Áp dụng PTG: \(x^2+\left(x+2\right)^2=6^2\Leftrightarrow2x^2+4x+4=36\)
\(\Leftrightarrow2x^2+4x-32=0\Leftrightarrow\left[{}\begin{matrix}x=-1+\sqrt{17}\left(tm\right)\\x=-1-\sqrt{17}\left(ktm\right)\end{matrix}\right.\)
Do đó \(S_{ABC}=\dfrac{1}{2}AB\cdot AC=\dfrac{1}{2}\left(-1+\sqrt{17}\right)\left(1+\sqrt{17}\right)\)
\(S_{ABC}=\dfrac{1}{2}\left(17-1\right)=\dfrac{1}{2}\cdot16=8\left(cm^2\right)\)