Cho hình chóp tứ giác đều S.ABCD có cạnh đáy bằng...

D

- Hướng dẫn giải

Đặt \(\left\{ \begin{array}{l}
{V_1} = {V_{SABIKN}}\\
{V_2} = {V_{NBCDIK}}
\end{array} \right. \to \frac{{{V_1}}}{{{V_2}}} = ?\)

\({V_{S.ABCD}} = \frac{1}{3}.\frac{{a\sqrt 6 }}{2}{a^2} = \frac{{\sqrt 6 }}{6}{a^3}\)

\(\begin{array}{l}
{V_{N.BMC}} = \frac{1}{3}.NH.{S_{\Delta BMC}} = \frac{1}{3}.\frac{{SO}}{2}.{S_{\Delta BMC}}\\
 = \frac{1}{3}\frac{{a\sqrt 6 }}{4}.\frac{1}{2}.a.2a = \frac{{\sqrt 6 }}{{12}}{a^3}
\end{array}\)

Nhận thấy K là trọng tâm của tam giác SMC \( \to \frac{{MK}}{{MN}} = \frac{2}{3}\)

\(\frac{{{V_{M.DIK}}}}{{{V_{M.CBN}}}} = \frac{{MD}}{{MC}}.\frac{{MI}}{{MB}}.\frac{{MK}}{{MN}} = \frac{1}{2}.\frac{1}{2}.\frac{2}{3} = \frac{1}{6}\)

\(\begin{array}{l}
 \to {V_2} = {V_{M.CBN}} - {V_{M.DIK}} = \frac{5}{6}{V_{M.CBN}}\\
 = \frac{5}{6}.\frac{{\sqrt 6 }}{{12}}{a^3} = \frac{{5\sqrt 6 }}{{72}}{a^3}
\end{array}\)

\(\begin{array}{l}
 \to {V_1} = {V_{S.ABCD}} - {V_2}\\
 = \frac{{\sqrt 6 }}{6}{a^3} - \frac{{5\sqrt 6 }}{{72}}{a^3} = \frac{{7\sqrt 6 }}{{72}}{a^3}\\
 \to \frac{{{V_1}}}{{{V_2}}} = \frac{{\frac{{7\sqrt 6 }}{{72}}{a^3}}}{{\frac{{5\sqrt 6 }}{{72}}{a^3}}} = \frac{7}{5}
\end{array}\)