NCERT Solutions for Class 9 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.

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Chapter 6 - 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 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

Chapter 6 - 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 4

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

Chapter 6 - Lines And Angles Exercise Ex. 6.3

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.