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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.
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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°
Question 2.
In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. 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. 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°
FREE CBSE Videos for Science, Maths, Mathematics, English Grammar for Class 9. Every class in your academics are important. The most important reason why class 9th is important is: that you have to pass class 9th to appear for the board exams in class 10th. The third reason why class 9th is important is: that the continuation of class 9th is in class 11th, so you will get the basics cleared in class 9th.
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.
Grand theft auto 4. 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.
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°
We hope the NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1, drop a comment below and we will get back to you at the earliest.
CBSE has released the revised syllabus for all subjects of class 9. Amid COVID-19 pandemic, the board has reduced the syllabus by 30%. We are providing here the CBSE syllabus of Class 9 Mathematics subject for the academic session 2020-21. This latest CBSE Class 9 Maths syllabus gives details of all topics and lessons to be prepared in the current academic session. It also mentions the unit-wise weightage and question paper design for the annual examination. All the CBSE affiliated schools are expected to follow the same pattern while preparing the question papers for the exam. With this article, students may read and download the CBSE Class 9 Maths Syllabus 2020-21 in its revised form.
Important* CBSE Class 9 Online Resources for Self-Study during COVID-19 Lockdown
Find below the complete syllabus for CBSE Class 9 Mathematics:
CBSE Class 9 Mathematics Unit-Wise Weightage
Units | Unit Name | Marks |
I | NUMBER SYSTEMS | 08 |
II | ALGEBRA | 17 |
III | COORDINATE GEOMETRY | 04 |
IV | GEOMETRY | 28 |
V | MENSURATION | 13 |
VI | STATISTICS & PROBABILTY | 10 |
Total | 80 |
Also Check: CBSE Class 9 Maths Deleted Portion of Syllabus for 2020-2021
You may also download the latest NCERT Book and Solutions for Class 9 Maths to help you effectively study throughout the year and make effective preparations for the periodic tests and the CBSE Annual Examinations.
Also Check: CBSE Class 9 Revised Syllabus of All Subjects for Academic Year 2020-2021
UNIT I: NUMBER SYSTEMS
1. Real Numbers (10 Periods)
1. Review of representation of natural numbers, integers, rational numbers on the number line. Rational numbers as recurring/ terminating decimals. Operations on real numbers.
2. Examples of non-recurring/non-terminating decimals. Existence of non-rational numbers (irrational numbers) such as and their representation on the number line.
3. Rationalization (with precise meaning) of real numbers of the type (and their combinations) where x and y are natural number and a and b are integers.
6. Recall of laws of exponents with integral powers. Rational exponents with positive real bases (to be done by particular cases, allowing learner to arrive at the general laws.)
UNIT II: ALGEBRA
1. Polynomials (15 Periods)
Definition of a polynomial in one variable, with examples and counter examples. Coefficients of a polynomial, terms of a polynomial and zero polynomial. Degree of a polynomial. Constant, linear, quadratic and cubic polynomials. Monomials, binomials, trinomials. Factors and multiples. Zeros of a polynomial. Factorization of ax2 + bx + c, a ≠ 0 where a, b and c are real numbers, and of cubic polynomials using the Factor Theorem.
Recall of algebraic expressions and identities. Verification of identities:
and their use in factorization of polynomials.
2. Linear Equations in Two Variables (10 Periods)
Recall of linear equations in one variable. Introduction to the equation in two variables. Focus on linear equations of the type ax + by + c = 0. Prove that a linear equation in two variables has infinitely many solutions and justify their being written as ordered pairs of real numbers, plotting them and showing that they lie on a line. Graph of linear equations in two variables. Examples, problems from real life, including problems on Ratio and Proportion and with algebraic and graphical solutions being done simultaneously.
UNIT III: COORDINATE GEOMETRY
1. Coordinate Geometry (6 Periods)
The Cartesian plane, coordinates of a point, names and terms associated with the coordinate plane, notations, plotting points in the plane.
UNIT IV: GEOMETRY
1. Lines and Angles (13 Periods)
1. (Motivate) If a ray stands on a line, then the sum of the two adjacent angles so formed is 180o and the converse.
2. (Prove) If two lines intersect, vertically opposite angles are equal.
3. (Motivate) Results on corresponding angles, alternate angles, interior angles when a transversal intersects two parallel lines.
4. (Motivate) Lines which are parallel to a given line are parallel.
5. (Prove) The sum of the angles of a triangle is 180o.
6. (Motivate) If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.
2. Triangles (20 Periods)
1. (Motivate) Two triangles are congruent if any two sides and the included angle of one triangle is equal to any two sides and the included angle of the other triangle (SAS Congruence).
2. (Motivate) Two triangles are congruent if the three sides of one triangle are equal to three sides of the other triangle (SSS Congruence).
3. (Motivate) Two right triangles are congruent if the hypotenuse and a side of one triangle are equal (respectively) to the hypotenuse and a side of the other triangle (RHS Congruence).
4. (Prove) The angles opposite to equal sides of a triangle are equal.
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5. (Motivate) The sides opposite to equal angles of a triangle are equal.
4. Quadrilaterals (10 Periods)
1. (Prove) The diagonal divides a parallelogram into two congruent triangles.
2. (Motivate) In a parallelogram opposite sides are equal, and conversely.
3. (Motivate) In a parallelogram opposite angles are equal, and conversely.
4. (Motivate) A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and equal.
5. (Motivate) In a parallelogram, the diagonals bisect each other and conversely.
6. (Motivate) In a triangle, the line segment joining the mid points of any two sides is parallel to the third side and in half of it and (motivate) its converse.
6. Circles (12 Periods)
Through examples, arrive at definition of circle and related concepts-radius, circumference, diameter, chord, arc, secant, sector, segment, subtended angle.
1. (Prove) Equal chords of a circle subtend equal angles at the center and (motivate) its converse.
2. (Motivate) The perpendicular from the center of a circle to a chord bisects the chord and conversely, the line drawn through the center of a circle to bisect a chord is perpendicular to the chord.
3. (Motivate) Equal chords of a circle (or of congruent circles) are equidistant from the center (or their respective centers) and conversely.
4. (Prove) The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
5. (Motivate) Angles in the same segment of a circle are equal.
6. (Motivate) The sum of either of the pair of the opposite angles of a cyclic quadrilateral is 180o and its converse.
7. Constructions (5 Periods)
1. Construction of bisectors of line segments and angles of measure 60o, 90o, 45o etc., equilateral triangles.
2. Construction of a triangle given its base, sum/difference of the other two sides and one base angle.
UNIT V: MENSURATION