Cho lăng trụ đứng tam giác ABC.A'B'C'. Gọi M, N, P...

B

- Hướng dẫn giải

Ta có \({V_1} = {V_{MNPQ}} = \frac{1}{3}d\left( {M;(NPQ)} \right).{S_{NPQ}},\) 

\({V_2} = {V_{ABC.A'B'C'}} = \frac{3}{2}{V_{A.BCC'B'}} = \frac{3}{2}.\frac{1}{3}d\left( {A;(BCC'B')} \right).{S_{BCC'B'}}.\) 

Ta có: \(d\left( {M;(NPQ)} \right) = d\left( {A;(BCC'B')} \right).\) 

Đặt \(BC = x,BB' = y\) ta có \({S_{BCC'B'}} = xy\) 

\(\begin{array}{l}
S{}_{BCPN} = \frac{{\left( {BN + CP} \right).BC}}{2} = \frac{{\left( {\frac{y}{3} + \frac{y}{4}} \right).x}}{2} = \frac{7}{{24}}xy\\
{S_{B'NQ}} = \frac{1}{2}B'N.B'Q = \frac{1}{2}.\frac{2}{3}y.\frac{4}{5}x = \frac{4}{{15}}xy\\
{S_{C'PQ}} = \frac{1}{2}C'P.C'Q = \frac{1}{2}.\frac{3}{4}y.\frac{1}{5}x = \frac{3}{{40}}xy\\
 \Rightarrow {S_{NPQ}} = xy - \frac{7}{{24}}xy - \frac{4}{{15}}xy - \frac{3}{{40}}xy = \frac{{11xy}}{{30}} = \frac{{11}}{{30}}{S_{BCC'B'}}\\
 \Rightarrow {V_1} = {V_{MNPQ}} = \frac{1}{3}d\left( {A;(BCC'B')} \right).\frac{{11}}{{30}}{S_{BCC'B'}} = \frac{{11}}{{90}}d\left( {A;(BCC'B')} \right).{S_{BCC'B'}}\\
 \Rightarrow \frac{{{V_1}}}{{{V_2}}} = \frac{{\frac{{11}}{{90}}d\left( {A;(BCC'B')} \right).{S_{BCC'B'}}}}{{\frac{3}{2}.\frac{1}{3}d\left( {A;(BCC'B')} \right).{S_{BCC'B'}}}} = \frac{{11}}{{45}}.
\end{array}\)