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

* Hướng dẫn giải

Chọn C

Ta xét lăng trụ tam giác $ABA\text{'}.DCD\text{'}$ có thể tích bằng $\frac{1}{2}V.$

Kéo dài D'N cắt A'B tại E

+) $\frac{EN}{ED\text{'}}=\frac{BN}{A\text{'}D\text{'}}=\frac{1}{4}\Rightarrow \frac{D\text{'}N}{D\text{'}E}=\frac{3}{4};\frac{A\text{'}B}{EA\text{'}}=\frac{D\text{'}N}{D\text{'}E}=\frac{3}{4}\iff \frac{EA\text{'}}{BA\text{'}}=\frac{4}{3}.$

+) $\frac{V_{D\text{'}.A\text{'}ME}}{V_{D\text{'}.A\text{'}AB}}=\frac{S_{\Delta MA\text{'}E}}{S_{AA\text{'}B}}=\frac{MB}{AB}.\frac{A\text{'}E}{A\text{'}B}=\frac{1}{2}.\frac{4}{3}=\frac{2}{3}$

$\Rightarrow V_{D\text{'}.A\text{'}ME}=\frac{2}{3}{V}_{D\text{'}.A\text{'}AB}=\frac{2}{3}.\frac{1}{3}{V}_{D\text{'}DC.A\text{'}AB}=\frac{2}{3}.\frac{1}{3}.\frac{1}{2}V=\frac{1}{9}V.$

Vậy $\frac{V_{D\text{'}.PMN}}{V_{D\text{'}.A\text{'}ME}}=\frac{D\text{'}P}{D\text{'}A\text{'}}.\frac{D\text{'}M}{D\text{'}M}.\frac{D\text{'}N}{D\text{'}E}=\frac{2}{3}\mathrm{.1.}\frac{3}{4}=\frac{1}{2}\Rightarrow V_{D\text{'}.PMN}=\frac{1}{2}{V}_{D\text{'}.A\text{'}ME}=\frac{1}{18}V.$