Đáp án:
$V_{A.BCD} = \dfrac{a^3\sqrt3}{12}$
Giải thích các bước giải:
$AB\perp (BCD) \quad (gt)$
$\Rightarrow AB\perp CD$
mà $CD\perp BC$
$\Rightarrow CD\perp (ABC)$
$\Rightarrow CD\perp AC$
Ta có:
$\begin{cases}(ACD)\cap (BCD) = CD\\AC\perp CD \quad (cmt)\\AC\subset (ACD)\\BC\perp CD\quad (gt)\\BC\subset (BCD)\end{cases}\Rightarrow \widehat{((ACD);(BCD))} = \widehat{ACB} = 60^o$
$\Rightarrow AB = BC.\tan60^o = a\sqrt3$
Ta được:
$V_{A.BCD} = \dfrac{1}{3}S_{BCD}.AB = \dfrac{1}{6}BC.CD.AB = \dfrac{1}{6}.a.\dfrac{a}{2}.a\sqrt3 = \dfrac{a^3\sqrt3}{12}\quad (đvtt)$
