NCERT Solutions for Class 9 Maths chapter-6 Lines And Angles
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NCERT Solutions for Class 9 Maths chapter-6 Lines And Angles
NCERT solutions for class 9 maths chapter 6 Lines and Angles is prepared by academic team of Physics Wallah. We have prepared NCERT Solutions for all exercise of chapter 6. Given below is step by step solutions of all questions given in NCERT textbook for chapter 6. Read chapter 6 theory make sure you have gone through the theory part of chapter-6 from NCERT textbook and you have learned the formula of the given chapter. Physics Wallah prepared a detail notes and additional questions for class 9 maths with short notes of all maths formula of class 9 maths. Do read these contents before moving to solve the exercise of NCERT chapter 6.
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 that
and 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.
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Additional Resource and Notes for class 9 Maths
4. Pair of Linear Equations in Two Variables
