First let us find the lengths of sides of the trinagle. Let a,b,c be sides of triangle
By triangle inequality we have as $a+b > c$ so $a+b+c > 2c$ or $8 > 2c$ or $ c< 4$
$c = 3 => a + b = 5$ giving a = 3 and b =2 or a =2 and b = 3
$c=2=> a+b=6$ giving a = b= 3
$c=1$ is not possible as it gives $a+b= 7$ give a a =3 , b= 4 invalid triangle
sides 3,3,2 say a = b= 3 and c = 2
and s (semiperimeter) = 8/2 = 4
so if A is area $A^2 = s(s-a)(s-b)(s-c) = 4 * 1 * 1 * 2 = 8$ or Area = $2\sqrt{2}$
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