Áp dụng định lý Pytago ta được:
$BD^2 = AB^2 + AD^2$
$\Rightarrow BD =\sqrt{AB^2 + AD^2} =\sqrt{4a^2 + 3a^2} = a\sqrt7$
Từ $S$ kẻ $SH\perp BD$ ta có:
$\begin{cases}(SBD)\perp (ABCD)\\(SBD)\cap (ABCD)=BD\\SH\subset (SBD);\, SH\perp BD\end{cases}$
$\Rightarrow SH\perp (ABCD)$
$\Rightarrow SH = SB.\sin\widehat{SBH}$
Ta lại có:
$\dfrac{SB}{SD}=\dfrac{1}{\sqrt3}$
$\Leftrightarrow \cot\widehat{SBD}=\dfrac{1}{\sqrt3}$
$\Rightarrow \widehat{SBD}=\widehat{SBH}=60^o$
$\Rightarrow SB = BD.\cos60^o = \dfrac{a\sqrt7}{2}$
$\Rightarrow SH = \dfrac{a\sqrt7}{2}.\sin60^o = \dfrac{a\sqrt{21}}{4}$
$\Rightarrow V_{S.ABCD}=\dfrac{1}{3}S_{ABCD}.SH=\dfrac{1}{3}.2a.a\sqrt3.\dfrac{a\sqrt{21}}{4} = \dfrac{a^3\sqrt7}{2}$
