Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  Short & Long Question Answer: Lines and Angle

Class 9 Maths Chapter 6 Question Answers - Lines & Angles

Q1: In the figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Class 9 Maths Chapter 6 Question Answers - Lines & AnglesSol:
From the given figure, we can see;
∠AOC, ∠BOE, ∠COE and ∠COE, ∠BOD,  ∠BOE form a straight line each.
So, ∠AOC + ∠BOE +∠COE = ∠COE +∠BOD + ∠BOE = 180°
Now, by substituting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° we get:
70° +∠COE = 180°
∠COE = 110°
Similarly,
110° +  40° + ∠BOE = 180°
∠BOE = 30°

Q2: In the Figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2(∠QOS – ∠POS).
Class 9 Maths Chapter 6 Question Answers - Lines & AnglesSol:

In the question, it is given that (OR ⊥ PQ) and ∠POQ = 180°
So, ∠POS + ∠ROS + ∠ROQ = 180°    (Linear pair of angles)
Now, ∠POS + ∠ROS = 180° – 90°     (Since ∠POR = ∠ROQ = 90°)
∴ ∠POS + ∠ROS = 90°
Now, ∠QOS = ∠ROQ + ∠ROS
It is given that ∠ROQ = 90°,
∴ ∠QOS = 90° + ∠ROS
Or, ∠QOS – ∠ROS = 90°
As ∠POS + ∠ROS = 90° and ∠QOS – ∠ROS = 90°, we get
∠POS + ∠ROS = ∠QOS – ∠ROS
⇒ 2 ∠ROS + ∠POS = ∠QOS
Or, ∠ROS = ½ (∠QOS – ∠POS) (Hence proved).

Q3: In the Figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Class 9 Maths Chapter 6 Question Answers - Lines & AnglesSol:

Since AB || CD GE is a transversal.
It is given that ∠GED = 126°
So, ∠GED = ∠AGE = 126° (alternate interior angles)
Also,
∠GED = ∠GEF + ∠FED
As
EF ⊥ CD, ∠FED = 90°
∴ ∠GED = ∠GEF + 90°
Or, ∠GEF = 126° – 90° = 36°
Again, ∠FGE + ∠GED = 180° (Transversal)
Substituting the value of ∠GED = 126° we get,
∠FGE = 54°
So,
∠AGE = 126°
∠GEF = 36° and
∠FGE = 54°

Q4: In Fig. 6.33, 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.
Class 9 Maths Chapter 6 Question Answers - Lines & AnglesSol:
First, draw two lines BE and CF such that BE ⊥ PQ and CF ⊥ RS.
Now, since PQ || RS,
So, BE || CF
Class 9 Maths Chapter 6 Question Answers - Lines & AnglesBE and CF are normals between the incident ray and reflected ray.As we know,
Angle of incidence = Angle of reflection (By the law of reflection)
So,
∠1 = ∠2 and
∠3 = ∠4
We also know that alternate interior angles are equal.
Here, BE ⊥ CF and the transversal line BC cuts them at B and C.
So, ∠2 = ∠3 (As they are alternate interior angles)
Now, ∠1 + ∠2 = ∠3 + ∠4
Or, ∠ABC = ∠DCB
So, AB ∥ CD (alternate interior angles are equal)

Q5: In the figure, if AB || CD || EF, PQ || RS, ∠RQD = 25° and ∠CQP = 60°, then find ∠QRS.
Class 9 Maths Chapter 6 Question Answers - Lines & AnglesSol:
According to the given figure, we have
AB || CD || EF
PQ || RS
∠RQD = 25°
∠CQP = 60°
PQ || RS.
As we know,
If a transversal intersects two parallel lines, then each pair of alternate exterior angles is equal.
Now, since, PQ || RS
⇒ ∠PQC = ∠BRS
We have ∠PQC = 60°
⇒ ∠BRS = 60° … eq.(i)
We also know that,
If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.
Now again, since, AB || CD
⇒ ∠DQR = ∠QRA
We have ∠DQR = 25°
⇒ ∠QRA = 25° … eq.(ii)
Using linear pair axiom,
We get,
∠ARS + ∠BRS = 180°
⇒ ∠ARS = 180° – ∠BRS
⇒ ∠ARS = 180° – 60° (From (i), ∠BRS = 60°)
⇒ ∠ARS = 120° … eq.(iii)
Now, ∠QRS = ∠QRA + ∠ARS
From equations (ii) and (iii), we have,
∠QRA = 25° and ∠ARS = 120°
Hence, the above equation can be written as:
∠QRS = 25° + 120°
⇒ ∠QRS = 145°

