# NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles| PDF Download

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**Class 9 Maths Chapter 6 Lines and Angles NCERT Solutions**that will help you in knowing the important concepts provided in the chapter and completing your homework on time.Â You can also Download PDF of Chapter 6 Lines and Angles NCERT Solutions Class 9 Maths to practice in a better manner.Â Through help of these**Chapter 6 NCERT Solutions**, you can raise the level of quality of your studies and at the same time it will also help in solving the difficulties that lie ahead with ease.There are variety of concepts given in this chapter 6 Class 9 Maths textbook that will develop your necessary skill to solve more and more questions not only in this class 9 but upcoming board exams also. It will introduce to a new set of topics through which you can understand the geometry better. These NCERT Solutions will provide good experience and provide opportunities to learn new things.

**Exercise 6.1**

1. In Fig. 6.13, lines AB and CD intersect at O. If âˆ AOC +Â âˆ BOE = 70Â° and âˆ BOD = 40Â°, find âˆ BOE and reflex âˆ COE.

**Answer**

Given,

âˆ AOC + âˆ BOE = 70Â° and âˆ BOD = 40Â°

A/q,

âˆ AOC + âˆ BOE +âˆ COE = 180Â° (Forms a straight line)

â‡’ 70Â° +âˆ COE = 180Â°

â‡’ âˆ COE = 110Â°

also,

âˆ COEÂ +âˆ BODÂ + âˆ BOE = 180Â° (Forms a straight line)

â‡’ 110Â° +40Â°Â + âˆ BOE = 180Â°

â‡’ 150Â° + âˆ BOE = 180Â°

â‡’ âˆ BOE = 30Â°

Page No: 97

2. In Fig. 6.14, lines XY and MN intersect at O. If âˆ POY = 90Â° and a : b = 2 : 3, find c.

**Answer**

Given,

âˆ POY = 90Â° and a : b = 2 : 3

A/q,

âˆ POYÂ + a + b = 180Â°

â‡’ 90Â° + a + b = 180Â°

â‡’ a + b = 90Â°

Let a be 2x then will be 3x

2x + 3x = 90Â°

â‡’ 5x = 90Â°

â‡’ x = 18Â°

âˆ´ a = 2Ã—18Â° = 36Â°

and b = 3Ã—18Â° = 54Â°

also,

b + c = 180Â° (Linear Pair)

â‡’ 54Â° + c = 180Â°

â‡’ c = 126Â°

3. In Fig. 6.15, âˆ PQR = âˆ PRQ, then prove that âˆ PQS = âˆ PRT.

**Answer**

Given,

âˆ PQR = âˆ PRQ

To prove,

âˆ PQS = âˆ PRT

A/q,

âˆ PQRÂ +âˆ PQS = 180Â° (Linear Pair)

â‡’Â âˆ PQS = 180Â° - âˆ PQR --- (i)

also,

âˆ PRQ +âˆ PRT = 180Â° (Linear Pair)

â‡’ âˆ PRT = 180Â° - âˆ PRQ

â‡’ âˆ PRQ = 180Â° - âˆ PQR --- (ii) (âˆ PQR = âˆ PRQ)

From (i) and (ii)

âˆ PQS = âˆ PRT = 180Â° - âˆ PQR

Therefore,Â âˆ PQS = âˆ PRT

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

**Answer**

Given,

x + y = w + z

To Prove,

AOB is a line or x + y = 180Â° (linear pair.)

A/q,

xÂ + y + w + z = 360Â° (Angles around a point.)

â‡’ (x + y) + Â (w + z) = 360Â°

â‡’ (x + y) + Â (x + y) = 360Â° (Given x + y = w + z)

â‡’ 2(x + y) = 360Â°

â‡’ (x + y) = 180Â°

Hence, x + y makes a linear pair. Therefore, AOB is a staright line.

5. 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 âˆ ROS = 1/2(âˆ QOS â€“ âˆ POS).

