Cho hình lăng trụ ABC.A'B'C'. Gọi M , N, P lần lượ...

C

- Hướng dẫn giải

Gọi V là thể tích khối lăng trụ ABC.A'B'C'.

Ta có \({V_1} = {V_{M.ABC}} + {V_{M.BCPN}}\)

\({V_{M.ABC}} = \frac{1}{3}d\left( {M;\left( {ABC} \right)} \right).{S_{ABC}} = \frac{1}{3}.\frac{2}{3}d\left( {A';\left( {ABC} \right)} \right).{S_{ABC}} = \frac{2}{9}V\)

\(\frac{{{V_{M.BCPN}}}}{{{V_{M.BCC'B'}}}} = \frac{{{S_{BCPN}}}}{{{S_{BCC'B'}}}} = \frac{{\frac{1}{2}d\left( {C;BB'} \right).\left( {BN + CP} \right)}}{{\frac{1}{2}d\left( {C;BB'} \right).\left( {BB' + CC'} \right)}} = \frac{{BN + CP}}{{BB' + CC'}} = \frac{{\frac{1}{3}BB' + \frac{1}{2}CC'}}{{BB' + CC'}} = \frac{5}{{12}}\left( {BB' = CC'} \right)\)

\( \Rightarrow {V_{M.BCPN}} = \frac{5}{{12}}{V_{M.BCC'B'}} = \frac{5}{{12}}.2{V_{ABCB'}} = \frac{5}{{12}}.2.\frac{1}{3}V = \frac{5}{{18}}V\)

\( \Rightarrow {V_1} = {V_{M.ABC}} + {V_{M.BCPN}} = \frac{2}{9}V + \frac{5}{{18}}V = \frac{1}{2}V \Rightarrow {V_2} = V - \frac{1}{2}V = \frac{1}{2}V \Rightarrow \frac{{{V_1}}}{{{V_2}}} = 1\)