Cho hình hộp ABCD. A'B'C'D' có thể tích V

* Hướng dẫn giải

Đáp án A

$\overrightarrow{D^{\text{'}}M}=2\overrightarrow{MD}\Rightarrow M$ nằm trên đoạn D'D và $D^{\text{'}}M=\frac{2}{3}{D}^{\text{'}}D.$

$\overrightarrow{C^{\text{'}}N}=2\overrightarrow{NC}\Rightarrow N$ nằm trên đoạn C'C và $C^{\text{'}}N=\frac{2}{3}{C}^{\text{'}}C.$

Trong (BB'C'C) qua N kẻ HK vuông với $BC,B^{\text{'}}{C}^{\text{'}}\left(H\in BC,K\in B^{\text{'}}{C}^{\text{'}}\right).$

$BC//B^{\text{'}}{C}^{\text{'}}\Rightarrow \frac{NK}{NH}=\frac{N{C}^{\text{'}}}{NC}=2\Rightarrow NK=2NH,NH=\frac{1}{3}HK.$

$\begin{array}{l}BC//B^{\text{'}}{C}^{\text{'}}\Rightarrow \frac{Q{C}^{\text{'}}}{BC}=\frac{C^{\text{'}}N}{CN}=2\Rightarrow Q{C}^{\text{'}}=2BC.\\ S_{Q{C}^{\text{'}}N}=\frac{1}{2}NK.Q{C}^{\text{'}}=\frac{1}{2}.2NH.2BC=4.\frac{1}{2}.\frac{1}{3}HK.BC=\frac{2}{3}{S}_{B{B}^{\text{'}}{C}^{\text{'}}C}.\\ \frac{V_{PQNM{D}^{\text{'}}{C}^{\text{'}}}}{V}=\frac{V_{NQ{C}^{\text{'}}.MP{D}^{\text{'}}}}{V}=\frac{S_{NQ{C}^{\text{'}}}}{S_{BC{C}^{\text{'}}{B}^{\text{'}}}}=\frac{2}{3}\Rightarrow V_{PQNM{D}^{\text{'}}{C}^{\text{'}}}=\frac{2}{3}V.\end{array}$