Question

STATEMENT 1: In a $\Delta ABC$, if $r_1,r_2,r_3$ are in $H.P$. then $\frac{r_2}{r}=\frac{1}{3}$

STATEMENT 2: In a $\Delta ABC,\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{2}{r}$

• A. Statement-1 is true, Statement-2 is true, Statement-2 is correct explanation of Statement- 1
• B. Statement-1 is true, Statement-2 is true, Statement-2 is not correct explanation for Statement-1
• C. Statement-1 is true, Statement-2 is false
• D. Statement-1 is false, Statement-2 is false

Answer 1

The correct option is D Statement-1 is false, Statement-2 is false

If $r_1,r_2,r_3$ are in G.P then

$\frac{2}{r_2}=\frac{1}{r_1}+\frac{1}{r_3}\frac{2(s-b)}{\Delta }=\frac{(s-a)}{\Delta }+\frac{(s-c)}{\Delta }2s-2b=s-a+s-ca+c=2b......\left(i\right)$

$\frac{r_2}{r}=\frac{\frac{\Delta }{s-b}}{\frac{\Delta }{s}}=\frac{s}{s-b}$

Substituting $\left(i\right)$

$\frac{r_2}{r}=\frac{a+c+b}{a+c-b}\frac{r_2}{r}=\frac{2b+b}{2b-b}=3$

So statement $\left(i\right)$ is false

$\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{s-a}{\Delta }+\frac{s-b}{\Delta }+\frac{s-c}{\Delta }=\frac{3s-a-b-c}{\Delta }=\frac{s}{\Delta }$

So statement $2$ is also false