Appying the Pythagoras theorem to the right \(\Delta ABC\),we have :

\(AB^2+AC^2=BC^2\Rightarrow BC=\sqrt{3^2+4^2}=\sqrt{25}=5\) (cm)

\(S_{ABC}=\dfrac{AB\times AC}{2}=\dfrac{AH\times BC}{2}\Rightarrow AB\times AC=AH\times BC\)

\(\Rightarrow AH=\dfrac{3\times4}{5}=2.4\) (cm)

Because \(\Delta ABC\) is a square triangle.

\(\Rightarrow\)\(AB^2+AC^2=BC^2\) (Pytago)

\(\Rightarrow BC=\sqrt{3^2+4^2}=5\)

We have:
\(AB\cdot AC=BC\cdot AH\)

\(\Rightarrow AH=\dfrac{AB\cdot AC}{BC}=\dfrac{3\cdot4}{5}=\dfrac{12}{5}=2.4\left(cm\right)\)

Answer: 2.4 cm