Question

In a triangle $\left(1-\frac{r_1}{r_2}\right)\left(1-\frac{r_1}{r_3}\right)=2$, then the triangle is

• A. Equilateral.
• B. Right angled.
• C. Isosceles.
• D. None of these.

Answer 1

The correct option is B

Right angled.

Explanation for the correct option:

In a triangle with sides $a,b,c$.

$$\begin{gathered} r_1=\frac{∆}{(s-a)} \\ {r}_2=\frac{∆}{(s-b)} \\ {r}_3=\frac{∆}{(s-c)} \end{gathered}$$

Where, $s=$semi perimeter of the triangle and $a,b,c$ are its three sides.

Given;

$\left(1-\frac{r_1}{r_2}\right)\left(1-\frac{r_1}{r_3}\right)=2$

Now, substitute the values of $r_1,r_2,r_3$ we get,

$\begin{array}{rcl}\left\{1-\frac{{\displaystyle \frac{∆}{(s-a)}}}{{\displaystyle \frac{∆}{(s-b)}}}\right\}\left\{1-\frac{{\displaystyle \frac{∆}{(s-a)}}}{{\displaystyle \frac{∆}{(s-c)}}}\right\}& =& 2\\ \left\{1-\frac{(s-b)}{(s-a)}\right\}\left\{1-\frac{(s-c)}{(s-a)}\right\}& =& 2\\ \frac{(b-a)(c-a)}{{(s-a)}^2}& =& 2\\ (b-a)(c-a)& =& 2{(s-a)}^2\\ (b-a)(c-a)& =& \frac{1}{2}{(b+c-a)}^2\\ b^2+c^2& =& a^2\end{array}$

Therefore, the triangle is right angled.

Hence, the correct option is B.