Giải thích các bước giải:
Ta có : $AC\cap BD=O\to SO\perp (ABCD)\to \widehat{SAO}=60^o\to SO=AO\sqrt{3}=\dfrac{a\sqrt{3}}{\sqrt{2}}$
$\to S_{SABCD}=\dfrac 13SO.S_{ABCD}=\dfrac 13.\dfrac{a\sqrt{3}}{\sqrt{2}}.a^2=\dfrac{a^3}{\sqrt{6}}$
Gọi $MB\cap AD=P, MN\cap SD=Q$
$\to P$ là trung điểm AD, Q là trọng tâm $\Delta SMC$
$\to\dfrac{V_{MPDQ}}{V_{MBCN}}=\dfrac{MP}{MB}.\dfrac{MD}{MC}.\dfrac{MQ}{MN}=\dfrac 12.\dfrac 12.\dfrac 23=\dfrac 16$
$\to V_{MBCN}=V_{MPDQ}+V_{PDQBCN}$
$\to V_{PDQBCN}=\dfrac 56V_{MBCN}$
Mà $S_{MBC}=S_{ABCD}, d(N,SBCD)=\dfrac 12 (S,ABCD)$
$\to V_{MBCN}=\dfrac 12V_{SABCD}$
$\to V_{PDQBCN}=\dfrac 5{12}V_{SABCD}=\dfrac{5a^3}{12\sqrt{6}}=\dfrac{5\sqrt{6}a^3}{72}\to C$
