NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1.
Check out the latest CBSE NCERT Class 9 Maths Syllabus.The syllabus is for the academic year 2020-21 session. First, of all check the CBSE Class 9 Mathematics Syllabus in PDF formate with exam pattern. Students are advised to check out the complete syllabus. Class 9 Maths Exam Pattern. Here in this Section, we have mentioned the Class 9 Maths Exam Pattern. If you are searching for NCERT Solutions for Class 9 Maths, you have reached the correct place.LearnCBSE.in has created most accurate and detailed solutions for Class 9 Maths NCERT solutions. Class 9 Maths NCERT Solutions includes all the questions provided as per new revised syllabus in Class 9 math NCERT textbook. Creating awareness among schools as to how mathematics laboratory will help in improving teaching and learning of the subject and providing general guidelines to school on setting up and using a mathematics laboratory. Besides, it also included a number of suggested hands-on activities related to concepts in mathematics for Class. Check out the latest CBSE NCERT Class 9 Maths Syllabus.The syllabus is for the academic year 2020-21 session. First, of all check the CBSE Class 9 Mathematics Syllabus in PDF formate with exam pattern. Students are advised to check out the complete syllabus. Tama serial number lookup. Aug 06, 2020 NCERT Solutions for Class 9 Maths are given for the students so that they can get to know the answers to the questions in case they are not able to find it.It is important for all the students who are in Class 9 currently. Here we are providing the solutions to all the chapters of NCERT Maths Class 9 Textbook for the students.
Class 9 Maths Ncert Lab Manual Pdf
Board | CBSE |
Textbook | NCERT |
Class | Class 9 |
Subject | Maths |
Chapter | Chapter 4 |
Chapter Name | Lines and Angles |
Exercise | Ex 4.1 |
Number of Questions Solved | 6 |
Category | NCERT Solutions |
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1
Question 1.
In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Solution:
Here, ∠AOC and ∠BOD are vertically opposite angles.
∴ ∠AOC = ∠BOD
⇒ ∠AOC = 40° [∵ ∠BOD = 40°(Given)] …(i)
We have, ∠AOC + ∠ BOE = 70° (Given)
40°+ ∠BOE = 70° [From Eq. (i)]
⇒ ∠BOE = 30°
Also, ∠AOC + ∠COE + ∠BOE = 180° (Linear pair axiom)
⇒ 40° + ∠COE + 30° = 180°
⇒ ∠COE = 110°
Now, ∠COE + reflex ∠COE = 360° (Angles at a point)
110°+reflex ∠COE = 360°
⇒ Reflex ∠COE = 250°
In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Solution:
Here, ∠AOC and ∠BOD are vertically opposite angles.
∴ ∠AOC = ∠BOD
⇒ ∠AOC = 40° [∵ ∠BOD = 40°(Given)] …(i)
We have, ∠AOC + ∠ BOE = 70° (Given)
40°+ ∠BOE = 70° [From Eq. (i)]
⇒ ∠BOE = 30°
Also, ∠AOC + ∠COE + ∠BOE = 180° (Linear pair axiom)
⇒ 40° + ∠COE + 30° = 180°
⇒ ∠COE = 110°
Now, ∠COE + reflex ∠COE = 360° (Angles at a point)
110°+reflex ∠COE = 360°
⇒ Reflex ∠COE = 250°
![Manual Manual](https://farm5.staticflickr.com/4665/26273086448_684d39f437_o.png)
Question 2.
In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. Hitman 2 silent assassin hd cheats ps3. find c.
Solution:
We have, ∠POY = 90°
⇒ ∠POY + ∠POX = 180° (Linear pair axiom)
⇒ ∠POX = 90°
⇒ a+b = 90°
Also, a : b = 2 : 3 (Given)
⇒ Let a = 2k,b = 3k
Now, from Eq. (j), we get
2k + 3k = 90°
⇒ 5k = 90°
⇒ k = 18°
∴ a = 2 x 18°=36°
and b=3 x 18°=54°
Now, ∠MOX + ∠XON = 1800 (Linear pair axiom)
b+ c = 180°
⇒ 540 + c= 180°
⇒ c = 126°
In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. Hitman 2 silent assassin hd cheats ps3. find c.
Solution:
We have, ∠POY = 90°
⇒ ∠POY + ∠POX = 180° (Linear pair axiom)
⇒ ∠POX = 90°
⇒ a+b = 90°
Also, a : b = 2 : 3 (Given)
⇒ Let a = 2k,b = 3k
Now, from Eq. (j), we get
2k + 3k = 90°
⇒ 5k = 90°
⇒ k = 18°
∴ a = 2 x 18°=36°
and b=3 x 18°=54°
Now, ∠MOX + ∠XON = 1800 (Linear pair axiom)
b+ c = 180°
⇒ 540 + c= 180°
⇒ c = 126°
Question 3.
