In a $\triangle A B C$, let $\angle C=\frac{\pi}{2}$. If $r$ and $R$ are respectively inradius and circumradius of $A B C$, then $R+r=$

• A

  $\frac{a-b}{2}$
• Correct answer
  B

  $\frac{a+b}{2}$
• C

  $a+b$
• D

  $a-b$

Solution

$$ x+y=2 R $$ Now, $$ \begin{array}{ll} & a+b=x+r+y+r=2 R+2 r \\ \therefore & R+r=\frac{a+b}{2} \end{array} $$

Hence, the correct answer is option: 'B'.