Giúp mình câu này với ạ toán hình 12

Gọi $M$ là trung điểm của $BC$

$\begin{array}{l}  \Rightarrow \left\{ \begin{array}{l} SM \bot AB\\ \left( {SAB} \right) \bot \left( {ABCD} \right)\\ \left( {SAB} \right) \cap \left( {ABCD} \right) = AB \end{array} \right. \Rightarrow SM \bot \left( {ABCD} \right)\\ \left\{ \begin{array}{l} BC \bot AB\\ SM \bot \left( {ABCD} \right) \Rightarrow SM \bot BC \end{array} \right. \Rightarrow BC \bot \left( {SAB} \right)\\  \Rightarrow BC \bot SB\\  \bullet SB = \dfrac{{AB}}{{\sqrt 2 }} = \dfrac{a}{{\sqrt 2 }} = SA\\ SC = \sqrt {S{B^2} + B{C^2}}  = \sqrt {{{\left( {\dfrac{a}{{\sqrt 2 }}} \right)}^2} + {{\left( {2a} \right)}^2}} \\  = \dfrac{{3a\sqrt 2 }}{2}\\  \bullet S{B^2} = SH.SC \Rightarrow SH = \dfrac{{S{B^2}}}{{SC}}\\  \Rightarrow \dfrac{{SH}}{{SC}} = {\left( {\dfrac{{SB}}{{SC}}} \right)^2} = {\left( {\dfrac{{\dfrac{a}{{\sqrt 2 }}}}{{\dfrac{{3a\sqrt 2 }}{2}}}} \right)^2} = \dfrac{1}{9}\\  \Rightarrow \dfrac{{{d_{\left( {H,\left( {ABCD} \right)} \right)}}}}{{{d_{\left( {S,\left( {ABCD} \right)} \right)}}}} = \dfrac{{HC}}{{SC}} = \dfrac{{SC - SH}}{{SH}} = \dfrac{{9 - 1}}{9} = \dfrac{8}{9}\\  \Rightarrow {V_{H.ABCD}} = \dfrac{8}{9}{V_{S.ABCD}} = \dfrac{8}{9}.\dfrac{1}{3}.SM.{S_{ABCD}}\\  = \dfrac{8}{9}.\dfrac{1}{3}.\dfrac{{AB}}{2}.AD.AB = \dfrac{8}{9}.\dfrac{1}{3}.\dfrac{a}{2}.2a.a\\  = \dfrac{{8{a^3}}}{{27}} \end{array}$