Class IX CBSE Mathematics Guide | Chapter 6 Lines and Angles | NCERT Solutions for Maths Exercise 6.1

CBSE Class 9 NCERT Mathematics Solutions

Lines and Angles

NCERT Mathematics Textbook Exercise 6.1 Solved

(Page 96, 97)

Q1: In the given figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70^O and ∠BOD = 40^O find ∠BOE and reflex ∠COE.

Solution:

Given that / BOD = 40^o, / AOC + / BOE = 70^o

To find / BOE and  / COE.

 / BOD = / COA = 40^o [vertically opposite angles]

As,  / AOC + / BOE = 70^o

Or, 40^o + / BOE = 70^o

Or, / BOE = 30^o

Also, / AOC + / COE + / EOB = 180^o [angle on same straight line having same vertex].

Or, 40^o + / COE + 30^o = 180^o

Or, / COE = 110^o

Or, Reflex / COE = 360^o - 180^o = 250^o

Q2: In the given figure, lines XY and MN intersect at O. If ∠POY = 90^O and a:b = 2 : 3, find c. 

 Answer:  Let us assume a = 2x and b = 3x

Then, / POX = 90^O [/ POY + / POX = 180^O, and / POY = 90^O given]   

Or, a + b = 90^O

Or, 2x + 3x = 90^O

Or, x = 18^O

a = 2x = 36^O and,

b = 3x = 54^O

Now, b + c = 180^O [linear pair]

Or, 54^O + c = 180^O

Or, c = 126^O

Q3: In the given figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Solution:

Given:  / PQR = / PRQ (/ 1 = / 2) ….. (i)

To prove: / PQS = / PRT (/ 4 = / 3)

Proof: / 1 + / 4 = 180^o and / 2 + / 3 = 180^o [linear pairs]

So, / 1 + / 4 = / 2 + / 3  

Or, / 2 + / 4 = / 2 + / 3 [from eq. (i)]

Or, / 4 = / 3

Or, / PQS = / PRT

Hence proved.

Q4: In the given figure, if x + y = w + z then prove that AOB is a line.

Solution:

Given that, x + y = w + z

To prove: AOB is a line.

Proof: (x + y) + (w + z) = 360^o [the sum of all angles round a point is equal to 360^o]

Or, (x + y) + (x + y) = 360^o

Or, (x + y) = 180^o

But we know that if the sum of two adjacent angles is 180^o then, the non-common arms of the angles form a straight line. Therefore, AOB is a line.

Q5: In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that 

Answer:

It is given that RO ⊥ PQ

∴ ∠ROP = ∠ROQ = 90º

Or, ∠POS + ∠SOR = ∠SOQ − ∠SOR

Or, 2∠ROS = ∠QOS − ∠POS

Or, ∠ROS = ½ (∠QOS − ∠POS)

Q6: It is given that ∠XYZ = 64^O and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.

Solution:

It can be observed that PX is a line. Ray YZ stands on it.

∴ ∠XYZ + ∠ZYP = 180º [linear pair]

⇒ 64º + ∠ZYP = 180º

⇒ ∠ZYP = 180º − 64º = 116º

Given that ray YQ bisects ∠YZP.

Hence, ∠QYP = ∠ZYQ = ½ ∠YZP = 58º

Reflex ∠QYP = 360º − 58º = 302º [the sum of all the angles round the point is 360^o].

Or, ∠XYQ = ∠XYZ + ∠ZYQ = 64º + 58º = 122º