NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (2024)

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1.

Ex 6.1 Class 9 MathsQuestion 1
In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (1)
Solution:
Since AB is a straight line,
∴ ∠AOC + ∠COE + ∠EOB = 180°
or (∠AOC + ∠BOE) + ∠COE = 180° or 70° + ∠COE = 180° [ ∵∠AOC + ∠BOE = 70° (Given)]
or ∠COE = 180° – 70° = 110°
∴ Reflex ∠COE = 360° – 110° = 250°
Also, AB and CD intersect at O.
∴∠COA = ∠BOD [Vertically opposite angles]
But ∠BOD = 40° [Given]
∴ ∠COA = 40°
Also, ∠AOC + ∠BOE = 70°
∴ 40° + ∠BOE = 70° or ∠BOE = 70° -40° = 30°
Thus, ∠BOE = 30° and reflex ∠COE = 250°.

Ex 6.1 Class 9 MathsQuestion 2.
In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (2)
Solution:
Since XOY is a straight line.
∴ b+a+∠POY= 180°
But ∠POY = 90° [Given]
∴ b + a = 180° – 90° = 90° …(i)
Also a : b = 2 : 3 ⇒ b = \(\frac { 3a }{ 2 }\) …(ii)
Now from (i) and (ii), we get
\(\frac { 3a }{ 2 }\) + A = 90°
⇒ \(\frac { 5a }{ 2 }\) = 90°
⇒ a = \(\frac { { 90 }^{ \circ } }{ 5 } \times 2\quad =\quad { 36 }^{ \circ }\) = 36°
From (ii), we get
b = \(\frac { 3 }{ 2 }\) x 36° = 54°
Since XY and MN interstect at O,
∴ c = [a + ∠POY] [Vertically opposite angles]
or c = 36° + 90° = 126°
Thus, the required measure of c = 126°.

Ex 6.1 Class 9 MathsQuestion 3.
In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (3)
Solution:
ST is a straight line.
∴ ∠PQR + ∠PQS = 180° …(1) [Linear pair]
Similarly, ∠PRT + ∠PRQ = 180° …(2) [Linear Pair]
From (1) and (2), we have
∠PQS + ∠PQR = ∠PRT + ∠PRQ
But ∠PQR = ∠PRQ [Given]
∴ ∠PQS = ∠PRT

Ex 6.1 Class 9 MathsQuestion 4.
In figure, if x + y = w + ⇒, then prove that AOB is a line.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (4)
Solution:
Sum of all the angles at a point = 360°
∴ x + y + ⇒ + w = 360° or, (x + y) + (⇒ + w) = 360°
But (x + y) = (⇒ + w) [Given]
∴ (x + y) + (x + y) = 360° or,
2(x + y) = 360°
or, (x + y) = \(\frac { { 360 }^{ \circ } }{ 2 }\) = 180°
∴ AOB is a straight line.

Ex 6.1 Class 9 MathsQuestion 5.
In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (5)
Solution:
rara POQ is a straight line. [Given]
∴ ∠POS + ∠ROS + ∠ROQ = 180°
But OR ⊥ PQ
∴ ∠ROQ = 90°
⇒ ∠POS + ∠ROS + 90° = 180°
⇒ ∠POS + ∠ROS = 90°
⇒ ∠ROS = 90° – ∠POS … (1)
Now, we have ∠ROS + ∠ROQ = ∠QOS
⇒ ∠ROS + 90° = ∠QOS
⇒ ∠ROS = ∠QOS – 90° ……(2)
Adding (1) and (2), we have
2 ∠ROS = (∠QOS – ∠POS)
∴ ∠ROS = \(\frac { 1 }{ 2 } (\angle QOS-\angle POS)\)

