Question

In a $∆ABC$, let $\angle C=\frac{\pi }{2}$, if $r$ is the inradius and $R$ is the circumradius of the $∆ABC$, then $2(r+R)$ equals

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

Answer 1

The correct option is C

$a+b$

Explanation for the correct option:

Step 1: Finding relation between circumradius and side of the triangle:

Given, $△ABC$ in which $\angle C=\frac{\pi }{2}$

$a,b,c$ are the sides of a given triangle. $R$ is circumradius of the triangle and $r$ is inradius.

Semi perimeter is denoted by, $s=\frac{a+b+c}{2}$

We know that,

$$\begin{gathered} c=2R\sin \left(C\right) \\ \Rightarrow c=2R\sin \left(\frac{\pi }{2}\right) \\ \Rightarrow c=2R\mathbf{}\left[\because \sin \left(\frac{\pi }{2}\right)=1\right] \\ \Rightarrow R=\frac{c}{2} \end{gathered}$$

Step 2: Finding relation between inradus and side of the triangle

And also, the inradius,

$$\begin{gathered} r=\left(s-c\right)\tan \left(\frac{C}{2}\right) \\ \Rightarrow r=\left(s-c\right)\tan \left(\frac{\pi }{4}\right)\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{C}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{]} \\ \Rightarrow r=s-c\mathbf{}\mathbf{[}\mathbf{t}\mathbf{a}\mathbf{n}\left(\frac{\pi }{4}\right)\mathbf{=}\mathbf{1}\mathbf{]} \end{gathered}$$

Step 3: Finding the value of $2\left(r+R\right)$

Now we have,

$$\begin{gathered} 2\left(r+R\right)=2r+2R \\ =2\left(s-c\right)+c \\ =2s-2c+c \\ =2s-c \\ =2\left(\frac{a+b+c}{2}\right)-c \\ =a+b+c-c \\ =a+b \end{gathered}$$

Therefore, $2\left(R+r\right)$ $=a+b$

Hence, the correct answer is option (C).