NCERT Solutions Class 7 Maths Chapter 5

NCERT Solutions Class 7 Mathematics Chapter 5

NCERT Solutions for Class 7 Mathematics Chapter 5 Lines and Angles 

Chapter 5 Lines and Angles is considered to be one of the most  important chapters in Class 7 Mathematics. To help the students, Extramarks offers NCERT Solutions for Class 7 Mathematics Chapter 5, which are detailed solutions to the textbook exercises in this chapter.

Students can now refer to  answers to all difficult questions. These solutions are prepared by subject matter experts. They have provided explanations for all of the questions, keeping in mind the importance of this chapter in the final exams and the best answer pattern for solving the textbook questions are provided. There are many exercise problems for students to solve. Going through these solutions will help students with their preparations, last-minute revisions, and also help them with their assignments.

NCERT Solutions for Class 7 Mathematics Chapter 5 Lines and Angles 

Access NCERT Solutions for Class 7 Mathematics Chapter 5 – Lines and Angles 

[Solutions]

NCERT Solutions for Class 7 Mathematics Chapter 5 

NCERT Class 7 Mathematics Chapter 5 is Lines and Angles. This chapter has four major topics. To learn and relate to this important chapter in Mathematics, students are advised to go through the subtopics that are there in  this chapter.

The topics that are discussed in NCERT Solutions for Class 7 Mathematics Chapter 5 lines and angles are as follows: 

  • Introduction
  • Related Angles 
  1. Complementary Angles 
  2. Supplementary Angles 
  3. Adjacent Angles
  4. Linear pairs 
  5. Vertically Opposite Angles 
  • Pairs of Lines 
  1. Intersecting Lines 
  2. Transversal
  3. Angles made by Transversal
  4. Transversal of Parallel Lines
  • Checking for Parallel Lines 

Extramarks recommends students study each of these topics carefully and then  turn to  the NCERT Solutions that are provided for the chapter on Lines and Angles.

List of Exercises in Class 7 Mathematics Chapter 5

Chapter 5 – Lines and Angles Exercises
Exercise 5.1 Questions & Solutions
Exercise 5.2 Questions & Solutions

NCERT Solutions for Class 7 Mathematics Chapter 5 Lines and Angles 

  • Introduction

Students are introduced to this chapter with a recap of ray, line, line segment, and angle. A ray has no endpoints.  If it has an endpoint in one direction, we call it a line. A line segment has endpoints on both sides. When two lines or line segments intersect each other, an angle is formed. 

Exercise 5.1 with 14 problems (4 are long answers, and 10 are short answers)

  • Related Angles 

Here, students will learn about different types of relative angles. Students can try to solve problems on each type of related angle so that it gives them knowledge about their properties too. 

Exercise 5.2 with six sums (2 are short answers, 4 are long answers)

  • Complementary Angles 

In this section, students will learn complementary angles, properties, and other kinds. The law is that the sum of two complementary angles is 90 degrees. If an angle between two lines is 90 degrees, we can say one angle is  complementary  to another. 

  • Supplementary Angles 

The next relative angle  is  supplementary angle. Contrary to the complementary aspects, the sum of the two angles is 180 degrees. Then they are called supplementary angles.

  • Adjacent angles 

If any two angles have a common vertex and arm on any one side, they are called adjacent angles. The pages in a book, and the car stereo are the best examples of adjacent angles.

  • Linear Pairs 

In this section, students will come across the concepts of Linear Pairs. A pair of adjacent angles having rays in their non-common sides are nothing but linear pairs. 

  • Vertically Opposite Angles  

The angles formed between the metal blades and handles of a pair of scissors, laundry stands, etc., are the examples given to learn about vertically opposite angles. Students will experiment with pencils to learn more about these angles. When the two lines are intersecting each other, two equal and opposite angles are formed in a vertical direction. 

  • Pairs of Lines 

In this segment, students of Class 7 will learn about lines, its types and properties with examples and explanations. 

  • Intersecting Lines

The first concept that the students will have to learn is intersecting lines. If two lines touch each other or meet at a certain point, those lines are called intersecting lines and the point where they meet is called the point of intersection.

  • Transversal 

It is an extension of intersecting lines. If two or more lines intersect at more than one point it is called a transversal. Students will understand them with examples like railway lines and cross multiple roads. 

