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In a $△ ABC, let ∠ C = \frac{π}{2}$ 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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$a + b$

$b + c$

$a + b + c$

$c + a$

answer is A.

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Detailed Solution

$∠ C = 90^{\circ} \Rightarrow C = 2 R \sin 90^{\circ} = 2 R$ .

and $r = (s - c) \tan \frac{c}{2} = s - c$

$\begin{array}{l} ∴ 2 (r + R) = 2 r + 2 R \\ = 2 s - 2 c + c = 2 s - c = a + b \end{array}$

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