Question

The relation $“$is congruent to$"$ on the set of all triangles in a plane is

• A. reflexive only
• B. symmetric only
• C. transitive only
• D. equivalence

Answer 1

The correct option is D equivalence

$\begin{array}{cc}Now,T=setofalltriangles& \\ R:T\to T& \\ R=\left\{\left(T_1,T_2\right):T_1\cong T_2\right\}& \\ & \\ \left(i\right)Forreflexive& \\ T_1\cong T_1& \left[alltriangles\cong toitself\right]\\ So,\left(T_1,T_1\right)\in R& \end{array}$

So it is a reflexive relation.

$\begin{array}{c}\left(ii\right)Forsymmetric\\ Let\left(T_1,T_2\right)\in R\Rightarrow T_1\cong T_2\\ \Rightarrow T_2\cong T_1\\ So,\left(T_2,T_1\right)\in R\end{array}$

Thus, it is a symmetric relation.

$\begin{array}{c}\left(iii\right)Fortransitive\\ Let\left(T_1,T_2\right)\in R\Rightarrow T_1\cong T_2\to \left(i\right)\\ \left(T_2,T_3\right)\in R\Rightarrow T_2\cong T_3\to \left(ii\right)\\ Fromequation\left(i\right)\&\left(ii\right),weget\\ T_1\cong T_2\cong T_3\\ T_1\cong T_3\Rightarrow \left(T_1,T_3\right)\in R\end{array}$

Thus, it is a transitive.

Hence, given relation is an equivalence.