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Câu hỏi: Cho khối lập phương $ABC\text{D}\text{.{A}'{B}'{C}'{D}'}$ cạnh a. Gọi M, N lần lượt là trung điểm của đoạn thẳng ${A}'{D}'$ và ${C}'{D}'$. Mặt phẳng $(BMN)$ chia khối lập phương thành hai phần, gọi V là thể tích phần chứa đỉnh ${B}'$. Tính V?
A. $\dfrac{25{{\text{a}}^{3}}}{72}$
B. $\dfrac{\text{7}{{\text{a}}^{3}}}{24}$
C. $\dfrac{25{{\text{a}}^{3}}}{24}$
D. $\dfrac{\text{7}{{\text{a}}^{3}}}{72}$
Ta có thể tích cần tính là ${{V}_{B{B}'E\text{{A}'}MN{C}'F}}$.
Mà ${{V}_{B{B}'E\text{{A}'MN{C}'F}}}={{V}_{B.E\text{{A}'}M}}+{{V}_{B{B}'{A}'MN{C}'}}+{{V}_{V.F{C}'N}}$.
Ta có: $\left\{ \begin{aligned}
& \Delta P{A}'M=\Delta N\text{{D}'}M \\
& \Delta Q{C}'N=\Delta M\text{{D}'}N \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& P{A}'=N{D}' \\
& Q{C}'=M{D}' \\
\end{aligned} \right.$.
Lại có: $M\text{{D}'}=N\text{{D}'}\Rightarrow P{A}'=N\text{{D}'}=M\text{{D}'}=Q{C}'=\dfrac{a}{2}$.
Mà: $\left\{ \begin{aligned}
& \Delta P{A}'E\#\Delta BA\text{E} \\
& \Delta Q{C}'F\#\Delta BCF \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \dfrac{{A}'E}{A\text{E}}=\dfrac{P{A}'}{BA}=\dfrac{1}{2} \\
& \dfrac{{C}'F}{CF}=\dfrac{Q{C}'}{BC}=\dfrac{1}{2} \\
\end{aligned} \right.\Rightarrow {A}'E=Q{C}'=\dfrac{a}{3}$.
Vậy ta có: ${{V}_{B.E\text{{A}'}M}}=\dfrac{1}{3}BA.{{S}_{E\text{{A}'}M}}=\dfrac{1}{3}BA.\dfrac{1}{2}{A}'M.{A}'E=\dfrac{{{a}^{3}}}{36};{{V}_{B.F{C}'N}}={{V}_{B.E\text{{A}'}M}}=\dfrac{{{a}^{3}}}{36}$.
Có: ${{V}_{B{B}'{A}'MN{C}'}}=\dfrac{1}{3}B{B}'.{{S}_{{B}'{A}'MN{C}'}}=\dfrac{1}{3}B{B}'\left( {{S}_{{A}'{B}'{C}'{D}'}}-{{S}_{M\text{{D}'}N}} \right)=\dfrac{7{{\text{a}}^{3}}}{24}$.
Vậy: ${{V}_{B{B}'E\text{{A}'}MN{C}'F}}=\dfrac{25{{\text{a}}^{3}}}{72}$.
A. $\dfrac{25{{\text{a}}^{3}}}{72}$
B. $\dfrac{\text{7}{{\text{a}}^{3}}}{24}$
C. $\dfrac{25{{\text{a}}^{3}}}{24}$
D. $\dfrac{\text{7}{{\text{a}}^{3}}}{72}$
Mà ${{V}_{B{B}'E\text{{A}'MN{C}'F}}}={{V}_{B.E\text{{A}'}M}}+{{V}_{B{B}'{A}'MN{C}'}}+{{V}_{V.F{C}'N}}$.
Ta có: $\left\{ \begin{aligned}
& \Delta P{A}'M=\Delta N\text{{D}'}M \\
& \Delta Q{C}'N=\Delta M\text{{D}'}N \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& P{A}'=N{D}' \\
& Q{C}'=M{D}' \\
\end{aligned} \right.$.
Lại có: $M\text{{D}'}=N\text{{D}'}\Rightarrow P{A}'=N\text{{D}'}=M\text{{D}'}=Q{C}'=\dfrac{a}{2}$.
Mà: $\left\{ \begin{aligned}
& \Delta P{A}'E\#\Delta BA\text{E} \\
& \Delta Q{C}'F\#\Delta BCF \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \dfrac{{A}'E}{A\text{E}}=\dfrac{P{A}'}{BA}=\dfrac{1}{2} \\
& \dfrac{{C}'F}{CF}=\dfrac{Q{C}'}{BC}=\dfrac{1}{2} \\
\end{aligned} \right.\Rightarrow {A}'E=Q{C}'=\dfrac{a}{3}$.
Vậy ta có: ${{V}_{B.E\text{{A}'}M}}=\dfrac{1}{3}BA.{{S}_{E\text{{A}'}M}}=\dfrac{1}{3}BA.\dfrac{1}{2}{A}'M.{A}'E=\dfrac{{{a}^{3}}}{36};{{V}_{B.F{C}'N}}={{V}_{B.E\text{{A}'}M}}=\dfrac{{{a}^{3}}}{36}$.
Có: ${{V}_{B{B}'{A}'MN{C}'}}=\dfrac{1}{3}B{B}'.{{S}_{{B}'{A}'MN{C}'}}=\dfrac{1}{3}B{B}'\left( {{S}_{{A}'{B}'{C}'{D}'}}-{{S}_{M\text{{D}'}N}} \right)=\dfrac{7{{\text{a}}^{3}}}{24}$.
Vậy: ${{V}_{B{B}'E\text{{A}'}MN{C}'F}}=\dfrac{25{{\text{a}}^{3}}}{72}$.
Đáp án A.