NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4

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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.4 is given in this article. Class 10 Maths Chapter 4 Exercise 4.4 has word problems and questions for solving quadratic equations.

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Check out the solutions of Class 10 Maths NCERT solutions chapter 4 Quadratic Equations 4.4

Check out other exercise solutions of Class 10 Maths Chapter 4

Class 10 Chapter 4 Topics:

Nature of Roots	Discriminant Formula	Quadratic Interpolation Formula
Quadratic Equations Formula	Quadratic Equations MCQ	Complex Numbers and Quadratic Equations
Arithmetic Sequence Formula	Fundamental Theorem of Arithmetic	Arithmetic Progression Formula

CBSE Class 10 Maths Study Guides:

Geometry	Number System	Class 10 Notes
Trigonometry	Math Study Notes	Math Formula
Calculus	Mensuration	Arithmetic
Difference between in Maths	NCERT Solutions for Class 10 Maths	Math MCQs

CBSE X Related Questions

1.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals
- \(2.5\) cm
- \(10.5\) cm
- \(3.5\) cm
- \(\frac{14}{3}\) cm
View Solution
2.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)
View Solution
3.
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
4.
The HCF of 960 and 432 is :
- 48
- 54
- 72
- 36
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

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NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3

CBSE X Syllabus

Mathematics Preparation Guides

Arithmetic Progressions: Properties, Formulas and Solved Examples

Distance Formula and Derivation of Coordinate Geometry

Probability- A Theoretical Approach: An Overview and Examples

Area of Circle: Area Formula, Derivation with Solved Examples

Mode of Grouped Data

Pythagoras Theorem: Formula, Theorem, Proof and Examples

Nature of Roots of Quadratic Equation

Probability: Concepts, Problems and Solved Examples

Trigonometry: Ratios, Identities & Trigonometric Table

Heights and Distances: Applications in Trigonometry

Real Numbers: Euclid’s Division Lemma, Properties & Sets

Surface Areas and Volumes Revision Notes

Circles: Definition, Properties, Parts, Theorems

Constructions Revision Notes

Pair of Linear Equations In Two Variables Important Questions

Circles Revision Notes: Theorems of Tangent to the Circle

Statistics Important Questions

Mathematics NCERT Solutions

Euclid's Division Lemma: Proof, Finding HCF & Examples

Area of a Sector: Definition, Formulae and Solved Examples

Real Numbers: Definition, Properties and Examples

Events in Probability: Definition, Types, Examples

Relation Between HCF and LCM: Definition, Methods and Formulas

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NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4

CBSE X Related Questions

1.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

2.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

3.
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.

4.
The HCF of 960 and 432 is :

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

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NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4

CBSE X Related Questions

1.In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

2.Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

3.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.

4.The HCF of 960 and 432 is :

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.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

2.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

3.
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.

4.
The HCF of 960 and 432 is :

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 :