Cho tứ diện đều ABCD có cạnh bằng a. Gọi M, N lần...

B

- Hướng dẫn giải

Ta có: \({V_{ACMNPQ}} = {V_{EAMNC}} - {V_{EACPQ}}\)

\(\begin{array}{l}\mathop V\nolimits_{EAMNC}  = \frac{1}{3}d(E,(AMNC)).\mathop S\nolimits_{AMNC}  = \frac{1}{3}d(E,(ABC)).\left( {\mathop S\nolimits_{\Delta ABC}  - \mathop S\nolimits_{\Delta BMN} } \right) = \frac{2}{3}d(D,(ABC)).\frac{3}{4}\mathop S\nolimits_{\Delta ABC} \\ = \frac{1}{2}d(D,(ABC)).\mathop S\nolimits_{\Delta ABC}  = \frac{3}{2}\mathop V\nolimits_{ABCD} \end{array}\)

\(\begin{array}{l}\mathop V\nolimits_{EACPQ}  = \frac{1}{3}d(E,(ACPQ)).\mathop S\nolimits_{ACPQ}  = \frac{1}{3}d(E,(ACD)).\left[ {{S_{ACD}} - {S_{DPQ}}} \right]\\ = \frac{1}{3}d(B,(ACD)).\left[ {{S_{ACD}} - \frac{1}{9}{S_{ACD}}} \right] = \frac{8}{{27}}d(B,(ACD)).{S_{ACD}} = \frac{8}{9}\mathop V\nolimits_{ABCD} \end{array}\)

( Vì P, Q là trọng tâm của \(\Delta BCE\) và \(\Delta ABE\))

Vậy \(\mathop V\nolimits_{ACMNPQ}  = \frac{{11}}{{18}}\mathop V\nolimits_{ABCD}  = \frac{{11}}{{18}}.\frac{{{a^3}\sqrt 2 }}{{12}} = \frac{{11\sqrt 2 {a^3}}}{{216}}.\)