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Câu hỏi: Cho tứ diện $ABCD$ có thể tích bằng $V$, hai điểm $M$ và $P$ lần lượt là trung điểm của $AB, CD$ ; điểm $N$ thuộc đoạn $AD$ sao cho $AD=3AN$. Tính thể tích tứ diện $BMNP$.
A. $\dfrac{V}{4}$.
B. $\dfrac{V}{12}$.
C. $\dfrac{V}{8}$.
D. $\dfrac{V}{6}$.
Ta có:
$MB=\dfrac{AB}{2},AN=\dfrac{AD}{3}\Rightarrow d\left( N,AB \right)=\dfrac{1}{3}d\left( D,AB \right)\Rightarrow {{S}_{\Delta NMB}}=\dfrac{1}{6}{{S}_{\Delta DAB}}$
$DP=\dfrac{CD}{2}\Rightarrow d\left( P,\left( MNB \right) \right)=\dfrac{1}{2}d\left( C,\left( ABD \right) \right)$ $\Rightarrow {{V}_{P.MNB}}=\dfrac{1}{3}d\left( P,\left( MNB \right) \right)\text{.}{{\text{S}}_{\Delta MNB}}=\dfrac{1}{3}.\dfrac{1}{2}d\left( C,\left( ABD \right) \right).\dfrac{1}{6}{{\text{S}}_{\Delta ABD}}=\dfrac{1}{12}V$
A. $\dfrac{V}{4}$.
B. $\dfrac{V}{12}$.
C. $\dfrac{V}{8}$.
D. $\dfrac{V}{6}$.
$MB=\dfrac{AB}{2},AN=\dfrac{AD}{3}\Rightarrow d\left( N,AB \right)=\dfrac{1}{3}d\left( D,AB \right)\Rightarrow {{S}_{\Delta NMB}}=\dfrac{1}{6}{{S}_{\Delta DAB}}$
$DP=\dfrac{CD}{2}\Rightarrow d\left( P,\left( MNB \right) \right)=\dfrac{1}{2}d\left( C,\left( ABD \right) \right)$ $\Rightarrow {{V}_{P.MNB}}=\dfrac{1}{3}d\left( P,\left( MNB \right) \right)\text{.}{{\text{S}}_{\Delta MNB}}=\dfrac{1}{3}.\dfrac{1}{2}d\left( C,\left( ABD \right) \right).\dfrac{1}{6}{{\text{S}}_{\Delta ABD}}=\dfrac{1}{12}V$
Đáp án B.