# NCERT Solutions for Class 9 Maths Chapter 6 – Lines And Angles

## NCERT Solutions Class 9 Maths Chapter 6 โ Free PDF Download

**NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles** deals with the questions and answers related to the chapter Lines and Angles. This topic introduces you to basic Geometry primarily focusing on the properties of the angles formed i) when two lines intersect each other and ii) when a line intersects two or more parallel lines at distinct points.

Now, you must be wondering why we are studying Lines and Angles. What are the real-life applications of it? The answer is that lines and angles are everywhere around us. The architecture uses lines and angles to design the structure of a building. When you stop at a signal and then move on when the signal light is green, then you either take a left angle turn or right-angle turn or move in a straight line. When you have to find the height of a tower or the location of an aircraft, then you need to know angles. In NCERT Solutions for Class 9 Maths Chapter 6, you will learn to solve the questions related to all the concepts of Lines and Angles. Students can avail a complete and free PDF of this chapterโs **NCERT Solutions for Class 9 Maths** from the link given below.Chapter 6 โ Lines And Angles

### Download PDF of NCERT Solutions for Class 9 Maths Chapter 6 โ Lines And Angles

### Access Answers of Maths NCERT Class 9 Maths Chapter 6 โ Lines And Angles

**List of Exercises in class 9 Maths Chapter 6**

Exercise 6.1 Solutions 6 Questions (5 Short Answer Questions, 1 Long Answer Question)

Exercise 6.2 Solutions 6 Questions (3 Short Answer Questions, 3 Long Answer Question)

Exercise 6.3 Solutions 6 Questions (5 Short Answer Questions, 1 Long Answer Question)

**Class 9 Maths Chapter 6 Exercise: 6.1 (Page No: 96)**

**Exercise: 6.1 (Page No: 96)**

**1. In Fig. 6.13, lines AB and CD intersect at O. If AOC +BOE = 70ยฐ and BOD = 40ยฐ, find BOE and reflex COE.**

**Solution:**

From the diagram, we have

(โ AOC +โ BOE +โ COE) and (โ COE +โ BOD +โ BOE) forms a straight line.

So, โ AOC+โ BOE +โ COE = โ COE +โ BOD+โ BOE = 180ยฐ

Now, by putting the values of โ AOC + โ BOE = 70ยฐ and โ BOD = 40ยฐ we get

โ COE = 110ยฐ and โ BOE = 30ยฐ

So, reflex โ COE = 360^{o} โ 110^{o} = 250^{o}

**2. In Fig. 6.14, lines XY and MN intersect at O. If POY = 90ยฐ and a : b = 2 : 3, find c.**

**Solution:**

We know that the sum of linear pair are always equal to 180ยฐ

So,

POY +a +b = 180ยฐ

Putting the value of POY = 90ยฐ (as given in the question) we get,

a+b = 90ยฐ

Now, it is given that a : b = 2 : 3 so,

Let a be 2x and b be 3x

โด 2x+3x = 90ยฐ

Solving this we get

5x = 90ยฐ

So, x = 18ยฐ

โด a = 2ร18ยฐ = 36ยฐ

Similarly, b can be calculated and the value will be

b = 3ร18ยฐ = 54ยฐ

From the diagram, b+c also forms a straight angle so,

b+c = 180ยฐ

c+54ยฐ = 180ยฐ

โด c = 126ยฐ

**3. In Fig. 6.15, PQR = PRQ, then prove that PQS = PRT.**

**Solution:**

Since ST is a straight line so,

**โ **PQS+**โ **PQR = 180ยฐ (linear pair) and

**โ **PRT+**โ **PRQ = 180ยฐ (linear pair)

Now, **โ **PQS + **โ **PQR = **โ **PRT+**โ **PRQ = 180ยฐ

Since **โ **PQR =**โ **PRQ (as given in the question)

**โ **PQS = **โ **PRT. (Hence proved).

**4. In Fig. 6.16, if x+y = w+z, then prove that AOB is a line.**

**Solution:**

For proving AOB is a straight line, we will have to prove x+y is a linear pair

i.e. x+y = 180ยฐ

We know that the angles around a point are 360ยฐ so,

x+y+w+z = 360ยฐ

In the question, it is given that,

x+y = w+z

So, (x+y)+(x+y) = 360ยฐ

2(x+y) = 360ยฐ

โด (x+y) = 180ยฐ (Hence proved).

