Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

Let $a^2+b^2=c^2$. If $R$ and $r$ are its circumradius and inradius respectively then $R+r$ is equal to?

• A. $AM$ of $a$ and $b$
• B. $AM$ of $b$ and $c$
• C. $AM$ of $c$ and $a$
• D. Perimeter of $\Delta ABC$

Answer 1

The correct option is A $AM$ of $a$ and $b$

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

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

Given $a^2+b^2=c^2$ means $\angle C=\frac{\pi }{2}$

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

Put eqn($3$) in eqn($1$) we get,

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

$\therefore R=\frac{c}{2}$ - - - - - - $4$

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

Put it in eqn($2$)

After solving eqn($2$) we will get,

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

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

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

Hence proved $(R+r)=A.M.of$ $a$ and $b$