Do \(AB||CD\Rightarrow AB||\left(SCD\right)\Rightarrow d\left(B;\left(SCD\right)\right)=d\left(A;\left(SCD\right)\right)\)
Từ A kẻ \(AH\perp SD\) (1)
\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp CD\\CD\perp AD\end{matrix}\right.\) \(\Rightarrow CD\perp\left(SAD\right)\)
\(\Rightarrow CD\perp AH\) (2)
(1);(2) \(\Rightarrow AH\perp\left(SCD\right)\Rightarrow AH=d\left(A;\left(SCD\right)\right)\)
Áp dụng hệ thức lượng:
\(\dfrac{1}{AH^2}=\dfrac{1}{SA^2}+\dfrac{1}{AD^2}\Rightarrow AH=\dfrac{SA.AD}{\sqrt{SA^2+AD^2}}=\dfrac{6\sqrt{13}}{13}\)
Đúng 0
Bình luận (0)
