NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles is curated and compiled by the expert team of professionals at GFG to assist students in resolving questions related to lines and angles they may have as they go through problems from the NCERT textbook.

This chapter Lines and Angles mainly covers the characteristics of the angles created when two or more lines cross one another. Additionally, it gives a fundamental comprehension of the definitions of various terminology used in geometry. The many kinds of angles, pairs of angles, transversals, parallel lines, the angle sum property of triangles, and axioms related to these ideas are all taught to students.

Class 9 Maths NCERT Solutions Chapter 6 Exercises:
NCERT Maths Solutions Class 9 Exercise 6.1 – 6 Questions & Solutions (5 Short Answers, 1 Long Answer)
NCERT Maths Solutions Class 9 Exercise 6.2 – 3 Questions & Solutions (3 Short Answers, 3 Long Answers)
NCERT Maths Solutions Class 9 Exercise 6.3 – 6 Questions & Solutions (5 Short Answers, 1 Long Answer)

Lines and Angles – Exercise 6.1

Question 1. In the given figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE? 

Solution:

Given, AB and CD are straight lines.

∠AOC + ∠BOE = 70°  —-eq(i)

∠BOD = 40° —-eq(ii)

Since, AB is a straight line, the sum of all angles made on it is 180°

⇒ ∠AOC  + ∠COE + ∠BOE = 180° —eq(iii)

We can rearrange this equation as,

⇒ ∠AOC + ∠BOE + ∠COE = 180°

⇒ 70° + ∠COE = 180°  —from eq(i)

⇒ ∠COE = 180° – 70° = 110°

⇒ ∠COE = 110° —eq(iv)

Reflex ∠COE = 360° – ∠COE = 360° – 110° = 250°

Now, it is also given that CD is also a straight line, so the sum of all angles made on it is 180°

⇒ ∠COE + ∠BOE + ∠BOD = 180° —eq(v)

We can rearrange this equation as,

⇒ ∠COE + ∠BOD + ∠BOE = 180°

⇒ 110° + 40° + ∠BOE = 180°  —from eq(ii) and eq(iv)

⇒ 150° + ∠BOE = 180°

⇒ ∠BOE = 180° – 150° = 30°

⇒ ∠BOE = 30°

Question 2. In the given figure, lines XY and MN intersect at O. If∠POY = 90° and a : b = 2 : 3, find c?

Solution:

Given, XY and  MN are straight lines.

∠POY = 90° –eq(i)

a : b = 2 : 3 –eq(ii)

∠POM = a

∠XOM = b

∠XON = c

Taking XY as a straight line, so the sum of all angles made on it is 180°

⇒ ∠XOM + ∠POM + ∠POY = 180° —eq(iii)

⇒ b + a + 90° = 180°

⇒ 3x + 2x + 90° = 180° from eq(i) and eq(ii)

⇒ 5x + 90° = 180°

⇒ 5x = 180° – 90° = 90°

⇒ 5x = 90°

⇒ x = 18°

a : b = 2x : 3x = 2×18 : 3×18 

a = 36°

b = 54°

Taking MN as a straight line so,the sum of all the angles made on it is 180°

⇒ ∠XOM  + ∠XON = 180°

⇒ 54° + ∠XON = 180°  from above finding value 

⇒ ∠XON = 126° or c = 126°

Question 3. In the given figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT?

Solution:

Given, ∠PQR = ∠PRQ 

Taking ST is a straight line, so the sum of all angles made on it is 180°

⇒ ∠PQS + ∠PQR = 180° —-eq(i)

also,  ∠PRQ +  ∠PRT = 180° —eq(ii)

By equating both the equations because RHS of both the equation is equal So, LHS will also be equal.

⇒ ∠PQS +  ∠PQR  = ∠PRQ  + ∠PRT

⇒ ∠PQS + ∠PQR = ∠PQR + ∠PRT  –[ Given in question ∠PQR = ∠PRQ ]

⇒ ∠PQS = ∠PRT 

Question 4. In the given figure, if x + y = w + z, then prove that AOB is a line?

Solution:

Given, x + y  = w + z  –eq(i)

We know that , sum of all angles made along  a point is 360°

So, Taking O as a point ∠AOC + ∠BOC + ∠BOD + ∠AOD = 360°

⇒ y + x + w + z = 360° from the given figure

⇒ (x + y) + (x + y) = 360°  from eq(i)

⇒ 2x + 2y = 360°

⇒ 2(x + y) = 360°

⇒ x + y=180°

From this statement it is proved that AOB is a straight line because the sum of angles made on the line is 180°. So, AOB is a straight 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∠ROS = (1/2) (∠QOS – ∠POS)?

