Bài 4:
\(S_{ABC}=\dfrac{1}{2}\cdot6\cdot9\cdot sin60=13.5\sqrt{3}\)
Xét ΔABC có \(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
=>\(\dfrac{6^2+9^2-BC^2}{2\cdot6\cdot9}=\dfrac{1}{2}\)
=>\(36+81-BC^2=54\)
=>BC^2=63
=>\(BC=3\sqrt{7}\left(cm\right)\)
\(\dfrac{1}{2}\cdot AH\cdot BC=S_{ABC}\)
=>\(AH\cdot\dfrac{3\sqrt{7}}{2}=13.5\sqrt{3}\)
=>\(AH=\dfrac{9}{7}\sqrt{21}\)