Áp dụng PTG: \(AC=\sqrt{BC^2-AB^2}=6\sqrt{6}\left(cm\right)\)
Áp dụng HTL: \(\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=\dfrac{75}{7}\left(cm\right)\\CH=\dfrac{AC^2}{BC}=\dfrac{72}{7}\left(cm\right)\\AH=\sqrt{BH\cdot CH}=\dfrac{5400}{49}\end{matrix}\right.\)