**Answer**

Given,

OR is perpendicular to line PQ

To prove,

âˆ ROS = 1/2(âˆ QOS â€“ âˆ POS)

A/q,

âˆ POR = âˆ ROQ = 90Â° (Perpendicular)

âˆ QOS = âˆ ROQ

âˆ POS = âˆ POR - âˆ ROS = 90Â° - âˆ ROS --- (ii)

Subtracting (ii) from (i)

âˆ QOS - âˆ POS = 90Â° + âˆ ROS - (90Â° - âˆ ROS)

â‡’ âˆ QOS - âˆ POS = 90Â° + âˆ ROS - 90Â° + âˆ ROS

â‡’ âˆ QOS - âˆ POS = 2âˆ ROS

â‡’ âˆ ROS = 1/2(âˆ QOS â€“ âˆ POS)

Hence, Proved.

6. It is given that âˆ XYZ = 64Â° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects âˆ ZYP, find âˆ XYQ and reflex âˆ QYP.

**Answer**

Given,

âˆ XYZ = 64Â°

YQ bisects âˆ ZYP

â‡’Â 64Â° +âˆ ZYP = 180Â°

â‡’ âˆ ZYP = 116Â°

also, âˆ ZYP = âˆ ZYQÂ + âˆ QYP

âˆ ZYQ = âˆ QYP (YQ bisects âˆ ZYP)

â‡’ âˆ ZYP = 2âˆ ZYQ

â‡’ 2âˆ ZYQ = 116Â°

â‡’ âˆ ZYQ = 58Â° = âˆ QYP

Now,

âˆ XYQ = âˆ XYZÂ + âˆ ZYQ

â‡’ âˆ XYQ = 64Â° + 58Â°

â‡’ âˆ XYQ = 122Â°

also,

reflex âˆ QYP = 180Â°Â + âˆ XYQ

âˆ QYP = 180Â°Â + 122Â°

â‡’ âˆ QYP = 302Â°

Page No: 103

**Exercise 6.2**

1. In Fig. 6.28, find the values of x and y and then show that AB || CD.

Â

**Answer**

x + 50Â° = 180Â° (Linear pair)

â‡’ x = 130Â°

also,

y = 130Â° (Vertically opposite)

Now,

x = y = 130Â° (Alternate interior angles)

Alternate interior angles are equal.

Therefore, AB || CD.

Page No: 104

2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

**Answer**

Given,

AB || CD and CD || EF

y : z = 3 : 7

Now,

xÂ + y = 180Â° (Angles on the same side of transversal.)

also,

âˆ O = z (Corresponding angles)

and, yÂ

â‡’ y + z = 180Â°

A/q,

y = 3w and z = 7w

3w + 7w = 180Â°

â‡’ 10 w = 180Â°

â‡’ w = 18Â°

âˆ´ y = 3Ã—18Â° = 54Â°

and, z = 7Ã—18Â° = 126Â°

Now,

xÂ + y = 180Â°

â‡’ x + 54Â° = 180Â°

â‡’ x = 126Â°

3. In Fig. 6.30, if AB || CD, EF âŠ¥ CD and âˆ GED = 126Â°, find âˆ AGE, âˆ GEF and âˆ FGE.

**Answer**

Given,

AB || CD

EF âŠ¥ CD

âˆ GED = 126Â°

A/q,

âˆ FED = 90Â° (EF âŠ¥ CD)

Now,

âˆ AGE = âˆ GED (Since, AB || CD and GE is transversal. Alternate interior angles.)

âˆ´ âˆ AGE = 126Â°

Also, âˆ GEF = âˆ GED - âˆ FED

â‡’ âˆ GEF = 126Â° - 90Â°

â‡’ âˆ GEF = 36Â°

Now,

âˆ FGEÂ +âˆ AGE = 180Â° (Linear pair)

â‡’ âˆ FGE = 180Â° - 126Â°

â‡’ âˆ FGE = 54Â°

4. In Fig. 6.31, if PQ || ST, âˆ PQR = 110Â° and âˆ RST = 130Â°, find âˆ QRS.

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

Â

Â

**Answer**
Given,

PQ || ST, âˆ PQR = 110Â° and âˆ RST = 130Â°

Construction,

A line XY parallel to PQ and ST is drawn.

