\(AM=\sqrt{\dfrac{AB^2+AC^2}{2}-\dfrac{BC^2}{4}}\)
=>\(\dfrac{3^2+5^2}{2}-\dfrac{BC^2}{4}=4\)
=>\(\dfrac{BC^2}{4}=13\)
=>BC^2=52
=>\(BC=2\sqrt{13}\left(cm\right)\)
Xét ΔABC có \(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{3^2+5^2-52}{2\cdot3\cdot5}=\dfrac{-3}{5}\)
=>\(sinA=\dfrac{4}{5}\)
\(S_{ABC}=\dfrac{1}{2}\cdot3\cdot5\cdot\dfrac{4}{5}=6\left(cm^2\right)\)