Most of the time students don’t ask their doubts in class due to shyness and fear of the teacher. Due to this, they don’t get to know how the questions are being solved by the teacher. This happens most often in Maths class. In other subjects, students can still find the answer if they re-read the chapter. But for mathematics, if they miss the concepts, then they would not be able to solve it. So, to help students here, we have provided the detailed NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles Exercise 6.1.
Students can download the NCERT Solutions of Class 9 Maths Chapter 6 Exercise 6.1 in PDF format by clicking on the link below. These solutions are prepared by subject experts and are described in detail. By going through NCERT Solutions PDF, students can easily understand the method of solving the questions.
Access Answers to NCERT Solutions for Class 9 Maths Chapter 6 – Lines And Angles Exercise 6.1
Access other Exercise Solutions of Class 9 Maths Chapter 6 – Lines and Angles
The NCERT Class 9 Chapter 6 Lines and Angles has two more exercises. Students can access the solutions of these exercises by visiting the link below.
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)
NCERT Solutions for Class 9 Maths Chapter 6 – Lines and Angles Exercise 6.1
1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC +∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
From the diagram, we have
(∠AOC +∠BOE +∠COE) and (∠COE +∠BOD +∠BOE) forms a straight line.
So, ∠AOC+∠BOE +∠COE = ∠COE +∠BOD+∠BOE = 180°
Now, by putting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° we get
∠COE = 110° and ∠BOE = 30°
So, reflex ∠COE = 360o – 110o = 250o
2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
We know that the sum of linear pair are always equal to 180°
∠POY +a +b = 180°
Putting the value of ∠POY = 90° (as given in the question) we get,
a+b = 90°
Now, it is given that a : b = 2 : 3 so,
Let a be 2x and b be 3x
∴ 2x+3x = 90°
Solving this we get
5x = 90°
So, x = 18°
∴ a = 2×18° = 36°
Similarly, b can be calculated and the value will be
b = 3×18° = 54°
From the diagram, b+c also forms a straight angle so,
b+c = 180°
c+54° = 180°
∴ c = 126°
3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Since ST is a straight line so,
∠PQS+∠PQR = 180° (linear pair) and
∠PRT+∠PRQ = 180° (linear pair)
Now, ∠PQS + ∠PQR = ∠PRT+∠PRQ = 180°
Since ∠PQR =∠PRQ (as given in the question)
∠PQS = ∠PRT. (Hence proved).
4. In Fig. 6.16, if x+y = w+z, then prove that AOB is a line.
For proving AOB is a straight line, we will have to prove x+y is a linear pair
i.e. x+y = 180°
We know that the angles around a point are 360° so,
x+y+w+z = 360°
In the question, it is given that,
x+y = w+z
So, (x+y)+(x+y) = 360°
2(x+y) = 360°
∴ (x+y) = 180° (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 = ½ (∠QOS – ∠POS).
In the question, it is given that (OR ⊥ PQ) and ∠POQ = 180°
So, ∠POS+∠ROS+∠ROQ = 180°
Now, ∠POS+∠ROS = 180°- 90° (Since ∠POR = ∠ROQ = 90°)
∴ ∠POS + ∠ROS = 90°
Now, ∠QOS = ∠ROQ+∠ROS
It is given that ∠ROQ = 90°,
∴ ∠QOS = 90° +∠ROS
Or, ∠QOS – ∠ROS = 90°
As ∠POS + ∠ROS = 90° and ∠QOS – ∠ROS = 90°, we get
∠POS + ∠ROS = ∠QOS – ∠ROS
2 ∠ROS + ∠POS = ∠QOS
Or, ∠ROS = ½ (∠QOS – ∠POS) (Hence proved).
6. It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.
Here, XP is a straight line
So, ∠XYZ +∠ZYP = 180°
Putting the value of ∠XYZ = 64° we get,
64° +∠ZYP = 180°
∴ ∠ZYP = 116°
From the diagram, we also know that ∠ZYP = ∠ZYQ + ∠QYP
Now, as YQ bisects ∠ZYP,
∠ZYQ = ∠QYP
Or, ∠ZYP = 2∠ZYQ
∴ ∠ZYQ = ∠QYP = 58°
Again, ∠XYQ = ∠XYZ + ∠ZYQ
By putting the value of ∠XYZ = 64° and ∠ZYQ = 58° we get.
∠XYQ = 64°+58°
Or, ∠XYQ = 122°
Now, reflex ∠QYP = 180°+XYQ
We computed that the value of ∠XYQ = 122°.
∠QYP = 180°+122°
∴ ∠QYP = 302°
This exercise will help students recall the types of angles (acute, obtuse, right, straight, reflex, complementary and supplementary) that they have studied in earlier classes. The new concepts in the exercise will start from the intersecting and non-intersecting lines, and then Pairs of Angles are explained in detail along with a theorem proof, as well as some examples in the NCERT textbook.
The questions in the exercise are based on finding the value of angles from the given diagram and proving the statements. In the NCERT Class 9 Maths Solutions, we have provided the step by step solutions along with the diagram for students convenience. Also, the questions are solved by the easiest method, so students can easily understand them.
We hope this information on “NCERT Solution for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.1” is useful for students. Stay tuned for further updates on CBSE and other competitive exams. To access interactive Maths and Science Videos download the BYJU’S App and subscribe to YouTube Channel.