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twierdzenie pitagorasa
5²+12²=c²
c²=169
c=13
OBW=4c
OBW = 52 [cm]
d2 - 24cm
Obw = 4a
(przekatne w rombie przecinaka sie pod katem prostym i w polowie tworzac 4 trojkaty prostokatne o boach a, 1/2d1, 1/2d2)
(1/2d1)² +(1/2d2)²=a²
5² + 12² =a²
a = 13cm
Obw= 4a
Obw= 52cm