Question

In $∆ABC$ with usual notation, observe the two statements given below

$$\begin{gathered} \text{I}.r{r}_1{r}_2{r}_3={∆}^2 \\ \text{II}.r_1{r}_2+r_2{r}_3+r_3{r}_1=s^2 \end{gathered}$$

Which of the following is correct?

• A. Both $\text{I}$ and $\text{II}$ are correct
• B. $\text{I}$ and $\text{II}$ are incorrect
• C. $\text{I}$ is incorrect, $\text{II}$ is correct
• D. $\text{I}$ is correct, $\text{II}$ is incorrect

Answer 1

The correct option is A

Both $\text{I}$ and $\text{II}$ are correct

Explanation For The Correct Option:

Explanation for statement $\text{I}$:

In $∆ABC$ inradius and ex-radius are as follows

$$\begin{gathered} r=\frac{∆}{s} \\ {r}_1=\frac{∆}{\left(s-a\right)} \\ {r}_2=\frac{∆}{\left(s-b\right)} \\ {r}_3=\frac{∆}{\left(s-c\right)} \\ \Rightarrow r{r}_1{r}_2{r}_3=\left(\frac{∆}{s}\right)\left(\frac{∆}{\left(s-a\right)}\right)\left(\frac{∆}{\left(s-b\right)}\right)\left(\frac{∆}{\left(s-c\right)}\right) \\ \Rightarrow r{r}_1{r}_2{r}_3=\frac{{∆}^4}{s(s-a)(s-b)(s-c)} \\ \Rightarrow r{r}_1{r}_2{r}_3=\frac{{∆}^4}{{∆}^2}\left[\because (s-a)(s-b)(s-c)={∆}^2\right] \\ \Rightarrow r{r}_1{r}_2{r}_3={∆}^2 \end{gathered}$$

Hence the statement $\text{I}$ is true.

Explanation for statement $\text{II}$:

$$\begin{gathered} r_1{r}_2+r_2{r}_3+r_3{r}_1=\frac{∆}{(s–a)}\times \frac{∆}{(s–b)}+\frac{∆}{(s–b)}\times \frac{∆}{(s–c)}+\frac{∆}{(s–c)}\times \frac{∆}{(s–a)} \\ =\frac{{∆}^2}{(s–a)(s–b)}+\frac{{∆}^2}{(s–b)(s–c)}+\frac{{∆}^2}{(s–c)(s–a)} \\ ={∆}^2\left[\frac{(s–c)+(s–a)+(s–b)}{(s–a)(s–b)(s–c)}\right] \\ ={∆}^2\left[\frac{(3s–(a+b+c\left)\right)}{(s–a)(s–b)(s–c)}\right] \\ ={∆}^2\left[\frac{(3s–2s)}{(s–a)(s–b)(s–c)}\right]\left[\because 2s=a+b+c\right] \\ ={∆}^2\left[\frac{s}{(s–a)(s–b)(s–c)}\right] \\ ={∆}^2\left[\frac{s^2}{s(s–a)(s–b)(s–c)}\right]\left[\because Multiplynumeratoranddenominatorbys\right] \\ ={∆}^2\left[\frac{s^2}{{∆}^2}\right]\left[\because s(s–a)(s–b)(s–c)={∆}^2\right] \\ =s^2 \end{gathered}$$

$\Rightarrow r_1{r}_2+r_2{r}_3+r_3{r}_{1=}{s}^2$

Hence the statement $\text{II}$ is true.

Hence, the correct answer is option (A).