Ta có :
\(S_{ABC}=\dfrac{1}{2}AH.BC\)
\(\Leftrightarrow BC=\dfrac{2S_{ABC}}{AH}=\dfrac{2.37,5}{6}=12,5\left(cm\right)\)
\(AB.AC=AH.BC=6.12,5=75\left(cm^2\right)\left(1\right)\)
\(AB^2+AC^2=BC^2\left(Pitago\right)\)
\(\Leftrightarrow\left(AB+AC\right)^2-2AB.AC=BC^2\)
\(\Leftrightarrow\left(AB+AC\right)^2=BC^2+2AB.AC\)
\(\Leftrightarrow\left(AB+AC\right)^2=156,25+2.75=306,25\)
\(\Leftrightarrow AB+AC=5\sqrt[]{12,5}\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow AB;AC\) là nghiệm phương trình:
\(x^2-5\sqrt[]{12,5}x+75=0\)
\(\Delta=312,5-300=12,5\)
Phương trình có 2 nghiệm \(AB< AC\)
\(\left[{}\begin{matrix}AB=\dfrac{5\sqrt[]{12,5}-\sqrt[]{12,5}}{2}=2\sqrt[]{12,5}\left(cm\right)\\AC=\dfrac{5\sqrt[]{12,5}+\sqrt[]{12,5}}{2}=3\sqrt[]{12,5}\left(cm\right)\end{matrix}\right.\)
