Đáp án:
$D.\,\dfrac{\sqrt6}{3}$
Giải thích các bước giải:
Kẻ $SH\perp BC\quad (H\in BC)$
Ta có:
$\begin{cases}(SBC)\cap (ABCD)=BC\\(SBC)\perp (ABCD)\\SH\perp BC\\SH\subset (SBC)\end{cases}$
$\Rightarrow SH\perp (ABCD)$
$\Rightarrow SH\perp CD$
mà $CD\perp BC$
nên $CD\perp (SBC)$
$\Rightarrow \widehat{(SD;(SBC))}=\widehat{DSC}=60^o$
$\Rightarrow SC =\dfrac{CD}{\tan60^o}=\dfrac{\sqrt3}{\sqrt3}=1$
Áp dụng định lý Pytago ta được:
$BC^2 = SC^2 + SB^2$
$\Rightarrow SB =\sqrt{BC^2 - SC^2}=\sqrt{3 -1} = \sqrt2$
Ta có:
$SC.SB = BC.SH=2S_{SBC}$
$\Rightarrow SH =\dfrac{SB.SC}{BC}=\dfrac{\sqrt2.1}{\sqrt3}=\dfrac{\sqrt6}{3}$
Do đó:
$V_{S.ABCD}=\dfrac13S_{ABCD}.SH =\dfrac13.(\sqrt3)^2\cdot\dfrac{\sqrt6}{3}=\dfrac{\sqrt6}{3}$