  • Angles made by Transversal 

In this section, students can learn the interrelation between lines and angles. When two lines act as transversal, students can observe eight kinds of different angles formed with these lines. All the angles are different from each other. Some are interior, exterior, adjacent angles, and even corresponding angles. 

  • Transversal of Parallel Lines  

In this topic, students will come to know the variations of parallel lines if they are transversal. If transversal cuts two parallel lines, each pair of corresponding angles and alternate interior angles are equal. Also, the pair of interior angles on the transversal side are supplementary to each other. So, a simple intersection can create so many angles and can change their properties. 

  • Checking for Parallel Lines 

So far, students have learned to find angles based on lines. Now, students can try to learn about the lines based on the angles formed, i.e., vice versa of the previous topic. If two lines are transversal and the corresponding angles are equal, then the lines are parallel to each other. 

Key Features of NCERT Class 7 Mathematics Chapter 5 

NCERT Solutions for Class 7 Mathematics Chapter 5 are easily accessible from Extramarks. The key features include:

  • The NCERT Solutions are well explained and provide detailed understanding of the topic by the experienced faculty who meticulously follow the CBSE examination guidelines. 
  • Our solutions are compiled in an easy-to-understand and simple language so as to help students prepare for their exams and by guiding them  to solve the different kinds of problems in a step-by-step manner.
  • It enhances students’ confidence to answer questions in the final exam. 
  • Students can also use these NCERT Solutions to prepare for other competitive exams.

NCERT Solutions for Class 7 Mathematics 

Students who find it difficult to answer the textbook questions can follow the NCERT Solutions provided by Extramarks. NCERT Solutions for Class 7 Mathematics has all the solved answers to the exercises in Class 7 Mathematics. The solutions are given in a   simple and comprehensive manner which makes it easier for students to understand and perform better.

NCERT Solutions for Class 7

Apart from Mathematics, Class 7 students can also find NCERT solutions for other subjects including Science, Social Science, and English on Extramarks official website. Students can use these resources  to prepare more effectively and efficiently. Since most of the final exam questions are based on the same pattern as NCERT questions,  these solutions will come handy before the exams and will guarantee students higher scores.

NCERT Solutions Class 7 Maths Chapter-wise List

Chapter 1 – Integers
Chapter 2 – Fractions and Decimals
Chapter 3 – Data Handling
Chapter 4 – Simple Equations
Chapter 5 – Lines and Angles
Chapter 6 – The Triangle and Its Properties
Chapter 7 – Congruence of Triangles
Chapter 8 – Comparing Quantities
Chapter 9 – Rational Numbers
Chapter 10 – Practical Geometry
Chapter 11 – Perimeter and Area
Chapter 12 – Algebraic Expressions
Chapter 13 – Exponents and Powers
Chapter 14 – Symmetry
Chapter 15 – Visualising Solid Shapes

Q.1

Find the complement of the following angles:

Ans.

