Cho khối hộp chữ nhật ABCD.A'B'...

A

- Hướng dẫn giải

Đáp án A

Cho lăng trụ $ABCD.A\text{'}B\text{'}C\text{'}D\text{'}$ có $AM=a,BN=b,CP=c,S=S_{ABC}.$ 

Khi đó $V_{MNP.ABC}=\frac{a+b+c}{3}.S$ 

Đặt $AA\text{'}=a\Rightarrow AM=\frac{a}{2},PC=\frac{2a}{3};$ 

Ta có $DN+BQ=2II\text{'}=MA+PC=\frac{7a}{6};S_{ABD}=S_{CBD}=S$ 

Áp dụng tính chất có

$\begin{array}{l}{V}_{MNPQ.ABCD}=V_{MNP.ADB}+V_{NQP.CBC}=\frac{1}{3}\left(AM+BQ+DN\right).S+\frac{1}{3}\left(DN+BQ+CP\right).S\\ =\frac{1}{3}\left(3AM+3PC\right).S=\frac{7}{6}a.S=\frac{7}{12}{V}_{ABCD.A\text{'}B\text{'}C\text{'}D\text{'}}\Rightarrow V_{A\text{'}B\text{'}C\text{'}D\text{'}.MNPQ}=\frac{5}{12}{V}_{ABCD.A\text{'}B\text{'}C\text{'}D\text{'}}=\frac{5}{12}.2018=\frac{5045}{6}\end{array}$