RBSE Solutions for Class 7 Maths Chapter 9 Congruence of Triangles In Text Exercise

RBSE Solutions for Class 7 Maths Chapter 9 Congruence of Triangles In Text Exercise is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 9 Congruence of Triangles In Text Exercise.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 7
Subject Maths
Chapter Chapter 9
Chapter Name Congruence of Triangles
Exercise In Text Exercise
Number of Questions 3
Category RBSE Solutions

Rajasthan Board RBSE Class 7 Maths Chapter 9 Congruence of Triangles In Text Exercise

Do and Learn

(Page 114)
Question 1
When two triangles ∆ABC and ∆PQR are given, then (RBSESolutions.com) six possible matching are there in these two. Find these matching by using cutout of two triangles.
Solution:
In two triangles ∆ABC and ∆PQR. The proper pair between their vertices are six pairs as:
RBSE Solutions for Class 7 Maths Chapter 9 Congruence of Triangles In Text Exercise
(i) We write ∆ABC ↔ ∆PQR for A ↔ P, B ↔ Q and C ↔ R
(ii) We write ∆ABC ↔ ∆QRP for A ↔ Q, B ↔ R and C ↔ P
(iii) We write ∆ABC ↔ ∆RPQ for A ↔ R, B ↔ P and C ↔ Q
(iv) We write ∆ABC ↔ ∆PRQ for A ↔ P B ↔ R and C ↔ Q
(v) We write ∆ABC ↔ ∆QPR for A ↔ Q, B ↔ P and C ↔ R
(vi) We write ∆ABC ↔ ∆RQP for A ↔ R, B ↔ Q and C ↔ P

RBSE Solutions

Question 2
Do all matching show congruency? Identify it (RBSESolutions.com) by superimposing.
Solution:
If ∆ABC and ∆PQR are equilateral triangles and congruent also then all pairs will show the congurency. we can find by overlapping all these pairs in figured given in Q. 1.

(Page 118)
Question 1
Some triangles are given in figure. Which triangles (RBSESolutions.com) are congruent by R.H.S.rule?
RBSE Solutions for Class 7 Maths Chapter 9 Congruence of Triangles In Text Exercise
Sol. :
∆ABC is not congruent to ∆DEF because in RHS rule, one side (RBSESolutions.com) and hypotenuse of one right angled triangle are equal to corresponding side and hypotenuse which is not in ∆ABC and ∆DEF
In right angle triangle ∆PRQ and ∆QSP
PR = SQ = 2 cm
PQ = PQ (common)
∠PRQ = ∠QSP = 90° (right angle)
∴ From R.H.S. congruency
∆PRQ ≅ ∆QSP

RBSE Solutions

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