Giả sử $AC=2BD=2a$
Gọi $H$ là trung điểm cạnh $AD$
$\Rightarrow SH\bot AD$
$SH\bot(ABCD)$
Gọi $O=AC\cap BD$
Áp dụng định lý Pitago vào $\Delta $ vuông $AOD$
$AD^2=AO^2+OD^2=a^2+(\dfrac{a}{2})^2=\dfrac{5a^2}{4}$
$\Rightarrow AH^2=\dfrac{5a^2}{16}$
$\Delta $ vuông $SAD$:
$SA^2+SD^2=AD^2$
$\Rightarrow 2SA^2=\dfrac{5a^2}{4}$
$\Rightarrow SA^2=\dfrac{5a^2}{8}$
$\Delta$ vuông $ SAH$
$SH^2=SA^2-AH^2=\dfrac{5a^2}{8}-\dfrac{5a^2}{16}=\dfrac{5a^2}{16}$
$\Rightarrow SH=\dfrac{a\sqrt5}{4}$
$\Rightarrow V_{SABCD}=\dfrac{1}{3}SH.S_{ABCD}=\dfrac{1}{3}\dfrac{a\sqrt5}{4}\dfrac{a.2a}{2}=\dfrac{a^3\sqrt5}{12}$.
