mng giúp em với ạ

Gọi F là trung điểm AD
Vì tam giác SAD đều
⟹ $\displaystyle SF\perp AD$
Có: $\displaystyle ( SAD) \perp ( ABCD) \Longrightarrow \ SF\perp ( ABCD)$
Vì ABCD là hình vuông
⟹ AD//BC⟹ AD//(SBD)
d(A,SBC)=d(F,SBC)
Gọi E là trung điểm BC
⟹ EF//AB//DC
Kẻ $\displaystyle FH\perp SE$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
SF\perp ( ABCD) \Longrightarrow \ SF\perp BC\\
\Longrightarrow \ BC\perp ( SFE) \Longrightarrow \ BD\perp FH
\end{array}$
Do $\displaystyle FH\perp SE$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Longrightarrow \ FH\perp ( SBC)\\
\Longrightarrow \ FH=d( F,SBC) =a\sqrt{3}
\end{array}$
Đặt AB=x
ABCD là hình vuông
⟹ EF=AB=CD=AD=BC=x
Tam giác SAD đều, $\displaystyle SF\perp AD$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Longrightarrow \ SF=\frac{x\sqrt{3}}{x}\\
SF\perp FE,\ FH\perp SE\\
\Longrightarrow \ \frac{1}{FH^{2}} =\frac{1}{FS^{2}} +\frac{1}{FE^{2}}\\
\Longrightarrow \ \frac{1}{3a^{2}} =\frac{1}{\left(\frac{x\sqrt{3}}{2}\right)^{2}} +\frac{1}{x^{2}}\\
\end{array}$

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Longrightarrow \ x=a\sqrt{7}\\
AB=BC=CD=DA=a\sqrt{7} ,\ SF=\frac{a\sqrt{21}}{2}\\
\Longrightarrow \ V=\frac{1}{3} .SF.S_{ABCD} =\frac{1}{3} .\frac{a\sqrt{21}}{2} .\left( a\sqrt{7}\right)^{2}\\
=\frac{7\sqrt{21} a^{3}}{6}
\end{array}$
Chọn A