Cho lăng trụ đứng ABCD.A'B'C'D' có đáy là hình bình hành. Các đường

* Hướng dẫn giải

Đáp án A

$\begin{array}{l}\widehat{\left(DB\text{'},\left(ABCD\right)\right)}=\widehat{BDB\text{'}}=60°\\ \Rightarrow BD=\frac{BB\text{'}}{\sqrt{3}}=\frac{2}{\sqrt{3}}\\ \widehat{\left(AC\text{'},\left(ABCD\right)\right)}=\widehat{CAC\text{'}}=60°\\ \Rightarrow AC=CC\text{'}=2\end{array}$

Áp dụng định lí Cosi ta có:

$$\begin{gathered} \begin{array}{l}\left\{\begin{array}{l}A{B}^2+A{D}^2-2AB.AD\cos \widehat{BAD}=B{D}^2\\ A{B}^2+A{D}^2-2AB.AD\cos \widehat{ABC}=A{C}^2\end{array}\right.\\ \iff \left\{\begin{array}{l}A{B}^2+A{D}^2-2AB.AD\sqrt{2}=\frac{4}{3}\\ A{B}^2+A{D}^2+2AB.AD\sqrt{2}=4\end{array}\right.\end{array} \\ \Rightarrow AB.AD=\frac{2\sqrt{2}}{3} \\ \Rightarrow V_{S.ABCD}=2S_{ABD}=AB.AD.\sin \widehat{BAD}=\frac{2}{3} \\ \Rightarrow V_{ABCD.A\text{'}B\text{'}C\text{'}D\text{'}}=S_{ABCD}.AA\text{'}=\frac{4}{3} \end{gathered}$$