Xét ∆AHB và ∆CHA có:

• \(\widehat{AHB} = \widehat{AHC} = 90^0\),
• \(\widehat{BAH} = \widehat{ACH}\) (cùng phụ $\widehat{HAC}$)

=> ∆AHB ∽ ∆CHA (góc - góc)

=> \(\frac{AH}{CH }= \frac{BH}{CH}\)

=>  $AH^2 = CH.BH = 25.36$

=> $AH^2 = 900 => AH = 30$

Vậy \(S_{ABC}= \frac{1}{2}.AH.BC = \frac{1}{2}.30.(25 + 26) = 915 cm^2\)

Xét ∆ABC và ∆HBA có: 

• $\widehat{B}$ chung
• $\widehat{BAC}=\widehat{BHA}=90^0$

=> ∆ABC ∽ ∆HBA (góc - góc)

=> $\frac{AB}{HB}=\frac{BC}{BA}=>AB^2=HB.HC=25.(25+36)=1525$

=> $AB=\sqrt{1525}\approx 39,05(cm)$

Xét ∆ABC và ∆HAC có:

• $\widehat{C}$ chung
• $\widehat{BAC}=\widehat{AHC}=90^0$

=> ∆ABC ∽ ∆HAC (góc - góc)

=> $\frac{AC}{HC}=\frac{BC}{AC}=>AC^2=HC.BC=36.(25+36)=2196$

=> $AC=\sqrt{2196}\approx 36,61(cm)$

=> Chu vi tam giác ABC là: $P=AB+AC+BC=146,91(cm)$