Q6: In the Figure, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
Class 9 Maths Chapter 6 Question Answers - Lines & AnglesSol: 
As we know, the sum of the linear pair is always equal to 180°
So,
∠POY + a + b = 180°
Substituting the value of ∠POY = 90° (as given in the question) we get,
a + b = 90°
Now, it is given that a : b = 2 : 3 so,
Let a be 2x and b be 3x.
∴ 2x + 3x = 90°
Solving this we get
5x = 90°
So, x = 18°
∴ a = 2 × 18° = 36°
Similarly, b can be calculated and the value will be
b = 3 × 18° = 54°
From the diagram, b + c also forms a straight angle so,
b + c = 180°
⇒ c + 54° = 180°
∴ c = 126°

Q7: 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.
Class 9 Maths Chapter 6 Question Answers - Lines & AnglesSol:
Here, XP is a straight line
So, ∠XYZ +∠ZYP = 180°
substituting the value of ∠XYZ = 64° we get,
64° +∠ZYP = 180°
∴ ∠ZYP = 116°
From the diagram, we also know that ∠ZYP = ∠ZYQ + ∠QYP
Now, as YQ bisects ∠ZYP,
∠ZYQ = ∠QYP
Or, ∠ZYP = 2∠ZYQ
∴ ∠ZYQ = ∠QYP = 58°
Again, ∠XYQ = ∠XYZ + ∠ZYQ
By substituting the value of ∠XYZ = 64° and ∠ZYQ = 58° we get.
∠XYQ = 64° + 58°
Or, ∠XYQ = 122°
Now, reflex ∠QYP = 180° + ∠XYQ
We computed that the value of ∠XYQ = 122°. So,
∠QYP = 180° + 122°
∴ ∠QYP = 302°

Q8: In the Figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
[Hint: Draw a line parallel to ST through point R.]

Class 9 Maths Chapter 6 Question Answers - Lines & AnglesSol: First, construct a line XY parallel to PQ.
Class 9 Maths Chapter 6 Question Answers - Lines & AnglesAs we know, the angles on the same side of the transversal are equal to 180°.So, ∠PQR + ∠QRX = 180°
Or,∠QRX = 180° – 110°
∴ ∠QRX = 70°
Similarly,
∠RST + ∠SRY = 180°
Or, ∠SRY = 180° – 130°
∴ ∠SRY = 50°
Now, for the linear pairs on the line XY-
∠QRX + ∠QRS + ∠SRY = 180°
Substituting their respective values we get,
∠QRS = 180° – 70° – 50°
Or, ∠QRS = 60°

Q9: In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of Δ XYZ, find ∠OZY and ∠YOZ.
Class 9 Maths Chapter 6 Question Answers - Lines & AnglesSol: 
As we know, the sum of the interior angles of the triangle is 180°.
So, ∠X +∠XYZ + ∠XZY = 180°
substituting the values as given in the question we get,
62° + 54° + ∠XZY = 180°
Or, ∠XZY = 64°
Now, As we know, ZO is the bisector so,
∠OZY = ½ ∠XZY
∴ ∠OZY = 32°
Similarly, YO is a bisector and so,
∠OYZ = ½ ∠XYZ
Or, ∠OYZ = 27° (As ∠XYZ = 54°)
Now, as the sum of the interior angles of the triangle,
∠OZY +∠OYZ + ∠O = 180°
Substituting their respective values we get,
∠O = 180° – 32° – 27°
Or, ∠O = 121°

The document Class 9 Maths Chapter 6 Question Answers - Lines & Angles is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9
48 videos|378 docs|65 tests
48 videos|378 docs|65 tests
Download as PDF
Explore Courses for Class 9 exam
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Download the FREE EduRev App
Track your progress, build streaks, highlight & save important lessons and more!
Related Searches

Objective type Questions

,

video lectures

,

past year papers

,

Class 9 Maths Chapter 6 Question Answers - Lines & Angles

,

pdf

,

Summary

,

Previous Year Questions with Solutions

,

Class 9 Maths Chapter 6 Question Answers - Lines & Angles

,

ppt

,

Viva Questions

,

MCQs

,

Exam

,

study material

,

Sample Paper

,

practice quizzes

,

Class 9 Maths Chapter 6 Question Answers - Lines & Angles

,

Free

,

shortcuts and tricks

,

Important questions

,

Extra Questions

,

Semester Notes

,

mock tests for examination

;