Let the length of the isosceles sides be s respectively,
the hypotenuse will be sqrt(2).s
so 2s+sqrt(s).s = 2P
-> s = 2P/(2+sqrt(2))
since its a isosceles right angled triangle, it's area is half of a square with the same side
area = s^2/2 = 2P^2/(2+sqrt(2))^2
