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

Ex. 6.1

Ex. 6.2

## 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

In the given figure, ST is a straight line and QP ray stand on it.

PQS + PQR = 180 (Linear Pair)

PQS + PQR = 180 (Linear Pair)

PQR = 180 - PQS (1)

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

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

We may observe that

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

It is given that

x + y = z + w

x + y + x + y = 360

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

Given that OR PQ

POR = 90

POS + SOR = 90

ROS = 90 - POS ... (1)

QOR = 90 (As OR PQ)

QOS - ROS = 90

ROS = QOS - 90 ... (2)

On adding equations (1) and (2), we have

2 ROS = QOS - POS

### Solution 6

Given that line YQ bisects PYZ.

Hence, QYP = ZYQ

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

Hence, QYP = ZYQ

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

XYZ + ZYQ + QYP = 180

64 + 2QYP = 180

2QYP = 180 - 64 = 116

QYP = 58

Also, ZYQ = QYP = 58

Reflex QYP = 360

^{o}- 58^{o}= 302^{o} XYQ = XYZ + ZYQ

= 64

^{o}+ 58^{o}= 122^{o}## Lines And Angles Exercise Ex. 6.2

### Solution 1

Given that AB || CD and CD || EF

AB || CD || EF (Lines parallel to a same line are parallel to each other)

Now we may observe that

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

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)

z + y = 180 [Using equation (1)]

7a + 3a = 180

10a = 180

a = 18

x = 7 a = 7 18 = 126

x = 7 a = 7 18 = 126

### Solution 2

It is given that

AB || CD

AB || CD

EF CD

GED = 126

GEF + FED = 126

GEF + FED = 126

GEF + 90 = 126

GEF = 36

Now, AGE and GED are alternate interior angles

GEF = 36

Now, AGE and GED are alternate interior angles

AGE = GED = 126

But AGE +FGE = 180 (linear pair)

126 + FGE = 180

FGE = 180 - 126 = 54

AGE = 126, GEF = 36, FGE = 54

### Solution 3

Let us draw a line XY parallel to ST and passing through point R.

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

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 4

APR = PRD (alternate interior angles)

50 + y = 127

y = 127 - 50

y = 77

Also APQ = PQR (alternate interior angles)

50 = x

50 + y = 127

y = 127 - 50

y = 77

Also APQ = PQR (alternate interior angles)

50 = x

x = 50 and y = 77

### Solution 5

Let us draw BM PQ and CN RS.

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.

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.

2 = 3 (alternate interior angles)

But 1 = 2 and 3 = 4 (By laws of reflection)

1 = 2 = 3 = 4

Now, 1 + 2 = 3 + 4

ABC = DCB

But, these are alternate interior angles

AB || CD