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.