**5. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ROS = ยฝ (QOS โ POS).**

**Solution:**

In the question, it is given that (OR โฅ PQ) and POQ = 180ยฐ

So, POS+ROS+ROQ = 180ยฐ

Now, POS+ROS = 180ยฐ- 90ยฐ (Since POR = ROQ = 90ยฐ)

โด POS + ROS = 90ยฐ

Now, QOS = ROQ+ROS

It is given that ROQ = 90ยฐ,

โด QOS = 90ยฐ +ROS

Or, QOS โ ROS = 90ยฐ

As POS + ROS = 90ยฐ and QOS โ ROS = 90ยฐ, we get

POS + ROS = QOS โ ROS

2 ROS + POS = QOS

Or, ROS = ยฝ (QOS โ POS) (Hence proved).

**6. It is given that XYZ = 64ยฐ and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ZYP, find XYQ and reflex QYP.**

**Solution:**

Here, XP is a straight line

So, XYZ +ZYP = 180ยฐ

Putting the value of XYZ = 64ยฐ we get,

64ยฐ +ZYP = 180ยฐ

โด ZYP = 116ยฐ

From the diagram, we also know that ZYP = ZYQ + QYP

Now, as YQ bisects ZYP,

ZYQ = QYP

Or, ZYP = 2ZYQ

โด ZYQ = QYP = 58ยฐ

Again, XYQ = XYZ + ZYQ

By putting the value of XYZ = 64ยฐ and ZYQ = 58ยฐ we get.

XYQ = 64ยฐ+58ยฐ

Or, XYQ = 122ยฐ

Now, reflex QYP = 180ยฐ+XYQ

We computed that the value of XYQ = 122ยฐ.

So,

QYP = 180ยฐ+122ยฐ

โด QYP = 302ยฐ

**Exercise: 6.2 (Page No: 103)**

**1. In Fig. 6.28, find the values of x and y and then show that AB CD.**

**Solution:**

We know that a linear pair is equal to 180ยฐ.

So, x+50ยฐ = 180ยฐ

โด x = 130ยฐ

We also know that vertically opposite angles are equal.

So, y = 130ยฐ

In two parallel lines, the alternate interior angles are equal. In this,

x = y = 130ยฐ

This proves that alternate interior angles are equal and so, AB CD.

**2. In Fig. 6.29, if AB CD, CD EF and y : z = 3 : 7, find x.**

**Solution:**

It is known that AB CD and CDEF

As the angles on the same side of a transversal line sums up to 180ยฐ,

x + y = 180ยฐ โโ(i)

Also,

O = z (Since they are corresponding angles)

and, y +O = 180ยฐ (Since they are a linear pair)

So, y+z = 180ยฐ

Now, let y = 3w and hence, z = 7w (As y : z = 3 : 7)

โด 3w+7w = 180ยฐ

Or, 10 w = 180ยฐ

So, w = 18ยฐ

Now, y = 3ร18ยฐ = 54ยฐ

and, z = 7ร18ยฐ = 126ยฐ

Now, angle x can be calculated from equation (i)

x+y = 180ยฐ

Or, x+54ยฐ = 180ยฐ

โด x = 126ยฐ

**3. In Fig. 6.30, if AB CD, EF โฅ CD and GED = 126ยฐ, find AGE, GEF and FGE.**

**Solution:**

Since AB CD, GE is a transversal.

It is given that GED = 126ยฐ

So, GED = AGE = 126ยฐ (As they are alternate interior angles)

Also,

GED = GEF +FED

As EFโฅ CD, FED = 90ยฐ

โด GED = GEF+90ยฐ

Or, GEF = 126ยฐ โ 90ยฐ = 36ยฐ

Again, FGE +GED = 180ยฐ (Transversal)

Putting the value of GED = 126ยฐ we get,

FGE = 54ยฐ

So,

AGE = 126ยฐ

GEF = 36ยฐ and

FGE = 54ยฐ

**4. In Fig. 6.31, if PQ ST, PQR = 110ยฐ and RST = 130ยฐ, find QRS.**

**[Hint : Draw a line parallel to ST through point R.]**

**Solution:**

First, construct a line XY parallel to PQ.

We know that the angles on the same side of transversal is equal to 180ยฐ.

So, PQR+QRX = 180ยฐ

Or,QRX = 180ยฐ-110ยฐ

โด QRX = 70ยฐ

Similarly,

RST +SRY = 180ยฐ

Or, SRY = 180ยฐ- 130ยฐ

โด SRY = 50ยฐ

Now, for the linear pairs on the line XY-

QRX+QRS+SRY = 180ยฐ

Putting their respective values, we get,

QRS = 180ยฐ โ 70ยฐ โ 50ยฐ

Hence, QRS = 60ยฐ

**5. In Fig. 6.32, if AB CD, APQ = 50ยฐ and PRD = 127ยฐ, find x and y.**

**Solution:**

From the diagram,

APQ = PQR (Alternate interior angles)

Now, putting the value of APQ = 50ยฐ and PQR = x we get,

x = 50ยฐ

Also,

APR = PRD (Alternate interior angles)

Or, APR = 127ยฐ (As it is given that PRD = 127ยฐ)

We know that

APR = APQ+QPR

Now, putting values of QPR = y and APR = 127ยฐ we get,

127ยฐ = 50ยฐ+ y

Or, y = 77ยฐ

Thus, the values of x and y are calculated as:

x = 50ยฐ and y = 77ยฐ

**6. In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB CD.**

**Solution:**

First, draw two lines BE and CF such that BE โฅ PQ and CF โฅ RS.

