Question

If in a $△ABC,r_1=2,r_2=3,$ and $r_3=6$, then $a$ equals

• A. $4$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D

$3$

Explanation for the correct option:

Step 1. Find the value of $s$.

It is known that $∆=r_1\left(s-a\right)=r_2\left(s-b\right)=r_3\left(s-c\right)$.

Now for $∆=r_1\left(s-a\right)$:

$$\begin{gathered} ∆=2\left(s-a\right) \\ \Rightarrow \frac{∆}{2}+a=s...\left(1\right) \end{gathered}$$

Similarly for $∆=r_2\left(s-b\right)$ it is found that $\frac{∆}{3}+b=s...\left(2\right)$.

And for $∆=r_3\left(s-c\right)$ it is found that $\frac{∆}{6}+c=s...\left(3\right)$.

Now, add the three equations $1,2,$ and $3$.

$$\begin{gathered} \frac{∆}{2}+a+\frac{∆}{3}+b+\frac{∆}{6}+c=s+s+s \\ \Rightarrow \left(\frac{∆}{2}+\frac{∆}{3}+\frac{∆}{6}\right)+\left(a+b+c\right)=3s \\ \Rightarrow \frac{3∆+2∆+∆}{6}+2s=3s\mathbf{[}\mathbf{s}\mathbf{=}\frac{\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{+}\mathbf{c}}{\mathbf{2}}\mathbf{]} \\ \Rightarrow \frac{6∆}{6}=s \\ \Rightarrow ∆=s \end{gathered}$$

Step 2. Find the value of $a$.

The Heron's formula $∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ can be simplified as:

$$\begin{gathered} ∆=\sqrt{∆·\frac{∆}{2}·\frac{∆}{3}·\frac{∆}{6}}\left[\mathit{s}\mathbf{=}\mathbf{∆}\mathbf{,}\mathbf{}\mathbf{∆}\mathbf{=}{\mathit{r}}_{\mathbf{1}}\left(s-a\right)\mathbf{=}{\mathit{r}}_{\mathbf{2}}\left(s-b\right)\mathbf{=}{\mathit{r}}_{\mathbf{3}}\left(s-c\right)\right] \\ \Rightarrow ∆=\sqrt{\frac{{∆}^4}{6^2}} \\ \Rightarrow ∆=\frac{{∆}^2}{6} \\ \Rightarrow 6=∆ \end{gathered}$$

Now, using $∆=6$ , $s=∆=6$ and equation $1$ the value of $a$ can be given as:

$\begin{array}{rcl}a& =& s-\frac{∆}{2}\\ & =& 6-\frac{6}{2}\\ & =& 6-3\\ & =& 3\end{array}$

So the value of $a$ is $3$.

Hence, the correct option is D.