Gọi O là tâm đáy \(\Rightarrow SO\perp\left(ABCD\right)\Rightarrow SO\perp BC\)
Mà \(BC\perp OM\Rightarrow BC\perp\left(SOM\right)\Rightarrow\widehat{SMO}=60^0\)
\(\Rightarrow SO=OM.tan60^0=a\sqrt{3}\)
Gọi N là trung điểm AD \(\Rightarrow AM//CN\Rightarrow AM//\left(SCN\right)\)
\(\Rightarrow d\left(AM;SC\right)=d\left(AM;\left(SCN\right)\right)=d\left(A;\left(SCN\right)\right)=2d\left(O;\left(SCN\right)\right)\)
Kẻ \(AH\perp CN\)
\(CN=\sqrt{CD^2+\left(\frac{AD}{2}\right)^2}=a\sqrt{5}\)
\(\Rightarrow cos\widehat{ACN}=\frac{AC^2+CN^2-AN^2}{2AC.CN}=\frac{3\sqrt{10}}{10}\Rightarrow sin\widehat{ACN}=\frac{\sqrt{10}}{10}\)
\(\Rightarrow OH=OC.sin\widehat{ACN}=\frac{a\sqrt{2}.\sqrt{10}}{10}=\frac{a\sqrt{5}}{5}\)
Từ O kẻ \(OK\perp SH\Rightarrow OK\perp\left(SCN\right)\Rightarrow OK=d\left(O;\left(SCN\right)\right)\)
\(\frac{1}{OK^2}=\frac{1}{SO^2}+\frac{1}{OH^2}\Rightarrow OK=\frac{SO.OH}{\sqrt{SO^2+OH^2}}=\frac{a\sqrt{3}}{4}\)
\(\Rightarrow d\left(AM;SC\right)=2OH=\frac{a\sqrt{3}}{2}\)
