Question

How many different (mutually non-congruent) trapezium s can be constructed using four distinct side lengths from the set $\left\{1, 2, 3, 4, 5, 6\right\}$ ?

• A. 5
• B. 11
• C. 15
• D. 30

Answer 1

Answer: (B)

Length of sides of trapezium from $\left\{1, 2, 3, 4, 5, 6\right\}$
From non-congruent trapezium
$|r-p|<\,q+s<\,r+p$
Possible combination are
$\left(r,p\right), \left(s,q\right) \equiv\left\{\left(5, 6\right), \left(1,3\right)\right\}, \left\{\left(5, 6\right), \left(2,4\right)\right\}$
$\left\{\left(5,6\right), \left(1,4\right)\right\}, \left\{\left(5,6\right), \left(3,4\right)\right\},\left\{\left(6,4\right), \left(1, 3\right)\right\}$,
$\left\{\left(6,4\right),\left(1,5\right)\right\}, \left\{\left(6,4\right), \left(2,3\right)\right\}, \left\{\left(6,4\right), \left(3,5\right)\right\}$,
$\left\{\left(4,5\right), \left(1,3\right)\right\} \left\{\left(4,5\right), \left(1,6\right)\right\}\left\{\left(4,5\right), \left(2,6\right)\right\}$
Total $11$ combination is possible