Question

In a triangle ABC, let $\angle C=\frac{\pi }{2}$. If $r$ is the in-radius 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 $a+b$

We know $R=\frac{abc}{4\Delta }$ - - - - - - - - $\left(1\right)$

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

Given $\angle C=\frac{\pi }{2}$

hence $\Delta =\frac{b\times a}{2}$ --------$3$

put eqn($3$) in eqn($1$)

we get $R=\frac{abc}{\frac{4\times b\times c}{2}}$

We will get $R=\frac{c}{2}$ - - - - - - $4$

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

Put it on eqn($2$)

After solving eqn($2$) we will get

$r=\frac{a+b-c}{2}$ - - - - - - - - $5$

now $2(R+r)=2(\frac{c}{2}+\frac{a+b-c}{2})$

Hence, $2(R+r)=a+b$