If an equivalence relation R defined in the set A of all polygons as$k=\left\{ \left( {{P}_{1}},{{P}_{2}} \right):{{P}_{1}}\,and\,{{P}_{2}}\,have\,same\,number\,of\,sides \right\}$. Then the set of all elements in A relation to the right-angle triangles T with sides 3,4 and 5 is
(a) Right angles triangle
(b) Triangle
(c) Square
(d) Circle

Hint: In this question, we will write the required set in set builder form and then step by step check all its conditions to find the correct option.

Complete step-by-step solution -
In a given question, we have set A of all polygons.
Polygon is a closed figure made with straight lines only.
On a set A, a relation is defined, which is given as $k=\left\{ \left( {{P}_{1}},{{P}_{2}} \right):{{P}_{1}}\,and\,{{P}_{2}}\,have\,same\,number\,of\,sides \right\}$
Now, we have a third set, let us name it as B, which is defined as
$B=\left\{ b:\left( b,T \right)\in R \right\}$where T is a right-angled triangle with side 3,4 and 5. That is, any element of A, which is related to T with relation R, is element of B.
So, for an element to belong to B, it must first belong to A, that is, it should be a polygon.
Here, a circle cannot be a polygon, as a polygon must be made with straight lines only, but the circle is a curved figure.
There, circle does not belong to A.
Hence, the circle does not belong to B.
Also, squares and all triangles are made of only straight lines and are also closed figures. So, squares and all triangles are polygons and hence belong to A.
Now, for the ordered pair $\left( b,T \right)$, where b is an element of A, to belong to relation R, the number of sides of b and T must be equal. Here, T is a right-angled triangle, So, number of sides of T are three. So, for $\left( b,T \right)$ to belong to R, b must also have three sides. Also, b is an element of A, that is, it is a polygon. So, b must be a polygon of three sides. Now, we know that, polygon of three sides is known as a triangle.
Here b is a set of triangles.
Therefore, the answer is option (b).

Note: In this question, we are asked all elements of A that belong to b and not some elements. So, even if option (a) will also belong to b, it will not be whole set b. So, it is not the correct option.