# NCERT SOLUTIONS FOR CLASS 9 MATHS CHAPTER 6 LINES AND ANGLES

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### NCERT Solutions for Class 9 Maths Exercise 6.1

Question 1. In Fig. 6.13, lines AB and CD intersect at O. If and, find and reflex.

Solution:
We are given thatand.

We need to find.

From the given figure, we can conclude thatform a linear pair.

We know that sum of the angles of a linear pair is.

or

Reflex

(Vertically opposite angles), or

But, we are given that

Therefore, we can conclude thatand.

Question 2. In Fig. 6.14, lines XY and MN intersect at O. If and a:b = 2 : 3, find c.

Solution:
We are given thatand.

We need find the value of c in the given figure.

Let a be equal to 2x and b be equal to 3x.

Therefore

Now[Linear pair]

Question 3. In the given figure,, then prove that .

Solution:
We need to prove that.

We are given that.

From the given figure, we can conclude thatform a linear pair.

We know that sum of the angles of a linear pair is.

and (i)

(ii)

From equations (i) and (ii), we can conclude that

Therefore, the desired result is proved.

Question 4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.

Solution:
We need to prove that AOB is a line.

We are given that.

We know that the sum of all the angles around a fixed point is.

Thus, we can conclude that

But, (Given).

From the given figure, we can conclude that y and x form a linear pair.

We know that if a ray stands on a straight line, then the sum of the angles of linear pair formed by the ray with respect to the line is.

.

Therefore, we can conclude that AOB is a line.

Question 5. 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

Solution:
We need to prove that.

We are given that OR is perpendicular to PQ, or

From the given figure, we can conclude that form a linear pair.

We know that sum of the angles of a linear pair is.

, or

.

From the figure, we can conclude that.

, or

.(i)

From the given figure, we can conclude thatform a linear pair.

We know that sum of the angles of a linear pair is.

, or

.(ii)

Substitute (ii) in (i), to get

Therefore, the desired result is proved.

Question 6. It is given thatand XY is produced to point P. Draw a figure from the given information. If ray YQ bisects, find

Solution:
We are given that, XY is produced to P and YQ bisects.

We can conclude the given below figure for the given situation:

We need to find.

From the given figure, we can conclude thatform a linear pair.

We know that sum of the angles of a linear pair is.

.

But.

Ray YQ bisects, or

.

Reflex

Therefore, we can conclude thatand Reflex

### NCERT Solutions for Class 9 Maths Exercise 6.2

Question 1. In figure, find the values of x and y and then show that AB || CD.

Solution:
In the figure, we have CD and PQ intersect at F.

∴ y = 130° …(1)
[Vertically opposite angles]
Again, PQ is a straight line and EA stands on it.
∠AEP + ∠AEQ = 180° [Linear pair]
or 50° + x = 180°
⇒ x = 180° – 50° = 130° …(2)
From (1) and (2), x = y
As they are pair of alternate interior angles.
∴ AB || CD

Question 2. In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Solution:
AB || CD, and CD || EF [Given]
∴ AB || EF
∴ x = z [Alternate interior angles] ….(1)
Again, AB || CD
⇒ x + y = 180° [Co-interior angles]
⇒ z + y = 180° … (2) [By (1)]
But y : z = 3 : 7
z = y = (180°- z) [By (2)]
⇒ 10z = 7 x 180°
⇒ z = 7 x 180° /10 = 126°
From (1) and (3), we have
x = 126°.

Question 3. In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Solution:
AB || CD and GE is a transversal.
∴ ∠AGE = ∠GED [Alternate interior angles]
But ∠GED = 126° [Given]
∴∠AGE = 126°
Also, ∠GEF + ∠FED = ∠GED
or ∠GEF + 90° = 126° [∵ EF ⊥ CD (given)]
x = z [Alternate interior angles]… (1) Again, AB || CD
⇒ x + y = 180° [Co-interior angles]
∠GEF = 126° -90° = 36°
Now, AB || CD and GE is a transversal.
∴ ∠FGE + ∠GED = 180° [Co-interior angles]
or ∠FGE + 126° = 180°
or ∠FGE = 180° – 126° = 54°
Thus, ∠AGE = 126°, ∠GEF=36° and ∠FGE = 54°.

Question 4. In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS.

Solution:
Draw a line EF parallel to ST through R.

Since PQ || ST [Given]
and EF || ST [Construction]
∴ PQ || EF and QR is a transversal
⇒ ∠PQR = ∠QRF [Alternate interior angles] But ∠PQR = 110° [Given]
∴∠QRF = ∠QRS + ∠SRF = 110° …(1)
Again ST || EF and RS is a transversal
∴ ∠RST + ∠SRF = 180° [Co-interior angles] or 130° + ∠SRF = 180°
⇒ ∠SRF = 180° – 130° = 50°
Now, from (1), we have ∠QRS + 50° = 110°
⇒ ∠QRS = 110° – 50° = 60°
Thus, ∠QRS = 60°.

