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CBSE Class 10 Maths Notes Chapter 13- Surface Area and Volumes

CBSE Notes for Class 10 Maths Chapter 13- Surface Area and Volumes

The topic experts make sure that the issues in the NCERT textbooks are solved according to the NCERT Syllabus and rules. It is more than just vital to practice the NCERT solutions for class 10 Math’s if students want to do well in the CBSE Class 10 2023 examinations.

The following are some of the subjects covered in Exercise 13.1 of 10th grade Math:

The solutions to the NCERT-based questions of Chapter 13 for Class 10 Maths, Surface Area and Volumes, is an essential study resource for Class 10 students. These assist students in comprehending or solving the sorts of problems that a paper setter asked in the first term examinations of CBSE Class 10 Maths. Furthermore, offering solutions to all relating to surface area and volumes assists students in effectively preparing for the Term I examinations.

 

CBSE Class 10 Maths Notes Chapter 13- Surface Area and Volumes | Free PDF Download

CBSE CLASS 10 MATHS NOTES 2022-23 | ALL CHAPTERS FREE PDF DOWNLOAD

 

The subject matter experts make sure that the errors in the NCERT textbooks are resolved according to the NCERT rules and syllabus.

 

The solutions to the NCERT-based questions of Chapter 13 for Grade 10 Maths are accessible at our website. These solutions will assist students in acing their first term examinations by preparing for the examinations ahead of time. Students may clarify all of their conceptual questions or doubts with the assistance of the notes of Maths NCERT Solutions for Class 10 of this chapter which we provide. Experts have created these NCERT Solutions based on the most recent update to the CBSE syllabus for 2022-23 to help students in Class 10 prepare well for their CBSE 2023 examinations.

 

The 10th-grade Maths Chapter 13 of NCERT Solutions includes topics related to surface area and volume of shapes like cylinder, sphere, cube, cone, cuboid, etc. It has formulas and theories related to area and volume of these shapes.

Chapter 13 Surface Area and Volumes is one of the parts of the Mensuration, and it has a total weightage of 10 points in the first term examinations. One question from this chapter is frequently asked in the first term test.

Parts of circles, their measurements, and the areas of planar figures are covered in Chapter 13 of Maths NCERT Solutions for Class 10. Our subject matter specialists have created solutions for each question based on the CBSE syllabus 2022-2023.

 

The following are some of the concepts covered in Chapter 13, Surface Area and Volumes are:

Surface Area: Surface area is a measurement of the entire area occupied by the object's surface.

 

Volume: Volume in math refers to the quantity of space in a 3D object. In this chapter we will learn how to calculate the volume of shapes like cylinder, sphere, cube, cone, cuboid, etc. with the help of some formulas. A fish tank, for example, is 3 feet long, 1 foot wide, and 2 feet tall. To calculate the volume, multiply the length by the breadth by the height, which is 3x1x2, six. As a result, the fish tank has six cubic feet.

  • Different shapes and their surface area and volume

Cuboid

  • It has six rectangular faces

Surface area = 2(lb + bh + lh) ( where l, b, h are length, breadth and      height respectively.

Volume = lbh

Cube

  • It has six square faces.

Surface area = 6a (where a is the side length of cube)

Volume = a3

Cylinder

It has 2 circular bases and are connected by surface area.

  • It has 2 disc and a curved surface

Surface area = 2 πrh (where r is the radius of the circular base and h is the height of the cylinder)

Total surface area=2 πrh+πr2

Volume = πr2h

 

Cone

  • It’s a 3D shape, the most common example of it is ice cream cone.

Surface area = 2 πrh (where r is the radius of the circular base and h is the height of the cone)

Total surface area= πrl + πr2 = πr (l + r)

Volume = (1/3) πr2h

Sphere

  • It’s a solid round shape whose every point is equidistant from the center.

Total surface area= 4πr2 (where r is the radius of the sphere)

Volume = (4/3) πr3

Hemisphere

  • It’s a half cut sphere.

Total surface area= 3πr2 (where r is the radius of the sphere)

Volume = (2/3) πr3

Frustum

  • When a cone is cut then the part without the vertex becomes a frustum.

Total surface area= π (r1+r2) l, (where r1 is the radius of the upper circular part and r2 is the radius of lower circular part)

Volume = (1/3) πh (r12 + r2+ r1r2)

 

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Frequently Asked Questions for Class 10 Maths Chapter 13 NCERT Solutions

  1. What are the benefits of studying NCERT Solutions for Class 10 Maths Chapter 13?

Answer: NCERT Class 10 Maths Chapter 13 solutions assist students in obtaining a summary of the question paper format, which includes a range of questions, such as usually repeated questions, brief and long response type questions, multiple-choice questions, marks, and so on. The more problems you solve, the more confident you will be in your achievement.

 

  1. Mention exam-relevant concepts in NCERT Solutions for Class 10 Maths Chapter 13?

Answer: The most important exam topics in NCERT Solutions for Class 10 Maths Chapter 13 are Introduction, Surface Area of a Combination of Solids, Volume of a Combination of Solids, Conversion of Solid from One Shape to Another, Frustum of a Cone and finally the Summary.

 

  1. What number of exercises in NCERT Solutions for Class 10 Maths Chapter 13?

Answer: It includes five exercises. The first exercise has eight questions. The second contains 8 questions. The third exercise contains 9. The fourth exercise contains 5 questions. The fifth exercise contains 7 questions based on introduction, Surface Area of a Combination of Solids, Volume of a Combination of Solids, Conversion of Solid from One Shape to Another and Frustum of a Cone respectively.

 

  1. What is the formula for the surface area and volume of the frustum?

Answer: When a cone is cut then the part without the vertex becomes a frustum.

Total surface area= π (r1+r2) l, (where r1 is the radius of the upper circular part and r2 is the radius of lower circular part)

Volume = (1/3) πh (r12 + r2+ r1r2)

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