6. Lines and Angles Mathematics Exercise - 6.1 class 9 Maths in English - CBSE Study
NCERT Solutions for Class 9 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 6. Lines and Angles with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 6.1 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 9 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.
Class 9 English Medium Mathematics All Chapters:
6. Lines and Angles
1. 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°
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