The area of the right angled triangle in terms of its circum radius and in radius is

$\begin{array}{l}\frac{c}{2}=R,\frac{a+b-c}{2}=r\\ \therefore \frac{a+b-c}{2}=r\therefore \frac{a+b-c}{2}=r\\ \Rightarrow a+b=2r+2R\\ \Rightarrow a^2+b^2+2ab=4r^2+4R^2+8rR\end{array}$

$\begin{array}{l}\Rightarrow c^2+2ab=4r^2+4R^2+8rR\\ \Rightarrow \frac{1}{4}2ab=r^2+2Rr\\ \Rightarrow \frac{1}{2}ab=r(r+2R)\end{array}$