Đáp án:
$\begin{array}{l}
a)\,\Delta ABB'\,vuông\,tại\,B\\
\Rightarrow BB' = \sqrt {AB{'^2} - A{B^2}} = \sqrt {5{a^2} - {a^2}} = 2a\\
\Rightarrow BM = a\\
\Delta ABC\,vuông\,tại\,B\\
\Rightarrow {S_{ABC}} = \frac{1}{2}.a.a\sqrt 2 = \frac{{{a^2}\sqrt 2 }}{2}\\
\Rightarrow {V_{ABC.A'B'C'}} = BB'.{S_{ABC}} = a.\frac{{{a^2}\sqrt 2 }}{2} = \frac{{{a^3}\sqrt 2 }}{2}\\
b)\\
{V_{B.AMC}} = \frac{1}{3}.{S_{ABC}}.BM = \frac{1}{3}.\frac{{{a^2}\sqrt 2 }}{2}.a = \frac{{{a^3}\sqrt 2 }}{6}\\
c)\\
AM = \sqrt {A{B^2} + B{M^2}} = \sqrt {{a^2} + {a^2}} = a\sqrt 2 \\
CM = \sqrt {B{C^2} + B{M^2}} = \sqrt {2{a^2} + {a^2}} = a\sqrt 3 \\
AC = a\sqrt 3 \\
\Rightarrow {S_{AMC}} = \frac{{\sqrt 5 {a^2}}}{2}\\
{V_{B.AMC}} = \frac{1}{3}.{d_{\left( {B,AMC} \right)}}.{S_{AMC}} = \frac{{{a^3}\sqrt 2 }}{6}\\
\Rightarrow {d_{\left( {B,AMC} \right)}} = \frac{{\sqrt {10} a}}{5}
\end{array}$
