Cho hình chóp\(S.ABCD\) có đáy \(ABCD\) là hình hì...

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- Hướng dẫn giải

Ta có \(\frac{{{V}_{S.MNC}}}{{{V}_{S.ABC}}}=\frac{SM}{SA}.\frac{SN}{SB}.\frac{SC}{SC}=\frac{1}{2}.\frac{1}{2}=\frac{1}{4}\Rightarrow {{V}_{S.MNC}}=\frac{1}{4}{{V}_{S.ABC}}=\frac{1}{8}{{V}_{S.ABCD}}\)

 \(\frac{{{V}_{S.MCD}}}{{{V}_{S.ACD}}}=\frac{SM}{SA}.\frac{SC}{SC}.\frac{SD}{SD}=\frac{1}{2}\Rightarrow {{V}_{S.MCD}}=\frac{1}{2}{{V}_{S.ACD}}=\frac{1}{4}{{V}_{S.ABCD}}\)

 \(\Rightarrow {{V}_{S.MNCD}}={{V}_{S.MNC}}+{{V}_{S.MCD}}=\frac{1}{8}{{V}_{S.ABCD}}+\frac{1}{4}{{V}_{S.ABCD}}=\frac{3}{8}{{V}_{S.ABCD}}\)

Vậy \({{V}_{ABCDMN}}={{V}_{S.ABCD}}-{{V}_{S.MNCD}}=\frac{5}{8}{{V}_{S.ABCD}}=\frac{5}{8}.18=\frac{45}{4}\)