Cho khối chóp S.ABCD có đáy ABCD là hình bình hành

* Ta có: \({V_{S.CDMN}} = {V_{S.CDM}} + {V_{S.CNM}}\left( 1 \right)\)

\(\frac{{{V_{S.CDM}}}}{{{V_{S.CDA}}}} = \frac{{SC}}{{SC}}.\frac{{SD}}{{SD}}.\frac{{SM}}{{SA}} = \frac{1}{2} \Rightarrow {V_{S.CDM}} = \frac{1}{2}{V_{S.CDA}} = \frac{1}{4}{V_{S.ABCD}}\),

\(\frac{{{V_{S.CNM}}}}{{{V_{S.CBA}}}} = \frac{{SC}}{{SC}}.\frac{{SN}}{{SB}}.\frac{{SM}}{{SA}} = \frac{1}{4} \Rightarrow {V_{S.CNM}} = \frac{1}{4}{V_{S.CBA}} = \frac{1}{8}{V_{S.ABCD}}\).

Thay vào (1), ta được: \({V_{S.CDMN}} = \frac{3}{8}{V_{S.ABCD}}\),

suy ra \({V_{ABCDMN}} = {V_{S.ABCD}} - {V_{S.CNMN}} = \frac{5}{8}{V_{S.ABCD}}\).

Vậy \(\frac{{{V_{S.CDMN}}}}{{{V_{ABCDMN}}}} = \frac{3}{5}\).