MCQ
How many different (mutually non-congruent) trapeziums can be constructed using four distinct side lengths from the set $\{1,2,3,4,5,6\}$ ?
- A$5$
- ✓$11$
- C$15$
- D$30$
✓
Answer
Correct option: B.
$11$
b
(b)
(b)
Length of sides of trapezium from $\{1,2,3$, $4,5,6\}$.
From non-congruent trapezium
$|a-c| < b+d < a+c$
Possible combination are
$(r, p),(s, q) \equiv\{(5,6),(1,3)\},\{(5,6),(2,4)\}$
$\{(5,6),(1,4)\},\{(5,6),(3,4)\},\{(6,4),(1,3)\}$,
$\{(6,4),(1,5)\},\{(6,4),(2,3)\},\{(6,4),(3,5)\}$
$\{(4,5),(1,3)\}\{(4,5),(1,6)\}\{(4,5),(2,6)\}$
Total $11$ combination is possible.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
For any set $A, (A')'$ is equal to:
→If $P(A) = P(B) = x$ and $P(A \cap B) = P(A' \cap B') = \frac{1}{3}$, then $x = $
→Find the Center of circle $x^2+ y^2 - 4x - 8x + 25 = 0:$
→A card from a pack of $52$ cards is lost. From the remaining cards of the pack, two cards are drawn. Then the probability that they both are diamonds is:
→If $4{x^2} + p{y^2} = 45$ and ${x^2} - 4{y^2} = 5$ cut orthogonally, then the value of $p$ is
→If $– 3x + 17 < – 13,$ then:
→In a class of $60$ students, $40$ opted for $NCC,\,30$ opted for $NSS$ and $20$ opted for both $NCC$ and $NSS.$ If one of these students is selected at random, then the probability that the student selected has opted neither for $NCC$ nor for $NSS$ is
→The coordinates of a point which divides the line joining the points $P(2, 3, 1)$ and $Q(5, 0, 4)$ in the ratio $1 : 2$ are:
→Let $\alpha > 0$, be the smallest number such that the expansion of $\left(x^{\frac{2}{3}}+\frac{2}{x^3}\right)^{30}$ has a term $\beta x^{-\alpha}, \beta \in N$. Then $\alpha$ is equal to $.............$.
→The number of different words can be formed from the letters of the word $'SUCCESS'$ in which the two $C$ are together but no two $S$ are together are :-
→