Đáp án:
Giải thích các bước giải:
a) Ta có: \(MN//AB\).
Dễ thấy:
\(\begin{array}{l}AB \bot PO,AB \bot SO\\ \Rightarrow AB \bot \left( {SPO} \right) \Rightarrow AB \bot SP\\ \Rightarrow MN \bot SP\end{array}\)
b) Do \(M\) là trung điểm \(SA\) nên \(d\left( {A,\left( {MNP} \right)} \right) = d\left( {S,\left( {MNP} \right)} \right) \Rightarrow {V_{A.MNP}} = {V_{S.MNP}}\)
Lại có: \(\frac{{{V_{S.MNP}}}}{{{V_{S.ABP}}}} = \frac{{SM}}{{SA}}.\frac{{SN}}{{SB}}.\frac{{SP}}{{SP}} = \frac{1}{4} \Rightarrow {V_{S.MNP}} = \frac{1}{4}{V_{S.ABP}}\)
Mà \(\frac{{{V_{S.ABP}}}}{{{V_{S.ABCD}}}} = \frac{{{S_{ABP}}}}{{{S_{ABCD}}}} = \frac{1}{2} \Rightarrow {V_{S.ABP}} = \frac{1}{2}{V_{S.ABD}}\)
Do đó \({V_{A.MNP}} = {V_{S.MNP}} = \frac{1}{4}.\frac{1}{2}{V_{S.ABCD}} = \frac{1}{8}{V_{S.ABCD}}\).
Có \({S_{ABCD}} = {a^2},SO = \sqrt {S{A^2} - A{O^2}} = \sqrt {2{a^2} - \frac{{{a^2}}}{2}} = \frac{{a\sqrt 6 }}{2}\)
\( \Rightarrow {V_{S.ABCD}} = \frac{1}{3}SO.{S_{ABCD}} = \frac{1}{3}.\frac{{a\sqrt 6 }}{2}.{a^2} = \frac{{{a^3}\sqrt 6 }}{6}\)
\( \Rightarrow {V_{A.MNP}} = \frac{1}{8}{V_{S.ABCD}} = \frac{1}{8}.\frac{{{a^3}\sqrt 6 }}{6} = \frac{{{a^3}\sqrt 6 }}{{48}}\)
