Question

In $\Delta ABC$, if $\sqrt{3}r+a=\sqrt{3}r+b=s$, then the triangle is

• A. a right angled triangle
• B. an isosceles triangle
• C. an equilateral triangle
• D. a scalene triangle

Answer 1

The correct option is C an equilateral triangle

Given, $\sqrt{3}r+a=\sqrt{3}r+b$
$\Rightarrow a=b$...............(1)

$\sqrt{3}r+a=s$
$\sqrt{3}r+b=s$

$2\sqrt{3}r+a+b=2s=a+b+c$
$c=2\sqrt{3}r$

$\frac{2s-(a+b)}{r}=2\sqrt{3}$

$\frac{2s-(a+a)}{r}=2\sqrt{3}$......(by 1)

$\frac{s-a}{r}=\sqrt{3}$
also, $r=(s-a)\tan \frac{A}{2}$

$\tan \frac{A}{2}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{A}{2}=\frac{\pi }{6}$$\Rightarrow A=\frac{\pi }{3}$

$\because A=B=\frac{\pi }{3}$
$\therefore \angle c=\frac{\pi }{3}$
$\therefore$ the triangle equilateral triangle.
Hence, option C.