How many different (mutually non-congruent) t | vedclass

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(B) Let the side lengths of the trapezium be $p, q, r, s$ where $p$ and $r$ are the parallel sides $(p > r)$ and $q, s$ are the non-parallel sides.For

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$11$

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(B) Let the side lengths of the trapezium be $p, q, r, s$ where $p$ and $r$ are the parallel sides $(p > r)$ and $q, s$ are the non-parallel sides.For a trapezium to exist with sides $p, q, r, s$,the condition $|p - r| We need to choose $4$ distinct lengths from $\{1, 2, 3, 4, 5, 6\}$.Let the parallel sides be $p$ and $r$. The number of ways to choose $p$ and $r$ is $\binom{6}{2} = 15$.For each pair $(p, r)$,we need to choose $q$ and $s$ from the remaining $4$ numbers such that $|p - r| By testing all combinations,we find there are $11$ such sets of four distinct side lengths that satisfy the condition for forming a non-congruent trapezium.Thus,the total number of such trapeziums is $11$.

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

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