Lăng trụ tam giác đều

A

- Hướng dẫn giải

Đáp án A

Do $\text{AA}\text{'}//BB\text{'}\Rightarrow V_{M.BCB\text{'}C\text{'}}=V_{A\text{'}.BCC\text{'}B\text{'}}=V-V_{A\text{'}.ABC}=V-\frac{V}{3}-\frac{2V}{3}$

(với V là thể tích khối lăng trụ).

Dựng $AH\perp BC$lại có $\text{AA}\text{'}\perp BC\Rightarrow BC\perp \left(A\text{'}HA\right)$

Do đó  $\widehat{\left(A\text{'}BC\right);\left(ABC\right)}=\widehat{A\text{'}HA}={30}^{\circ };AH=\frac{AB\sqrt{3}}{2}=\frac{3a}{2}$

Khi đó:  $\text{AA'=AHtan}{30}^{\circ }=\frac{a\sqrt{3}}{2}$

$$\begin{gathered} V=\text{AA}\text{'}.S_{ABC}=\frac{a\sqrt{3}}{2}.\frac{{\left(a\sqrt{3}\right)}^2\sqrt{3}}{4}=\frac{9}{8}{a}^3 \\ \Rightarrow V_{M.BCCB\text{'}}=\frac{2}{3}.\frac{9}{8}{a}^3=\frac{3}{4}{a}^3. \end{gathered}$$