(i) 20° Since the sum of complementary angle is 90° So, we have Complement=90°20°= 70° (ii) 63° Since the sum of complementary angle is 90° So, we have Complement=90°63°= 27° (iii) 57° Since the sum of complementary angle is 90° So, we have Complement=90°57°= 33° MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcacaqGGaGaaeOm aiaabcdacqGHWcaSaeaacaqGtbGaaeyAaiaab6gacaqGJbGaaeyzai aabccacaqG0bGaaeiAaiaabwgacaqGGaGaae4CaiaabwhacaqGTbGa aeiiaiaab+gacaqGMbGaaeiiaiaabogacaqGVbGaaeyBaiaabchaca qGSbGaaeyzaiaab2gacaqGLbGaaeOBaiaabshacaqGHbGaaeOCaiaa bMhacaqGGaGaaeyyaiaab6gacaqGNbGaaeiBaiaabwgacaqGGaGaae yAaiaabohacaqGGaGaaeyoaiaabcdacqGHWcaSaeaacaqGtbGaae4B aiaabYcacaqGGaGaae4DaiaabwgacaqGGaGaaeiAaiaabggacaqG2b GaaeyzaaqaaiaaboeacaqGVbGaaeyBaiaabchacaqGSbGaaeyzaiaa b2gacaqGLbGaaeOBaiaabshacqGH9aqpcaqG5aGaaeimaiabgclaWk abgkHiTiaaikdacaaIWaGaeyiSaaRaeyypa0ZaauIhaeaacaaI3aGa aGimaiabgclaWcaaaeaacaqGOaGaaeyAaiaabMgacaqGPaGaaeiiai aabAdacaqGZaGaeyiSaalabaGaae4uaiaabMgacaqGUbGaae4yaiaa bwgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabohacaqG1bGaae yBaiaabccacaqGVbGaaeOzaiaabccacaqGJbGaae4Baiaab2gacaqG WbGaaeiBaiaabwgacaqGTbGaaeyzaiaab6gacaqG0bGaaeyyaiaabk hacaqG5bGaaeiiaiaabggacaqGUbGaae4zaiaabYgacaqGLbGaaeii aiaabMgacaqGZbGaaeiiaiaabMdacaqGWaGaeyiSaalabaGaae4uai aab+gacaqGSaGaaeiiaiaabEhacaqGLbGaaeiiaiaabIgacaqGHbGa aeODaiaabwgaaeaacaqGdbGaae4Baiaab2gacaqGWbGaaeiBaiaabw gacaqGTbGaaeyzaiaab6gacaqG0bGaeyypa0JaaeyoaiaabcdacqGH WcaScqGHsislcaaI2aGaaG4maiabgclaWkabg2da9maaL4babaGaaG OmaiaaiEdacqGHWcaSaaaabaGaaeikaiaabMgacaqGPbGaaeyAaiaa bMcacaqGGaGaaeynaiaabEdacqGHWcaSaeaacaqGtbGaaeyAaiaab6 gacaqGJbGaaeyzaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaae4C aiaabwhacaqGTbGaaeiiaiaab+gacaqGMbGaaeiiaiaabogacaqGVb GaaeyBaiaabchacaqGSbGaaeyzaiaab2gacaqGLbGaaeOBaiaabsha caqGHbGaaeOCaiaabMhacaqGGaGaaeyyaiaab6gacaqGNbGaaeiBai aabwgacaqGGaGaaeyAaiaabohacaqGGaGaaeyoaiaabcdacqGHWcaS aeaacaqGtbGaae4BaiaabYcacaqGGaGaae4DaiaabwgacaqGGaGaae iAaiaabggacaqG2bGaaeyzaaqaaiaaboeacaqGVbGaaeyBaiaabcha caqGSbGaaeyzaiaab2gacaqGLbGaaeOBaiaabshacqGH9aqpcaqG5a GaaeimaiabgclaWkabgkHiTiaaiwdacaaI3aGaeyiSaaRaeyypa0Za auIhaeaacaaIZaGaaG4maiabgclaWcaaaaaa@25ED@

Q.2 Find the supplement of the following angles:

Ans.

