$\ \Rightarrow \left\{ \begin{array}{l}
MK \bot \left( {ABCD} \right)\\
KH = KD
\end{array} \right.$
- Trong (ABCD) dựng $\ KL \bot AC\left( {L \in AC} \right).$
- Do $\ \left\{ \begin{array}{l}
KL \bot AC\\
MK \bot AC
\end{array} \right. \Rightarrow ML \bot AC.$
$\ \Rightarrow \widehat {\left( {\left( {AMC} \right),\left( {ABCD} \right)} \right)} = \widehat {MLK} = {30^0}.$
- Xét hình thoi ABCD ta thấy:
$\ AC = AB = 2a;BD = 2a\sqrt 3 .$
$\ \Rightarrow {S_{ABCD}} = AC.BD = 4{a^2}\sqrt 3 .$
- Trong $\ \Delta BHD:OK// = \frac{1}{2}BH.$
$\ \left\{ \begin{array}{l}
\widehat {KOD} = \widehat {ABD} = {30^0}\\
OK = \frac{{BH}}{2} = \frac{a}{2}
\end{array} \right. \Rightarrow \widehat {KOL} = {60^0}.$
$\ \Rightarrow KL = OK\sin {60^0} = \frac{{a\sqrt 3 }}{4}.$
- Trong tam giác vuông MKL có:
$\ \frac{h}{2} = \frac{{SH}}{2} = MK = LK\tan {30^0} = \frac{a}{4}$$\ \Rightarrow h = \frac{a}{2} \Rightarrow {V_{S.ABCD}} = \frac{1}{3}.\frac{a}{2}.4{a^2}\sqrt 3 = \frac{{2{a^3}\sqrt 3 }}{3}.$
- Do OM là đường trung bình của tam giác SBD nên MO//SB.
Vậy$\ d\left( {CM,SB} \right) = d\left( {SB,\left( {AMC} \right)} \right) = d\left( {S,\left( {AMC} \right)} \right).$
Mặ khác, $\ \left\{ \begin{array}{l}
d\left( {S,\left( {AMC} \right)} \right) = \frac{{3{V_{S.AMC}}}}{{{S_{\Delta AMC}}}}\\
\frac{{{V_{S.AMC}}}}{{{V_{S.ADC}}}} = \frac{{SA}}{{SA}}.\frac{{SM}}{{SD}}.\frac{{SC}}{{SC}} = \frac{1}{2} \Rightarrow {V_{S.AMC}} = \frac{{{V_{S.ADC}}}}{2} = \frac{{{V_{S.ABCD}}}}{4} = \frac{{{a^3}\sqrt 3 }}{6}\\
{S_{\Delta AMC}} = \frac{{ML.AC}}{2} = \frac{{AC\sqrt {M{K^2} + L{K^2}} }}{2} = \frac{{{a^2}}}{2}
\end{array} \right. \Rightarrow d\left( {S,\left( {AMC} \right)} \right) = a\sqrt 3 .$
Vậy: $\ {V_{S.ABCD}} = \frac{{2{a^3}\sqrt 3 }}{3};\,d\left( {SB,CM} \right) = a\sqrt 3 .$
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