+)Qua $M$ ke $MN//OA(N\in BC),MQ//SB(Q\in SA),NP//MQ//SB(P\in SC)$$\Rightarrow (\beta )\equiv (MNPQ)$Ta co: $NP//MQ$( theo cach dung) $\Rightarrow MNPQ$ la hinh thang Mat khac: $\left\{ \begin{array}{l} SB AO\\ MN//AO\\NP//SB \end{array} \right.\Rightarrow MN |NP$Vay$MNPQ$la hinh thang vuong+)Ta co:$BC=\frac{2a}{\sqrt3}\Rightarrow BO=OC=OA=\frac{a}{\sqrt3} $$MN//AO\Rightarrow \frac{BN}{BO}=\frac{MN}{OA}\Rightarrow MN=BN=x$$ MQ//SB\Rightarrow \frac{MQ}{SB}=\frac{AM}{AB}=\frac{ON}{OB}=1-\frac{BN}{OB} $$\Rightarrow \frac{MQ}{a}=1-\frac{x}{a/\sqrt3}\Rightarrow MQ=a-\sqrt3x$$ NP//SB\Rightarrow \frac{NP}{SB}=\frac{CN}{BC}=1-\frac{BN}{BC}\Leftrightarrow \frac{NP}{a}=1-\frac{x\sqrt3}{2a}\Rightarrow NP=a-\frac{x\sqrt3}{2}$$S_{MNPQ}=(NP+MQ)MN/2$ ban tu tinh tiep nha! :)

+)Qua $M$ ke $MN//OA(N\in BC),MQ//SB(Q\in SA),NP//MQ//SB(P\in SC)$$\Rightarrow (\beta )\equiv (MNPQ)$Ta co: $NP//MQ$( theo cach dung) $\Rightarrow MNPQ$ la hinh thang Mat khac: $\left\{ \begin{array}{l} SB \perp AO\\ MN//AO\\NP//SB \end{array} \right.\Rightarrow MN \perp NP$Vay$MNPQ$la hinh thang vuong+)Ta co:$BC=\frac{2a}{\sqrt3}\Rightarrow BO=OC=OA=\frac{a}{\sqrt3} $$MN//AO\Rightarrow \frac{BN}{BO}=\frac{MN}{OA}\Rightarrow MN=BN=x$$ MQ//SB\Rightarrow \frac{MQ}{SB}=\frac{AM}{AB}=\frac{ON}{OB}=1-\frac{BN}{OB} $$\Rightarrow \frac{MQ}{a}=1-\frac{x}{a/\sqrt3}\Rightarrow MQ=a-\sqrt3x$$ NP//SB\Rightarrow \frac{NP}{SB}=\frac{CN}{BC}=1-\frac{BN}{BC}\Leftrightarrow \frac{NP}{a}=1-\frac{x\sqrt3}{2a}\Rightarrow NP=a-\frac{x\sqrt3}{2}$$S_{MNPQ}=(NP+MQ)MN/2$ ban tu tinh tiep nha! :)