Show thatin a right triangle the inradius circumcenter and s

Show thatin a right triangle, the inradius, circumcenter, and semiperimeter are related by the formula s = r + 2R.

Solution

<C = 90 deg and the opposite side AB = c,
which equals to a + b - 2r.
So, r = (a + b - c)/2

2) In a right triangle, center of the circumscribed circle is the mid point of the hypotenuse. So its radius R = c/2 in this figure; ==> 2R = c

3) From the above two relations, r + 2R = (a + b - c)/2 + c =

==> (a + b - c + 2c)/2 = (a + b + c)/2 = s [by definition of semi perimeter of a triangle]

Thus it is proved that in a right triangle, s = r + 2R