Ex 6.1 Class 9 MathsQuestion 6.
It is given that ∠XYZ = 64° 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:
XYP is a straight line.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (6)
∴ ∠XYZ + ∠ZYQ + ∠QYP = 180°
⇒ 64° + ∠ZYQ + ∠QYP = 180°
[∵ ∠XYZ = 64° (given)]
⇒ 64° + 2∠QYP = 180°
[YQ bisects ∠ZYP so, ∠QYP = ∠ZYQ]
⇒ 2∠QYP = 180° – 64° = 116°
⇒ ∠QYP = \(\frac { { 116 }^{ \circ } }{ 2 }\) = 58°
∴ Reflex ∠QYP = 360° – 58° = 302°
Since ∠XYQ = ∠XYZ + ∠ZYQ
⇒ ∠XYQ = 64° + ∠QYP [∵∠XYZ = 64°(Given) and ∠ZYQ = ∠QYP]
⇒ ∠XYQ = 64° + 58° = 122° [∠QYP = 58°]
Thus, ∠XYQ = 122° and reflex ∠QYP = 302°.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (रेखाएँ और कोण) (Hindi Medium) Ex 6.1

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (7)
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (8)
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (9)
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (10)
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (11)
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (12)

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2

Ex 6.2 Class 9 MathsQuestion 1.
In figure, find the values of x and y and then show that AB || CD.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (13)
Solution:
In the figure, we have CD and PQ intersect at F.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (14)
∴ 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

Ex 6.2 Class 9 MathsQuestion 2.
In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (15)
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 = \(\frac { 7 }{ 3 }\) y = \(\frac { 7 }{ 3 }\)(180°- z) [By (2)]
⇒ 10z = 7 x 180°
⇒ z = 7 x 180° /10 = 126°
From (1) and (3), we have
x = 126°.

Ex 6.2 Class 9 MathsQuestion 3.
In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (16)
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°.

Ex 6.2 Class 9 MathsQuestion 4.
In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (17)
Solution:
Draw a line EF parallel to ST through R.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (18)
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°.

Ex 6.2 Class 9 MathsQuestion 5.
In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (19)
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°.

Ex 6.2 Class 9 MathsQuestion 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.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (20)
Solution:
Draw ray BL ⊥PQ and CM ⊥ RS
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (21)
∵ 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 Chapter 6 Lines and Angles Ex 6.3

Ex 6.3 Class 9 MathsQuestion 1.
In figure, sides QP and RQ of ∆PQR are produced to points S and T, respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (22)
Solution:
We have, ∠TQP + ∠PQR = 180°
[Linear pair]
⇒ 110° + ∠PQR = 180°
⇒ ∠PQR = 180° – 110° = 70°
Since, the side QP of ∆PQR is produced to S.
⇒ ∠PQR + ∠PRQ = 135°
[Exterior angle property of a triangle]
⇒ 70° + ∠PRQ = 135° [∠PQR = 70°]
⇒ ∠PRQ = 135° – 70° ⇒ ∠PRQ = 65°

Ex 6.3 Class 9 MathsQuestion 2.
In figure, ∠X = 62°, ∠XYZ = 54°, if YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (23)
Solution:
In ∆XYZ, we have ∠XYZ + ∠YZX + ∠ZXY = 180°
[Angle sum property of a triangle]
But ∠XYZ = 54° and ∠ZXY = 62°
∴ 54° + ∠YZX + 62° = 180°
⇒ ∠YZX = 180° – 54° – 62° = 64°
YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively.
∴ ∠OYZ = \(\frac { 1 }{ 2 } \angle XYZ\) = \(\frac { 1 }{ 2 }\)(54°) = 27°
and ∠OZY = \(\frac { 1 }{ 2 } \angle YZX\) = \(\frac { 1 }{ 2 }\)(64°) = 32°
Now, in ∆OYZ, we have
∠YOZ + ∠OYZ + ∠OZY = 180°
[Angle sum property of a triangle]
⇒ ∠YOZ + 27° + 32° = 180°
⇒ ∠YOZ = 180° -27° – 32° = 121°
Thus, ∠OZY = 32° and ∠YOZ = 121°

Ex 6.3 Class 9 MathsQuestion 3.
In figure, if AB || DE, ∠BAC = 35° and ∠CDE = 539 , find ∠DCE.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (24)
Solution:
AB || DE and AE is a transversal.
So, ∠BAC = ∠AED
[Alternate interior angles]
and ∠BAC = 35° [Given]
∴ ∠AED = 35°
Now, in ∆CDE, we have ∠CDE + ∠DEC + ∠DCE = 180°
{Angle sum property of a triangle]
∴ 53° + 35° + ∠DCE =180°
[∵ ∠DEC = ∠AED = 35° and∠CDE = 53° (Given)]
⇒ ∠DCE = 180° – 53° – 35° = 92°
Thus, ∠DCE = 92°