â‡’ 110Â° + âˆ QRX = 180Â°

â‡’ âˆ QRX = 70Â°

Also,

âˆ RST + âˆ SRY = 180Â° (Angles on the same side of transversal.)

â‡’ 130Â° + âˆ SRY = 180Â°

â‡’ âˆ SRY = 50Â°

Now,

âˆ QRXÂ +âˆ SRYÂ

â‡’ 70Â°Â + 50Â°Â + âˆ QRS = 180Â°

â‡’ âˆ QRS = 60Â°

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

**Answer**

Given,

AB || CD, âˆ APQ = 50Â° and âˆ PRD = 127Â°

A/q,

x = 50Â° (Alternate interior angles.)

âˆ PRDÂ + âˆ RPB = 180Â° (Angles on the same side of transversal.)

â‡’ 127Â° + âˆ RPB = 180Â°

â‡’ âˆ RPB = 53Â°

Now,

yÂ + 50Â° + âˆ RPB = 180Â° (AB is a straight line.)

â‡’ y + 50Â° + 53Â° = 180Â°

â‡’ y + 103Â° = 180Â°

â‡’ y = 77Â°

6. In Fig. 6.33, 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.

Â

**Answer**

Let us draw BE âŸ‚ PQ and CF âŸ‚ RS.

Â As PQ || RSSo, BE || CF

By laws of reflection we know that,

Angle of incidence = Angle of reflection

Thus, âˆ 1 = âˆ 2 and âˆ 3 = âˆ 4Â --- (i)

also, âˆ 2 = âˆ 3 Â Â (alternate interior angles because BE || CF and a transversal line BC cuts them at B and C)Â Â Â --- (ii)

From (i) and (ii),
âˆ 1 + âˆ 2 = âˆ 3 + âˆ 4

â‡’ âˆ ABC = âˆ DCB
â‡’ AB || CDÂ Â Â Â Â (alternate interior angles are equal)

**Exercise 6.3**

1. In Fig. 6.39, sides QP and RQ of Î”PQR are produced to points S and T respectively. If âˆ SPR = 135Â° and âˆ PQT = 110Â°, find âˆ PRQ.

**Answer**

Given,

âˆ SPR = 135Â° and âˆ PQT = 110Â°

A/q,

âˆ SPRÂ +âˆ QPR = 180Â° (SQ is a straight line.)

â‡’ 135Â° +âˆ QPR = 180Â°

â‡’ âˆ QPR = 45Â°

also,

âˆ PQTÂ +âˆ PQR = 180Â° (TR is a straight line.)

â‡’ 110Â° +âˆ PQR = 180Â°

â‡’ âˆ PQR = 70Â°

Now,

âˆ PQRÂ +âˆ QPRÂ + âˆ PRQ = 180Â° (Sum of the interior angles of the triangle.)

â‡’ 70Â°Â + 45Â° + âˆ PRQ = 180Â°

â‡’ 115Â° + âˆ PRQ = 180Â°

â‡’ âˆ PRQ = 65Â°

2. In Fig. 6.40, âˆ X = 62Â°, âˆ XYZ = 54Â°. If YO and ZO are the bisectors of âˆ XYZ and âˆ XZY respectively of Î” XYZ, find âˆ OZY and âˆ YOZ.

**Answer**

Given,

âˆ X = 62Â°, âˆ XYZ = 54Â°

YO and ZO are the bisectors of âˆ XYZ and âˆ XZY respectively.

A/q,

âˆ XÂ +âˆ XYZÂ

â‡’ 62Â°Â + 54Â° + âˆ XZY = 180Â°

â‡’ 116Â° + âˆ XZY = 180Â°

â‡’ âˆ XZY = 64Â°

Now,

âˆ OZY = 1/2âˆ XZY (ZO is the bisector.)

