Đáp án:
${V_{A.MNP}} = 2{a^3}$
Giải thích các bước giải:
Sửa đề tính ${V_{AMNP}}$
Gọi $E,F,G$ lần lượt là trung điểm của $BD,BC,CD$
Ta có:
$\dfrac{{{V_{A.MNP}}}}{{{V_{A.EFG}}}} = \dfrac{{AP}}{{AE}}.\dfrac{{AM}}{{AF}}.\dfrac{{AN}}{{AG}} = \dfrac{2}{3}.\dfrac{2}{3}.\dfrac{2}{3} = \dfrac{8}{{27}}\left( 1 \right)$
Lại có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\dfrac{{{S_{BEF}}}}{{{S_{BCD}}}} = \dfrac{{BE}}{{BD}}.\dfrac{{BF}}{{BC}} = \dfrac{1}{2}.\dfrac{1}{2} = \dfrac{1}{4}\\
\dfrac{{{S_{DEG}}}}{{{S_{BCD}}}} = \dfrac{{DE}}{{DB}}.\dfrac{{DG}}{{DC}} = \dfrac{1}{2}.\dfrac{1}{2} = \dfrac{1}{4}\\
\dfrac{{{S_{CFG}}}}{{{S_{BCD}}}} = \dfrac{{CF}}{{CB}}.\dfrac{{CG}}{{CD}} = \dfrac{1}{2}.\dfrac{1}{2} = \dfrac{1}{4}
\end{array} \right.\\
\Rightarrow \dfrac{{{S_{EFG}}}}{{{S_{BCD}}}} = \dfrac{1}{4}\\
\Rightarrow \dfrac{{{V_{A.EFG}}}}{{{V_{A.BCD}}}} = \dfrac{{\dfrac{1}{3}.d\left( {A,\left( {EFG} \right)} \right).{S_{EFG}}}}{{\dfrac{1}{3}.d\left( {A,\left( {BCD} \right)} \right).{S_{BCD}}}} = \dfrac{{{S_{EFG}}}}{{{S_{BCD}}}} = \dfrac{1}{4}\left( 2 \right)
\end{array}$
Mặt khác:
${V_{A.BCD}} = \dfrac{1}{3}AB.{S_{ACD}} = \dfrac{1}{3}.AB.\dfrac{1}{2}AC.AD = \dfrac{1}{6}.6a.9a.3a = 27{a^3}\left( 3 \right)$
Từ (1),(2),(3) ta có:
${V_{A.MNP}} = \dfrac{8}{{27}}{V_{A.EFG}} = \dfrac{8}{{27}}.\dfrac{1}{4}{V_{A.BCD}} = \dfrac{8}{{27}}.\dfrac{1}{4}.27{a^3} = 2{a^3}$
Vậy ${V_{A.MNP}} = 2{a^3}$
