=\dfrac{a^3\sqrt{2}}{12}\)
