Answer
$b\sqrt{4-b}$
where $0\lt b\lt 4$.
Work Step by Step
Let the other two sides have length $x$. Then we also know that $2x+b=8$ because the perimeter is $8$. Thus, we can solve for $x$:
$x=(8-b)/2$.
Then the area of the triangle can be calculated with Heron's formula:
$A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{4(4-(8-b)/2)(4-b)(4-(8-b)/2)}=\sqrt{4(b/2)(4-b)(b/2)}=b\sqrt{4-b}$
and
$0\lt b\lt 4$.
Where $s=(a+b+c)/2=8/2=4$.
