If $r_{1}^{2} = r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1},$ then the triangle is:

Equilateral triangle

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Isosceles triangle

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Right angled triangle

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Scalene triangle

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Solution

The correct option is C Right angled triangle
Given, $r_{1}^{2} = r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1}$
We know, $r_{1} = s tan \frac{A}{2}$
Also, we know $r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = s^{2}$
$∴ r_{1}^{2} = s^{2}$
$\Rightarrow r_{1} = s$
$\Rightarrow s tan \frac{A}{2} = s$
$\Rightarrow tan \frac{A}{2} = 1$
$\Rightarrow A = 90^{\circ}$
Hence, given triangle is a right angled triangle.

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Similar questions

A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} =

\frac{r_{1} (r_{2} + r_{3})}{\sqrt{r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1}}} =

In a $△ A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} =$

a (r r_{1} + r_{2} r_{3}) = b (r r_{2} + r_{3} r_{1}) = c (r r_{3} + r_{1} r_{2}) = a b c

\frac{r_{1} (r_{2} + r_{3})}{\sqrt{r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1}}} = a

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