Cho khối lăng trụ ABCA'B'C' có thể tích bằng 1. Gọi M, N lần

* Hướng dẫn giải

Ta có A' là trung điểm PC'; B' là trung điểm QC'.

Do đó $V_{C.C^{\text{'}}PQ}=\frac{S_{C^{\text{'}}PQ}}{S_{C^{\text{'}}{A}^{\text{'}}{B}^{\text{'}}}}.V_{C.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}=4V_{C.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}=4\left(\frac{1}{3}{V}_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}\right)=\frac{4}{3}$ 

Mặt khác$V_{A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}.MNC}=\frac{\frac{A^{\text{'}}M}{A^{\text{'}}A}+\frac{B^{\text{'}}N}{B^{\text{'}}B}+\frac{C^{\text{'}}C}{C^{\text{'}}C}}{3}.V_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}=\frac{\frac{1}{2}+\frac{1}{2}+1}{3}.V_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}=\frac{2}{3}$

Do đó .$V_{A^{\text{'}}M{B}^{\text{'}}NQ}=V_{C.C^{\text{'}}PQ}-V_{A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}.MNC}=\frac{4}{3}-\frac{2}{3}=\frac{2}{3}$