In triangle `A B C ,` let `/_c=pi/2dot` If `r` is the inradius and `R` is circumradius of the triangle, then `2(r+R)` is equal to `a+b` (b) `b+c` `c+a` (d) `a+b+c`

Explanation

Here, `C = pi/2`

Now, `c = 2RsinC = 2R`

`=>R = c/2`

`=>2R = c->(1)`

`r = (s-c)tan(C/2) = (s-c)tan(pi/4)`

`=>r = s-c`

`=>2r = 2s-2c = a+b+c-2c = a+b-c`

From (1),

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