SOS!! Mình cần gấp! Mọi người giúp mình với

Ta có $SA\bot (ABCD)$ nên $SA\bot AB$ nên theo hệ thức lượng trong tam giác vuông ta có:

$\begin{array}{l}
S{A^2} = SH.SB = SH.\sqrt {S{A^2} + A{B^2}} \\
 \Rightarrow SH = \dfrac{{S{A^2}}}{{\sqrt {S{A^2} + A{B^2}} }} = \dfrac{{3{a^2}}}{{\sqrt {3{a^2} + {a^2}} }} = \dfrac{{3a}}{2}\\
 \Rightarrow \dfrac{{SH}}{{SB}} = \dfrac{{SH}}{{\sqrt {S{A^2} + A{B^2}} }} = \dfrac{{\dfrac{{3a}}{2}}}{{2a}} = \dfrac{3}{4}\\
 \Rightarrow \dfrac{{HB}}{{SB}} = \dfrac{1}{4}\\
{V_{H.ABC}} = \dfrac{1}{3}{d_{\left( {H,\left( {ABC} \right)} \right)}}.{S_{ABC}} = \dfrac{1}{3}.\dfrac{1}{4}{d_{\left( {S,\left( {ABC} \right)} \right)}}.{S_{ABC}}\\
 = \dfrac{1}{4}{V_{S.ABC}} = \dfrac{1}{8}{V_{S.ABCD}} = \dfrac{1}{8}SA.S = \dfrac{1}{8}a\sqrt 3 .{a^2} = \dfrac{{{a^3}\sqrt 3 }}{8}\\
\dfrac{{{d_{\left( {H,\left( {SCD} \right)} \right)}}}}{{{d_{\left( {B,\left( {SCD} \right)} \right)}}}} = \dfrac{{HS}}{{BS}} = \dfrac{3}{4} \Rightarrow {d_{\left( {H,\left( {SCD} \right)} \right)}} = \dfrac{3}{4}{d_{\left( {B,\left( {SCD} \right)} \right)}}\\
AB//CD \subset \left( {SCD} \right) \Rightarrow {d_{\left( {B,\left( {SCD} \right)} \right)}} = {d_{\left( {A,\left( {SCD} \right)} \right)}}\\
 \Rightarrow {d_{\left( {H,\left( {SCD} \right)} \right)}} = \dfrac{3}{4}{d_{\left( {A,\left( {SCD} \right)} \right)}}
\end{array}$

Kẻ $AK\bot SD$

$\begin{array}{l}
\left\{ \begin{array}{l}
AK \bot SD\\
CD \bot AD\\
SA \bot CD
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
AK \bot SD\\
CD \bot \left( {SAD} \right)
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
AK \bot SD\\
CD \bot AK
\end{array} \right.\\
 \Rightarrow AK \bot \left( {SCD} \right) \Rightarrow {d_{\left( {A,\left( {SCD} \right)} \right)}} = AK\\
AK = \dfrac{{SA.AD}}{{\sqrt {S{A^2} + A{D^2}} }} = \dfrac{{{a^2}\sqrt 3 }}{{\sqrt {3{a^2} + {a^2}} }} = \dfrac{{a\sqrt 3 }}{2}\\
 \Rightarrow {d_{\left( {H,\left( {SCD} \right)} \right)}} = \dfrac{3}{4}.\dfrac{{a\sqrt 3 }}{2} = \dfrac{{3a\sqrt 3 }}{8}
\end{array}$