In tr dreptunghic PAO aplic T. lui Pitagora si avem PA^2=35^2-27^2=(35-27)(35+27)=8*62=16*31
PA=4[tex] \sqrt{31} [/tex]
In tr PAB, PB^2=54^2+16*31=3412
PB=2[tex] \sqrt{853} [/tex]
P=4[tex] \sqrt{31} [/tex]+54+2[tex] \sqrt{853} [/tex]
Aria A=54*4[tex] \sqrt{31} [/tex]/2=108[tex] \sqrt{31} [/tex]
Desi eu cred ca OP era 45 nu 35 . In aceasta situatie AP=36 (27,36,45 numere pitagorice , PB=54^2+36^2=18^2(3^2+2^2)=18^2*13, PB =18[tex] \sqrt{13} [/tex]