Now, since PQ RS,

So, BE CF

We know that,

Angle of incidence = Angle of reflection (By the law of reflection)

So,

1 = 2 and

3 = 4

We also know that alternate interior angles are equal. Here, BE โฅ CF and the transversal line BC cuts them at B and C

So, 2 = 3 (As they are alternate interior angles)

Now, 1 +2 = 3 +4

Or, ABC = DCB

So, AB CD alternate interior angles are equal)

**Exercise: 6.3 (Page No: 107)**

**1. In Fig. 6.39, sides QP and RQ of ฮPQR are produced to points S and T respectively. If SPR = 135ยฐ and PQT = 110ยฐ, find PRQ.**

**Solution:**

It is given the TQR is a straight line and so, the linear pairs (i.e. TQP and PQR) will add up to 180ยฐ

So, TQP +PQR = 180ยฐ

Now, putting the value of TQP = 110ยฐ we get,

PQR = 70ยฐ

Consider the ฮPQR,

Here, the side QP is extended to S and so, SPR forms the exterior angle.

Thus, SPR (SPR = 135ยฐ) is equal to the sum of interior opposite angles. (Triangle property)

Or, PQR +PRQ = 135ยฐ

Now, putting the value of PQR = 70ยฐ we get,

PRQ = 135ยฐ-70ยฐ

Hence, PRQ = 65ยฐ

**2. In Fig. 6.40, X = 62ยฐ, XYZ = 54ยฐ. If YO and ZO are the bisectors of XYZ and XZY respectively of ฮ XYZ, find OZY and YOZ.**

**Solution:**

We know that the sum of the interior angles of the triangle.

So, X +XYZ +XZY = 180ยฐ

Putting the values as given in the question we get,

62ยฐ+54ยฐ +XZY = 180ยฐ

Or, XZY = 64ยฐ

Now, we know that ZO is the bisector so,

OZY = ยฝ XZY

โด OZY = 32ยฐ

Similarly, YO is a bisector and so,

OYZ = ยฝ XYZ

Or, OYZ = 27ยฐ (As XYZ = 54ยฐ)

Now, as the sum of the interior angles of the triangle,

OZY +OYZ +O = 180ยฐ

Putting their respective values, we get,

O = 180ยฐ-32ยฐ-27ยฐ

Hence, O = 121ยฐ

**3. In Fig. 6.41, if AB DE, BAC = 35ยฐ and CDE = 53ยฐ, find DCE.**

**Solution:**

We know that AE is a transversal since AB DE

Here BAC and AED are alternate interior angles.

Hence, BAC = AED

It is given that BAC = 35ยฐ

AED = 35ยฐ

Now consider the triangle CDE. We know that the sum of the interior angles of a triangle is 180ยฐ.

โด DCE+CED+CDE = 180ยฐ

Putting the values, we get

DCE+35ยฐ+53ยฐ = 180ยฐ

Hence, DCE = 92ยฐ

**4. In Fig. 6.42, if lines PQ and RS intersect at point T, such that PRT = 40ยฐ, RPT = 95ยฐ and TSQ = 75ยฐ, find SQT.**

**Solution:**

Consider triangle PRT.

PRT +RPT + PTR = 180ยฐ

So, PTR = 45ยฐ

Now PTR will be equal to STQ as they are vertically opposite angles.

So, PTR = STQ = 45ยฐ

Again, in triangle STQ,

TSQ +PTR + SQT = 180ยฐ

Solving this we get,

SQT = 60ยฐ

**5. In Fig. 6.43, if PQ โฅ PS, PQ SR, SQR = 28ยฐ and QRT = 65ยฐ, then find the values of x and y.**

**Solution:**

x +SQR = QRT (As they are alternate angles since QR is transversal)

So, x+28ยฐ = 65ยฐ

โด x = 37ยฐ

It is also known that alternate interior angles are same and so,

QSR = x = 37ยฐ

Also, Now,

QRS +QRT = 180ยฐ (As they are a Linear pair)