In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Solution:
∵ ∠PQS+ ∠PQR = 180° (Linear pair axiom) ,.(i)
and ∠PRT + ∠PRQ = 180° (Linear pair axiom). .(ii)
From Eqs. (i) and (ii), we get
∠PQS + ∠PQR =∠PRT + ∠PRQ
∠PQS + ∠PRQ =∠PRT + ∠PRQ
[Given, ∠PQR = ∠PRQ]
⇒ ∠PQS = ∠PRT
In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Solution:
∵ ∠PQS+ ∠PQR = 180° (Linear pair axiom) ,.(i)
and ∠PRT + ∠PRQ = 180° (Linear pair axiom). .(ii)
From Eqs. (i) and (ii), we get
∠PQS + ∠PQR =∠PRT + ∠PRQ
∠PQS + ∠PRQ =∠PRT + ∠PRQ
[Given, ∠PQR = ∠PRQ]
⇒ ∠PQS = ∠PRT
Question 4.
In figure, if x + y = w + z, then prove that AOB is a line.
Solution:
∵ x+ y+w+ z = 360° (Angle at a point)
x + y = w + z (Given)…(i)
∴ x+ y+ x+ y = 360° [From Eq. (i)]
2(x + y) = 360°
⇒ x + y = 180° (Linear pair axiom)
Hence, AOB is a straight line.
In figure, if x + y = w + z, then prove that AOB is a line.
Solution:
∵ x+ y+w+ z = 360° (Angle at a point)
x + y = w + z (Given)…(i)
∴ x+ y+ x+ y = 360° [From Eq. (i)]
2(x + y) = 360°
⇒ x + y = 180° (Linear pair axiom)
Hence, AOB is a straight line.
Question 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
Solution:
We have,
∠POR = ∠ROQ = 90° (∵ Given that, OR is perpendicular to PQ)
∴ ∠POS + ∠ROS = 90°
⇒ ∠ROS = 90° – ∠POS
On adding ∠ROS both sides, we get
2 ∠ROS = 90° – ∠POS + ∠ROS
⇒ 2 ∠ROS = (90° + ∠ROS) – ∠POS
⇒ 2∠ROS = ∠QOS – ∠POS (∵ ∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS)
⇒ ∠ROS = (frac { 1 }{ 2 }) (∠QOS – ∠POS)
Hence proved.
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
Solution:
We have,
∠POR = ∠ROQ = 90° (∵ Given that, OR is perpendicular to PQ)
∴ ∠POS + ∠ROS = 90°
⇒ ∠ROS = 90° – ∠POS
On adding ∠ROS both sides, we get
2 ∠ROS = 90° – ∠POS + ∠ROS
⇒ 2 ∠ROS = (90° + ∠ROS) – ∠POS
⇒ 2∠ROS = ∠QOS – ∠POS (∵ ∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS)
⇒ ∠ROS = (frac { 1 }{ 2 }) (∠QOS – ∠POS)
Hence proved.
![Lab Lab](https://www.raajkart.com/media/catalog/product/cache/1/image/800x800/9df78eab33525d08d6e5fb8d27136e95/1/2/128.jpg)
Class 9 Maths Ncert Lab Manual Class 12 Biology
Question 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:
Here, YQ bisects ∠ZYP.
Hence, ∠ZYQ = ∠QYP = (frac { 1 }{ 2 }) ∠ZYP …….(i)
Given, ∠XYZ = 64° ….(ii)
∵ ∠XYZ + ∠ZYQ + ZQYP = 180° (Linear pair axiom)
⇒ 64° + ∠ZYQ + ∠ZYQ = 180° [From Eqs. (i) and (ii)]
⇒ 2 ∠ZYQ = 180° – 64°
⇒ ∠ZYQ = (frac { 1 }{ 2 }) x 116°
⇒ ∠ZYQ = 58°
∴ ∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58° = 122°
Now, ∠QYP + reflex ∠QYP = 360°
58° + reflex ∠QYP = 360°
⇒ reflex ∠ QYP = 302°
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:
Here, YQ bisects ∠ZYP.
Hence, ∠ZYQ = ∠QYP = (frac { 1 }{ 2 }) ∠ZYP …….(i)
Given, ∠XYZ = 64° ….(ii)
∵ ∠XYZ + ∠ZYQ + ZQYP = 180° (Linear pair axiom)
⇒ 64° + ∠ZYQ + ∠ZYQ = 180° [From Eqs. (i) and (ii)]
⇒ 2 ∠ZYQ = 180° – 64°
⇒ ∠ZYQ = (frac { 1 }{ 2 }) x 116°
⇒ ∠ZYQ = 58°
∴ ∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58° = 122°
Now, ∠QYP + reflex ∠QYP = 360°
58° + reflex ∠QYP = 360°
⇒ reflex ∠ QYP = 302°
Ncert Maths Solutions Of Class 9
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