Show that the relation $R$ defined in the set $A$ of all triangles as $R = \{(T_{1}, T_{2}) : T_{1} \text{ is similar to } T_{2}\}$,is an equivalence relation. Consider three right-angled triangles $T_{1}$ with sides $3, 4, 5$,$T_{2}$ with sides $5, 12, 13$,and $T_{3}$ with sides $6, 8, 10$. Which triangles among $T_{1}, T_{2}$,and $T_{3}$ are related?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(A) The relation is defined as $R = \{(T_{1}, T_{2}) : T_{1} \text{ is similar to } T_{2}\}$.
$1$. Reflexivity: Every triangle $T_{1}$ is similar to itself. Therefore,$(T_{1}, T_{1}) \in R$. Thus,$R$ is reflexive.
$2$. Symmetry: If $(T_{1}, T_{2}) \in R$,then $T_{1}$ is similar to $T_{2}$. This implies $T_{2}$ is similar to $T_{1}$. Therefore,$(T_{2}, T_{1}) \in R$. Thus,$R$ is symmetric.
$3$. Transitivity: If $(T_{1}, T_{2}) \in R$ and $(T_{2}, T_{3}) \in R$,then $T_{1}$ is similar to $T_{2}$ and $T_{2}$ is similar to $T_{3}$. This implies $T_{1}$ is similar to $T_{3}$. Therefore,$(T_{1}, T_{3}) \in R$. Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.
Regarding the triangles:
For $T_{1}$ (sides $3, 4, 5$) and $T_{3}$ (sides $6, 8, 10$):
$\frac{3}{6} = \frac{4}{8} = \frac{5}{10} = \frac{1}{2}$.
Since the ratios of corresponding sides are equal,$T_{1}$ is similar to $T_{3}$.
Thus,$T_{1}$ and $T_{3}$ are related.

Explore More

Similar Questions

In the set $N$,a relation $R$ is defined as $aRb \Leftrightarrow b$ is divisible by $a$. Then $R$ is:

The number of reflexive relations on a set $A$ of $n$ elements is equal to

Let $R$ be the real line. Let the relations $S$ and $T$ on $R$ be defined by $S = \{(x, y) : y = x + 1, 0 < x < 2\}$ and $T = \{(x, y) : (x - y) \text{ is an integer}\}$. Then:

Let $L$ be the set of all straight lines in the Euclidean plane. Two lines $l_1$ and $l_2$ are said to be related by the relation $R$ if and only if $l_1$ is parallel to $l_2$. Then the relation $R$ is

On $R$, the relation $\rho$ is defined by '$x \rho y$ holds if and only if $x-y$ is zero or irrational'. Then,

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo