NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 Solutions

Content Curator

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volume Exercise 13.2 Solutions are based on Volume of Combination of Solids.

Download PDF NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 Solutions

Check out the solutions of NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 Solutions

Exercise Solutions of Class 10 Maths Chapter 13 Surface Areas and Volume

Also check other Exercise Solutions of Class 10 Maths Chapter 13 Surface Areas and Volume

Also Check:

Chapter Related Topics
Surface Areas And Volumes Revision Notes	Area Of Prism	Volume Of A Rectangular Prism Formula
Volume Of A Square Pyramid Formula	Volume Of A Triangular Prism Formula	Surface Area Of A Prism Formula
Surface Area Of A Pyramid Formula	Surface Area Of A Rectangular Prism Formula	Surface Area Of A Square Pyramid Formula
Surface Area of a Cylinder	Area of Hollow Cylinder	Surface Area of a Prism Formula

Also check:

Math Study Guides
Maths Important Formulas	Maths MCQs	Comparison Articles in Maths
Difference Between Topics in Maths	Math Study Material	Arithmetic
Class 10 Maths Notes	Permutations and Combinations	Mensuration
Trigonometry	Number Systems	Geometry
Math Formula	NCERT Solutions for Class 10 Maths	Algebra

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.
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
3.
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
4.
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
5.
If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).
View Solution
6.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.
View Solution

View All

Similar Mathematics Concepts

Surface Areas and Volumes Revision Notes

Area of Hollow Cylinder: Total Surface Area & Lateral Surface Area

Surface Area of a Cylinder Formula: Definition and Solved Examples

Area of Square: Formula & How to Find Area of Square?

Volume of a Sphere: Formula, Derivation & Solved Examples

NCERT Solutions for class 10 Mathematics Chapter 13: Surface Areas and Volumes

NCERT Solutions for Class 9 Maths Chapter 13 : Surface Areas And Volumes

Surface Area of Right Circular Cone: Formula, Examples and Characteristics

MCQ On Surface Area And Volume

Surface Areas and Volumes MCQs

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

Geometric Probability: Formula, Illustration, Model

Algebra Formula: Important Formulas in Algebra & Solved Examples

Angle of Depression: Definition, Formula and Solved Examples

Cyclic Quadrilateral: Properties, Theorems and Formulas

Polynomial Notation: Types, Remainder Theorem, Linear and Quadratic Equation

You can also check

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 Solutions

Exercise Solutions of Class 10 Maths Chapter 13 Surface Areas and Volume

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.
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.

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

5.
If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).

6.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 Solutions

Exercise Solutions of Class 10 Maths Chapter 13 Surface Areas and Volume

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.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.

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

5.If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).

6.D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

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.
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.

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

5.
If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).

6.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.