â‡’ âˆ OZY = 32Â°

also,

âˆ OYZ = 1/2âˆ XYZ (YO is the bisector.)

â‡’ âˆ OYZ = 27Â°

Now,

âˆ OZYÂ +âˆ OYZÂ

â‡’ 32Â°Â + 27Â° + âˆ O = 180Â°

â‡’ 59Â° + âˆ O = 180Â°

â‡’ âˆ O = 121Â°

3. In Fig. 6.41, if AB || DE, âˆ BAC = 35Â° and âˆ CDE = 53Â°, find âˆ DCE.

**Answer**

Given,

AB || DE, âˆ BAC = 35Â° and âˆ CDE = 53Â°

A/q,

âˆ BAC = âˆ CED (Alternate interior angles.)

âˆ´ âˆ CED = 35Â°

Now,

âˆ DCEÂ +âˆ CEDÂ + âˆ CDE = 180Â° (Sum of the interior angles of the triangle.)

â‡’ âˆ DCEÂ + 35Â° + 53Â° = 180Â°

â‡’ âˆ DCEÂ + 88Â° = 180Â°

â‡’ âˆ DCE = 92Â°

4. In Fig. 6.42, if lines PQ and RS intersect at point T, such that âˆ PRT = 40Â°, âˆ RPT = 95Â° and âˆ TSQ = 75Â°, find âˆ SQT.

**Answer**

Given,

âˆ PRT = 40Â°, âˆ RPT = 95Â° and âˆ TSQ = 75Â°

A/q,

âˆ PRTÂ +âˆ RPTÂ + âˆ PTR = 180Â° (Sum of the interior angles of the triangle.)

â‡’ 40Â°Â + 95Â°Â + âˆ PTR = 180Â°

â‡’ 40Â°Â + 95Â°Â + âˆ PTR = 180Â°

â‡’ 135Â°Â + âˆ PTR = 180Â°

â‡’ âˆ PTR = 45Â°

âˆ PTR = âˆ STQ = 45Â° (Vertically opposite angles.)

Now,

âˆ TSQÂ +âˆ PTR

â‡’ 75Â°Â + 45Â° + âˆ SQT = 180Â°

â‡’ 120Â° + âˆ SQT = 180Â°

â‡’ âˆ SQT = 60Â°

Page No: 108

5. In Fig. 6.43, if PQ âŠ¥ PS, PQ || SR, âˆ SQR = 28Â° and âˆ QRT = 65Â°, then find the values of x and y.

**Answer**

Given,

PQ âŠ¥ PS, PQ || SR, âˆ SQR = 28Â° and âˆ QRT = 65Â°

A/q,

x +âˆ SQR = âˆ QRT (Alternate anglesÂ as QR is transveersal.)

â‡’ x + 28Â° = 65Â°

â‡’ x = 37Â°

also,

âˆ QSR = x

â‡’ âˆ QSR = 37Â°

also,

âˆ QRSÂ +âˆ QRT = 180Â° (Linea pair)

â‡’ âˆ QRSÂ + 65Â° = 180Â°

â‡’ âˆ QRS = 115Â°

Now,

âˆ PÂ

â‡’ 90Â°Â + 65Â°Â + 115Â°Â + âˆ S = 360Â°

â‡’ 270Â° + yÂ + âˆ QSR = 360Â°

â‡’ 270Â° + yÂ + 37Â° = 360Â°

â‡’ 307Â° + y = 360Â°

â‡’ y = 53Â°

6. In Fig. 6.44, the side QR of Î”PQR is produced to a point S. If the bisectors of âˆ PQR and âˆ PRS meet at point T, then prove that âˆ QTR = 1/2âˆ QPR.

**Answer**

Given,

Bisectors of âˆ PQR and âˆ PRS meet at point T.

To prove,

âˆ QTR = 1/2âˆ QPR.

