Question

Let $R$ be the relation “is congruent to” on set of all triangles in a plane. Then $R$ is

[1 mark]

• A. reflexive and symmetric only
• B. symmetric only
• C. transitive only
• D. an equivalence relation

Answer 1

The correct option is D an equivalence relation

For reflexive :
Since every triangle is congruent to itself,
$\therefore R$ is reflexive.

For symmetric :
Let $(T_1,T_2)\in R$
$\Rightarrow T_1\cong T_2$
$\Rightarrow T_2\cong T_1$
So, $(T_2,T_1)\in R$
$\therefore R$ is symmetric.

For transitive :
Let $(T_1,T_2)\in R$ and $(T_2,T_3)\in R$
$\Rightarrow T_1\cong T_2\cdots \left(1\right)$
and $T_2\cong T_3\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we get
$T_1\cong T_2\cong T_3$
$T_1\cong T_3\Rightarrow (T_1,T_3)\in R$
$\therefore R$ is transitive.

Hence, given relation $R$ is an equivalence relation.