Cho hình hộp ABCD.A'B'C'D' có thể tích bằng V. Gọi G là trọng

* Hướng dẫn giải

Chọn C.

Gọi I là trung điểm đoạn BC

Ta có

$$\begin{gathered} S_{\Delta B\text{'}C\text{'}I\text{'}}=S_{\Delta A\text{'}B\text{'}C\text{'}}=\frac{1}{2}{S}_{A\text{'}B\text{'}C\text{'}D\text{'}}=\frac{1}{2}B \\ \frac{d\left(G;\left(A\text{'}B\text{'}C\text{'}D\text{'}\right)\right)}{d\left(I;\left(A\text{'}B\text{'}C\text{'}D\text{'}\right)\right)}=\frac{GA\text{'}}{IA\text{'}}=\frac{2}{3}\Rightarrow d\left(G;\left(A\text{'}B\text{'}C\text{'}D\text{'}\right)\right)=\frac{2}{3}d\left(I;\left(A\text{'}B\text{'}C\text{'}D\text{'}\right)\right)=\frac{2}{3}h \\ \Rightarrow V_{GB\text{'}C\text{'}I\text{'}}=\frac{1}{3}d\left(G;\left(A\text{'}B\text{'}C\text{'}D\text{'}\right)\right).S_{\Delta B\text{'}C\text{'}I\text{'}}=\frac{1}{3}.\frac{2}{3}h.\frac{1}{2}B=\frac{1}{9}B.h \\ \Rightarrow V_{GB\text{'}C\text{'}I\text{'}}=\frac{1}{9}V \end{gathered}$$