Ex 6.3 Class 9 MathsQuestion 4.
In figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠TSQ = 75°, find ∠ SQT.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (25)
Solution:
In ∆PRT, we have ∠P + ∠R + ∠PTR = 180°
[Angle sum property of a triangle]
⇒ 95° + 40° + ∠PTR =180°
[ ∵ ∠P = 95°, ∠R = 40° (given)]
⇒ ∠PTR = 180° – 95° – 40° = 45°
But PQ and RS intersect at T.
∴ ∠PTR = ∠QTS
[Vertically opposite angles]
∴ ∠QTS = 45° [ ∵ ∠PTR = 45°]
Now, in ∆ TQS, we have ∠TSQ + ∠STQ + ∠SQT = 180°
[Angle sum property of a triangle]
∴ 75° + 45° + ∠SQT = 180° [ ∵ ∠TSQ = 75° and ∠STQ = 45°]
⇒ ∠SQT = 180° – 75° – 45° = 60°
Thus, ∠SQT = 60°

Ex 6.3 Class 9 MathsQuestion 5.
In figure, if PQ ⊥ PS, PQ||SR, ∠SQR = 2S° and ∠QRT = 65°, then find the values of x and y.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (26)
Solution:
In ∆ QRS, the side SR is produced to T.
∴ ∠QRT = ∠RQS + ∠RSQ
[Exterior angle property of a triangle]
But ∠RQS = 28° and ∠QRT = 65°
So, 28° + ∠RSQ = 65°
⇒ ∠RSQ = 65° – 28° = 37°
Since, PQ || SR and QS is a transversal.
∴ ∠PQS = ∠RSQ = 37°
[Alternate interior angles]
⇒ x = 37°
Again, PQ ⊥ PS ⇒ AP = 90°
Now, in ∆PQS,
we have ∠P + ∠PQS + ∠PSQ = 180°
[Angle sum property of a triangle]
⇒ 90° + 37° + y = 180°
⇒ y = 180° – 90° – 37° = 53°
Thus, x = 37° and y = 53°

Ex 6.3 Class 9 MathsQuestion 6.
In figure, the side QR of ∆PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (27)
Solution:
In ∆PQR, side QR is produced to S, so by exterior angle property,
∠PRS = ∠P + ∠PQR
⇒ \(\frac { 1 }{ 2 }\)∠PRS = \(\frac { 1 }{ 2 }\)∠P + \(\frac { 1 }{ 2 }\)∠PQR
⇒ ∠TRS = \(\frac { 1 }{ 2 }\)∠P + ∠TQR …(1)
[∵ QT and RT are bisectors of ∠PQR and ∠PRS respectively.]
Now, in ∆QRT, we have
∠TRS = ∠TQR + ∠T …(2)
[Exterior angle property of a triangle]
From (1) and (2),
we have ∠TQR + \(\frac { 1 }{ 2 }\)∠P = ∠TQR + ∠T
⇒ \(\frac { 1 }{ 2 }\)∠P = ∠T
⇒ \(\frac { 1 }{ 2 }\)∠QPR = ∠QTR or ∠QTR = \(\frac { 1 }{ 2 }\)∠QPR

NCERT Solutions for Class 9 Maths All Chapters

  1. Chapter 1 Number systems
  2. Chapter 2 Polynomials
  3. Chapter 3 Coordinate Geometry
  4. Chapter 4 Linear Equations in Two Variables
  5. Chapter 5 Introduction to Euclid Geometry
  6. Chapter 6 Lines and Angles
  7. Chapter 7 Triangles
  8. Chapter 8 Quadrilaterals
  9. Chapter 9 Areas of Parallelograms and Triangles
  10. Chapter 10 Circles
  11. Chapter 11 Constructions
  12. Chapter 12 Heron’s Formula
  13. Chapter 13 Surface Areas and Volumes
  14. Chapter 14 Statistics
  15. Chapter 15 Probability
  16. Class 9 Maths (Download PDF)

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NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (2024)

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