The circumradius of the triangle is the radius of a circle that passes through all vertices of the triangle.

Let, we have a triangle $∆ABC$,

whose sides $AB, BC$ and $CA$ are  $3, 4$ and $5$ respectively.

As, they are Pythagorean triplets, hence, $∆ABC$ is a right-angled triangle,with $\angle B=90°$

Let, the vertices are, $A\left(0, 3\right)$, $B\left(0, 0\right)$ and $C\left(4, 0\right)$

So, $AC$ is the diameter. $\left( \because \angle ABC=90°\right)$

Hence radius$=\frac{AC}{2}=\frac{5}{2}$