Xét \(ΔHBA\) và \(ΔHAC\):
\(\widehat{HBA}=\widehat{HAC}\) (cùng phụ \(\widehat{C}\) )
\(\widehat{BHA}=\widehat{AHC}(=90^\circ)\)
\(→\dfrac{BH}{AH}=\dfrac{AH}{CH}\)
\(↔AH^2=BH.CH\) hay \(AH^2=4.9=36\)
\(↔AH=6(AH>0)\)
\(S_{ΔABC}=\dfrac{1}{2}.AH.BC=\dfrac{1}{2}.6.(4+9)=39(cm^2)\)
Vậy \(S_{ΔABC}=39(cm^2)\)