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.

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