Solution:

Given POQ is a straight line 

So, the sum of all angles made on it is 180°

⇒ ∠POS + ∠ROS + ∠ROQ = 180°

⇒ ∠POS + ∠ROS + 90° = 180° [given ∠ROQ = 90°]

⇒ ∠POS + ∠ROS = 90°

⇒ ∠ROS = 90°  – ∠POS –eq(i)

Now, ∠ROS + ∠ROQ = ∠QOS   [from figure]

⇒ ∠ROS + 90° = ∠QOS 

⇒ ∠ROS = ∠QOS – 90° –eq(ii)

Now Adding both the equations eq(i) + eq(ii)

⇒ ∠ROS + ∠ROS = 90° – ∠POS + ∠QOS – 90°

⇒ 2∠ROS =(∠QOS  – ∠POS)

⇒ ∠ROS = (1/2) (∠QOS – ∠POS)     

Hence Verified!!!

Question 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?

Solution:

From the drawn figure, it is clearly shown that XYP is a straight line.

So, ∠XYZ + ∠ZYQ + ∠QYP = 180°

⇒ 64°+ ∠ZYQ + ∠QYP = 180°  [ given ∠XYZ = 64°]

⇒ 64° + 2∠QYP = 180°  [ YQ bisect ∠ZYP so, ∠QYP = ∠ZYQ]

⇒ 2∠QYP = 180° – 64° = 116°

⇒ ∠QYP = 58°

So, Reflex ∠QYP = 360° – 58° = 302°

Since ∠XYQ = ∠XYZ + ∠ZYQ

⇒ ∠XYQ = 64° + ∠QYP  [ given ∠XYZ = 64° and ∠ZYQ = ∠QYP]

⇒ ∠XYQ =64° + 58° = 122° 

Thus, ∠XYQ = 122° and Reflex ∠QYP = 302°

Lines and Angles – Exercise 6.2

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

Solution:

After given names to the remaining vertices we get,

Now, Given ∠AEP = 50°, ∠CFQ = 130° 

⇒ ∠EFD = ∠CFQ  [vertically opposite angles are equal]

⇒ y = 130°   [Given ∠CFQ = 130°]

⇒ y = 130° —eq(i)

Now, PQ is taking as straight line so, sum of all angles made on it is 180°

⇒ ∠AEP  + ∠AEQ = 180°

⇒ 50° + x = 180°

⇒ x = 180° – 50° = 130°

⇒ x = 130° –eq(ii) 

Now, from eq(i) and eq(ii) We conclude that x = y

As they are pair of alternate interior angles

So, AB || CD

Hence proved!!!

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

Solution:

Given AB || CD and CD || EF  y : z = 3 : 7

⇒ AB || CD || EF

⇒ AB || EF  

So, x=z [alternate interior angles] –eq(i)

Again AB || CD 

⇒ x + y = 180° [Co-interior angles]

⇒ z + y = 180°  –eq(ii) [from eq(i)]

But given that y : z = 3 : 7

⇒ z = (7/3) y = (7/3)(180° – z) [from eq(ii)]

⇒ 10z = 7 * 180°

⇒ z = (7 * 180°)/10 =126°

⇒ z  = 126° –eq(iii)

from eq(i) and eq(iii) we have

⇒ x = 126° 

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

Solution:

Given AB || CD, EF ⊥ CD, ∠GED = 126° and ∠FED =90° 

⇒ ∠GED =  ∠GEF + ∠FED

⇒ 126° = ∠GEF + 90°  [Given]

⇒ ∠GEF = 36°

As, AB || CD and GE is a transversal 

So, ∠FGE + ∠GED = 180° [ sum of Co- interior angles is 180 ]

⇒  ∠FGE + 126° = 180° [ Given ]

⇒  ∠FGE = 54°

As, AB || CD and GE is a transversal 

So, ∠AGE = ∠GED [ alternate angles are equal ]

⇒ ∠AGE = 126°

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

[Hint: Draw a line parallel to ST through point R.]

