Giải:
Áp dụng định lí Py-ta-go vào \(\Delta ABC\left(\widehat{A}=90^o\right)\) ta có:
\(AB^2+AC^2=BC^2\)
\(\Rightarrow AB^2+AC^2=10^2=100\)
Ta có: \(\frac{AB}{AC}=\frac{3}{4}\Rightarrow\frac{AB}{3}=\frac{AC}{4}\)
Đặt \(\frac{AB}{3}=\frac{AC}{4}=k\left(k>0\right)\Rightarrow\left\{\begin{matrix}AB=3k\\AC=4k\end{matrix}\right.\)
Mà \(AB^2+AC^2=100\)
\(\Rightarrow9k^2+16k^2=100\)
\(\Rightarrow25k^2=100\)
\(\Rightarrow k^2=4\)
\(\Rightarrow k=2\)
\(\Rightarrow AB=4\left(cm\right);AC=8\left(cm\right)\)
Lại có: \(S_{\Delta ABC}=\frac{AB.AC}{2}=\frac{6.8}{2}=24\left(cm^2\right)\)
\(S_{\Delta ABC}=\frac{AH.BC}{2}=5AH\)
\(\Rightarrow24=5AH\)
\(\Rightarrow AH=4,8\left(cm\right)\)
Vậy AH = 4,8 cm
