Đáp án:
$d(A;(SBC)) = \dfrac{a\sqrt3}{2}$
Giải thích các bước giải:
Ta có:
$ΔABC$ vuông cân tại $A$ cạnh $AB = a$
$\Rightarrow S_{ABC} = \dfrac{a^2}{2}$
Bên cạnh đó:
$SA\perp (ABC)$
$\Rightarrow \widehat{(SB;(ABC))} = \widehat{SBA} = 60^o$
$\Rightarrow \begin{cases}SA = AB.\tan60^o = a\sqrt3\\SB = \dfrac{AB}{\cos60^o} = 2a\end{cases}$
$\Rightarrow V_{S.ABC} = \dfrac{1}{3}S_{ABC}.SA = \dfrac{1}{3}\cdot\dfrac{a^2}{2}\cdot a\sqrt3 = \dfrac{a^3\sqrt3}{6}$
Mặt khác:
$CB\perp AB$
$CB\perp SA \quad (SA\perp (ABC))$
$\Rightarrow CB\perp (SAB)$
$\Rightarrow CB\perp SB$
$\Rightarrow S_{SCB} = \dfrac{1}{2}CB.SB = \dfrac{1}{2}.a.2a = a^2$
Ta có:
$V_{A.SBC} = \dfrac{1}{3}S_{SBC}.d(A;(SBC))$
$\Leftrightarrow d(A;(SBC)) = \dfrac{3V_{A.SBC}}{S_{SBC}} = \dfrac{3V_{S.ABC}}{S_{SBC}}$
$\Leftrightarrow d(A;(SBC)) = \dfrac{3\cdot\dfrac{a^3\sqrt3}{6}}{a^2} = \dfrac{a\sqrt3}{2}$
