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Câu hỏi: Cho tứ diện $ABCD$. Gọi $M$, $N$, $P$, $Q$, $R$, $S$ lần lượt là các điểm thuộc các cạnh $AB$, $BC$, $CD$, $AD$, $BD$, $AC$ sao cho $AM=MB$ ; $BN=NC$ ; $DQ=QA$ ; $BR=2021RD$ ; $AS=\dfrac{1}{2022}SC$ (tham khảo hình vẽ bên).
Thể tích của khối bát diện $MNPQRS$ bằng
A. $\dfrac{1}{4}V$.
B. $\dfrac{1}{3}V$.
C. $\dfrac{1011}{2021}V$.
D. $\dfrac{1}{2}V$.
Thể tích của khối bát diện $MNPQRS$ bằng
A. $\dfrac{1}{4}V$.
B. $\dfrac{1}{3}V$.
C. $\dfrac{1011}{2021}V$.
D. $\dfrac{1}{2}V$.
Ta có: ${{V}_{MNPRRS}}={{V}_{A.BCD}}-{{V}_{C.SNP}}-{{V}_{D.PQR}}-{{V}_{A.MQS}}-{{V}_{B.MNR}}$
Đặt ${{V}_{A.BCD}}=V$, ta được: $\dfrac{{{V}_{C.NPS}}}{V}=\dfrac{CN.CS.CP}{CB.CA.CD}=\dfrac{1.2022.1}{2.2023.2}$ ; $\dfrac{{{V}_{A.MQS}}}{V}=\dfrac{AM.AS.AQ}{AB.AC.AD}=\dfrac{1.1.1}{2.2023.2}$
$\dfrac{{{V}_{D.PQR}}}{V}=\dfrac{DP.DQ.DR}{DC.DA.DB}=\dfrac{1.1.1}{2.2.2022}$ ; $\dfrac{{{V}_{B.MNR}}}{V}=\dfrac{BM.BN.BR}{BA.BC.BD}=\dfrac{1.1.2021}{2.2.2022}$
$\Rightarrow {{V}_{MNPQRS}}=V-\dfrac{2022}{2.2023.2}V-\dfrac{1}{2.2023.2}V-\dfrac{1}{2.2.2022}V-\dfrac{2021}{2.2.2022}V=\dfrac{V}{2}$
Đặt ${{V}_{A.BCD}}=V$, ta được: $\dfrac{{{V}_{C.NPS}}}{V}=\dfrac{CN.CS.CP}{CB.CA.CD}=\dfrac{1.2022.1}{2.2023.2}$ ; $\dfrac{{{V}_{A.MQS}}}{V}=\dfrac{AM.AS.AQ}{AB.AC.AD}=\dfrac{1.1.1}{2.2023.2}$
$\dfrac{{{V}_{D.PQR}}}{V}=\dfrac{DP.DQ.DR}{DC.DA.DB}=\dfrac{1.1.1}{2.2.2022}$ ; $\dfrac{{{V}_{B.MNR}}}{V}=\dfrac{BM.BN.BR}{BA.BC.BD}=\dfrac{1.1.2021}{2.2.2022}$
$\Rightarrow {{V}_{MNPQRS}}=V-\dfrac{2022}{2.2023.2}V-\dfrac{1}{2.2023.2}V-\dfrac{1}{2.2.2022}V-\dfrac{2021}{2.2.2022}V=\dfrac{V}{2}$
Đáp án D.