a) Ta có: $\widehat{A} = 60^o$
$\Rightarrow AB=BD=DA=a$
$\Rightarrow AB=BC=CD=DA=SO=a$
$\Rightarrow OB = \dfrac{BD}{2} = \dfrac{a}{2}; \, \, OC = BC\dfrac{\sqrt{3}}{2} = \dfrac{a\sqrt{3}}{2}$
Áp dụng định lý Pytago, ta được:
$SB^2 = SO^2 + OB^2 = a^2 + \dfrac{a^2}{4} = \dfrac{5a^2}{4}$
$\Rightarrow SB = \dfrac{a\sqrt{5}}{2}$
$SC^2 = SO^2 + OC^2 = a^2 + \dfrac{3a^2}{4} = \dfrac{7a^2}{4}$
$\Rightarrow SC = \dfrac{a\sqrt{7}}{4}$
$\Rightarrow p_{SBC} = \dfrac{SB+SC+BC}{2} = \dfrac{a\sqrt{5} + a\sqrt{7} + 2a}{2}$
$\Rightarrow S_{SBC} = \sqrt{p(p-SB)(p-SC)(p-BC)} = \dfrac{a^2\sqrt{7}}{8}$
Ta có:
$3V_{S.OCB} = S_{OCB}.SO = \dfrac{1}{2}OB.OC.SO = \dfrac{1}{2}.\dfrac{a}{2}.\dfrac{a\sqrt{3}}{2}.a = \dfrac{a^3\sqrt{3}}{8}$
mà $3V_{S.OCB} = 3V_{O.SBC} = S_{SBC}.d(O;(SBC))$
$\Rightarrow d(O;(SBC)) = \dfrac{3V_{S.OCB}}{S_{SBC}} = \dfrac{\dfrac{a^3\sqrt{3}}{8}}{ \dfrac{a^2\sqrt{7}}{8}}= \dfrac{a\sqrt{21}}{7}$
b) Ta có: $AD//CB$
$\Rightarrow AD//(SBC)$
$\Rightarrow d(AD;SB) = d(AD;(SBC)) = d(D;(SBC))$
Gọi $DH = d(D;(SBC))$
$OI = d(O;(SBC))$
$\Rightarrow DH//OI \, (\perp (SBC))$
mà $DO = OB$
$\Rightarrow DH = 2OI$
$\Rightarrow DH = \dfrac{2a\sqrt{21}}{7}$
Vậy $d(AD;SB) = \dfrac{2a\sqrt{21}}{7}$
