**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 +__/__3Or,

__/__2 +__/__4 =__/__2 +__/__3 [from eq. (i)]Or,

__/__4 =__/__3Or,

__/__PQS =__/__PRTHence 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

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