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}

Step 1: Without losing generality, assume $a$ > $c$ and $d$ > $b$ and sides AB || CD

Step 2: d2 – x2 = b2 – (c – $a$ + x)2, Similarly d2 – x2 = b2 – (a – $c$ – x)2

Step 3: (a − $c$ )2 + $d$ 2 − $b$ 2 (a − $c$ )2 + $d$ 2 − $b$ 2

Step 4: => x= => x=

Step 5: 2(a − $c$ ) 2(a − $c$ )

Step 6: If $x$ in (0, d), then there will be unique trapezoid

Step 7: (a − $c$ )2 + $d$ 2 − $b$ 2

Step 8: => in (0, $d$ )

Step 9: 2c (c − $a$ )

Step 10: => (a – c)2 + d2 – b2 – 2d(a – c) < 0

Step 11: => (a – $c$ – d)2 – b2 < 0 => (a – $c$ – $d$ – b)(a – $c$ – $d$ + b) < 0

Step 12: => (a – $c$ – $d$ + b) > 0 => $a$ + $b$ > $c$ + d

Step 13: And $a$ > c, $d$ > b

Step 14: Using these inequality, numerate these pairs (a, b, c, d)

Step 15: Case I : $a$ = 10 => total no. of cases = 16

Step 16: Case II : $a$ = 9 => total no. of cases = 9

Step 17: Case III : $a$ = 8 => total no. of cases = 4

Step 18: Case (IV) : $a$ = 7 => (7, 9, 5, 10) and (7, 8, 5, 9) => 2 cases

Step 19: => Total = 31