(i) 105° Since the sum of supplementary angle is 180° So, we have Complement=180°105°= 75° (ii) 87° Since the sum of Supplementary angle is 180° So, we have Complement=180°87°= 93° (iii) 154° Since the sum of Supplementary angle is 180° So, we have Complement=180°154°= 26° MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcacaqGGaGaaeym aiaabcdacaqG1aGaeyiSaalabaGaae4uaiaabMgacaqGUbGaae4yai aabwgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabohacaqG1bGa aeyBaiaabccacaqGVbGaaeOzaiaabccacaqGZbGaaeyDaiaabchaca qGWbGaaeiBaiaabwgacaqGTbGaaeyzaiaab6gacaqG0bGaaeyyaiaa bkhacaqG5bGaaeiiaiaabggacaqGUbGaae4zaiaabYgacaqGLbGaae iiaiaabMgacaqGZbGaaeiiaiaabgdacaqG4aGaaeimaiabgclaWcqa aiaabofacaqGVbGaaeilaiaabccacaqG3bGaaeyzaiaabccacaqGOb GaaeyyaiaabAhacaqGLbaabaGaae4qaiaab+gacaqGTbGaaeiCaiaa bYgacaqGLbGaaeyBaiaabwgacaqGUbGaaeiDaiabg2da9iaabgdaca qG4aGaaeimaiabgclaWkabgkHiTiaaigdacaaIWaGaaGynaiabgcla Wkabg2da9maaL4babaGaaG4naiaaiwdacqGHWcaSaaaabaGaaeikai aabMgacaqGPbGaaeykaiaabccacaqG4aGaae4naiabgclaWcqaaiaa bofacaqGPbGaaeOBaiaabogacaqGLbGaaeiiaiaabshacaqGObGaae yzaiaabccacaqGZbGaaeyDaiaab2gacaqGGaGaae4BaiaabAgacaqG GaGaae4uaiaabwhacaqGWbGaaeiCaiaabYgacaqGLbGaaeyBaiaabw gacaqGUbGaaeiDaiaabggacaqGYbGaaeyEaiaabccacaqGHbGaaeOB aiaabEgacaqGSbGaaeyzaiaabccacaqGPbGaae4CaiaabccacaqGXa GaaeioaiaabcdacqGHWcaSaeaacaqGtbGaae4BaiaabYcacaqGGaGa ae4DaiaabwgacaqGGaGaaeiAaiaabggacaqG2bGaaeyzaaqaaiaabo eacaqGVbGaaeyBaiaabchacaqGSbGaaeyzaiaab2gacaqGLbGaaeOB aiaabshacqGH9aqpcaqGXaGaaeioaiaabcdacqGHWcaScqGHsislca aI4aGaaG4naiabgclaWkabg2da9maaL4babaGaaGyoaiaaiodacqGH WcaSaaGaaeikaiaabMgacaqGPbGaaeyAaiaabMcacaqGGaGaaeymai aabwdacaqG0aGaeyiSaalabaGaae4uaiaabMgacaqGUbGaae4yaiaa bwgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabohacaqG1bGaae yBaiaabccacaqGVbGaaeOzaiaabccacaqGtbGaaeyDaiaabchacaqG WbGaaeiBaiaabwgacaqGTbGaaeyzaiaab6gacaqG0bGaaeyyaiaabk hacaqG5bGaaeiiaiaabggacaqGUbGaae4zaiaabYgacaqGLbGaaeii aiaabMgacaqGZbGaaeiiaiaabgdacaqG4aGaaeimaiabgclaWcqaai aabofacaqGVbGaaeilaiaabccacaqG3bGaaeyzaiaabccacaqGObGa aeyyaiaabAhacaqGLbaabaGaae4qaiaab+gacaqGTbGaaeiCaiaabY gacaqGLbGaaeyBaiaabwgacaqGUbGaaeiDaiabg2da9iaabgdacaqG 4aGaaeimaiabgclaWkabgkHiTiaaigdacaaI1aGaaGinaiabgclaWk abg2da9maaL4babaGaaGOmaiaaiAdacqGHWcaSaaaaaaa@2D1D@

Q.3

Identify which of the following pairs of angles arecomplementary and which are supplementary.(i) 65°, 115° (ii) 63°, 27° (iii) 112°, 68°(iv)130°, 50° (v) 45°, 45° (vi) 80°, 10°

Ans.

(i) 65°, 115°Since, the sum of complementary angle is 90°and sum ofsupplementary angle is 180°.So, 65°+115°=180°Therefore, given pair is supplemenatry.(ii) 63°, 27°Since, the sum of complementary angle is 90°and sum ofsupplementary angle is 180°.So, 63°+27°=90°Therefore, given pair is complementary.(iii) 112°, 68°Since, the sum of complementary angle is 90°and sum ofsupplementary angle is 180°.So, 112°+68°=180°Therefore, given pair is supplemenatry.(iv) 130°, 50°Since, the sum of complementary angle is 90°and sum ofsupplementary angle is 180°.So, 130°+50°=180°Therefore, given pair is supplemenatry.(v) 45°, 45°Since, the sum of complementary angle is 90°and sum ofsupplementary angle is 180°.So, 45°+45°=90°Therefore, given pair is complementary.(vi) 80°, 10°Since, the sum of complementary angle is 90°and sum ofsupplementary angle is 180°.So, 80°+10°=90°Therefore, given pair is complementary.

Q.4

Find the angle which is equal to its complement.

Ans.

Let the angle be x.Since, it also equal to its complement.So, complement angle = x.Sum of complementary angle is 90°So, we getx+x=90°2x=90°x=90°2=45°Thus, the angle be 45°.

Q.5

Find the angle which is equal to its supplement.

Ans.

