Question

Answer 1

\(\Delta ABC\) đều \(\Rightarrow BC\perp AM\) (trung tuyến đồng thời là đường cao) (1)

\(SA\perp\left(ABC\right)\Rightarrow SA\perp BC\) (2)

(1);(2) \(\Rightarrow BC\perp\left(SAM\right)\)

Từ A kẻ \(AH\perp SM\Rightarrow AH\perp\left(SBC\right)\)

\(\Rightarrow AH=d\left(A;\left(SBC\right)\right)\)

\(AM=\dfrac{a\sqrt{3}}{2}\)

\(\dfrac{1}{AH^2}=\dfrac{1}{SA^2}+\dfrac{1}{AM^2}\Rightarrow AH=\dfrac{SA.AM}{\sqrt{SA^2+AM^2}}=\dfrac{a\sqrt{15}}{5}\)