\(SA\perp\left(ABCD\right)\Rightarrow AB\) là hình chiếu vuông góc của SB lên (ABCD)
\(\Rightarrow\widehat{SBA}\) là góc giữa SB avf (ABCD)
\(\Rightarrow\widehat{SBA}=60^0\)
\(\Rightarrow SA=AB.tan60^0=a\sqrt{3}\)
a/
\(BC//AD\Rightarrow BC//\left(SAD\right)\Rightarrow d\left(BC;SD\right)=d\left(BC;\left(SAD\right)\right)=d\left(B;\left(SAD\right)\right)=AB=a\)
b/ \(AB//DC\Rightarrow AB//\left(SCD\right)\)
\(\Rightarrow d\left(AB;SC\right)=d\left(AB;\left(SCD\right)\right)=d\left(A;\left(SCD\right)\right)\)
\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp CD\\CD\perp AD\end{matrix}\right.\) \(\Rightarrow CD\perp\left(SAD\right)\)
Từ A kẻ \(AH\perp SD\Rightarrow AH\perp\left(SCD\right)\)
\(\Rightarrow d\left(A;\left(SCD\right)\right)=AH\)
\(\frac{1}{AH^2}=\frac{1}{SA^2}+\frac{1}{AD^2}\Rightarrow AH=\frac{SA.AD}{\sqrt{SA^2+AD^2}}=\frac{a\sqrt{3}}{2}\)
