Q. 125.0( 1 Vote )

# Mark the tick against the correct answer in the following:Let S be the set of all triangles in a plane and let R be a relation on S defined by ∆1 S ∆2⇔ ∆1 ≡ A2. Then, R isA. reflexive and symmetric but not transitiveB. reflexive and transitive but not symmetricC. symmetric and transitive but not reflexiveD. an equivalence relation

According to the question ,

Given set S = {…All triangles in plane….}

And R = {(∆1 , ∆2) : ∆1 , ∆2 S and 1 ≡ ∆2}

Formula

For a relation R in set A

Reflexive

The relation is reflexive if (a , a) R for every a A

Symmetric

The relation is Symmetric if (a , b) R , then (b , a) R

Transitive

Relation is Transitive if (a , b) R & (b , c) R , then (a , c) R

Equivalence

If the relation is reflexive , symmetric and transitive , it is an equivalence relation.

Check for reflexive

Consider , (1, 1)

We know every triangle is congruent to itself.

(1, 1) R all 1 S

Therefore , R is reflexive ……. (1)

Check for symmetric

(∆1 , ∆2) R then 1 is congruent to ∆2

(∆2 , ∆1) R then 2 is congruent to ∆1

Both the equation are the same and therefore will always be true.

Therefore , R is symmetric ……. (2)

Check for transitive

Let 1, 2, 3 S such that (1, 2) R and (2, 3) R

Then (1, 2)R and (2, 3)R

1 is congruent to 2, and 2 is congruent to 3

1 is congruent to 3

(1, 3) R

Therefore , R is transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (D)

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