`a,` V_(S.ABCD) = 1/3 .SA.AB.AD=1/3 .a sqrt2 .a . a= sqrt2/3 a^3`

`=>` Đúng

`b,SA⊥(ABCD)=>(SC,(ABCD))=hat(SCA)`

`=>` Đúng

`c,BD⊥AC,BD⊥SA=>BD⊥(SAB)=>(SDB)⊥(SAC)`

`=>((SDB),(SAC))=90^o`

`=>` Sai

`d,` Ta có:

`d(O,(SCD)) = 1/2 d(A,(SCD))`

Trong (SAD), kẻ AH⊥SD

CD⊥AD,CD⊥SA `=>` CD⊥AH

AH⊥CD,AH⊥SD `=>` AH⊥(SCD)

`=>d(A,(SCD))=AH`

Ta có:

`1/(SA^2) +1/(AD^2) = 1/(AH^2)`

`=> 1/((a sqrt2)^2) +1/(a^2) = 1/(AH^2)`

`=>AH = sqrt6/3`

`=> `d(O,(SCD)) =1/2 AH = sqrt6/6`

`=>` Sai