NIMCET 2017 — Mathematics PYQ

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Question

NIMCET 2017 — Mathematics PYQ

NIMCET | Mathematics | 2017

In a triangle ABC, Let $C = \frac{π}{2}$ , If $r$ is the inradius and $R$ is circumradius of the triangle ABC, then $2 (r + R)$ equals

Choose the correct answer:

Explanation

Concept:

Sine law:
$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = 2 R$ ,
where $a, b, c$ are sides and $R$ is circumradius.

Inradius:
$r = (s - a) tan (\frac{A}{2}) = (s - b) tan (\frac{B}{2}) = (s - c) tan (\frac{C}{2})$

where $a, b, c$ are sides, $r$ is inradius, and $s$ is semiperimeter

$s = \frac{a + b + c}{2}$

Calculation:
Here, in triangle ABC, Let $C = \frac{π}{2}$ ,

$\frac{c}{sin C} = 2 R$

$\Rightarrow 2 R = \frac{c}{sin ( π /2 )} = c$

$R = \frac{C}{2}$

Now, inradius $r = (s - c) tan (\frac{C}{2})$

$= (s - c) tan (\frac{π}{4})$

$= (s - c)$

So, $2 (r + R) = 2 (s - c + \frac{C}{2}) = 2 (s - \frac{C}{2})$

$= a + b + c - c$

$= a + b$

Hence, option (1) is correct.