Question

In $△ABC$, the incircle touches the sides $BC,CA$ and $AB$, respectively, $D,E,$ and $F$. If the radius of the incircle is $4$ units and $BD,CE$ and $AF$ are consecutive integers, then the value of $\frac{s}{3}$, where $s$ is a semi-perimeter of triangle, is

Answer 1

Let $BD=x,CE=x+1$ and $AF=x+2$ Then
$CD=CE=x+1=s-c$
$AE=AF=x+2=s-a$
$BF=BD=x=s-b$
On adding, we get
$3s-(a+b+c)=3x+3\therefore s=3x+3\Rightarrow s-a=x+2,s-b=x,s-c=x+1$
Now, $r=\frac{△}{s}=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{△}$
Or, $4=\sqrt{\frac{(x+2)x(x+1)}{3(x+1)}}$
Or, $x^2+2x=48$
$\Rightarrow x=6$
Hence, $s=21\Rightarrow \frac{s}{3}=7$