Đáp án:
Gọi AB=3x, AC=4x (x>0)
$\begin{array}{l}
Theo\,Pytago:\\
B{C^2} = A{B^2} + A{C^2} = {\left( {3x} \right)^2} + {\left( {4x} \right)^2} = 25{x^2}\\
\Rightarrow BC = 5x\\
{S_{ABC}} = \dfrac{1}{2}.AB.AC = \dfrac{1}{2}.AH.BC\\
\Rightarrow AH = \dfrac{{AB.AC}}{{BC}}\\
\Rightarrow 4,8 = \dfrac{{3x.4x}}{{5x}} = 2,4.x\\
\Rightarrow x = 2\\
\Rightarrow \left\{ \begin{array}{l}
AB = 6\\
AC = 8\\
BC = 10
\end{array} \right.\\
Trong\,\Delta ABH \bot tai\,H\\
\Rightarrow A{B^2} = A{H^2} + B{H^2}\\
\Rightarrow B{H^2} = {6^2} - 4,{8^2} = 12,96\\
\Rightarrow BH = 3,6\left( {cm} \right)\\
\Rightarrow CH = BC - BH = 6,4\left( {cm} \right)
\end{array}$
