The National Council of Educational Research and Training (NCERT) is an autonomous body of the Indian government that formulates the curricula for schools in India that are governed by the Central Board of Secondary Education (CBSE) and certain state boards. Therefore, students who will be taking the Class 10 tests administered by various boards should consult this NCERT Syllabus in order to prepare for those examinations, which in turn will assist those students get a passing score.
When working through the exercises in the NCERT textbook, if you run into any type of difficulty or uncertainty, you may use the swc NCERT Solutions for class 9 as a point of reference. While you are reading the theory form textbook, it is imperative that you always have notes prepared. You should make an effort to understand things from the very beginning so that you may create a solid foundation in the topic. Use the NCERT as your parent book to ensure that you have a strong foundation. After you have finished reading the theoretical section of the textbook, you should go to additional reference books.
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 that
and
.
We need to find
.
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
Reflex 
(Vertically opposite angles), or
But, we are given that
Therefore, we can conclude that
and
.
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 that
and
.
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 that
form 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 that
form 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 that
form 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 that
and 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 of
in the figure given below.
From the figure, we can conclude that
form 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 that
and 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 that
and
.
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 of
in 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 that
in 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.
