Cho hình chóp S.ABCD có đáy ABCD là hình vuông và...

B

- Hướng dẫn giải

Gọi \(M = SD \cap S'A\) 

Trong (S'AB) kẻ \(MN//AB\left( {N \in SC} \right)\) ta có:

\(MN \cap S'B = P \Rightarrow MP = \left( {S'AB} \right) \cap \left( {SCD} \right)\) 

Áp dụng định lí Ta-lét ta có: \(\frac{{MD}}{{MS}} = \frac{{S'D}}{{SA}} = \frac{1}{2} = \frac{{NC}}{{NS}}\) 

Ta có:

\(\begin{array}{l}
\frac{{{V_{S.AMN}}}}{{{V_{S.ADC}}}} = \frac{{SM}}{{SD}}.\frac{{SN}}{{SC}} = \frac{4}{9} \Rightarrow {V_{S.AMN}} = \frac{4}{9}{V_{S.ADC}} \Rightarrow {V_{S.AMN}} = \frac{2}{9}{V_2}\\
\frac{{{V_{S.ANB}}}}{{{V_{S.ACB}}}} = \frac{{SN}}{{SC}} = \frac{2}{3} \Rightarrow {V_{S.ANB}} = \frac{2}{3}{V_{S.ACB}} \Rightarrow {V_{S.ANB}} = \frac{1}{3}{V_2}\\
 \Rightarrow {V_{S.AMNB}} = \frac{2}{9}{V_2} + \frac{1}{3}{V_2} = \frac{5}{9}{V_2} \Rightarrow {V_{MN.ABCD}} = \frac{4}{9}{V_2}
\end{array}\) 

Áp dụng định lí Ta-lét ta có: \(\frac{{MP}}{{AB}} = \frac{{S'M}}{{S'A}} = \frac{1}{3} \Rightarrow MP = \frac{1}{3}AB = \frac{1}{3}MN\) 

\( \Rightarrow {V_1} = {V_{M.ABCD}} - {V_{P.NBC}} = \frac{4}{9}{V_2} - \frac{1}{9}{V_2} = \frac{1}{3}{V_2} \Rightarrow \frac{{{V_1}}}{{{V_2}}} = \frac{1}{3}\) 

\(\begin{array}{l}
 \Rightarrow PN = \frac{2}{3}MN = \frac{2}{3}AB;\,\,\frac{{{S_{NBC}}}}{{{S_{SBC}}}} = \frac{{NC}}{{SC}} = \frac{{MD}}{{SD}} = \frac{1}{3}\\
 \Rightarrow \frac{{{V_{P.NBC}}}}{{{V_{A.SBC}}}} = \frac{1}{3}.\frac{1}{3} = \frac{2}{9} \Rightarrow {V_{P.NBC}} = \frac{2}{9}{V_{A.SBC}} = \frac{1}{9}{V_2}
\end{array}\)