A trapezium in the plane is a quadrilateral in which a pair of opposite sides are parallel. A trapezium is said to be non-degenerate if it has positive area. Find the number of mutually non-congruent, non-degenerate trapeziums whose sides are four distinct integers from the set $$\{5,6,7,8,9,10\}$$.

Let the six given integral lengths be arranged in increasing order as
$$5,\;6,\;7,\;8,\;9,\;10.$$

We have to pick four distinct numbers and arrange them as sides of a non-degenerate trapezium (one pair of opposite sides parallel) in such a way that no two admissible trapezia are congruent.

Step-1 : Combinatorics of choosing sides
Choosing any four different numbers out of the six can be done in $$\binom{6}{4}=15$$ ways.
Fix such a 4-tuple and denote its members (in increasing order) by $$a\lt b\lt c\lt d.$$

Step-2 : Which two of the four will be the parallel bases?
Out of the four numbers, any unordered pair can be declared to be the bases; the remaining two then become the legs.
Number of unordered pairs of a 4-set = $$\binom{4}{2}=6.$$
Hence each 4-tuple offers 6 possible $$(\text{bases},\text{legs})$$ assignments.

Step-3 : Existence condition for a trapezium
Let $$B_1\;(\gt B_2)$$ be the two bases and $$L_1,\,L_2$$ the two legs. Draw $$B_1B_2$$ perpendiculars to obtain two triangles glued along a segment of length $$B_1-B_2$$. Non-degeneracy is possible ⇔ that segment can really be spanned by $$L_1,L_2$$;
in other words

$$\lvert L_1-L_2\rvert \; \lt\; B_1-B_2\; \lt\; L_1+L_2\qquad -(1)$$

When either inequality becomes equality the height of the trapezium is zero, i.e. it degenerates into a straight line. Therefore strict inequalities are needed.

Step-4 : Exhaustive check for every 4-tuple
Because the numbers are only from 5 to 10, the verification can be done case-wise very quickly. The table below lists, for every 4-tuple, those unordered pairs that satisfy (1). (The larger of the two numbers of a pair is written first; this does not double count anything.)

5 6 7 8 : (8,5)   (1)
5 6 7 9 : (9,7), (9,6), (9,5)   (3)
5 6 7 10: (10,7), (10,6), (10,5)  (3)
5 6 8 9 : (9,5)   (1)
5 6 8 10: (10,8), (10,6), (10,5)  (3)
5 6 9 10: (10,5)   (1)
5 7 8 9 : (9,5), (8,5), (7,5)  (3)
5 7 8 10: (10,5)   (1)
5 7 9 10: (10,5), (9,5), (7,5)  (3)
5 8 9 10: (10,5), (9,5), (8,5)  (3)
6 7 8 9 : (9,6)   (1)
6 7 8 10: (10,8), (10,7), (10,6)  (3)
6 7 9 10: (10,6)   (1)
6 8 9 10: (10,6), (9,6), (8,6)  (3)
7 8 9 10: (10,7)   (1)

The right-hand margin of each line shows how many of the 6 unordered pairs fulfil condition (1). Adding them up

$$1+3+3+1+3+1+3+1+3+3+1+3+1+3+1 = 31.$$

Step-5 : Congruence check
All the counted trapezia have four mutually different side lengths. Interchanging the two legs merely produces a mirror image, which is congruent. Any two trapezia appearing in different rows or with different pairs of parallel sides have different ordered 4-tuples of side lengths and therefore are not congruent. Thus every entry above corresponds to exactly one mutually non-congruent non-degenerate trapezium.

Hence the required number of mutually non-congruent non-degenerate trapeziums is 31.