fie ABCD-dreptunghiul cu diagonala AC
ΔABC-dreptunghic,iar BH⊥AC
BH²=AH×HC=9×16=144 (Teorema inaltimii)
BH=√144=12cm
din ΔAHB-dreptunghic
AB²=AH²+BH²=9²+12=81+144=225(teotema lui Pitagora)
AB=√225=15cm
din ΔBHC-dreptunghic
BC²=BH²+HC²=12²+16²=144+259=400(teorema lui Pitagora)
BC=√400=20 cm
P=2(AB+BC)=2(15+20)=70 cm