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Câu hỏi: Cho khối hộp $ABCD.{A}'{B}'{C}'{D}'$ có diện tích đáy bằng ${{a}^{2}}$ và chiều cao bằng $2a$. Lấy điểm $M$ thuộc đoạn $C{D}'$ sao cho $MC=3M{D}'$, lấy điểm $N$ thuộc đoạn $C{B}'$ sao cho $CN=2N{B}'$. Thể tích $V$ của khối đa diện $A{B}'{C}'{D}'MN$ là
A. $V=\dfrac{{{a}^{3}}}{6}$.
B. $V=\dfrac{{{a}^{3}}}{3}$.
C. $V=\dfrac{{{a}^{3}}}{4}$.
D. $V=\dfrac{{{a}^{3}}}{2}$.
$V={{V}_{A{D}'{B}'NM}}+{{V}_{NM{D}'{B}'{C}'}}$.
${{V}_{A{D}'{B}'NM}}={{V}_{CA{D}'{B}'}}-{{V}_{CAMN}}=\dfrac{1}{3}{{V}_{ABCD.{A}'{B}'{C}'{D}'}}-\dfrac{2}{3}.\dfrac{3}{4}{{V}_{CA{B}'{D}'}}$
$=\dfrac{1}{3}.2{{a}^{3}}-\dfrac{1}{2}.\dfrac{1}{3}{{V}_{ABCD.{A}'{B}'{C}'{D}'}}=\dfrac{1}{3}.2{{a}^{3}}-\dfrac{1}{2}.\dfrac{1}{3}2{{a}^{3}}=\dfrac{{{a}^{3}}}{3}$.
${{V}_{NM{D}'{B}'{C}'}}={{V}_{C{B}'{D}'{C}'}}-{{V}_{CNM{C}'}}=\dfrac{1}{6}{{V}_{ABCD{A}'{B}'{C}'{D}'}}-\dfrac{2}{3}.\dfrac{3}{4}{{V}_{C{B}'{D}'{C}'}}=\dfrac{1}{6}.2{{a}^{3}}-\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{1}{6}.2{{a}^{3}}=\dfrac{{{a}^{3}}}{6}$.
$\Rightarrow V={{V}_{A{D}'{B}'NM}}+{{V}_{NM{D}'{B}'{C}'}}=\dfrac{{{a}^{3}}}{3}+\dfrac{{{a}^{3}}}{6}=\dfrac{{{a}^{3}}}{2}$.
A. $V=\dfrac{{{a}^{3}}}{6}$.
B. $V=\dfrac{{{a}^{3}}}{3}$.
C. $V=\dfrac{{{a}^{3}}}{4}$.
D. $V=\dfrac{{{a}^{3}}}{2}$.
$V={{V}_{A{D}'{B}'NM}}+{{V}_{NM{D}'{B}'{C}'}}$.
${{V}_{A{D}'{B}'NM}}={{V}_{CA{D}'{B}'}}-{{V}_{CAMN}}=\dfrac{1}{3}{{V}_{ABCD.{A}'{B}'{C}'{D}'}}-\dfrac{2}{3}.\dfrac{3}{4}{{V}_{CA{B}'{D}'}}$
$=\dfrac{1}{3}.2{{a}^{3}}-\dfrac{1}{2}.\dfrac{1}{3}{{V}_{ABCD.{A}'{B}'{C}'{D}'}}=\dfrac{1}{3}.2{{a}^{3}}-\dfrac{1}{2}.\dfrac{1}{3}2{{a}^{3}}=\dfrac{{{a}^{3}}}{3}$.
${{V}_{NM{D}'{B}'{C}'}}={{V}_{C{B}'{D}'{C}'}}-{{V}_{CNM{C}'}}=\dfrac{1}{6}{{V}_{ABCD{A}'{B}'{C}'{D}'}}-\dfrac{2}{3}.\dfrac{3}{4}{{V}_{C{B}'{D}'{C}'}}=\dfrac{1}{6}.2{{a}^{3}}-\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{1}{6}.2{{a}^{3}}=\dfrac{{{a}^{3}}}{6}$.
$\Rightarrow V={{V}_{A{D}'{B}'NM}}+{{V}_{NM{D}'{B}'{C}'}}=\dfrac{{{a}^{3}}}{3}+\dfrac{{{a}^{3}}}{6}=\dfrac{{{a}^{3}}}{2}$.
Đáp án D.