NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3 Solutions

Content Curator

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.3 Solutions are based on the graph of linear equations.

Download PDF: NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3 Solutions

Check out NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3 Solutions

Exercise Solutions of Class 9 Maths Chapter 4 Linear Equation in Two Variables

Also check other Exercise Solutions of Class 9 Maths Chapter 4 Linear Equation in Two Variables

Also Check:

Chapter Related Topics
Pair of Linear Equations in Two Variables Formula	Pair of Linear Equations in Two Variables Revision Notes	Linear Equation in Two Variables

Also check:

Math Study Guides
Maths Important Formulas	Maths MCQs	Comparison Articles in Maths
Difference Between Topics in Maths	Math Study Material	Algebra
Trigonometry	Number Systems	Mensuration
Math Formula	NCERT Solutions for Class 9 Maths	Geometry

CBSE X Related Questions

1.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.
View Solution
2.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).
View Solution
3.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.
View Solution
4.
In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.
View Solution
5.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is
- \(4 \text{ cm}\)
- \(4\sqrt{2} \text{ cm}\)
- \(8 \text{ cm}\)
- \(2\sqrt{2} \text{ cm}\)
View Solution
6.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :
- \( 130^\circ \)
- \( 50^\circ \)
- \( 120^\circ \)
- \( 90^\circ \)
View Solution

View All

Similar Mathematics Concepts

Linear Equation in Two Variables: Solutions & Examples

NCERT Solutions for Class 9 Maths Chapter 4 : Linear Equations In Two Variables

Consistent and Inconsistent System of Equations

Linear Equations in Two Variables MCQs

Important Questions for Class 12 Maths Linear Equations in Two Variables

NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.1 Solutions

NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.2 Solutions

NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.4 Solutions

Linear Equations Formula: Derivation & Examples

Mathematics Preparation Guides

Statistics: Mean, Median, Variance and Cumulative Frequency Curve

Remainder Theorem: Definition, Steps & Examples

Angle Formula: Definition, Different Types, Solved Example

Trapezoids: Isosceles, Scalene Properties, Area, Properties, Examples

Quadrilateral Angle Sum Property: Theorem, Types & Example

Frequency Polygons: Steps, Formula & Examples

Area of Parallelogram: How to Calculate Area of Parallelogram?

Congruence of Triangles: SSS, SAS, ASA & RHS Rules

Quadrilateral Formulas: Types and Properties

Properties of Triangle: Types, Formulas & Examples

Class 9 Maths MCQ: Chapter-wise with Solutions

Properties of Parallel Lines: Transitive & Symmetric Properties

Isosceles triangle: Types, Properties, Angles & Examples

Operations on Rational Numbers:Rules, Properties & Examples

Addition and Subtraction of Integers: Rules & Examples

Scalene Triangle: Types, Formulas, Properties & Examples

Quadrant: Types, Sign Convention & Examples

Rationalize the Denominator: Explanation and Solved Examples

Area of a Kite: Formula, Derivation & Examples

Volume of a Pyramid: Formula and Solved Example

Mathematics NCERT Solutions

Linear Equation in Two Variables: Graph, Elimination and Substitution

Types of Triangles: Important Terms, Scalene, Acute, and Obtuse Triangle

Formula Of Perimeter Shapes

Sphere Formula: Diameter, Surface Area and Volume

Ordinate: Abscissa, Plotting Ordinates, Relation, Solved Examples

Integers as Exponents: Explanation, Rules and Method of Solving

Central Tendency: Definition, Methods and Sample Questions

Line Segment: Definition, Measurement and Examples

Line, Line Segment and Ray: Differentiation and Sample Questions

Area of Equilateral Triangle, Perimeter & Altitude Formulas

You can also check

NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3 Solutions

Exercise Solutions of Class 9 Maths Chapter 4 Linear Equation in Two Variables

CBSE X Related Questions

1.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

2.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

3.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

4.
In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.

5.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

6.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3 Solutions

Exercise Solutions of Class 9 Maths Chapter 4 Linear Equation in Two Variables

CBSE X Related Questions

1.PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

2.In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

3.In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

4.In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.

5.If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

6.In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

2.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

3.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

4.
In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.

5.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

6.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :