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Câu hỏi: Cho khối lăng trụ đứng $ABC.ABC$ có đáy $ABC$ là tam giác vuông tại $A,AB=a,$ $\widehat{ACB}={{30}^{0}}$ $BC'=a\sqrt{7}$ . Lấy hai điểm $M,N$ lần lượt trên hai cạnh AB' và AC' sao cho $\overrightarrow{AB'}=3\overrightarrow{AM},\overrightarrow{NC}=2\overrightarrow{A'N}$. Thể tích khối đa diện $BMNCC$ bằng
A. $\dfrac{8}{9}{{a}^{3}}.$
B. $\dfrac{4}{9}{{a}^{3}}.$
C. $\dfrac{1}{2}{{a}^{3}}.$
D. $\dfrac{3}{2}{{a}^{3}}.$
+ Ta có $BC=2a\Rightarrow \left\{ \begin{aligned}
& CC'=\sqrt{BC{{'}^{2}}-B{{C}^{2}}}=\sqrt{7{{a}^{2}}-4{{a}^{2}}}=a\sqrt{3} \\
& AC=\sqrt{B{{C}^{2}}-A{{B}^{2}}}=\sqrt{4{{a}^{2}}-{{a}^{2}}}=a\sqrt{3} \\
\end{aligned} \right..$
+) ${{S}_{\Delta ABC}}=\dfrac{1}{2}AB.AC=\dfrac{1}{2}a.a\sqrt{3}={{a}^{2}}\sqrt{3}\Rightarrow {{V}_{ABC.A'B'C'}}=CC'.{{S}_{\Delta ABC}}=a\sqrt{3}.{{a}^{2}}\sqrt{3}=3{{a}^{3}}.$
+) $\dfrac{{{V}_{B.ACC'A'}}}{{{V}_{ABC.A'B'C'}}}=\dfrac{2}{3}\Rightarrow {{V}_{B.ACC'A'}}=2{{a}^{3}}.$
+) $\dfrac{{{V}_{B.SCC'}}}{{{V}_{B.ACC'A'}}}=\dfrac{{{S}_{\Delta SCC'}}}{{{S}_{ACC'A'}}}=\dfrac{1}{2}\Rightarrow {{V}_{B.SCC'}}={{a}^{3}}.$
+) $\dfrac{{{V}_{S.MNC}}}{{{V}_{B.SCC'}}}=\dfrac{SN}{SC'}.\dfrac{SM}{SB}.\dfrac{SC}{SC}=\dfrac{1}{3}.\dfrac{1}{3}.\dfrac{1}{1}=\dfrac{1}{9}\Rightarrow {{V}_{BMNCC'}}=\dfrac{8}{9}{{a}^{3}}.$
A. $\dfrac{8}{9}{{a}^{3}}.$
B. $\dfrac{4}{9}{{a}^{3}}.$
C. $\dfrac{1}{2}{{a}^{3}}.$
D. $\dfrac{3}{2}{{a}^{3}}.$
& CC'=\sqrt{BC{{'}^{2}}-B{{C}^{2}}}=\sqrt{7{{a}^{2}}-4{{a}^{2}}}=a\sqrt{3} \\
& AC=\sqrt{B{{C}^{2}}-A{{B}^{2}}}=\sqrt{4{{a}^{2}}-{{a}^{2}}}=a\sqrt{3} \\
\end{aligned} \right..$
+) ${{S}_{\Delta ABC}}=\dfrac{1}{2}AB.AC=\dfrac{1}{2}a.a\sqrt{3}={{a}^{2}}\sqrt{3}\Rightarrow {{V}_{ABC.A'B'C'}}=CC'.{{S}_{\Delta ABC}}=a\sqrt{3}.{{a}^{2}}\sqrt{3}=3{{a}^{3}}.$
+) $\dfrac{{{V}_{B.ACC'A'}}}{{{V}_{ABC.A'B'C'}}}=\dfrac{2}{3}\Rightarrow {{V}_{B.ACC'A'}}=2{{a}^{3}}.$
+) $\dfrac{{{V}_{B.SCC'}}}{{{V}_{B.ACC'A'}}}=\dfrac{{{S}_{\Delta SCC'}}}{{{S}_{ACC'A'}}}=\dfrac{1}{2}\Rightarrow {{V}_{B.SCC'}}={{a}^{3}}.$
+) $\dfrac{{{V}_{S.MNC}}}{{{V}_{B.SCC'}}}=\dfrac{SN}{SC'}.\dfrac{SM}{SB}.\dfrac{SC}{SC}=\dfrac{1}{3}.\dfrac{1}{3}.\dfrac{1}{1}=\dfrac{1}{9}\Rightarrow {{V}_{BMNCC'}}=\dfrac{8}{9}{{a}^{3}}.$
Đáp án A.