Or, QRS+65ยฐ = 180ยฐ

So, QRS = 115ยฐ

Now, we know that the sum of the angles in a quadrilateral is 360ยฐ. So,

P +Q+R+S = 360ยฐ

Putting their respective values, we get,

S = 360ยฐ-90ยฐ-65ยฐ-115ยฐ

In ฮ SPQ

โ SPQ + x + y = 180^{0}

90^{0} + 37^{0} + y = 180^{0}

y = 180^{0} โ 127^{0} = 53^{0}

Hence, y = 53ยฐ

**6. In Fig. 6.44, the side QR of ฮPQR is produced to a point S. If the bisectors of PQR and PRS meet at point T, then prove that QTR = ยฝ QPR.**

**Solution:**

Consider the ฮPQR. PRS is the exterior angle and QPR and PQR are interior angles.

So, PRS = QPR+PQR (According to triangle property)

Or, PRS -PQR = QPR โโโโ(i)

Now, consider the ฮQRT,

TRS = TQR+QTR

Or, QTR = TRS-TQR

We know that QT and RT bisect PQR and PRS respectively.

So, PRS = 2 TRS and PQR = 2TQR

Now, QTR = ยฝ PRS โ ยฝPQR

Or, QTR = ยฝ (PRS -PQR)

From (i) we know that PRS -PQR = QPR

So, QTR = ยฝ QPR (hence proved).

### NCERT Solutions for Class 9 Maths Chapter 6 โ Lines And Angles

The Class 9 Maths theory paper is of 80 marks. Out of which Geometry constitute a total of 22 marks which includes Introduction to Euclidโs Geometry, Lines and Angles, Triangles, Quadrilaterals, Areas, Circles, Constructions. As you can see that it constitutes approximately 27% of weightage. So, using the **NCERT Solutions for Class 9**, students can easily score high marks if by having a thorough understanding of this topic.

Solve all the exercise problems of Lines and Angles. Refer to the NCERT Solutions of Class 9 provided by our Experts below. It will help you to solve the questions in an easy way.

### Summary of NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles

Before starting to solve the exercise problems, you must first read the theory part and get to know the basic terms, definitions and theorems. After that go through the solved examples of Lines and Angles that are given in the **Class 9 NCERT Book**. It will make your concepts more clear. Then you can start solving the exercise problems with the help of NCERT Solutions.

**Key Points of NCERT Class 9 Maths Chapter 6 โ Lines And Angles**

After solving the Line and Angles chapter of Class 9 Maths, you will get to know the following points:

- Basic terms and definitions related to a line segment, ray, collinear points, non-collinear points, intersecting and non-intersecting lines.
- Pair of angles (reflex, complementary, supplementary, adjacent, vertical opposite, linear pair).
- Parallel and Transversal Lines and theorems related to them.
- Angle sum property of a triangle.

We hope this information on โ**NCERT Solutions for Class 9 Maths Chapter 6** Lines and Anglesโ is useful for students. Stay tuned for further updates on CBSE and other competitive exams. To access interactive Maths and Science Videos download BYJUโS App and subscribe to YouTube Channel.

## Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 6

### How NCERT Solutions for Class 9 Maths Chapter 6 is helpful for exam preparation?

NCERT Solutions for Class 9 Maths Chapter 6 are created by the BYJUโS expert faculty to help students in the preparation of their examinations. These expert faculty solve and provide the NCERT Solution for Class 9, which would help students to solve the problems comfortably. They give a detailed and stepwise explanation to the problems given in the exercises in the NCERT Solutions for Class 9. These solutions help students prepare for their upcoming Board Exams by covering the whole syllabus, in accordance with the NCERT guidelines.

### How many exercises are present in NCERT Solutions for Class 9 Maths Chapter 6?

There are 3 exercises present in NCERT Solutions for Class 9 Maths Chapter 6. Viz,

Exercise 6.1 Solutions 6 Questions (5 Short Answer Questions, 1 Long Answer Question)

Exercise 6.2 Solutions 6 Questions (3 Short Answer Questions, 3 Long Answer Question)

Exercise 6.3 Solutions 6 Questions (5 Short Answer Questions, 1 Long Answer Question)

### Is BYJUโS providing answers for all questions present in NCERT Solutions for Class 9 Maths Chapter 6?

NCERT Solutions for Class 9 Maths Chapter 6 are useful for students as it helps them to score well in the class exams. We, in our aim to help students, have devised detailed chapter wise solutions for them to understand the concepts easily. We followed the latest Syllabus, while creating the NCERT Solutions and it is framed in accordance with the exam pattern of the CBSE Board. These solutions are designed by subject matter experts who have assembled model questions covering all the exercise questions from the textbook.