a: Vì SA\(\perp\)(ABCD)
nên \(d\left(S;\left(ABCD\right)\right)=SA\)
SA\(\perp\)(ABCD)
=>SA\(\perp\)AD
=>ΔSAD vuông tại A
=>\(SA^2+AD^2=SD^2\)
=>\(SA^2=\left(a\sqrt{5}\right)^2-a^2=4a^2=\left(2a\right)^2\)
=>SA=2a
=>d(S;(ABCD))=2a
b: ta có: BA\(\perp\)AD
BA\(\perp\)SA(SA\(\perp\)(ABCD))
AD,SA cùng thuộc mp(SAD)
Do đó: BA\(\perp\)(SAD)
=>\(d\left(B;\left(SAD\right)\right)=BA=a\)