Circumradius and inradius

Let $a=BC$ , $b=CA$ , $c=AB$ . Furthermore, let $s$ be the semiperimeter of triangle $ABC$ . The first equality becomes $abc=4350$ . From the metric identity $[ABC]=rs=\dfrac{abc}{4R}$ , we deduce that $s=\dfrac{abc}{4rR}=\dfrac{4350}{\left(4\cdot \dfrac{145}{4}\right)}=30$ so the perimeter of triangle $ABC$ is $\boxed{60}$ .

Let R be the circumradius r be the inradius,a=BC,b=AC and c=AB .Given that r * R=145/4 or Area of the triangle/{(a+b+c)/2} * abc/(4 * Area of the triangle)=145/4 or abc/[4 * {(a+b+c)/2}]=145/4 or 4350/{2 * (a+b+c)}=145/4 or 2175/(a+b+c)=145/4 or a+b+c=(2175*4)/145 or a+b+c=60 so AB+BC+CA=60(ans)