Kẻ \(HK\perp BC\Rightarrow\widehat{SKH}=60^o\) Ta có \(S_{ABCD}=3m^2\)
\(S_{HBC}=S_{ABCD}-S_{HDC}-S_{HAB}=\dfrac{3m^2}{2}\)
\(BC=m\sqrt{5}\) \(\Rightarrow HK=\dfrac{S_{HBC}}{\dfrac{1}{2}BC}=\dfrac{3m\sqrt{5}}{5}\)
\(\Rightarrow SH=HK.tan60^o=\dfrac{3m\sqrt{15}}{5}\)
Vậy \(V_{S.ABCD}=\dfrac{1}{3}.SH.S_{ABCD}=\dfrac{3m^3\sqrt{15}}{5}\)