I need to find number of mutually non congruent trapeziums with side lengths from {1,3,4,5,6}.
I know for a trapezium with a,c as non parallel sides the condition for my question is that
$$|a-c|<b+d<a+c$$
But i cant apply it to count the cases. Can someone help in how to get the number of cases?
If $a$ and $c$ are the nonparallel sides then the correct (necessary and sufficient) conditions for the existence of a nondegenerate trapezium are $$|a-c|<|b-d|<a+c\ .\tag{1}$$ There is no condition on $b+d$.
From $\{1,3,4,5,6\}$ you can choose $b>d$ in ${5\choose2}=10$ ways, and from the remaining lengths you can choose $a>c$ in ${3\choose2}=3$ ways. Therefore you obtain $30$ quadruples $(a,b,c,d)$, which you have to test for $(1)$. All admissible quadruples give rise to two mirror symmetric trapezoids.