#### Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 10 Maths Textbook Solution.

Answer: R is an equivalence relation.
The set of all elements in A related to triangle T is the set of all triangles.
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
$Given: \! R= \left \{ \left ( P_{1}, P_{2}\right ):P_{1}\, and\, P_{2} \: have\: same\: number\: o\! f\: sides \right \}$
Explanation:
Reflexive:
$R\: is\: re\! f\! lexive\: \text{ since } (P_{1}, P_{1})\epsilon \: R \: \text{ is }\: the\: same \: polygon\: as\: itsel\! f.$
Symmetric:
$Let (P_{1}, P_{2}) \epsilon \: R$
$P_{1}\, \text{ and }\, P_{2}\: have \: the \: same\: numbers\: o\! f\: sides.$
$P_{2}\, \text{ and }\, P_{1} \: \text{ have }\: the\: same \: number \: o\! f \: side.$
$(P_{2}, P_{1}) \epsilon \: R$
R is symmetric
Transitive:
$Let\:\left ( P_{1}, P_{2} \right ) , (P_{2}, P_{3})\: \epsilon \: R$
$P_{1} \text{ and } P_{2}\: have \: the\: same\: number \: o\! f \: sides$
$Also\: P_{2}\, and\, P_{3}\: have\: the\: same\: number\: o\! f\: sides.$
$P_{1}\, and\, P_{3} \: have \: the\: same\: number\: o\! f \: sides$.
$(P_{1}, P_{3}) \: \epsilon \: R$
R is transitive.
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation.
The elements in A related to the right-angle triangle (T) with sides 3, 4, and 5 are those polygons that have 3 sides (since T is a polygon with 3 sides).
Hence, the set of all elements in A related to triangle T is the set of all triangles.