\(SA\perp\left(ABCD\right)\Rightarrow\widehat{SCA}\) là góc giữa SC và đáy \(\Rightarrow\widehat{SCA}=60^0\)
\(AC=AD\sqrt{2}\Rightarrow SA=AC.tan60^0=AD\sqrt{6}\)
\(AB||CD\Rightarrow d\left(B;\left(SCD\right)\right)=d\left(A;\left(SCD\right)\right)\)
Từ A kẻ \(AH\perp SD\Rightarrow AH\perp\left(SCD\right)\Rightarrow AH=d\left(A;\left(SCD\right)\right)\)
\(\Rightarrow AH=\dfrac{a\sqrt{42}}{7}\)
Áp dụng hệ thức lượng:
\(\dfrac{1}{AH^2}=\dfrac{1}{SA^2}+\dfrac{1}{AD^2}\Rightarrow\dfrac{49}{42a^2}=\dfrac{1}{6AD^2}+\dfrac{1}{AD^2}\)
\(\Rightarrow AD=a\)
\(\Rightarrow CD=a;AB=2a\); \(SA=a\sqrt{6}\)
\(V=\dfrac{1}{6}SA.AD\left(AB+CD\right)=\dfrac{a^3\sqrt{6}}{2}\)
