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P =(AC+CB+BA) -----?
CH² =AH*BH ;
CH =√AH*BH ;
CH =√18*32 =√9*2*2*16 =3*2*4 =24;
ΔACH :
AC =√(AH² +CH²) =√(18² +24²) =2√(9² +12²) =2*15 =30;
ΔBCH:
BC =√(BH² +CH²) =√(32² +24²) =40 .
P =AC+CB+BA =AC+CB+(AH +BH) = 30+40+(18+32) =30+40+50=120.