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
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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.
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-1](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-1.jpeg)
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 = 360o โ 110o = 250o
2. In Fig. 6.14, lines XY and MN intersect at O. If POY = 90ยฐ and a : b = 2 : 3, find c.
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-2](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-2.png)
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.
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-3](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-3.png)
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.
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-4](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-4.png)
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).
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-5](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-5.jpeg)
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:
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-6](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-6.png)
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.
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-7](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-7.jpeg)
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.
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-8](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-8.png)
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.
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-9](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-9.png)
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.]
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-10](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-10.png)
Solution:
First, construct a line XY parallel to PQ.
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-11](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-11.png)
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.
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-12](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-12.jpeg)
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.
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-13](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-13.jpeg)
Solution:
First, draw two lines BE and CF such that BE โฅ PQ and CF โฅ RS.
Now, since PQ RS,
So, BE CF
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-14](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-14.jpeg)
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.
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-15](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-15.png)
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.
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-16](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-16.jpeg)
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.
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-17](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-17.jpeg)
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.
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-18](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-18.jpeg)
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.
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-19](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-19.png)
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 = 1800
900 + 370 + y = 1800
y = 1800 โ 1270 = 530
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.
![Ncert solutions class 9 chapter 6 Ncert solutions class 9 chapter 6-20](https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-6-20.png)
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 โNCERT 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.