Question

In a triangle $ABC$, the incircle touches the sides $BC,CA$ and $AB$ at $D,E,F$ respectively. If the radius of incircle is $\sqrt{33}$ units and $BD,CE$ and $AF$ are consecutive natural numbers, then the perimeter of the triangles is

• A. $30$
• B. $40$
• C. $20$
• D. $60$

Answer 1

The correct option is D $60$

Let $BD=x$, $CE=x+1$, $AF=x+2$

$\therefore$ $AB=AF+BF=2x+2$ $[\because \stackrel{BF=BD}{\underset{AE=AF}{CD=CE}}]$

$BC=BD+CD=2x+1$

$AC=AE+CE=2x+3$

area of $ABC=\sqrt{(3x+3)(x+1)(x+2)\left(x\right)}$

Radius of incentre $=\frac{\Delta }{5}$

$\Rightarrow \sqrt{33}=\sqrt{\frac{(3x+3)(x+1)(x+2)\left(x\right)}{{(3x+3)}^2}}$

$\Rightarrow \sqrt{33}=\frac{(x+1)(x+2)\left(x\right)}{3(x+1)}$

$\Rightarrow 99:x^2+2x$

$\Rightarrow x^2+2x-99=0$

$\Rightarrow (x+11)(x-9)=0$

$\Rightarrow x=9$ ($\because$ $x$ cannot be negative)

$\therefore$ $AB=20$, $BC=19$, $AC=21$

$\therefore$ Perimeter $=60$ [D]