\(\Delta AHB\) right at H has HA = HB,so \(\Delta AHB\) is an isosceles triangle

\(\Rightarrow AB=\sqrt{2}BH=20\sqrt{2}\) (cm)

\(\Delta AHC\) right at H has : AH² + HC² = AC²

\(\Rightarrow AC=\sqrt{20^2+25^2}=\sqrt{1025}=5\sqrt{41}\) (cm)

BC = BH + HC = 20 + 25 = 45 (cm)

\(\Rightarrow P_{\Delta ABC}=45+20\sqrt{2}+5\sqrt{41}\) cm