Cho hình lăng trụ đứng ABC.A’B’C’ có đáy ABC...

A

- Hướng dẫn giải

Đáp án A.

Dựng $B\text{'}M\perp A\text{'}C\text{'}\Rightarrow B\text{'}M\perp \left(ACC\text{'}A\text{'}\right)$

Dựng $MN\perp AC\text{'}\Rightarrow AC\text{'}\perp \left(MNB\text{'}\right)$

Khi đó $\widehat{\left(\left(AB\text{'}C\text{'}\right);\left(AC\text{'}A\text{'}\right)\right)}=\widehat{MNB\text{'}}={60}^0$

Ta có:

$$\begin{gathered} B\text{'}M=\frac{a\sqrt{2}}{2} \\ \Rightarrow MN=\frac{B\text{'}M}{\tan \widehat{MNB\text{'}}}=\frac{a\sqrt{6}}{6} \end{gathered}$$

Mặt khác $\tan \widehat{AC\text{'}A\text{'}}=\frac{MN}{C\text{'}N}=\frac{\text{AA}\text{'}}{A\text{'}C\text{'}}$

Trong đó:

$$\begin{gathered} MN=\frac{a\sqrt{6}}{6};MC\text{'}=\frac{a\sqrt{2}}{2} \\ \Rightarrow C\text{'}N=\sqrt{C\text{'}{M}^2-M{N}^2} \\ =\frac{a\sqrt{3}}{3} \end{gathered}$$

Suy ra $\text{AA}\text{'}=a$

Thể tích lăng trụ:

$$\begin{gathered} V=\frac{A{B}^2}{2}.h=\frac{a^3}{2} \\ \Rightarrow V_{B\text{'}.ACC\text{'}A\text{'}}=V-V_{B\text{'}.BAC} \\ =V-\frac{V}{3}=\frac{2}{3}V=\frac{a^3}{3}. \end{gathered}$$