Let the angle be x. Since, it also equal to its supplement.So, supplement angle = x.Sum of supplementary angle is 180°So, we getx+x=180°2x=180°x=180°2=90°Thus, the angle be 90°.

Q.6

In the given figure, 1 and 2 are supplementary angles.If 1 is decreased, what changes should take place in 2 sothat both angles still remain supplementary.

Ans.

Since,1 and 2are supplementary angles. If 1 reduced, then 2 should be increased by the same measure so that this angle remain supplementary. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGtbGaaeyAaiaab6gacaqGJbGaaeyz aiaabYcacaaMe8UaeyiiIaTaaGymaiaabccacaqGHbGaaeOBaiaabs gacaqGGaGaeyiiIaTaaGOmaiaaysW7caqGHbGaaeOCaiaabwgacaqG GaGaae4CaiaabwhacaqGWbGaaeiCaiaabYgacaqGLbGaaeyBaiaabw gacaqGUbGaaeiDaiaabggacaqGYbGaaeyEaiaabccacaqGHbGaaeOB aiaabEgacaqGSbGaaeyzaiaabohacaqGUaaabaGaaeysaiaabAgaca qGGaGaeyiiIaTaaGymaiaaysW7caqGGaGaaeOCaiaabwgacaqGKbGa aeyDaiaabogacaqGLbGaaeizaiaabYcacaqGGaGaaeiDaiaabIgaca qGLbGaaeOBaiaabccacqGHGic0caaIYaGaaeiiaiaabohacaqGObGa ae4BaiaabwhacaqGSbGaaeizaiaabccacaqGIbGaaeyzaiaabccaca qGPbGaaeOBaiaabogacaqGYbGaaeyzaiaabggacaqGZbGaaeyzaiaa bsgacaqGGaGaaeOyaiaabMhacaqGGaGaaeiDaiaabIgacaqGLbGaae iiaiaabohacaqGHbGaaeyBaiaabwgacaqGGaaabaGaaeyBaiaabwga caqGHbGaae4CaiaabwhacaqGYbGaaeyzaiaabccacaqGZbGaae4Bai aabccacaqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabshacaqGObGa aeyAaiaabohacaqGGaGaaeyyaiaab6gacaqGNbGaaeiBaiaabwgaca qGGaGaaeOCaiaabwgacaqGTbGaaeyyaiaabMgacaqGUbGaaeiiaiaa bohacaqG1bGaaeiCaiaabchacaqGSbGaaeyzaiaab2gacaqGLbGaae OBaiaabshacaqGHbGaaeOCaiaabMhacaqGUaaaaaa@C186@

Q.7

Can two angles be supplementary if both of them are(i) acute?(ii) obtuse?(iii) right?

Ans.

(i) No, if both angles are acute, that means that both angles are less than 90°. In that case, their sum can not be equal to 180°.(ii) No, if both angles are obtuse, that means that both angles greater than 90°. In that case, their sum will exceed 180°.(iii) Yes, if both angles are right angles, that is, 90°,then their sum will be exact 180°.

Q.8 An angle is greater than 45°. Is its complementary angle greater than 45° or equal to 45° or less than 45°?

Ans.

Let x and y be two angles having complementary angle pairand x is greater than 45°.Then,x+y=90°y=90°xThus, y will be less than 45°.

Q.9 

In the adjoining figure:(i) Is 1 adjacent to 2?(ii)Is AOC adjacent to AOE?(iii) Do COE andEODform a linear pair?(iv) Are BOD and DOAsupplementary?(v) Is 1 vertically opposite to 4?(v) What is the vertically opposite angle of 5?

Ans.

(i) Yes, since they have a common vetex O and also a common arm OC. Also, their non-common arms, OA and OB are on either side of the common arm. (ii) No. they have a common vertex O and also a common arm OA. However, their non common arms, OC and OE are on the same side of the common arm. Therefore, theses are not adjacent to each other. (iii) Yes, since they have a common vertex O and a common arm OE. Also, their non common arms OC and OD, are opposite rays. (iv) Yes, since BOD and DOA have a common vertex O and their non-common arms opposite to each other. (v) Yes, since these are fromed dure to the intersection of two straight lines (AB and CD) (vi) COB is the vertically opposite angle of 5 as these are formed due to the intersection of two straight lines AB and CD. 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Q.10

Indicate which pairs of angles are:(i) Vertically opposite angles.(ii) Linear pairs.

