V_ABCD = \(\frac{1}{3}.6a.\frac{1}{2}.7a.4a = 28{a^3}\)

\[\frac{{{V_{APMN}}}}{{{V_{ABCD}}}} = \frac{{\frac{1}{3}.d\left( {A,\left( {PMN} \right)} \right).{S_{PMN}}}}{{\frac{1}{3}.d\left( {A,\left( {BCD} \right)} \right).{S_{BCD}}}}\] (do mặt phẳng (PMN) chính là mặt phẳng (BCD) nên d(A,(PMN) = d(A,(BCD))

⇒ \[\frac{{{V_{APMN}}}}{{{V_{ABCD}}}} = \frac{{\frac{1}{3}.d\left( {A,\left( {PMN} \right)} \right).{S_{PMN}}}}{{\frac{1}{3}.d\left( {A,\left( {BCD} \right)} \right).{S_{BCD}}}} = \frac{{{S_{PMN}}}}{{{S_{BCD}}}}\]

Mà \[\frac{{{S_{PMN}}}}{{{S_{BCD}}}} = \frac{{\frac{1}{2}.PM.PN.\sin \left( {\widehat {PM,PN}} \right)}}{{\frac{1}{2}.DC.BC.\sin \left( {\widehat {DC,BC}} \right)}}\] (do PM, MN là đường trung bình của tam giác BCD, PNCM là hình bình hành nên \(\widehat {NPM} = \widehat {BCD}\)

Suy ra: \[\frac{{{V_{APMN}}}}{{{V_{ABCD}}}} = \frac{{{S_{PMN}}}}{{{S_{BCD}}}} = \frac = \frac{1}{2}.\frac{1}{2} = \frac{1}{4}\]

V_AMNP = \(\frac{1}{4}.28{a^3} = 7{a^3}\).