NCERT Solution for Class 9 Maths Chapter 6: Lines and Angles
Important questions and exercises from the chapter are covered in the solutions, along with definitions, properties, and theorems. Since each answer is provided simply, students will find it easy to fully understand the chapter's primary ideas and concepts.
Access Answers to NCERT Solution for Class 9 Maths Chapter 6: Lines and Angles
Students can access the NCERT Solution for Class 9 Maths Chapter 6: Lines and Angles. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise 6.1
In Fig. 6.14, lines XY and MN intersect at O. If 
We are given that
We need find the value of c in the given figure.
Let a be equal to 2x and b be equal to 3x.
Therefore
Now
Question 1. In Fig. 6.13, lines AB and CD intersect at O. If 
We are given that
We need to find
From the given figure, we can conclude that
We know that sum of the angles of a linear pair is
Reflex
But, we are given that
Therefore, we can conclude that
In the given figure,
We need to prove that
We are given that
From the given figure, we can conclude that
We know that sum of the angles of a linear pair is.
From equations (i) and (ii), we can conclude that
Therefore, the desired result is proved.
In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.
We need to prove that AOB is a line.
We are given that
We know that the sum of all the angles around a fixed point is
Thus, we can conclude that
But, 
From the given figure, we can conclude that y and x form a linear pair.
We know that if a ray stands on a straight line, then the sum of the angles of linear pair formed by the ray with respect to the line is
Therefore, we can conclude that AOB is a line.
In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that
We need to prove that
We are given that OR is perpendicular to PQ, or
From the given figure, we can conclude that
We know that sum of the angles of a linear pair is
From the figure, we can conclude that
From the given figure, we can conclude that
We know that sum of the angles of a linear pair is
Substitute (ii) in (i), to get
Therefore, the desired result is proved.
It is given that
We are given that
We can conclude the given below figure for the given situation:
We need to find
From the given figure, we can conclude that
We know that sum of the angles of a linear pair is
But
Ray YQ bisects
Reflex
Therefore, we can conclude that
Exercise 6.2
In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS.
Draw a line EF parallel to ST through R.
Since PQ || ST [Given]
and EF || ST [Construction]
∴ PQ || EF and QR is a transversal
⇒ ∠PQR = ∠QRF [Alternate interior angles] But ∠PQR = 110° [Given]
∴∠QRF = ∠QRS + ∠SRF = 110° …(1)
Again ST || EF and RS is a transversal
∴ ∠RST + ∠SRF = 180° [Co-interior angles] or 130° + ∠SRF = 180°
⇒ ∠SRF = 180° – 130° = 50°
Now, from (1), we have ∠QRS + 50° = 110°
⇒ ∠QRS = 110° – 50° = 60°
Thus, ∠QRS = 60°.
In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
We have AB || CD and PQ is a transversal.
∴ ∠APQ = ∠PQR
[Alternate interior angles]
⇒ 50° = x [ ∵ ∠APQ = 50° (given)]
Again, AB || CD and PR is a transversal.
∴ ∠APR = ∠PRD [Alternate interior angles]
⇒ ∠APR = 127° [ ∵ ∠PRD = 127° (given)]
⇒ ∠APQ + ∠QPR = 127°
⇒ 50° + y = 127° [ ∵ ∠APQ = 50° (given)]
⇒ y = 127°- 50° = 77°
Thus, x = 50° and y = 77°.
In figure, find the values of x and y and then show that AB || CD.
In the figure, we have CD and PQ intersect at F.
∴ y = 130° …(1)
[Vertically opposite angles]
Again, PQ is a straight line and EA stands on it.
∠AEP + ∠AEQ = 180° [Linear pair]
or 50° + x = 180°
⇒ x = 180° – 50° = 130° …(2)
From (1) and (2), x = y
As they are pair of alternate interior angles.
∴ AB || CD
In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.
AB || CD, and CD || EF [Given]
∴ AB || EF
∴ x = z [Alternate interior angles] ….(1)
Again, AB || CD
⇒ x + y = 180° [Co-interior angles]
⇒ z + y = 180° … (2) [By (1)]
But y : z = 3 : 7
z = y = (180°- z) [By (2)]
⇒ 10z = 7 x 180°
⇒ z = 7 x 180° /10 = 126°
From (1) and (3), we have
x = 126°.
