Question

Statement$-1$: In a $\Delta ABC$, if $a<b<c$ and $r$ is in-radius and $r_1,r_2,r_3$ are the exradii opposite to angle $A,B,C$ respectively, then $r<r_1<r_2<r_3$.

Statement$-2$ : For, $\Delta ABC$
$r_1{r}_2+r_2{r}_3+r_3{r}_1=\frac{r_1{r}_2{r}_3}{r}$

• A. Both statements are True and statement$-2$ is the correct explanantion of Statement$-1$
• B. Both statements are True but statement$-2$ is NOT the correct explanantion of Statement$-1$
• C. Statement$-1$ is True and statement$-2$ is False.
• D. Statement$-1$ is False and statement$-2$ is True.

Answer 1

The correct option is B Both statements are True but statement$-2$ is NOT the correct explanantion of Statement$-1$

Statement$-1$ :
Given,$a<b<c\Rightarrow \frac{\Delta }{s}<\frac{\Delta }{s-a}<\frac{\Delta }{s-b}<\frac{\Delta }{s-c}$
We know, $r=\frac{\Delta }{s},r_1=\frac{\Delta }{s-a},r_2=\frac{\Delta }{s-b},r_3=\frac{\Delta }{s-c}$
$\Rightarrow r<r_1<r_2<r_3$
Statement$-2$ :
Clearly, from the relations of inradius and exradii, we have:$\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{1}{r}$
$\Rightarrow r_1{r}_2+r_2{r}_3+r_3{r}_1=\frac{r_1{r}_2{r}_3}{r}$
Clearly, both statements are True but statement$-2$ is NOT the correct explanantion of Statement$-1$