Proof,

âˆ TRS = âˆ TQRÂ

â‡’ âˆ QTR = âˆ TRS - âˆ TQR --- (i)

also,

âˆ SRP = âˆ QPR

â‡’ 2âˆ TRS = âˆ QPR + 2âˆ TQR

â‡’ âˆ QPR =Â 2âˆ TRS - 2âˆ TQR

â‡’ 1/2âˆ QPR =Â âˆ TRS - âˆ TQR --- (ii)

Equating (i) and (ii)

âˆ QTR - âˆ TQR = 1/2âˆ QPR

Hence proved.

**Go Back to NCERT Solutions for Class 9 Maths**

## NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

Chapter 6 Lines and Angles is a great chapter which must be in your strategy to improve your marks in geometry section. You will learn variety of new terms and definitions which are going to help in solving questions.

â€¢Â Basic Terms and Definitions: A part of a line with two end points is called a line-segment and a part of a line with one end point is called a ray. If three or more points lie on the same line, they are called collinear points otherwise they are called non-collinear points. An angle is formed when two rays originate from the same end point. The rays making an angle are called the arms of the angle and the end point is called the vertex of the angle.

An acute angle measures between 0Â° and 90Â°, whereas a right angle is exactly equal to 90Â°. An angle greater than 90Â° but less than 180Â° is called an obtuse angle. A straight angle is equal to 180Â°. An angle which is greater than 180Â° but less than 360Â° is called a reflex angle. Two angles whose sum is 90Â° are called complementary angles, and two angles whose sum is 180Â° are called supplementary angles. Two angles are adjacent, if they have a common vertex, a common arm and their non-common arms are on different sides of the common arm.Â

â€¢Â Intersecting Lines and Non-intersecting Lines: If two lines intersect each other, then the vertically opposite angles are equal.

â€¢Â Pairs of Angles: There is a very important theorem is given in this section. On the basis of which you have to solve questions given in the exercise 6.1 If two lines intersect each other, then the vertically opposite angles are equal.

â€¢Â Parallel Lines and a Transversal: There are four theorems given in this section which is to going to help you in solving questions effectively.

(i) If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.

(ii) If a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel.

(iii) If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.

(iv) If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel.

â€¢ Lines Parallel to the Same Line: Lines which are parallel to the same line are parallel to each other.

â€¢ Angle Sum Property of a Triangle: (i) The sum of the angles of a triangle is 180Âº. (ii) 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.

Three exercises are given in

**Chapter 6 Lines and Angles NCERT Solutions**which will improve your knowledge of Geometry. Every students must try to solve each questions given in the exercise that is why we have also provided exercise wise solutions of every problem that you can find below.Studyrankers experts at every step, they have tried to prepare these

**Class 9 Maths NCERT Solutions**in such a way that you can easily understand even the most difficult problems. You can always clear related to this chapter just by visiting this page.### NCERT Solutions for Class 9 Maths Chapters:

**FAQ onÂ**

**ChapterÂ**

**6 Lines and Angles**

#### How many exercises in Chapter 6 Lines and Angles?

How many exercises in Chapter 6 Lines and Angles?

There are only three exercise in Chapter 6 Lines and Angles NCERT Solutions which are also important for competitive exams and higher grades. We have detailed every step through which one can always clear their doubts.

#### If the supplement of an angle is 4 times of its complement, find the angle.

Let the required angle be x

âˆ´Â (180Â°- x) = 4 (90Â° - x)

â‡’ x = 60Â°

#### What is the measure of an angle whose measure is 32Â° less than its supplement?

Let the required angle be x

âˆ´ x = (180Â°- x) - 32Â°

â‡’ x = 74Â°

#### Angles âˆ P and 100Â° form a linear pair. What is the measure of âˆ P?

Since, the sum of the angles of a linear pair equal to 180Â°.

âˆ´ âˆ P + 100Â° = 180Â°

â‡’ âˆ P = 180Â° - 100 = 80Â°.