# Class 9 NCERT Solutions Maths Chapter 6 - Lines And Angles

Refer to NCERT Solutions for CBSE Class 9 Mathematics Chapter 6 Lines and Angles at TopperLearning for thorough Maths learning. Get clarity on concepts like linear pairs, vertically opposite angles, co-interior angles, alternate interior angles etc. While practising the model solutions from this chapter, you will also learn to use the angle sum property of a triangle while solving problems.

Grasp the properties of angles and lines by observing the steps used by experts in the NCERT textbook solutions. The skills gained through our CBSE Class 9 Maths chapter resources can benefit you while preparing for your Class 10, Class 11 and Class 12 exams too.

## Lines And Angles Exercise Ex. 6.1

### Solution 1

### Solution 2

Let common ratio between a and b is x, a = 2x and b = 3x.

XY is a straight line, OM and OP rays stands on it.

XOM + MOP + POY = 180 b + a + POY = 180

3x + 2x + 90 = 180

5x = 90

x = 18

a = 2x

= 2 * 18

= 36

b = 3x

= 3 * 18

= 54

Now, MN is a straight line. OX ray stands on it.

*b* + *c* = 180

54 + c = 180

*c* = 180 54 = 126** **

*c* = 126

### Solution 3

PQS + PQR = 180 (Linear Pair)

PRT + PRQ = 180 (Linear Pair)

PRQ = 180 - PRT (2)

Given that PQR = PRQ. Now, equating equations (1) and (2), we have

180 - PQS = 180 - PRT

PQS = PRT

### Solution 4

x + y + z + w = 360 (Complete angle)

It is given that

x + y = z + w

x + y + x + y = 360

2(x + y) = 360

x + y = 180

Since x and y form a linear pair, thus AOB is a line.

### Solution 5

### Solution 6

Hence, QYP = ZYQ

Now we may observe that PX is a line. YQ and YZ rays stand on it.

^{o}- 58

^{o}= 302

^{o}

^{o}+ 58

^{o}= 122

^{o}

## Lines And Angles Exercise Ex. 6.2

### Solution 1

50 + x = 180 (Linear pair)

x = 130 ... (1)

Also, y = 130 (vertically opposite angles)

As x and y are alternate interior angles for lines AB and CD and also measures of these angles are equal to each other, so line AB || CD

### Solution 2

x = z (alternate interior angles) ... (1)

Given that y: z = 3: 7

Let common ratio between y and z be a

y = 3a and z = 7a

Also x + y = 180 (co-interior angles on the same side of the transversal)

x = 7 a = 7 18 = 126

### Solution 3

AB || CD

GEF + FED = 126

GEF = 36

Now, AGE and GED are alternate interior angles

### Solution 4

PQR + QRX = 180 (co-interior angles on the same side of transversal QR)

110 + QRX = 180

QRX = 70

Now,

RST +SRY = 180 (co-interior angles on the same side of transversal SR)

130 + SRY = 180

SRY = 50

XY is a straight line. RQ and RS stand on it.

QRX + QRS + SRY = 180

70 + QRS + 50 = 180

QRS = 180 - 120 = 60

### Solution 5

50 + y = 127

y = 127 - 50

y = 77

Also APQ = PQR (alternate interior angles)

50 = x

### Solution 6

As PQ || RS

So, BM || CN

Thus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.

## Lines And Angles Exercise Ex. 6.3

### Solution 1

SPR = 135 and PQT = 110

Now, SPR + QPR = 180 (linear pair angles)

135 + QPR = 180

QPR = 45

Also, PQT + PQR = 180 (linear pair angles)

110 + PQR = 180

PQR = 70

As we know that sum of all interior angles of a triangle is 180, so, for PQR

QPR + PQR + PRQ = 180

45 + 70 + PRQ = 180

PRQ = 180 - 115

PRQ = 65

### Solution 2

62 + 54 + XZY = 180

XZY = 180 - 116

XZY = 64

Similarly, OYZ = = 27

Using angle sum property for OYZ, we have

OYZ + YOZ + OZY = 180º

27 + YOZ + 32 = 180

YOZ = 180 - 59

YOZ = 121

### Solution 3

BAC =CED (alternate interior angle)

CED = 35

In CDE,

CDE + CED + DCE = 180 (angle sum properly of a triangle)

53 + 35 + DCE = 180

DCE = 180 - 88

DCE = 92

### Solution 4

PRT + RPT + PTR = 180

40 + 95 + PTR = 180

PTR = 180 - 135

PTR = 45

STQ = PTR = 45 (vertically opposite angles)

STQ = 45

By using angle sum property for STQ, we have

STQ + SQT + QST = 180

45 + SQT + 75 = 180

SQT = 180 - 120

SQT = 60

### Solution 5

PQR = QRT (alternate interior angles)

x + 28 = 65

x = 65 - 28

x = 37

By using angle sum property for SPQ, we have

SPQ + x + y = 180

90 + 37 + y = 180

y = 180 - 127

y = 53

x = 37 and y = 53.

### Solution 6

QTR + TQR = TRS

QTR = TRS - TQR (1)

For PQR, PRS is external angle

QPR = 2(TRS - TQR)

QPR = 2QTR [By using equation (1)]

QTR = QPR