Question

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

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

Answer 1

Answer: (B)

$a+b$

The given triangle is right angled at $C.$

Circumradius $R=\frac{abc}{4\Delta }$

For right angle triangle $\Delta =\frac{1}{2}ab$

So, $R=\frac{abc}{4\frac{ab}{2}}=\frac{c}{2}$

Now inradius $r=\frac{\Delta }{s}$

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

So, $r=\frac{\frac{ab}{2}}{\frac{a+b+c}{2}}=\frac{ab}{a+b+c}$

Now we need value of $2(r+R)$

$2(r+R)=\frac{2ab}{a+b+c}+c=\frac{2ab+ac+bc+c^2}{a+b+c}$

Apply Pythagores Theorem to get,

$\Rightarrow 2(r+R)=\frac{2ab+ac+bc+a^2+b^2}{a+b+c}=\frac{{(a+b)}^2+c(a+b)}{a+b+c}=\frac{(a+b)(a+b+c)}{a+b+c}=a+b$