Question

If two sides (in $\text{units}$) of a $△ABC$ are $3$ and $9,$ then the number of possible triangles whose third side is also integral, is

Answer 1

Let $a=3,b=9$
we know, $c<a+b$ and $c>|b-a|$
$\Rightarrow c<12$ and $c>6$
So, possible integral values of third side are : $7,8,9,10,11$
$\therefore$ Number of possible triangles $=5$