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c=dł. przeciwprostokatnej
c=a√2
a=2+√2
c=(2+√2)√2=2√2+2
obwód=2a+c=2(2+√2)+2√2+2=4+2√2+2√2+2=6+4√2
teraz z pitagorasa;
a²+a²=c²
2a²=c²
2(2+√2)²=c²
2(4+4√2+2)=c²
8√2+12=c²
c=√[8√2+12]=√[2√2+2]²=2√2+2
dalej, jak wyżej