RBSE Solutions for Class 7 Maths Chapter 9 Congruence of Triangles Ex 9.2 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 9 Congruence of Triangles Exercise 9.2.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 7 |
| Subject | Maths |
| Chapter | Chapter 9 |
| Chapter Name | Congruence of Triangles |
| Exercise | Ex 9.2 |
| Number of Questions | 7 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 7 Maths Chapter 9 Congruence of Triangles Ex 9.2
Question 1
Find the value of the (RBSESolutions.com) following if ∆ABC = ∆PRQ as given below :
(i) Side PR
(ii) Side QR
(iii) Side PQ
(iv) ∠P
(v) ∠Q
(vi) ∠R
Solution:
In ∆ABC
∠A + ∠B + ∠C = 180°
45° + 65° + ∠C = 180°
110° + ∠C = 180°
∠C = 180° – 110° = 70°
In ∆ABC and ∆PRQ
(i) side PR = side AB = 3.5 cm
(ii) side QR = side CB = 2 cm
(iii) side PQ = side AC = 2.9 cm
(iv) ∠P =∠A = 45°
(v) ∠Q =∠C = 70°
(vi) ∠R = ∠B = 65°
Question 2
Which condition applies of congruence of triangles in the (RBSESolutions.com) diagrams given below ? Write the congruent triangles in notations.
Solution:
(i) In ∆PQT and ∆SRT ,
side QT = side RT = 2.8 cm
∠PQT = ∠SRT = 70°
∠PTQ= ∠STR (Vertically opposite angle)
∴ A.S.A. condition
∆PQT ≅ ∆SRT
(ii) In ∆ABD and ∆CDB
∠ABD = ∠CDB = 35°
AB = CD = 4.5 cm
BD = BD (common side)
∴ S.A.S. condition
∆ABD ≅ ∆CDB,
(iii) In ∆ABC and ∆ADC
AB = AD = 2.5 cm
BC = CD = 3.5 cm
CA = CA (common)
S.S.S. condition
∆ABC ≅ ∆ADC
Question 3
Which pairs are congruent (RBSESolutions.com) in the pairs of triangle given below :
Solution:
(i) In ∆PRQand ∆QSP
PS = OR = 3 cm
PQ = PQ = 5 cm (common)
∠PQR =∠QPS = 50°
S.A.S. condition
∆PRO ≅ ∆QSP
(ii) In ∆MXZand ∆NYZ
XZ = YZ = 2.5 cm
∠X =∠Y = 110°
∠MXZ = ∠NTZ = 40°
A.S.A. condition
∆MXZ ≅ NYZ
(iii) In ∆QRPand ∆TSQ
PR = QS = 3 cm
PQ = QT = 5 cm
∠QRP = ∠TSQ = 90° (right angle)
S.A.S. condition
∆QRP ≅ ∆TSQ
Question 4
Complete the (RBSESolutions.com) statement :
(i) ∆ADB ≅ ?
(ii) ∆PQR ≅ ?
Solution:
(i) In ∆ABD and ∆ACD
AB = AC, BD = CD and AD = AD (common)
∴ S.S.S. condition
∆ADB ≅ ∆ADC
(ii) In ∆PQR and ∆PSR
QR = RS, PQ = PS and RP = RP (common)
By S.S.S. condition
∆PQR ≅ ∆PSR
Question 5
In the diagram given (RBSESolutions.com) below MNOL is a rectangle. Is ∆NOL ≅ ∆LMN? If yes, then give reason.
Solution:
∆NOL ≅ ∆LMN because
∵ OL = NM, ON= LM and NL = NL (common)
∴ By S.S.S. condition,
∆NOL ≅ ∆LMN
Question 6
In the given diagram in ∆PQR and ∆PQS, side PR = side QS and side RQ = side PS, then point out (RBSESolutions.com) which statement is correct.
(i) ∆PQR ≅ ∆PQS
(ii) ∆PQR ≅ ∆QPS
(ii) ∆PQR ≅ ∆QSP
Solution:
In ∆PQR and ∆QPS
PR ⊁ QS, RQ ⊁ PS and PQ is common to both triangles
∴ By S.S.S. condition ∆PQR ≅ ∆QPS
∴ Statement (ii) is true.
Question 7
In ∆ABC in given (RBSESolutions.com) diagram ∠A = 40°, ∠C = 35° and side AB = 2.5 cm and in ADEF ∠F = 35°, ∠E = 105° and side DE = 2.5 cm, is ∆ABC ≅ ∆DEF?
Solution:
In ∆ABC
∠A + B+ ∠C = 180°
⇒ 40° + ∠B + 359 = 180°
⇒ ∠B + 75° = 180°
∠ B = 180° – 750 = 105°
In ∆DEF
∠D + ∠E + ∠F = 180°
⇒ ∠D + 105° + 35° = 180°
⇒ ∠D + 140° = 180°
⇒ ∠D = 180° – 140° = 40°
In ∆ABC and ∆DEF
∠A = ∠D = 40°,
AB = DE = 2.5
∠B = ∠E = 105°
∴ By A.S.A. condition ∆ABC ≅ ∆DEF
We hope the RBSE Solutions for Class 7 Maths Chapter 9 Congruence of Triangles Ex 9.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 9 Congruence of Triangles Exercise 9.2, drop a comment below and we will get back to you at the earliest.