NCERT Solutions for Class 9 Maths chapter 6 Lines And Angles

NCERT solutions for class 9 maths  chapter 6 Lines and Angles are given here. We have prepared NCERT Solutions  for all exercise of chapter 6. Given below is step by step solutions of all questions given in NCERT textbook for chapter 6.

Read chapter 6 theory make sure you have gone through the theory part of chapter-6 from NCERT textbook and you have learned the formula of the given chapter. Do read these contents before moving to solve the exercise of Maths NCERT Solutions for Class 9 Chapter 6.

NCERT Solutions for Class 9 Maths Exercise 6.1

Students who have studied acute, obtuse, right, straight, reflex, complementary, and supplementary angles in previous sessions can discover this exercise useful in remembering these types of angles. The NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 's creative concepts will begin with the intersecting and non-intersecting lines. Next, pairs of angles are thoroughly discussed, supported by a theorem proof and several instances from the NCERT textbook.

Question 1. In Fig. 6.13, lines AB and CD intersect at O. If and , find and reflex .

Solution:
We are given that and .

We need to find .

From the given figure, we can conclude that form a linear pair.

We know that sum of the angles of a linear pair is .

or

Reflex

(Vertically opposite angles), or

But, we are given that

Therefore, we can conclude that and .

Question 2. In Fig. 6.14, lines XY and MN intersect at O. If and a:b = 2 : 3, find c.

Solution:
We are given that and .

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 [Linear pair]

Question 3. In the given figure, , then prove that .

Solution:
We need to prove that .

We are given that .

From the given figure, we can conclude that form a linear pair.

We know that sum of the angles of a linear pair is .

and (i)

(ii)

From equations (i) and (ii), we can conclude that

Therefore, the desired result is proved.

Question 4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.

Solution:
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, (Given).

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.

Question 5. 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

Solution:
We need to prove that .

We are given that OR is perpendicular to PQ, or

From the given figure, we can conclude that form a linear pair.

We know that sum of the angles of a linear pair is .

, or

.

From the figure, we can conclude that .

, or

.(i)

From the given figure, we can conclude that form a linear pair.

We know that sum of the angles of a linear pair is .

, or

.(ii)

Substitute (ii) in (i), to get

Therefore, the desired result is proved.

Question 6. It is given that and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects , find

Solution:
We are given that , XY is produced to P and YQ bisects .

We can conclude the given below figure for the given situation:

We need to find .

From the given figure, we can conclude that form a linear pair.

We know that sum of the angles of a linear pair is .

.

But .

Ray YQ bisects , or

.

Reflex

Therefore, we can conclude that and Reflex

NCERT Solutions for Class 9 Maths Exercise 6.2

NCERT Solutions for Class 9 Maths Exercise 6.2 are discussed here. We have included a diagram and step-by-step solutions in the NCERT Class 9 Maths Solutions for the benefit of the students. Additionally, the questions are answered in the simplest way possible so that students may grasp them with ease.

Question 1. In figure, find the values of x and y and then show that AB || CD.


Solution:
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

Question 2. In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.


Solution:
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°.

Question 3. In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.


Solution:
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°.

Question 4. In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS.


Solution:
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°.

Question 5. In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.


Solution:
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°.

Question 6. 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.


Solution:
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.

NCERT Solutions for Class 9 Maths Exercise 6.3

Chapter 6 of NCERT Solutions Maths for Class 9 provides a comprehensive guide on analyzing the characteristics of lines and angles. This exercise focuses on the significant characteristics of parallel lines when a transversal cuts them.

Question 1. 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.

Solution:
We are given that and .

We need to find the value of in the figure given below.

From the figure, we can conclude that form 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 .

Question 2. In the given figure, X = 62º, XYZ = 54º. If YO and ZO are the bisectors of XYZ and XZY respectively of ∆XYZ, find OZY and YOZ.

Solution:
We are given that and 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 , respectively.

and

From the figure, we can conclude that in

(Angle sum property)

Therefore, we can conclude that and .

Question 3. In the given figure, if AB || DE, BAC = 35º and CDE = 53º, find DCE.

Solution:
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 .

Question 4. In the given figure, if lines PQ and RS intersect at point T, such that PRT = 40º, RPT = 95º and TSQ = 75º, find SQT.

Solution:
We are given that .

We need to find the value of in the figure.

From the figure, we can conclude that in

(Angle sum property)

From the figure, we can conclude that

(Vertically opposite angles)

From the figure, we can conclude that in

(Angle sum property)

Therefore, we can conclude that .

Question 5. In the given figure, if , PQ || SR, , then find the values of x and y.

Solution:
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 .

Question 6. 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 .

Solution:
We need to prove that in 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 , is an exterior angle

…(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.

NCERT Solutions for Class 9 Maths Chapter 6 PDF download

Students can click the link below to obtain the NCERT Solutions of Class 9 Maths Chapter 6 Exercise 6.1 in PDF format. These solutions are well explained and created by subject matter specialists. Students who study the NCERT Solutions PDF can quickly grasp how to answer the problems.

NCERT Solutions for Class 9 Maths Chapter 6 PDF download

Tips for CBSE Class 9 Math Preparation

Below are the CBSE Class 9 Math Preparation Tips.

• To study the CBSE Class 9 syllabus systematically, divide the concepts of algebra, geometry, mensuration, and statistics.
  
  

• While revising each topic, focus on chapters that carry more importance to build a strong basis for Class 10.
  
  

• Examine the CBSE Class 9 Maths NCERT Solutions for more assistance after completing all of the NCERT textbook exercises.
  
  

• Review important NCERT theorems and formulas regularly.
  
  

• Use CBSE Class 9 Sample Papers to improve time management and understand exam patterns.
  

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