Ans.

(i) 1 , 4and 5, ( 2+3 ) are vertically opposite angles as these formed due to the intersection of straight lines (ii) 1 and 5, 5and, 4 as these have common vertex and also have non-common arms opposite to each other. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcacaqGGaGaeyii IaTaaGymaiaabccacaqGSaGaaeiiaiabgcIiqlaaisdacaaMe8Uaae yyaiaab6gacaqGKbGaaeiiaiabgcIiqlaaiwdacaGGSaGaaeiiamaa bmaabaGaeyiiIaTaaGOmaiabgUcaRiabgcIiqlaaiodaaiaawIcaca GLPaaacaqGGaGaaeyyaiaabkhacaqGLbGaaeiiaiaabAhacaqGLbGa aeOCaiaabshacaqGPbGaae4yaiaabggacaqGSbGaaeiBaiaabMhaca qGGaGaae4BaiaabchacaqGWbGaae4BaiaabohacaqGPbGaaeiDaiaa bwgacaqGGaGaaeyyaiaab6gacaqGNbGaaeiBaiaabwgacaqGZbaaba GaaeyyaiaabohacaqGGaGaaeiDaiaabIgacaqGLbGaae4Caiaabwga caqGGaGaaeOzaiaab+gacaqGYbGaaeyBaiaabwgacaqGKbGaaeiiai aabsgacaqG1bGaaeyzaiaabccacaqG0bGaae4BaiaabccacaqG0bGa aeiAaiaabwgacaqGGaGaaeyAaiaab6gacaqG0bGaaeyzaiaabkhaca qGZbGaaeyzaiaabogacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabohacaqG0bGaaeOCaiaabggacaqGPbGaae 4zaiaabIgacaqG0bGaaeiiaiaabYgacaqGPbGaaeOBaiaabwgacaqG ZbaabaGaaeikaiaabMgacaqGPbGaaeykaiaabccacqGHGic0caaIXa GaaeiiaiaabggacaqGUbGaaeizaiaabccacqGHGic0caqG1aGaaeil aiaabccacqGHGic0caaI1aGaaGjbVlaabggacaqGUbGaaeizaiaabY cacaqGGaGaeyiiIaTaaGinaiaabccacaqGHbGaae4CaiaabccacaqG 0bGaaeiAaiaabwgacaqGZbGaaeyzaiaabccacaqGObGaaeyyaiaabA hacaqGLbGaaeiiaiaabogacaqGVbGaaeyBaiaab2gacaqGVbGaaeOB aiaabccacaqG2bGaaeyzaiaabkhacaqG0bGaaeyzaiaabIhacaqGGa aabaGaaeyyaiaab6gacaqGKbGaaeiiaiaabggacaqGSbGaae4Caiaa b+gacaqGGaGaaeiAaiaabggacaqG2bGaaeyzaiaabccacaqGUbGaae 4Baiaab6gacaqGTaGaae4yaiaab+gacaqGTbGaaeyBaiaab+gacaqG UbGaaeiiaiaabggacaqGYbGaaeyBaiaabohacaqGGaGaae4Baiaabc hacaqGWbGaae4BaiaabohacaqGPbGaaeiDaiaabwgacaqGGaGaaeiD aiaab+gacaqGGaGaaeyzaiaabggacaqGJbGaaeiAaiaabccacaqGVb GaaeiDaiaabIgacaqGLbGaaeOCaiaab6caaaaa@04E1@

Q.11 In the following figure, is

1

adjacent to

2

? Give reasons.

Ans.

1

and

2

are not adjacent angles because they don’t have common vertex.

Q.12 Find the values of the angles x, y and z in each of the following:

Ans.

(i) Since x and 55° are vertically opposite angles.So, x=55°x°+y°=180°55°+y°=180°y°=180°55°     =125°y°=z°(vertically opposite angles)So,z°=125°(ii) z°=40°(vertically opposite angles)y°+z°=180°(Linear pair)y°=180°40°     =140°40°+x°+25°=180°x°+65°=180°x°=180°65°     =115°

Q.13

Fill in the blanks:(i) If two angles are complementary, then the sum of theirmeasuresis______.(ii) If two angles are supplementary, then the sum of theirmeasuresis_____.(iii) Two angles forming a linear pair are_______________.(iv) If two adjacent angles are supplementary, they form a__________.(v) If two lines intersect at a point, then the vertically oppositeangles are always_____________.(vi) If two lines intersect at a point, and if one pair ofvertically opposite angles are acute angles, then theother pair of vertically opposite angles are__________.

