Maximum number of non-congruent triangles with side lengths integers and less than $9$ 
Find the maximum number of non-congruent triangles whose side lengths are integers and less than $9$.

I tried using brute force to count all the triangles but there are so many triangles that can be formed and I counted about $60$ of them but after that I gave up. Is there any systematic approach to question?

We do some systematic counting.
First the isosceles (including equilateral) ones. If the base is $1$, other two sides can be any from $1$ to $8$. If base is $2$, other two can be any from $2$ to $8$. If base is $3$ - $2$ to $8$. Base $4$ - $3$ to $8$. Base $5$ - $3$ to $8$. Base $6$ - $4$ to $8$. Base $7$ - $4$ to $8$. Base $8$ - $5$ to $8$. Total $$8+7+7+6+6+5+5+4=48$$
For scalene, we write the triplet of side lengths in decreasing order respecting triangle inequality.

*

*$[8,7,(6 \,\text{to}\, 2)]. [8,6,(5\,\text{to}\, 3)]. [8, 5, 4]. [8,4,✗].$

*$[7,6,(5 \,\text{to}\, 2)]. [7,5,(4 \,\text{to}\, 3)]. [7, 4, ✗].$

*$[6,5,(4 \,\text{to}\, 2)]. [6,4,3]. [6, 3, ✗].$

*$[5,4,(3 \,\text{to}\, 2)]. [5,3,✗].$

*$[4,3,2]. [4,2,✗].$

*$[3,2,✗].$
Sum : $5+3+1+4+2+3+1+2+1=22$
Grand total : $\boxed{48+22=70}$
