Let the upper left corner be point p.

Then AB is the hypotenuse of the right triangle ABP.

AP=3，BP= 1

So AB = √ (AP 2+BP 2) = √ (9+1) = √10.

Similarly, AC = √ (3 2+2 2) = √13.

AD=√(3^2+3^2)=√ 18=3√2

AE=√(3^2+4^2)=5

AF=√(4^2+2^2)=√20=2√5

So the length is irrational, including AB, AC, AD and AF.