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# 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 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)
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

### 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

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.
XYZ + ZYQ + QYP = 180

64 + 2QYP = 180

2QYP = 180 - 64 = 116

QYP = 58

Also, ZYQ = QYP = 58

Reflex QYP = 360o - 58o = 302o

XYQ = XYZ + ZYQ

= 64o + 58o = 122o

## Lines And Angles Exercise Ex. 6.2

### Solution 1

We may observe that
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

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

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

### Solution 3

It is given that
AB || CD
EF    CD
GED = 126
GEF + FED = 126
GEF + 90 = 126
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 4

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

### Solution 5

APR = PRD                 (alternate interior angles)
50 + y = 127
y = 127 - 50
y = 77
Also APQ = PQR         (alternate interior angles)
50 = x
x = 50 and y = 77

### Solution 6

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.
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

## Lines And Angles Exercise Ex. 6.3

### Solution 1

Given that
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

As we know that sum of all interior angles of a triangle is 180, so for XYZ
X + XYZ + XZY = 180
62 + 54 + XZY = 180
XZY = 180 - 116
XZY = 64
OZY =   = 32         (OZ is angle bisector of XZY)
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

AB || DE and AE is a transversal
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

Using angle sum property for PRT, we have
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

Given that PQ || SR and QR is a transversal line
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

In QTR, TRS is an exterior angle.
QTR + TQR = TRS
QTR = TRS - TQR        (1)
For PQR, PRS is external angle
QPR + PQR = PRS
QPR + 2TQR = 2TRS    (As QT and RT are angle bisectors)
QPR = 2(TRS - TQR)
QPR = 2QTR            [By using equation (1)]
QTR =  QPR