Ans.

Fill in the blanks: ( i ) If two angles are complementary, then the sum of their measures is 90° . ( ii ) If two angles are supplementary, then the sum of their measures is 180° . ( iii ) Two angles forming a linear pair are supplementary . ( iv ) If two adjacent angles are supplementary, they form a Linear pair . ( v ) If two lines intersect at a point, then the vertically opposite angles are always equal . ( vi ) If two lines intersect at a point, and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are obtuse angles . 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Q.14

In the adjoining figure, name the following pairs of angle.(i) Obtuse vertically opposite angles(ii) Adjacent complementary angles(iii) Equal supplementary angles(iv) Unequal supplementary angles(v) Adjacent angles that do not form a linear pair.

Ans.

(i) AOD,BOC(ii) EOA,AOB(iii)EOB,EOD(iv)EOA,EOC(v)AOBandAOE,AOE  andEOD,EODandCOD.

Q.15

State the property that isused in each of the following statements?(i) If a b, then 1 = 5.(ii) If 4 = 6, then ab.(iii) If 4+5=180°, then ab.

Ans.

(i) Corresponding angles property(ii) Alternate interior angles property(ii) Interior angles on the same side of transversal are supplementary.

Q.16 In the adjoining figure, identify
(i) the pairs of corresponding angles.
(ii) the pairs of alternate interior angles.
(iii) the pairs of interior angles on the same side of the transversal.
(
iv)
the vertically opposite angles.

Ans.

(i) 1 and 5, 2 and 6, 3 and 7, 4 and 8.(ii) 2and8,3and5(iii) 2and5,3and8(iv) 1and3,2and4,5and7,6and8.

Q.17

In the adjoining figure, pq. Find the unknown angles.

Ans.

d=125°(Corresponding angles)e=180°125°=55°(Linear pair)f=e=55°(Vertically opposite angles)c=f=55°(Corresponding angles)a=e=55°(Corresponding angles)b=d=125°(Vertically opposite angles)

Q.18 Find the value of x in each of the following figures if

l m

.

Ans.

(i)y=110° (corresponding angles)x+y=180°(linear pair)So,x=180°110°=70°(ii)x=100°(corresponding angles)

Q.19

In the given figure, the arms of two angles are parallel.IfABC=70°, then find(i) DGC(ii) DEF

Ans.

(i) Consdier that ABDG and a transerval line BC isintersecting them.DGC=ABC (Corresponding angles)SO,DGC=70°(ii) Consdier that BCEF and a transerval line DE isintersecting them.DEF=DGC (Corresponding angles)DEF=70°

Q.20 In the given figures below, decide whether l is parallel to m.

Ans.

(i)Consider two lines, l and m, and a transversal line n which is intersecting them. Sum of the interior angles on the same sideof transersal=126°+44°=170°.As the sum of interior angles on the same side of transervsal is not 180°, Therefore l is not parallel to m.(ii) x+75°=180° (linear pair)x=180°75°=105°For l and m to be parallel to each other, correspondingangles (ABC and x) should be equal. However, here theyare 75° and 105°.Hence these lines are not parallel to each other.(iii) x+123°=180°x=180°123°=57°For l and m to be parallel to each other, correspondingangles (ABC and x) should be equal. However, here theyare 57° and 57°.Hence these lines are parallel to each other.(iv) 98°+x=180°x=180°98°=82°For l and m to be parallel to each other, correspondingangles (ABC and x) should be equal. However, here theyare 72° and 82°.Hence these lines are not parallel to each other

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The most important topics covered in NCERT Solutions for Class 7 Mathematics Chapter 5 are Relative Angles and their properties, Complementary Angles, Supplementary Angles, Adjacent Angles, Linear Pairs, Vertically Opposite Angles, Pairs of Lines, Intersecting Lines, Transversal, the Angle made by a Transversal, a Transversal of Parallel lines, and Checking for Parallel Lines.

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