Lời giải
Giả sử \(V = {V_{ABC.A'B'C'}} = 2020\).
Cách 1
Ta có \({V_{C'.ABC}} = \frac{1}{3}d\left( {C';\left( {ABC} \right)} \right).{S_{\Delta ABC}} = \frac{V}{3} \Rightarrow {V_{C'.ABB'A'}} = \frac{2}{3}V\).
Lại có \(\frac{{{V_{P.ABC}}}}{{{V_{C'.ABC}}}} = \frac{{\frac{1}{3}.d\left( {P;\left( {ABC} \right)} \right).{S_{\Delta ABC}}}}{{\frac{1}{3}.d\left( {C';\left( {ABC} \right)} \right).{S_{\Delta ABC}}}} = \frac{{d\left( {P;\left( {ABC} \right)} \right)}}{{d\left( {C';\left( {ABC} \right)} \right)}} = \frac{{PC}}{{CC'}} = \frac{3}{4} \Rightarrow {V_{P.ABC}} = \frac{1}{4}V\).
Ta có \(\frac{{{V_{P.ABNM}}}}{{{V_{C'.ABB'A'}}}} = & \frac{{\frac{1}{3}.d\left( {P;\left( {ABB'A'} \right)} \right).{S_{ABNM}}}}{{\frac{1}{3}.d\left( {C;\left( {ABB'A'} \right)} \right).{S_{ABB'A'}}}}\).
Mà \(d\left( {P;\left( {ABB'A'} \right)} \right) = d\left( {C;\left( {ABB'A'} \right)} \right)\)và \({S_{ABNM}} = \frac{1}{2}{S_{ABB'A'}}\).
Suy ra \(\frac{{{V_{P.ABNM}}}}{{{V_{C'.ABB'A'}}}} = \frac{1}{2} \Rightarrow {V_{P.ABNM}} = \frac{1}{3}V\).
Vậy \({V_{ABC.MNP}} = {V_{P.ABNM}} + {V_{P.ABC}} = \frac{7}{{12}}V = \frac{{3535}}{3}\).
Cách 2: Dùng công thức giải nhanh
Ta có: \(\frac{{{V_{ABC.MNP}}}}{{{V_{ABC.A'B'C'}}}} = \frac{1}{3}\left( {\frac{{AM}}{{AA'}} + \frac{{BN}}{{BB'}} + \frac{{CP}}{{CC'}}} \right)\)\( \Rightarrow {V_{ABC.MNP}} = \frac{{2020}}{3}\left( {\frac{1}{2} + \frac{1}{2} + \frac{3}{4}} \right) = \frac{{3535}}{3}\).
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