Problema 2
in ΔPMQ⇒T.Pitagora⇒PM²=PQ²-QM²
PM²=400-256
PM²=144
PM=12
In ΔPMR⇒T.Pitagora⇒PR²=PM²+MR²
PR²=144+81
PR²=225
PR=15
In ΔPQR verificam relatia intre laturi
QR²=QP²+PR², unde QR=QM+MR=16+9=25
625=400+225
625=625⇒Reciproca T.Pitagora⇒ΔPQR este dreptunghic in ∡QPR