The altitude AH is also a median since \(\Delta ABC\) isosceles at A

\(\Delta AHB\) right at H has :

 \(BH=\sqrt{AB^2-AH^2}=\sqrt{13^2-5^2}=12\) (cm)

The answer is :

\(\dfrac{AH.BC}{2}=AH.BH=5.12=60\) (cm²)