Cho hình lăng trụ ABC.A'B'C' Gọi E, F lần lượt là trung điểm của BB và

* Hướng dẫn giải

Đáp án A

$$\begin{gathered} V_1=\frac{1}{3}d\left(A;\left(BCC\text{'}B\text{'}\right)\right).S_{BEFC} \\ =\frac{1}{3}d\left(A;\left(BCC\text{'}B\text{'}\right)\right).S_{BCC\text{'}B\text{'}} \\ =\frac{1}{2}{V}_{ABCC\text{'}B\text{'}} \end{gathered}$$

Mà:

$$\begin{gathered} V_{ABC.A\text{'}B\text{'}C\text{'}}=V_{A.A\text{'}B\text{'}C\text{'}}+V_{A.BCC\text{'}B\text{'}} \\ \Rightarrow V_{ABC\text{'}C\text{'}B\text{'}}=\frac{2}{3}{V}_{ABC.A\text{'}B\text{'}C\text{'}} \\ \Rightarrow V_1=\frac{1}{2}.\frac{2}{3}{V}_{ABC.A\text{'}B\text{'}C\text{'}} \\ =\frac{1}{3}{V}_{ABC.A\text{'}B\text{'}C\text{'}} \end{gathered}$$

Mặt khác:

$$\begin{gathered} V_1+V_2=V_{ABC.A\text{'}B\text{'}C\text{'}} \\ \to V_2=\frac{2}{3}{V}_{ABC.A\text{'}B\text{'}C\text{'}} \\ \Rightarrow \frac{V_1}{V_2}=\frac{1}{3}:\frac{2}{3}=\frac{1}{2} \end{gathered}$$