rs=Δ
let sides be a and b
and let angle B =90
now, 2008(a+b+√a2+b2)/2=ab/2
next use r=4R sinA/2SinB/2SinC/2 SinB/2=1/√2 manipulate something with remaining sinC/2 and sinB/2
The number of rightangled triangles with integer sides and inradius 2008 is
rs=Δ
let sides be a and b
and let angle B =90
now, 2008(a+b+√a2+b2)/2=ab/2
next use r=4R sinA/2SinB/2SinC/2 SinB/2=1/√2 manipulate something with remaining sinC/2 and sinB/2
i am not interested to do. i think if we try we can get it via the method i said.