jmc

Let the sides of the pentagon be $a,$ $b,$ $c,$ $d$ and $e,$ and let $r$ and $s$ be the legs of the triangular region cut off, as shown. By the Pythagorean theorem, $r^2+s^2=e^2.$ Furthermore, we have $r=b-d$ and $s=c-a,$ which are integers because $a,b,c,$ and $d$ are integers. Thus, $e$ must be the hypotenuse of some Pythagorean triple. The possibilities for that triple are $$\{5,12,13\},\{12,16,20\},\{15,20,25\},\{7,24,25\}.$$Also, the leg lengths $r=b-d$ and $s=c-a$ must be among the pairwise differences of the given numbers. Since $16,$ $15$ and $24$ do not appear among any of the pairwise differences of $\{13,19,20,25,31\},$ the only possible triple is $\{5,12,13\}.$ Then we may take $r=b-d=5$ and $s=c-a=12,$ and this forces $a=19,$ $b=25,$ $c=31,$ $d=20$ and $e=13.$ Hence, the area of the pentagon is $$bc-\frac{1}{2}rs=31\cdot 25-\frac{1}{2}(12\cdot 5)=775-30=\boxed{745}.$$