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6. Lines and Angles Mathematics class 9 in English Medium ncert book solutions Exercise 6.1
6. Lines and Angles Exercise 6.1 – Complete NCERT Book Solutions for Class 9 Mathematics (English Medium). Get all chapter explanations, extra questions, solved examples and additional practice questions for 6. Lines and Angles Exercise 6.1 to help you master concepts and score higher.
6. Lines and Angles Mathematics class 9 in English Medium ncert book solutions Exercise 6.1
NCERT Solutions for Class 9 Mathematics play an important role in helping students understand the concepts of the chapter 6. Lines and Angles clearly. This chapter includes the topic Exercise 6.1, which is essential from both academic and examination point of view. The solutions provided here are prepared strictly according to the latest NCERT syllabus and follow the guidelines of CBSE to ensure accuracy and relevance. Each question is explained in a simple and student-friendly manner so that learners can grasp the concepts without confusion. These NCERT Solutions are useful for regular study, homework help, and exam preparation. All textbook questions are solved step by step to improve problem-solving skills and conceptual clarity. Students of Class 9 studying Mathematics can use these solutions to revise important topics, understand difficult questions, and practise effectively before examinations. The chapter 6. Lines and Angles is explained in a structured way, making it easier for students to connect the theory with the topic Exercise 6.1. By studying these updated NCERT Solutions for Class 9 Mathematics, students can build a strong foundation, boost their confidence, and score better marks in school and board exams.
6. Lines and Angles
Exercise 6.1
Exercise 6.1
Q1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠ BOE = 70° and ∠BOD = 40° find ∠BOE and reflex ∠COE.
Solution:
∠BOD = 40°
∠AOC = ∠BOD (Vertically opposite Angle)
∠AOC = 40°
∠AOC + ∠ BOE = 70° (Given)
∠BOE = 70°
∠BOE = 70° - 40°
∠BOE = 30°
AOB is straight line
∠AOC + ∠COE +∠BOE = 180° (linear pair)
⇒ 70° + ∠COE = 180°
⇒ ∠COE = 180° - 70°
⇒ ∠COE = 110°
Reflex ∠COE = 360 - 110°
= 250°
Q2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
Solution:
∠POY=90° (given)
Let ∠a and ∠b = 2x and 3x
XOY is a straight line
∠a + ∠b + ∠POY = 180°
⇒2x + 3x + 90°= 180°
⇒5x = 180° - 90°
⇒5x = 90°
⇒x = 90°/5
⇒x = 18°
Now ∠a = 2 x 18°
= 36°
∠b =3 x 18°
= 54°
MON is a straight line
∠b + ∠c = 180°(linear pair)
∠54° + ∠c = 180°
⇒∠c = 180°- 54°
=126°
Q3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT
Solution :
Given : ∠PQR = ∠PRQ
To prove : ∠PQS = ∠PRT
Proof :
∠PQS + ∠PQR = 180° .................. (1) Linear pair
∠PRT + ∠PRQ = 180° .................. (2) Linear pair
From equation (1) and (2)
∠PQS + ∠PQR = ∠PRT + ∠PRQ
Or, ∠PQS + ∠PQR = ∠PRT + ∠PQR (∠PQR = ∠PRQ given)
Or, ∠PQS = ∠PRT Proved
Q4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.
Solution:
Given : x + y = w + z
To prove : AOB is a line.
Proof :
We know that;
x + y + w + z = 360०
(Angle Sustained on centre)
x + y + x + y = 360० (x + y = w + z given)
2x + 2y = 360०
2 (x + y) = 360०
x + y = 180० (linear pair)
Therefore, AOB is a line
Hence, Proved
Q5. 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
Solution:
Given:
POQ is a straight line. OR ⊥ PQ and OS is
another ray lying between rays OP and OR.
To prove:
Proof: OR ⊥ PQ (given)
∴ ∠QOR = 90० …………… (1)
POQ is straight line
∴ ∠POR + ∠QOR = 180० (linear pair)
⇒ ∠POR + 90० = 180०
⇒ ∠POR = 180०– 90०
⇒ ∠POR = 90०…………… (2)
Now, ∠ROS + ∠QOR = ∠QOS
Or, ∠ROS = ∠QOS – ∠QOR ……………. (3)
Again, ∠ROS + ∠POS = ∠POR
Or, ∠ROS = ∠POR – ∠POS ……………. (4)
Adding equation (1) and (2)
∠ROS + ∠ROS = ∠QOS – ∠QOR + ∠POR – ∠POS
2 ∠ROS = ∠QOS – 90०+ 90०– ∠POS
2 ∠ROS = (∠QOS – ∠POS)
Hence Proved
Q6. 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.
Solution:
Given: ∠ XYZ = 64°and XY is produced to point P. YQ bisects ∠ ZYP.
To Find: ∠XYQ and reflex ∠QYP.
YQ bisects ∠ZYP
∴ ∠ZYQ = ∠QYP ................. (1)
∵ XY is produced to point P.
∴ PX is a straight line.
Now, ∠ XYZ + ∠ZYQ + ∠QYP = 180° (linear pair)
Or, 64° + ∠ZYQ + ∠QYP = 180°
⇒ ∠ZYQ + ∠QYP = 180° - 64°
⇒ ∠ZYQ + ∠ZYQ = 116° [Using equation (1) ]
⇒ 2∠ZYQ = 116°
⇒ ∠ZYQ = 116°/2
⇒ ∠ZYQ = 58°
∠ZYQ = ∠QYP = 58°
∠XYQ = ∠XYZ + ∠ZYQ
= 64° + 58°
= 122°
∵ ∠QYP = 58°
∴ Reflex ∠QYP = 360° - 58°
= 302°
∠XYQ = 122°, Reflex ∠QYP = 302°