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Câu hỏi: Cho tứ diện ABCD và M, N, P lần lượt thuộc BC, BD, AC sao cho $BC=4BM$, $BD=2BN$, $AC=3AP$. Mặt phẳng $\left( MNP \right)$ cắt AD tại Q. Tính tỷ số thể tích hai phần khối tứ diện ABCD bị chia bởi mặt phẳng $\left( MNP \right)$.
A. $\dfrac{2}{3}$.
B. $\dfrac{7}{13}$.
C. $\dfrac{5}{13}$.
D. $\dfrac{1}{3}$.
A. $\dfrac{2}{3}$.
B. $\dfrac{7}{13}$.
C. $\dfrac{5}{13}$.
D. $\dfrac{1}{3}$.
Gọi $I=MN\cap CD,Q=PI\cap AD$.
Kẻ $DH\text{//}BC\left( H\in IM \right)$ và $DK\text{//}AC\left( K\in IP \right)$.
$\Delta NMB=\Delta NDH\Rightarrow \dfrac{ID}{IC}=\dfrac{DH}{CM}=\dfrac{BM}{CM}=\dfrac{1}{3}$.
$\dfrac{IK}{IP}=\dfrac{DK}{CP}=\dfrac{ID}{IC}=\dfrac{1}{3}\Rightarrow \dfrac{DK}{2AP}=\dfrac{1}{3}\Rightarrow DK=\dfrac{2}{3}$
$\Delta APQ\Delta DKQ\Rightarrow \dfrac{AQ}{DQ}=\dfrac{AP}{DK}=\dfrac{2}{3}\Rightarrow \dfrac{AQ}{AD}=\dfrac{3}{5}$.
Đặt $V={{V}_{ABCD}}$.
Ta có: $\dfrac{{{V}_{ANPQ}}}{{{V}_{ANCD}}}=\dfrac{AP}{AC}.\dfrac{AQ}{AD}=\dfrac{1}{5}$
$\dfrac{{{V}_{ANCD}}}{{{V}_{ABCD}}}=\dfrac{{{V}_{DACN}}}{{{V}_{DABC}}}=\dfrac{DN}{DB}=\dfrac{1}{2}\Rightarrow {{V}_{ANPQ}}=\dfrac{1}{10}V$.
$\dfrac{{{V}_{CDMP}}}{{{V}_{CDBA}}}=\dfrac{CM}{CB}.\dfrac{CP}{CA}=\dfrac{1}{2}\Rightarrow {{V}_{CDMP}}=\dfrac{1}{2}V\Rightarrow {{V}_{V.ABMP}}=\dfrac{1}{2}{{V}_{DABMP}}=\dfrac{1}{2}V-{{V}_{CDMP}}=\dfrac{1}{4}V$
$\Rightarrow {{V}_{ABMNQP}}={{V}_{ANPQ}}+{{V}_{N.ABMP}}=\dfrac{7}{20}V\Rightarrow \dfrac{{{V}_{ABMNQP}}}{{{V}_{CDMNQP}}}=\dfrac{7}{13}$.
Vậy mặt phẳng $\left( MNP \right)$ chia khối chóp thành hai phần với tỉ lệ thể tích $\dfrac{7}{13}$.
Kẻ $DH\text{//}BC\left( H\in IM \right)$ và $DK\text{//}AC\left( K\in IP \right)$.
$\Delta NMB=\Delta NDH\Rightarrow \dfrac{ID}{IC}=\dfrac{DH}{CM}=\dfrac{BM}{CM}=\dfrac{1}{3}$.
$\dfrac{IK}{IP}=\dfrac{DK}{CP}=\dfrac{ID}{IC}=\dfrac{1}{3}\Rightarrow \dfrac{DK}{2AP}=\dfrac{1}{3}\Rightarrow DK=\dfrac{2}{3}$
$\Delta APQ\Delta DKQ\Rightarrow \dfrac{AQ}{DQ}=\dfrac{AP}{DK}=\dfrac{2}{3}\Rightarrow \dfrac{AQ}{AD}=\dfrac{3}{5}$.
Đặt $V={{V}_{ABCD}}$.
Ta có: $\dfrac{{{V}_{ANPQ}}}{{{V}_{ANCD}}}=\dfrac{AP}{AC}.\dfrac{AQ}{AD}=\dfrac{1}{5}$
$\dfrac{{{V}_{ANCD}}}{{{V}_{ABCD}}}=\dfrac{{{V}_{DACN}}}{{{V}_{DABC}}}=\dfrac{DN}{DB}=\dfrac{1}{2}\Rightarrow {{V}_{ANPQ}}=\dfrac{1}{10}V$.
$\dfrac{{{V}_{CDMP}}}{{{V}_{CDBA}}}=\dfrac{CM}{CB}.\dfrac{CP}{CA}=\dfrac{1}{2}\Rightarrow {{V}_{CDMP}}=\dfrac{1}{2}V\Rightarrow {{V}_{V.ABMP}}=\dfrac{1}{2}{{V}_{DABMP}}=\dfrac{1}{2}V-{{V}_{CDMP}}=\dfrac{1}{4}V$
$\Rightarrow {{V}_{ABMNQP}}={{V}_{ANPQ}}+{{V}_{N.ABMP}}=\dfrac{7}{20}V\Rightarrow \dfrac{{{V}_{ABMNQP}}}{{{V}_{CDMNQP}}}=\dfrac{7}{13}$.
Vậy mặt phẳng $\left( MNP \right)$ chia khối chóp thành hai phần với tỉ lệ thể tích $\dfrac{7}{13}$.
Đáp án B.