Đáp án:
\({V_{A'ABCD}} = \frac{{2a^3\sqrt 6 }}{{27}}\)
Giải thích các bước giải:
\(\begin{array}{l}
Ta\,\,co:\\
AC{'^2} = A'{C^2} = A{C^2} + AA{'^2} = A{B^2} + A{D^2} + AA{'^2}\\
\Leftrightarrow 2{a^2} = 3AA{'^2}\\
\Leftrightarrow AA{'^2} = \frac{{2{a^2}}}{3}\\
\Leftrightarrow AA' = \frac{{a\sqrt 6 }}{3}.\\
\Rightarrow {V_{hlp}} = {\left( {\frac{{a\sqrt 6 }}{3}} \right)^3} = \frac{{2a^3\sqrt 6 }}{9}.\\
\Rightarrow {V_{A'ABCD}} = \frac{1}{3}{V_{hlp}} = \frac{1}{3}.\frac{{2a^3\sqrt 6 }}{9} = \frac{{2a^3\sqrt 6 }}{{27}}.
\end{array}\)
