In a right-angled isosceles triangle, the ratio of the circumradius and inradius is `2(sqrt(2)+1):1` (b) `(sqrt(2)+1):1` `2:1` (d) `sqrt(2):1`

Explanation

Sides of right angled isoceles triangle will be , `a,a and sqrt2a.`

`:. Delta = 1/2*a*a = a^2/2`

`s = (a+a+asqrt2)/2 = (a(2+sqrt2) )/2`

`:.` Circumradius `(R) = (abc)/(4Delta) = (a*a*sqrt2a)/(4(a^2/2)) = a/sqrt2`

Inradius `(r) = Delta/s = (a^2/2)/((a(2+sqrt2) )/2) = a/(2+sqrt2)`

`:. R:r = a/sqrt2 : a/(2+sqrt2) = 2+sqrt2:sqrt2 = sqrt2+1:1`

So, option `b` is the correct option.