Question

In a triangle $ABC$, let $\angle C=\frac{\pi }{2}$.

If $r$ is the in-radius and $R$ is the circum-radius 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 $a+b$

We have, $r=\frac{\Delta }{s}$ , where s is the semi-perimeter
$\Rightarrow r=\frac{ab}{a+b+c}$ (It is a right angled triangle)
$\therefore R=\frac{c}{2}$
Now, $(R+r)$ $=\frac{ab}{a+b+c}+\frac{c}{2}$
$=\frac{2ab+ac+bc+c^2}{2(a+b+c)}$
$=\frac{2ab+ac+bc+a^2+b^2}{2(a+b+c)}$
$=\frac{a^2+ab+ab+b^2+ac+bc}{2(a+b+c)}$
$=\frac{(a+b+c)(a+b)}{2(a+b+c)}$
$=\frac{(a+b)}{2}$

Thus $2(R+r)=a+b$