Question 5. In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Solution:
We have AB || CD and PQ is a transversal.
∴ ∠APQ = ∠PQR
[Alternate interior angles]
⇒ 50° = x [ ∵ ∠APQ = 50° (given)]
Again, AB || CD and PR is a transversal.
∴ ∠APR = ∠PRD [Alternate interior angles]
⇒ ∠APR = 127° [ ∵ ∠PRD = 127° (given)]
⇒ ∠APQ + ∠QPR = 127°
⇒ 50° + y = 127° [ ∵ ∠APQ = 50° (given)]
⇒ y = 127°- 50° = 77°
Thus, x = 50° and y = 77°.

Question 6. In figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

Solution:
Draw ray BL ⊥PQ and CM ⊥ RS

∵ PQ || RS ⇒ BL || CM
[∵ BL || PQ and CM || RS]
Now, BL || CM and BC is a transversal.
∴ ∠LBC = ∠MCB …(1) [Alternate interior angles]
Since, angle of incidence = Angle of reflection
∠ABL = ∠LBC and ∠MCB = ∠MCD
⇒ ∠ABL = ∠MCD …(2) [By (1)]
Adding (1) and (2), we get
∠LBC + ∠ABL = ∠MCB + ∠MCD
⇒ ∠ABC = ∠BCD
i. e., a pair of alternate interior angles are equal.
∴ AB || CD.

### NCERT Solutions for Class 9 Maths Exercise 6.3

Question 1. In the given figure, sides QP and RQ of ∆PQR are produced to points S and T respectively. If SPR = 135º and PQT = 110º, find PRQ.

Solution:
We are given that and .

We need to find the value ofin the figure given below.

From the figure, we can conclude thatform a linear pair.

We know that the sum of angles of a linear pair is.

and

and

Or,

From the figure, we can conclude that

(Angle sum property)

Therefore, we can conclude that.

Question 2. In the given figure, X = 62º, XYZ = 54º. If YO and ZO are the bisectors of XYZ and XZY respectively of ∆XYZ, find OZY and YOZ.

Solution:
We are given thatand YO and ZO are bisectors of, respectively.

We need to find in the figure.

From the figure, we can conclude that in

(Angle sum property)

We are given that OY and OZ are the bisectors of, respectively.

and

From the figure, we can conclude that in

(Angle sum property)

Therefore, we can conclude thatand.

Question 3. In the given figure, if AB || DE, BAC = 35º and CDE = 53º, find DCE.

Solution:
We are given that,.

We need to find the value of in the figure given below.

From the figure, we can conclude that

(Alternate interior)

From the figure, we can conclude that in

(Angle sum property)

Therefore, we can conclude that.

Question 4. In the given figure, if lines PQ and RS intersect at point T, such that PRT = 40º, RPT = 95º and TSQ = 75º, find SQT.

Solution:
We are given that.

We need to find the value ofin the figure.

From the figure, we can conclude that in

(Angle sum property)

From the figure, we can conclude that

(Vertically opposite angles)

From the figure, we can conclude that in

(Angle sum property)

Therefore, we can conclude that.

Question 5. In the given figure, if, PQ || SR,, then find the values of x and y.

Solution:
We are given that.

We need to find the values of x and y in the figure.

We know that “If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.”

From the figure, we can conclude that

, or

From the figure, we can conclude that

(Alternate interior angles)

From the figure, we can conclude that

(Angle sum property)

Therefore, we can conclude that.

Question 6. In the given figure, the side QR of ∆PQR is produced to a point S. If the bisectors of meet at point T, then prove that.

Solution:
We need to prove thatin the figure given below.

We know that “If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.”

From the figure, we can conclude that in,is an exterior angle

…(i)

From the figure, we can conclude that in,is an exterior angle

We are given that are angle bisectors of

We need to substitute equation (i) in the above equation, to get

Therefore, we can conclude that the desired result is proved.

## Conclusions for NCERT SOLUTIONS FOR CLASS 9 MATHS CHAPTER 6 LINES AND ANGLES

SWC academic staff has developed NCERT answers for this chapter of the ninth grade mathematics curriculum. We have solutions prepared for all of the exercises of this chapter. The answers, broken down into steps, to all of the questions included in the NCERT textbook’s chapter are provided here. Read this chapter on theory. Be certain that you have read the theory section of this chapter of the NCERT textbook and that you have learnt the formulas for the chapter that you are studying.