In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
AB || CD and GE is a transversal.
∴ ∠AGE = ∠GED [Alternate interior angles]
But ∠GED = 126° [Given]
∴∠AGE = 126°
Also, ∠GEF + ∠FED = ∠GED
or ∠GEF + 90° = 126° [∵ EF ⊥ CD (given)]
x = z [Alternate interior angles]… (1) Again, AB || CD
⇒ x + y = 180° [Co-interior angles]
∠GEF = 126° -90° = 36°
Now, AB || CD and GE is a transversal.
∴ ∠FGE + ∠GED = 180° [Co-interior angles]
or ∠FGE + 126° = 180°
or ∠FGE = 180° – 126° = 54°
Thus, ∠AGE = 126°, ∠GEF=36° and ∠FGE = 54°.
In figure, 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.
Draw ray BL ⊥PQ and CM ⊥ RS
∵ PQ || RS ⇒ BL || CM
[∵ BL || PQ and CM || RS]
Now, BL || CM and BC is a transversal.
∴ ∠LBC = ∠MCB …(1) [Alternate interior angles]
Since, angle of incidence = Angle of reflection
∠ABL = ∠LBC and ∠MCB = ∠MCD
⇒ ∠ABL = ∠MCD …(2) [By (1)]
Adding (1) and (2), we get
∠LBC + ∠ABL = ∠MCB + ∠MCD
⇒ ∠ABC = ∠BCD
i. e., a pair of alternate interior angles are equal.
∴ AB || CD.
Exercise 6.3
In the given figure, sides QP and RQ of ∆PQR are produced to points S and T respectively. If SPR = 135º and PQT = 110º, find PRQ.
We are given that and .
We need to find the value ofin the figure given below.
From the figure, we can conclude thatform a linear pair.
We know that the sum of angles of a linear pair is.
and
and
Or,
From the figure, we can conclude that
(Angle sum property)
Therefore, we can conclude that.
In the given figure, 
We are given thatand YO and ZO are bisectors of, respectively.
We need to find in the figure.
From the figure, we can conclude that in
(Angle sum property)
We are given that OY and OZ are the bisectors of
From the figure, we can conclude that in
(Angle sum property)
Therefore, we can conclude that
In the given figure, if AB || DE, 
We are given that,
We need to find the value of in the figure given below.
From the figure, we can conclude that
(Alternate interior)
From the figure, we can conclude that in
(Angle sum property)
Therefore, we can conclude that.
In the given figure, if lines PQ and RS intersect at point T, such that PRT = 40º, RPT = 95º and TSQ = 75º, find 
We are given that.
We need to find the value ofin the figure.
From the figure, we can conclude that in
From the figure, we can conclude that
From the figure, we can conclude that in
(Angle sum property)
Therefore, we can conclude that.
In the given figure, if, PQ || SR,, then find the values of x and y.
We are given that.
We need to find the values of x and y in the figure.
We know that “If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.”
From the figure, we can conclude that
, or
From the figure, we can conclude that
(Alternate interior angles)
From the figure, we can conclude that
(Angle sum property)
Therefore, we can conclude that.
In the given figure, the side QR of ∆PQR is produced to a point S. If the bisectors of meet at point T, then prove that.
We need to prove thatin the figure given below.
We know that “If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.”
From the figure, we can conclude that in
…(i)
From the figure, we can conclude that in,is an exterior angle
We are given that are angle bisectors of
We need to substitute equation (i) in the above equation, to get
Therefore, we can conclude that the desired result is proved.
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Frequently Asked Questions
The NCERT solution for Class 9 Chapter 6: Lines and Angles is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.
Yes, the NCERT solution for Class 9 Chapter 6: Lines and Angles is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. They can solve the practice questions and exercises that allow them to get exam-ready in no time.
You can get all the NCERT solutions for Class 9 Maths Chapter 6 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for Class 9 Maths Chapter 6: Lines and Angles as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for Class 9 Maths Chapter 6 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.