Solution:

Firstly we have drawn a line EF parallel to ST (EF || ST)

Since, PQ || ST [Given] and EF || ST [ Construction ]

So, PQ || EF and QR is a transversal 

⇒ ∠PQR = ∠QRF [ Alternate interior angles ] 

⇒ ∠QRF = 110°   [Given ∠PQR =110° ]

⇒ ∠QRF = ∠QRS + ∠SRF 

⇒ ∠QRS + ∠SRF = 110°  –eq(i)

Again ST || EF and RS is a transversal 

⇒∠RST + ∠SRF = 180° [ sum of  Co-interior angles is 180° ]

⇒130° + ∠SRF =180° [Given]

⇒∠SRF =50°

Now , from eq(i) 

⇒ ∠QRS + ∠SRF = 110°

⇒ ∠QRS +50 = 110°

⇒ Thus, ∠QRS = 60°

Question 5. In Fig. 6.32, if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, find x and y.

Solution: 

Given  AB || CD and PQ is a transversal

⇒ ∠APQ = ∠PQR [ Alternate interior angles ]

⇒ x= 50° [ Given ∠APQ = 50° ]

⇒ x = 50°

Again, AB || CD and PR is a transversal 

⇒∠APR = ∠PRD [ Alternate interior angles ]

⇒ ∠APR = 127° [ Given ∠PRD = 127° ]

⇒ 50° + y =127° [Given ∠APQ = 50° ]

⇒ y =127° – 50° = 77°

⇒  y = 77°

Thus,  x = 50° and y = 77°

Question 6. In given 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

Since, PQ || RS ⇒ BL || CM 

⇒[ So, BL || PQ and CM || RS ]

Now, BL || CM and BC is a transversal 

 ⇒ ∠LBC = ∠MCB –eq(i) [ Alternate interior angles ]

Since, angle of incidence = angle of reflection 

⇒  ∠ABL = ∠LBC  and  ∠MCB = ∠MCD 

⇒ ∠ABL = ∠MCD –eq(ii) [By eq(i)]

⇒Adding eq(i) and eq(ii) we get

⇒  ∠LBC + ∠ABL = ∠MCB + ∠MCD 

⇒ ∠ABC = ∠BCD 

i.e, a pair of alternate angles are equal

Thus, AB || CD

Hence, Proved !!!

Lines and Angles – Exercise 6.3

Question 1: In Figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Solution:

Given: ∠TQP = 110°, ∠SPR = 135°

TQR is a Straight line as we can see in the figure

As we have studied in this chapter, TQP and PQR will form a linear pair 

⇒ ∠TQP + ∠PQR = 180°  ———-(i)

Putting the value of ∠TQP = 110° in Equation (i) we get,

⇒ 110° + ∠PQR = 180° 

⇒ ∠PQR = 70°

Consider the ΔPQR,

Here, the side QP is extended to S and so, SPR forms the exterior angle.

Thus, ∠SPR (∠SPR = 135°) is equal to the sum of interior opposite angles. (Triangle property)

Or, ∠PQR + ∠PRQ = 135° ———(ii)

Now, putting the value of PQR = 70° in equation (ii) we get,

∠PRQ = 135° – 70°

Hence, ∠PRQ = 65°

Question 2: In Figure, ∠X = 62°, ∠ XYZ = 54°. If YO and ZO are the bisectors of XYZ and XZY respectively of Δ XYZ, find OZY and YOZ.

Solution:

Given: ∠X = 62°, ∠XYZ = 54°

As we have studied in this chapter,

We know that the sum of the interior angles of the triangle is 180°.

So, ∠X +∠XYZ +∠XZY = 180°

Putting the values as given in the question we get,

62°+54° + ∠XZY = 180°

Or, ∠XZY = 64°

Now, we know that ZO is the bisector so,

∠OZY = ½ XZY

∴ ∠OZY = 32°

Similarly, YO is a bisector and so,

∠OYZ = ½ XYZ

Or, ∠OYZ = 27° (As XYZ = 54°)

Now, as the sum of the interior angles of the triangle,

∠OZY +∠OYZ +O = 180°

Putting their respective values, we get,

∠O = 180°-32°-27°

Hence, ∠O = 121°

Question 3: In Figure, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Solution:

Given: AB || DE, ∠BAC = 35° and ∠CDE = 53°

Since, we know that AE is a transversal of AB and DE

Here, BAC and AED are alternate interior angles.

Hence, ∠BAC = ∠AED

 ∠BAC = 35° (Given)

∠AED = 35°

Now consider the triangle CDE. We know that the sum of the interior angles of a triangle is 180°.

