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IACI=√[(1+7)²+(5+3)²]=√[64+64]=√128=8√2
IBCI=√[(5-1)²+(5-3)²]=√[16+4]=√20=2√5
OBWÓD=6√5+8√2+2√5=8√5+8√2
kazdy trojkat link na profilu
w zalaczniku kopia wynikow.
Nie widac to pierwiastkow
ale mozna policzy a+b+c=13,416+11,314+4,472=29,202
Pozdr
Hans