Lời giải chi tiết:

\(\begin{align}  & +)\,\widehat{ASB}=\widehat{ASC}={{60}^{o}}\Rightarrow AB=AC=a \\  & +)\,\widehat{BSC}={{90}^{o}}\Rightarrow BC=a\sqrt{2} \\ \end{align}\)

+) Nhận xét: D ABC có \(A{{B}^{2}}+A{{C}^{2}}=2{{a}^{2}}=B{{C}^{2}}\Rightarrow \) D ABC vuông ở A

 \(\begin{align} & +)\,\,{{R}_{\mathbf{}}}=\frac{BC}{2}=\frac{a\sqrt{2}}{2} \\  & +)\,\,h=\sqrt{S{{A}^{2}}-{{R}_{\mathbf{}}}^{2}}=\sqrt{{{a}^{2}}-\frac{2{{a}^{2}}}{4}}=\frac{a\sqrt{2}}{2} \\ & +)\,{{S}_{_{\mathbf{}}}}=\frac{1}{2}AB.AC=\frac{{{a}^{2}}}{2} \\ & +)\,\,{{V}_{SABC}}=\frac{1}{3}.h.{{S}_{_{\mathbf{}}}}=\frac{1}{3}.\frac{a\sqrt{2}}{2}.\frac{{{a}^{2}}}{2}=\frac{\sqrt{2}{{a}^{3}}}{12} \\ \end{align}\)

Chọn đáp án A