In a $\triangle A B C$, let $\angle C=\frac{\pi}{2}$. If $r$ and $R$ are respectively inradius and…
- $\frac{a-b}{2}$
- $\frac{a+b}{2}$
- $a+b$
- $a-b$
Solution

$ x+y=2 R $ Now, $ \begin{array}{ll} & a+b=x+r+y+r=2 R+2 r \\ \therefore & R+r=\frac{a+b}{2} \end{array} $
Asked in: AP EAMCET 2022 (07 Jul Shift 2)