Question

In $ \Delta ABC $ , if $ r{{r}_{1}}={{r}_{2}}{{r}_{3}} $ , then the triangle is
A. equilateral
B. isosceles
C. right-angled
D. scalene

Answer 1

Hint: We first try to use the properties of triangles theorem involving the area of the triangle and the semi-perimeter of the triangle. Putting the values we get the particular condition about the triangle.

Complete step by step solution:
It’s given that for $ \Delta ABC $ , if $ r{{r}_{1}}={{r}_{2}}{{r}_{3}} $ .
We can rewrite the given conditions using the theorems where $ r=\dfrac{\Delta }{s},{{r}_{1}}=\dfrac{\Delta }{s-a},{{r}_{2}}=\dfrac{\Delta }{s-b},{{r}_{3}}=\dfrac{\Delta }{s-c} $ . Here $ \Delta $ represents the area and $ s $ represents the semi-perimeter of $ \Delta ABC $ .
We put these values in $ r{{r}_{1}}={{r}_{2}}{{r}_{3}} $ to find $ \dfrac{\Delta }{s}\times \dfrac{\Delta }{s-a}=\dfrac{\Delta }{s-b}\times \dfrac{\Delta }{s-c} $ .
We simplify the equation to get $ \left( s-b \right)\left( s-c \right)=s\left( s-a \right) $ .
Therefore,
 $
   \left( s-b \right)\left( s-c \right)=s\left( s-a \right) \\
  \Rightarrow {{s}^{2}}-s\left( b+c \right)+bc={{s}^{2}}-as \\
  \Rightarrow bc=s\left( b+c-a \right) \\
  \Rightarrow 2bc=2s\left( b+c-a \right) \;
 $
Now we know that $ 2s=a+b+c $ . Putting the value, we get
 $
   2bc=2s\left( b+c-a \right) \\
  \Rightarrow 2bc=\left( a+b+c \right)\left( b+c-a \right) \;
 $
Now we use the identity where $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ .
Therefore, $ 2bc=\left( a+b+c \right)\left( b+c-a \right)={{\left( b+c \right)}^{2}}-{{a}^{2}} $ .
Simplifying we get
 $
   2bc=\left( a+b+c \right)\left( b+c-a \right)={{\left( b+c \right)}^{2}}-{{a}^{2}} \\
  \Rightarrow {{b}^{2}}+{{c}^{2}}-{{a}^{2}}=0 \\
  \Rightarrow {{a}^{2}}={{b}^{2}}+{{c}^{2}} \;
 $
So, the triangle is right-angled by the converse of Pythagoras. The correct option is C.
So, the correct answer is “Option C”.

Note: We can also use the circle circumscribing the triangle to find the relation between the sides with the area. Using those relations, we can find the relations between the sides directly.