Cho hình lăng trụ ABCD.A'B'C'D' đáy là hình bình hành.

Đặt AA' = x (x > 0)

Xét tam giác ACD có $A{C}^2+A{D}^2=2a^2=C{D}^2\Rightarrow \Delta ACD$ vuông tại A (định lí Pytago đảo).

Ta có: $\left\{\begin{array}{l}AD\perp AC\\ AD\perp CD\text{'}\left(doC\text{'}\perp A\text{'}D\text{'}\right)\end{array}\right.\Rightarrow AD\perp \left(ACD\text{'}\right)\Rightarrow AD\perp AD\text{'}.$

$\Rightarrow AD{\text{'}}^2=DD{\text{'}}^2-A{D}^2=x^2-a^2.$

Ta lại có $A\text{'}{D}^2=AD{\text{'}}^2+{\left(2AD\right)}^2$

$\Rightarrow A\text{'}{D}^2=\left(x^2-a^2\right)+4a^2=x^2+3a^2\left(1\right).$

Ta có: $\left\{\begin{array}{l}A\text{'}C\perp A\text{'}B\text{'}\left(gt\right)\\ A\text{'}B\text{'}//CD\end{array}\right.\Rightarrow A\text{'}C\perp CD.$

$\Rightarrow A\text{'}{D}^2=A\text{'}{C}^2+C{D}^2.$

Ta lại có: $A\text{'}{C}^2+AC{\text{'}}^2=2\left(AA{\text{'}}^2+A{C}^2\right)\Rightarrow A\text{'}{C}^2+3a^2=2\left(x^2+a^2\right)\Rightarrow A\text{'}{C}^2=2x^2-a^2$

$\Rightarrow x^2+3a^2=2x^2-a^2+2a^2$

$\Rightarrow x^2=2a^2\Rightarrow x=a\sqrt{2}$

$\Rightarrow A\text{'}C=\sqrt{2.2a^2-a^2}=a\sqrt{3},CD\text{'}=\sqrt{A\text{'}{C}^2-A\text{'}D{\text{'}}^2}=\sqrt{3a^2-a^2}=a\sqrt{2},AD{\text{'}}^2=\sqrt{x^2-a^2}=a.$

$\Rightarrow \Delta ACD\text{'}$ vuông cân tại A

Vậy $V_{BCDA\text{'}}=V_{A\text{'}.BCD}=V_{D\text{'}.ACD}=V_{D.ACD\text{'}}=\frac{1}{3}AD.S_{ACD\text{'}}=\frac{1}{3}.\frac{1}{2}.a.a.a=\frac{a^3}{6}.$

Chọn A.