Đáp án A

Dễ dàng dựng được thiết diện như hình vẽ.

Ta có: \(\frac{{SA'}}{{SA}} = \frac{{SM}}{{SI}} = \frac{{SP}}{{SD}} = \frac{{AM}}{{AI}} = \frac{1}{4}\) suy ra \(\frac{{{V_{S.AMP}}}}{{{V_{S.ADI}}}} = \frac{1}{{64}} \Rightarrow {V_{AMP.ADI}} = \frac{{63}}{{64}}{V_{S.ADI}}\)

\({V_{S.ADI}} = \frac{1}{3}.\frac{1}{2}.AD.AI.SA = \frac{1}{3}.\frac{1}{2}.a.2a.\frac{{4a}}{3} = \frac{{4{a^3}}}{9} \Rightarrow {V_{AMP.ADI}} = \frac{{63}}{{64}}{V_{S.ADI}} = \frac{{63}}{{64}}.\frac{{4{a^3}}}{9} = \frac{{7{a^3}}}{{16}}\)

\({V_{IPBN}} = \frac{1}{6}.BN.BI.BP = \frac{1}{6}.\frac{a}{2}.a.\frac{{2a}}{3} = \frac{{{a^3}}}{{18}} \Rightarrow {V_{\left( H \right)}} = {V_{AMP.ADI}} - {V_{IPBN}} = \frac{{7{a^3}}}{{16}} - \frac{{{a^3}}}{{18}} = \frac{{55{a^3}}}{{144}}\)

\( \Rightarrow {V_{\left( H \right)}} = {V_{k/p}} - {V_{\left( H \right)}} = {a^3} - \frac{{55{a^3}}}{{144}}\) suy ra \(\frac{{{V_{\left( H \right)}}}}{{{V_{\left( {H'} \right)}}}} = \frac{{55}}{{89}}\).