In a Delta ABC, let angle C = frac{pi}{2}. If | vedclass

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(A) In a right-angled triangle $\Delta ABC$ with $\angle C = \frac{\pi}{2},$ the hypotenuse is $c = AB.$The circumradius $R$ is half of the hypotenuse

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$a + b$

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(A) In a right-angled triangle $\Delta ABC$ with $\angle C = \frac{\pi}{2},$ the hypotenuse is $c = AB.$The circumradius $R$ is half of the hypotenuse,so $R = \frac{c}{2}.$The inradius $r$ is given by $r = \frac{\Delta}{s},$ where $\Delta = \frac{1}{2}ab$ and $s = \frac{a + b + c}{2}.$Thus,$r = \frac{\frac{1}{2}ab}{\frac{1}{2}(a + b + c)} = \frac{ab}{a + b + c}.$Now,$r + R = \frac{ab}{a + b + c} + \frac{c}{2} = \frac{2ab + c(a + b + c)}{2(a + b + c)}.$Since $c^2 = a^2 + b^2,$ we have $2ab + c(a + b + c) = 2ab + ca + cb + c^2 = 2ab + ca + cb + a^2 + b^2 = (a + b)^2 + c(a + b) = (a + b)(a + b + c).$Therefore,$r + R = \frac{(a + b)(a + b + c)}{2(a + b + c)} = \frac{a + b}{2}.$Hence,$2(r + R) = a + b.$

In a $\Delta ABC,$ let $\angle C = \frac{\pi}{2}.$ If $r$ and $R$ are the inradius and the circumradius respectively of the triangle,then $2(r + R)$ is equal to

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