1.

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 is A. reflexive and symmetric but not transitive B. reflexive and transitive but not symmetric C. symmetric and transitive but not reflexive D. an equivalence relation

Answer»

Correct Answer is (D) an equivalence relation

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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