In a $△ A B C$ , If $∠ C = 90^{o}$ , r and R are the inradius and circumradius of the $△ A B C$ respectively, then $2 (r + R)$ is equal to

b + c

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c + a

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a + b

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a + b + c

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Solution

The correct option is D $a + b$
Tangents drawn from external points are of equal length.
$\Rightarrow B D = B E = a - r$ $[∵ C D = r]$
and $A F = A E = b - r$ $[∵ C F = r]$
$∵ A E + B E = A B$
$\Rightarrow b - r + a - r = 2 R$ $[∵$ AB is a diameter of circumcircle $]$
$\Rightarrow b + a = 2 R + 2 r$ $2 (r + R) = a + b$

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