All I'm given is the perimeter?

From Pythagorean theorem, we know that for a right triangle, $a^2+b^2=c^2$ . Given this is an isosceles triangle, $a=b$ , therefore $a^2+a^2=c^2$ , $2a^2=c^2$

We solve for the sides which satisfy the condition for a right triangle first:

$2a+c=2008$

$2a^2=c^2$

$c=\sqrt{2}a$

substitute $c$ into the first equation,

$2a+\sqrt{2}a=2008$

$a\approx588.13$ ; $c\approx831.74$

From this we can infer that for obtuse triangles, the hypotenuse must be longer then the condition for right triangle, so in this case, to get an obtuse triangle, the hypotenuse must be $>831.74$ . We need integer sides, so the first set of sides are :

$832 ; 588 ; 588$

Also note that for the sides to be integer, the hypotenuse must increase by 2 and the two sides decrease by 1, so every even hypotenuse above 832 satisfy the condition.

As soon as we reach hypotenuse length of 1004, the two other sides are no longer long enough to form a triangle, therefore 1004 is our upper limit.

$1004-832 = 172$ This is the number of integer hypotenuses within the range, however, only the even-numbered hypotenuses satisfy our condition, therefore

$172\div2=86$

86 is the correct answer.

$For~ a~ right~ angled~ triangle,~ with~~ legs=x,~ we~ have ~Hypothesis,~h~=x\sqrt2.\\ So~x=\dfrac{2008}{2+\sqrt2}=588.12...and~h=831.55...\\ So ~upper~ limit~for~x ~is ~588~~and~~for~base~b~lower~limit~is~832>h.\\ If~all~three~are~in~line,~we~have~502-502-1004.\\ So~ b=1003~would~ be~ OK. But~x~is~not~an~integer.\\ So~b=1002~is~higher~limit~and~x=503~lower~limit.\\ Number~of~solutions-moving~ in~ steps~ of~2~to~keep~x~as~an~integer~is~\dfrac{1002-832} 2+1=86.\\ Check~for~x~Number~of~solutions-moving~ in ~steps~ of~1~~is~~(588 -503)+1=\Large~~\color{#D61F06}{86,~checked.}$