∴ ∠DCE + ∠CED + ∠CDE = 180°

Putting the values, we get

∠DCE + 35° + 53° = 180°

Hence, ∠DCE = 92°

Question 4: In Figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Solution:

Given: ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°

In △PRT.

∠PRT +∠RPT + ∠PTR = 180° (The Sum of all the angles of Triangle is 180°)

⇒ ∠PTR = 45°

Now ∠PTR will be equal to STQ as they are vertically opposite angles.

⇒ ∠PTR = ∠STQ = 45°

Again, in triangle STQ,

⇒ ∠TSQ +∠PTR + ∠SQT = 180°  (The Sum of all the angles of Triangle is 180°)

Solving this we get,

⇒ ∠SQT = 60°

Question 5: In Figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Solution:

Given: PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°

x + SQR = QRT (As they are alternate angles since QR is transversal)

Now, Putting the value of ∠SQR = 28° and ∠QRT = 65°

⇒ x + 28° = 65°

∴ x = 37°

It is also known that alternate interior angles are same and 

⇒ QSR = x = 37°

Also, 

⇒ QRS + QRT = 180° (As they form a Linear pair)

Putting the value of ∠QRT = 65° we get,

⇒ QRS + 65° = 180°

⇒ QRS = 115°

As we know that the sum of the angles in a quadrilateral is 360°. 

⇒ P + Q + R + S = 360°

Putting their respective values, we get,

⇒ S = 360° – 90° – 65° – 115° = 900

In Δ SPQ

⇒ ∠SPQ + x + y = 1800

⇒ 900 + 370 + y = 1800

⇒ y = 1800 – 1270 = 530

Hence, y = 53°

Question 6: In Figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = ½ ∠QPR.

Solution:

Given: T is the bisector of ∠PQR and ∠PRS, 

To Prove: ∠QTR = ½ ∠QPR

Proof:

Consider the ΔPQR. 

∠PRS is an exterior angle.

∠QPR and ∠PQR are interior angles.

⇒ ∠PRS = ∠QPR + ∠PQR (According to triangle property)

⇒ ∠PRS – ∠PQR = ∠QPR  ————(i)

Now, consider the ΔQRT,

∠TRS = ∠TQR + ∠QTR (Since exterior angle are equal)

⇒ ∠QTR = ∠TRS – ∠TQR

We know that QT and RT bisect ∠PQR and ∠PRS respectively.

So, ∠PRS = 2 ∠TRS and ∠PQR = 2∠TQR

Now,

⇒ ∠QTR = ½ ∠PRS – ½ ∠PQR

⇒ ∠QTR = ½ ∠(PRS – PQR)

From equation (i) we know that ∠PRS – ∠PQR = ∠QPR,

⇒ ∠QTR = ½ ∠QPR 

Hence, Proved.

Important Points to Remember:

• NCERT Solutions provides accurate and complete solutions for problems given in NCERT textbooks.
• These solutions are given in step by step manner for a better understanding of the problems.
• NCERT Solutions for Class 9 will help students to learn the solution for all the NCERT Problems.

FAQs on NCERT Solutions for Class 9 Maths Chapter 6 – Lines and Angles

1. Why is it important to learn about the lines and angles?

Lines and Angles are fundamental concepts in geometry. They form the building blocks for understanding the properties and relationships of shapes and figures. By studying lines and angles, students develop a strong geometric foundation and gain insights into the properties of various geometric shapes, such as triangles, quadrilaterals, and circles.

2. What topics are covered in NCERT Solutions for Chapter 6 – Lines and Angles?

NCERT Solutions for Class 9 Maths Chapter 6 – Lines and Angles covers topics such as characteristics of the angles, kinds of angles, pairs of angles, transversals, parallel lines, the angle sum property of triangles, and axioms.

3. How can NCERT Solutions for Class 9 Maths Chapter 6 – Lines and Angles help me?

NCERT Solutions for Class 9 Maths Chapter 6 – Lines and Angles can help you solve the NCERT exercise without any limitations. If you are stuck on a problem you can find its solution in these solutions and free yourself from the frustration of being stuck on some question.

4. How many exercises are there in Class 9 Maths Chapter 6 – Lines and Angles?

There are 3 exercises in the Class 9 Maths Chapter 6 – Lines and Angles which covers all the important topics and sub-topics.

5. Where can I find NCERT Solutions for Class 9 Maths Chapter 6 – Lines and Angles?

You can find these NCERT Solutions in this article created by our team of experts at GeeksforGeeks.