Dựng $IH\bot CD$, mặt khác $SI\bot \left( ABCD \right)$ .

Do đó $CD\bot BC\Rightarrow CD\bot \left( SHI \right)\Rightarrow \widehat{SHI}={{60}^{0}}$

Ta có: $BC=\sqrt{A{{D}^{2}}+\left( AB-C{{D}^{2}} \right)}=a\sqrt{5}$

${{S}_{ICD}}={{S}_{ABCD}}-{{S}_{ABI}}-{{S}_{CDI}}=\frac{3{{a}^{2}}}{2}$

Do vậy $IH=\frac{2{{S}_{ICD}}}{BC}=\frac{3a}{\sqrt{5}}\Rightarrow SI=\frac{3a\sqrt{3}}{\sqrt{5}}$

$\Rightarrow V=\frac{1}{3}SI.{{S}_{ABCD}}=\frac{3{{a}^{3}}\sqrt{15}}{5}$ .