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Câu hỏi: Cho hình hộp $ABCD.{A}'{B}'{C}'{D}'$ có thể tích ${{V}_{1}}$. Gọi ${{O}_{1}}, {{O}_{2}}, {{O}_{3}}, {{O}_{4}}$ lần lượt là tâm các mặt bên $AB{B}'{A}', BC{C}'{B}', CD{D}'{C}', DA{A}'{D}'$. Gọi ${{V}_{2}}$ là thể tích khối đa diện $ABCD.{{O}_{1}}{{O}_{2}}{{O}_{3}}{{O}_{4}}$. Tỷ số $\dfrac{{{V}_{1}}}{{{V}_{2}}}$ bằng
A. $\dfrac{13}{5}$.
B. $\dfrac{12}{5}$.
C. $\dfrac{6}{11}$.
D. $\dfrac{11}{6}$.
Ta có ${{V}_{B{B}'{{O}_{1}}{{O}_{2}}}}={{V}_{A{A}'{{O}_{1}}{{O}_{4}}}}={{V}_{C{C}'{{O}_{2}}{{O}_{3}}}}={{V}_{D{D}'{{O}_{3}}{{O}_{4}}}}={{V}_{3}}$ ; ${{V}_{2}}={{V}_{{A}'{B}'{C}'{D}'{{O}_{1}}{{O}_{2}}{{O}_{3}}{{O}_{4}}}}\Rightarrow {{V}_{2}}=\dfrac{{{V}_{1}}-4{{V}_{3}}}{2}$
Mặt khác, $\dfrac{{{V}_{3}}}{{{V}_{{B}'BAC}}}=\dfrac{{{V}_{{B}'B{{O}_{1}}{{O}_{2}}}}}{{{V}_{{B}'BAC}}}=\dfrac{1}{4}\Rightarrow {{V}_{3}}=\dfrac{1}{4}{{V}_{{B}'BAC}}=\dfrac{1}{4}.\dfrac{1}{6}{{V}_{1}}=\dfrac{{{V}_{1}}}{24}$.
Do vậy, ta được: ${{V}_{2}}=\dfrac{{{V}_{1}}-4\dfrac{{{V}_{1}}}{24}}{2}=\dfrac{5}{12}{{V}_{1}}\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{12}{5}$.
A. $\dfrac{13}{5}$.
B. $\dfrac{12}{5}$.
C. $\dfrac{6}{11}$.
D. $\dfrac{11}{6}$.
Mặt khác, $\dfrac{{{V}_{3}}}{{{V}_{{B}'BAC}}}=\dfrac{{{V}_{{B}'B{{O}_{1}}{{O}_{2}}}}}{{{V}_{{B}'BAC}}}=\dfrac{1}{4}\Rightarrow {{V}_{3}}=\dfrac{1}{4}{{V}_{{B}'BAC}}=\dfrac{1}{4}.\dfrac{1}{6}{{V}_{1}}=\dfrac{{{V}_{1}}}{24}$.
Do vậy, ta được: ${{V}_{2}}=\dfrac{{{V}_{1}}-4\dfrac{{{V}_{1}}}{24}}{2}=\dfrac{5}{12}{{V}_{1}}\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{12}{5}$.
Đáp án B.