Area=Perimeter? Recalling Pythagoras

Let $x,y,z$ are the sides of the triangle

We know two equations

$x^2+y^2=z^2$

$x+y+z=\frac{1}{2}xy$

From the second equation we get

$z=\frac{1}{2}xy-(x+y)$

Putting the value of $x$ on the first we get

$x^2+y^2=[\frac{1}{2}xy-(x+y)]^2$

or, $xy-4(x+y)+8=0$

or, $xy-4(x+y)+16=8$

or $(x-4)(y-4)=8$

From this we get the two triplets $6, 8, 10$ and $5,12,13$

So the answer is $24+30=\boxed{54}$

Only two Pythagorean triples have perimeter equal to area :

$6, 8, 10$ and $5,12,13$

Thus $24 + 30 = 54$