You can put this solution on YOUR website! Call the length of the leg of that triangle x.
Thus the perimeter is
P = x + x + x(sqrt(2)) = 2p
The area would be
A = (1/2)bh = 1/2 (x^2)
so we need to solve for x above and plug it in here.
2x + x(sqrt(2)) = 2p
Factor out x and divide by what's left and get
x(2 + sqrt(2)) = 2p
x = 2p / (2 + sqrt(2))
Now plug in and get
A = (1/2)[2p / (2 + sqrt(2))]^2
A = (1/2)(4p^2 / (2 + sqrt(2))^2)
A = 2p^2 / (2 + sqrt(2))^2
or if you have to simplify further
A = 2p^2 / (6 + 4sqrt(2)) = p^2 / (3 + 2sqrt(2))
You may still have to rationalize the denominator.
