Cho hình chóp \(S.ABCD\) có đáy \(ABCD\) là hình vuông cạnh \(a\),\(SA = a\) và \(SA\) vuông góc với đáy. Gọi \(M\)là trung điểm \(SB\),\(N\)là điểm thuộc cạnh \(SD\) sao cho \(SN = 2ND\).

a. Sai.

Ta có: \({V_{S.ABCD}} = \frac{1}{3}SA.{S_{ABCD}} = \frac{{{a^3}}}{3}\).

b. Đúng.

Vì \(ABCD\) là hình vuông nên \({S_{\Delta ABC}} = {S_{\Delta BCD}} \Rightarrow {V_{S.ABC}} = {V_{S.BCD}}\).

c. Sai.

Ta có: \(\frac{{{V_{S.AMC}}}}{{{V_{S.ABC}}}} = \frac{{SM}}{{SB}} = \frac{1}{2} \Rightarrow {V_{S.AMC}} = \frac{1}{2}{V_{S.ABC}} = \frac{1}{4}{V_{S.ABCD}}\).

d. Đúng.

Ta có: \({V_{S.ABCD}} = \frac{1}{3}SA.{S_{ABCD}} = \frac{{{a^3}}}{3}\).

Vì \(\frac{{ND}}{{SD}} = \frac{1}{3}\,\,\, \Rightarrow \,\,\,d\left( {N,\left( {ABCD} \right)} \right) = \frac{1}{3}SA = \frac{a}{3}\).

Do \(\frac{{MB}}{{SB}} = \frac{1}{2}\,\,\, \Rightarrow \,\,\,d\left( {M,\left( {ABCD} \right)} \right) = \frac{1}{2}SA = \frac{a}{2}\).

Mà \({V_{ACMN}} = {V_{S.ABCD}} - {V_{S.AMN}} - {V_{S.CMN}} - {V_{M.ABC}} - {V_{N.ADC}}\)

Mặt khác \({V_{S.ABD}} = {V_{S.BCD}} = \frac{1}{2}{V_{S.ABCD}} = \frac{{{a^3}}}{6}\).

\(\frac{{{V_{S.AMN}}}}{{{V_{S.ABD}}}} = \frac{{SM}}{{SB}}.\frac{{SN}}{{SD}} = \frac{1}{2}.\frac{2}{3} = \frac{1}{3}\)\( \Rightarrow {V_{S.AMN}} = \frac{1}{3}{V_{S.ABD}} = \frac{1}{3}.\frac{{{a^3}}}{6} = \frac{{{a^3}}}{{18}}\).

\(\frac{{{V_{S.CMN}}}}{{{V_{S.BCD}}}} = \frac{{SM}}{{SB}}.\frac{{SN}}{{SD}} = \frac{1}{2}.\frac{2}{3} = \frac{1}{3}\)\( \Rightarrow {V_{S.CMN}} = \frac{1}{3}{V_{S.BCD}} = \frac{1}{3}.\frac{{{a^3}}}{6} = \frac{{{a^3}}}{{18}}\).

\({V_{M.ABC}} = \frac{1}{3}d\left( {M,\left( {ABCD} \right)} \right).{S_{ABC}} = \frac{1}{3}.\frac{a}{2}.\frac{1}{2}{a^2} = \frac{{{a^3}}}{{12}}\).

\({V_{N.ADC}} = \frac{1}{3}d\left( {N,\left( {ABCD} \right)} \right).{S_{ADC}} = \frac{1}{3}.\frac{a}{3}.\frac{1}{2}{a^2} = \frac{{{a^3}}}{{18}}\).

Vậy \({V_{ACMN}} = \frac{{{a^3}}}{3} - \frac{{{a^3}}}{{18}} - \frac{{{a^3}}}{{18}} - \frac{{{a^3}}}{{12}} - \frac{{{a^3}}}{{18}} = \frac{{{a^3}}}{{12}}\).