Phương pháp giải:

Lời giải chi tiết:

+) Vẽ M, N lần lượt là trung điểm của AB, CD $\Rightarrow \left\{ \begin{align}  & AN\bot CD \\ & BN\bot CD \\\end{align} \right.\Rightarrow CD\bot \left( NAB \right)$

$\begin{align}  & +)\Delta ACD\text{ }=\Delta BCD\text{ }\left( c.c.c \right)\Rightarrow AN\text{ }=\text{ }BN\Rightarrow MN\bot AB \\ & +)\,\,{{V}_{ABCD}}={{V}_{DNAB}}+{{V}_{CNAB}}=2{{V}_{DNAB}}=2.\frac{1}{3}.ND.{{S}_{\Delta NAB}} \\ & +)\,{{\Delta }_{v}}BNC:\,\,BN=\sqrt{{{a}^{2}}-\frac{{{y}^{2}}}{4}} \\ & +)\,{{\Delta }_{v}}BMN:\,\,MN=\sqrt{{{a}^{2}}-\frac{{{y}^{2}}}{4}-\frac{{{x}^{2}}}{4}} \\ & +)\,{{S}_{\Delta NAB}}=\frac{1}{2}MN.AB=\frac{1}{2}.\sqrt{{{a}^{2}}-\frac{{{y}^{2}}}{4}-\frac{{{x}^{2}}}{4}}.x=\frac{x}{2}.\sqrt{{{a}^{2}}-\frac{{{y}^{2}}}{4}-\frac{{{x}^{2}}}{4}} \\ & +)\,{{V}_{ABCD}}=2.\frac{1}{3}.ND.{{S}_{\Delta NAB}}=2.\frac{1}{3}.\frac{y}{2}.\frac{x}{2}.\sqrt{{{a}^{2}}-\frac{{{y}^{2}}}{4}-\frac{{{x}^{2}}}{4}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{xy}{12}.\sqrt{4{{a}^{2}}-{{x}^{2}}-{{y}^{2}}} \\\end{align}$

Chọn đáp án B