Đáp án: $\dfrac{{\sqrt 6 a}}{3}$
Giải thích các bước giải:
AM//CD
=> d (M,SCD) = d (A,SCD)
$\begin{array}{l}
CD \bot AD;CD \bot SA\\
\Rightarrow \left( {SAD} \right) \bot CD\\
\Rightarrow \left( {SAD} \right) \bot \left( {SCD} \right)\\
ke:AH \bot SD = H\\
\Rightarrow AH \bot \left( {SCD} \right)\\
\Rightarrow {d_{A - \left( {SCD} \right)}} = AH\\
\dfrac{1}{{A{H^2}}} = \dfrac{1}{{S{A^2}}} + \dfrac{1}{{A{D^2}}} = \dfrac{1}{{{{\left( {a\sqrt 2 } \right)}^2}}} + \dfrac{1}{{{a^2}}}\\
\Rightarrow AH = \dfrac{{\sqrt 6 a}}{3}\\
\Rightarrow {d_{M - \left( {SCD} \right)}} = \dfrac{{\sqrt 6 a}}{3}
\end{array}$
