Đáp án:
$d(D,(SBC))=\dfrac{a\sqrt 3}{2}$
Lời giải:
$\widehat{SBA}=(\widehat{SB,AB})=(\widehat{SB,(ABCD)})\ (=60^\circ)$
$→ SA=a\sqrt 3$
Ta có: $\left.\begin{matrix}AD\ //\ BC\\AD\ //\ (SBC)\end{matrix}\right\} \to d(D;(SBC))=d(A,(SBC))$
Nối $AH ⊥SB$
$\left.\begin{matrix}BC⊥SA\\BC⊥AB\end{matrix}\right\} \to BC ⊥ (SAB) \to BC⊥AH$
$\left.\begin{matrix}AH⊥SB\\BC⊥AH\end{matrix}\right\} \to AH⊥(SBC)$
$→ d(D,(SBC))=AH=\dfrac{SA.AB}{\sqrt{SA^2+AB^2}}=\dfrac{a\sqrt 3}{2}$
