Cho hình hộp chữ nhật ABCD.A'B'C'D' có AB=2a, AD=3a, AA'=3a

* Hướng dẫn giải

Chọn C.

$$\begin{gathered} V_{ABCD.A\text{'}B\text{'}C\text{'}D\text{'}}=2a.3a.3a=18a^3. \\ {V}_{E.BCD}=\frac{1}{3}d\left(E;\left(BCD\right)\right).S_{BCD}. \end{gathered}$$

Vì $B\text{'}C\text{'}//\left(ABCD\right)$ nên $d\left(E;\left(BCD\right)\right)=d\left(B\text{'};\left(BCD\right)\right)=d\left(B\text{'};\left(ABCD\right)\right).$

$S_{BCD}=\frac{1}{2}{S}_{ABCD}.$

Do đó: $V_{E.BCD}=\frac{1}{3}d\left(B\text{'};\left(ABCD\right)\right).\frac{1}{2}.S_{ABCD}=\frac{1}{2}{V}_{B\text{'}.ABCD}=\frac{1}{2}.\frac{1}{3}{V}_{ABCD.A\text{'}B\text{'}C\text{'}D\text{'}}$

$\Rightarrow V_{E.BCD}=\frac{1}{6}.18a^3=3a^3.$