Question

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

• A. $a+b$
• B. $b+c$
• C. $c+a$
• D. $a+b+c$

Answer 1

The correct option is A $a+b$

Given that

$\Rightarrow \angle C={90}^0$

We know that $r=(s-c)\tan \frac{C}{2}$

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

$\Rightarrow r=(s-c)\tan \left(\frac{{90}^0}{2}\right)$

$\Rightarrow r=(s-c)\tan \left({45}^0\right)$

$\Rightarrow r=(s-c)$

According to sine rule

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

$\Rightarrow \frac{c}{\sin {90}^0}=2R$

$\Rightarrow 2R=c$

Now, $2(r+R)=2r+2R$

$\Rightarrow 2(r+R)=2(s-c)+c$

$\Rightarrow 2(r+R)=2s-c$

$\Rightarrow 2(r+R)=a+b+c-c$

$\Rightarrow 2(r+R)=a+b$