Cho hình chóp đều S.ABC có đáy ABC là tam giác đều...

A

- Hướng dẫn giải

Vì \(MN//BC \Rightarrow \frac{{AM}}{{AB}} = \frac{{AN}}{{AC}} = \frac{2}{3}\)

\(\begin{array}{l}
 \Rightarrow \frac{{{V_{ASMN}}}}{{{V_{ASBC}}}} = \frac{{AS}}{{AS}}.\frac{{AM}}{{AB}}.\frac{{AN}}{{AC}} = \frac{2}{3}.\frac{2}{3} = \frac{4}{9} \to {V_{ASMN}} = \frac{4}{9}{V_{ASBC}}\\
 \Rightarrow {V_{SMNCB}} = \frac{5}{9}{V_{SABC}}.\\
{V_{SABC}} = \frac{1}{3}.SG.{S_{ABC}} = \frac{1}{3}.a.\frac{{{a^2}\sqrt 3 }}{4} = \frac{{{a^3}\sqrt 3 }}{{12}}\\
 \Rightarrow {V_{SMNCB}} = \frac{5}{9}.\frac{{{a^3}\sqrt 3 }}{{12}} = \frac{{5{a^3}\sqrt 3 }}{{108}}
\end{array}\)