NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes (Ex 13.1)
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Mathematics is an important topic for pupils, both in school and after. Many professional paths and fields still require Mathematical ability after high school and college. Students need to get a solid understanding of the topic in order to make their daily lives easier. They can only do well on tests and provide quick calculations for answers if they have a thorough comprehension of the material. For the majority of college courses in engineering, economics, commerce, or mathematics honours, students must have taken mathematics as a mandatory subject in Class 12 and may also need to pass an admission exam in order to be accepted into the programme. For competitive exams, it must be given top priority because it is an important topic. Concepts and chapters covered in later classes are introduced to Class 10 mathematics students.
By giving students basic knowledge, assisting them in understanding theorems and concepts, and properly evaluating them through exercises and examples in the textbook, the Class 10 Mathematics curriculum aims to help students build a solid foundation in these areas. The exercises in the NCERT textbook are designed to help students get the most out of their skill and topic knowledge. Additionally, sufficient examples are provided to the students before each exercise, so they can refer to them and understand the questions with the help of the thorough explanations provided in the textbooks.
This page has the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1, which include all the questions and solutions for the exercise. These answers have been created by qualified teachers with years of expertise in instructing Mathematics to Class 10 students. The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1 are accessible in PDF format and include stepbystep explanations. For a seamless learning experience, students may download the 12.3 Class 10 Maths NCERT Solutions PDFs from the Extramarks website or Learning App.
Every question from Class 10 Maths Ex 12.3 is covered in the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1. Students should download the solutions for Class 10 Maths Chapter 12 Exercise 12.3 from reliable resources like Extramarks. The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1 are created in accordance with the requirements of the CBSE test in order to improve students’ comprehension of the material and aid in their academic success. The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1 are available as PDFs, which can aid students in better understanding the material as well as preparing them for their final exams. Class 10 students must excel in both Science and Mathematics, and Extramarks has precise and dependable sources like the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1.
NCERT Solutions for Class 10 Maths Chapter 12.3 and all the other exercises of the chapter are available for PDF download. The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1 are created by a professional instructor in accordance with the NCERT (CBSE) book rules. To help in reviewing the entire syllabus and getting more marks, Extramarks provides the questions and solutions for Class 10 Mathematics Chapter 13 Surface Areas and Volumes Exercise 13.1. Students can also download the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1 to help them review the entire syllabus and perform better on test day. On Extramarks, learners may discover NCERT Solutions for Class 10 Science in addition to NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1. Surface areas and volumes are covered in the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1. This chapter has six main sections that make it easier to comprehend the basics of surface areas and volumes. The six significant subjects addressed in Class 10 Maths Ex 13.1 Surface Areas and Volumes are mentioned below:
An Introduction
Surface Area of a Combination of Solids
The volume of a Combination of Solids
Conversion of Solid From One Shape to Another
Frustum of a Cone
A Summary
The Value of Understanding Surface Areas and Volumes
Calculating surface areas and volumes is very important in developing and determining the dimensions of any given thing. Mathematics strongly relies on measurements. Surface area is the twodimensional area defined by lengths, widths, and edges, whereas volume is the amount of space occupied by any given threedimensional object. By resolving and practising the sums presented in this chapter, students will be able to calculate surface areas and volumes of various figures efficiently in exams as well as in everyday situations.
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes (Ex 13.1) Exercise 13.1
Experienced academics in different domains assembled the knowledge of solid geometry in the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1 in a style that is simply understood by everyone. Additionally, the NCERT Textbooks are a trusted resource for students to practice and excel in geometry because the CBSE board strongly advises studying from them. If they rely on the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1 then Mathematics questions will become easy to cover.
Many crucial formulas are included in the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1, which the students should pay attention to. It is crucial that they do not mix up the formulas because each geometric figure has a specific formula. Some important formulas are given below:
Surface Area of a Cuboid = 2[(length x breadth) + (breadth x height) + (height x length)]
Volume of a Cylinder = πr2h
The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1 are available on the Extramarks website or mobile version. According to their needs, students can download the solution anytime. The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1 will be helping guide them during their problemsolving phase.
Access NCERT Solutions for Class 10 Maths Chapter 13 – Surface Areas and Volumes
The two categories of geometry are solid geometry and plane geometry. Solid geometry is the study of prisms, cylinders, cubes, pyramids, spheres, and other threedimensional objects, whereas plane geometry, also known as twodimensional geometry, is the study of lines and shapes such as squares, triangles, rectangles, and hexagons. Students may find more information on the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1 Surface Areas and Volumes in Chapter 13.
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1
For CBSE students preparing for exams, using the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1 is thought to be the best alternative. There are numerous exercises in this chapter. On this page, in a PDF format, The Extramarks offer the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1. Students can study this answer straight from the website or mobile app, or anyone can download it as needed.
The Extramarks internal subject matter experts carefully and in accordance with all CBSE regulations solved the problems and questions from the exercise. Any student in class 10 who is thoroughly familiar with all the ideas in the Subject Surface Areas and Volumes textbook and sufficiently knowledgeable about all the exercises provided in it can easily earn the highest possible score on the final examination. Students can discover the pattern of questions that may be asked in the test from this chapter and the chapter’s weight in marks with the aid of the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1. So that they can adequately study for the final exam.
There are numerous exercises in this chapter that contain numerous questions in addition to the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1. They are all guaranteed to be of the highest quality, and anyone can use them to study for exams. It is crucial to comprehend all the concepts in the textbooks and work through the exercises that are provided in order to receive the greatest grades possible in the class. For better test preparation, student should download the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1 from the Extramarks website. If students already have the Extramarks app on their phones, they may also download it there. The best feature of these solutions is that they can be used offline and online.
NCERT Solutions for Class 10 Maths Chapter 13 Exercises
Exercise 13.1 of Class 10 Mathematics has problems based on the following topics:
The surface area of a given cuboid
The surface area of a cylindrical vessel
The total surface area of a coneshaped object mounted on a hemisphere
The surface area of solid surmounted by a hemisphere
The surface area of the remained solid when a coneshaped is from a solid cylinder
Students can check the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1 for the entire Class 10 Maths Chapter 13 Exercise 13.1. They can also obtain additional study materials, past year question papers, revision notes, and study tips and tricks, to help them adequately prepare for the Mathematics exam. The questions in Maths Class 10 Chapter 13 Exercise 13.1 are developed utilising the guidelines provided in the chapter that comes before this exercise, as well as certain situations where the questions are already addressed. While learning how to calculate the surface areas of various forms, students may easily relate to the generic approaches that are employed in the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1.
The area or region that an object’s surface occupies is known as its surface area. Volume, on the other hand, refers to how much room an object has. There are numerous shapes and dimensions in geometry, including spheres, cubes, cuboids, cones, cylinders, etc. Each form has its volume and surface area. The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1 have all the simplified concepts to learn.
NCERT Solutions for Class 10 Maths Chapter 13 Exercises
Chapter 13 Ex 13.1 – 8 Questions
Chapter 13 Ex 13.2 – 11 Questions
Chapter 13 Ex 13.3 – 8 Questions
Chapter 13 Ex 13.4 – 9 Questions
Chapter 13 Ex 13.5 – 9 Questions
Chapter 13 Ex 13.6 – 8 Questions
Chapter 13 Ex 13.7 – 9 Questions
Chapter 13 Ex 13.8 – 10 Questions
Chapter 13 Ex 13.9 – 3 Questions
All the exercises are solved systematically in the NCERT Solution for Class 10 Mathematics. For a quick review, students can review the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1. The Extramarks come up with all these NCERT solutions For every class.
Q.1 Two cubes each of volume 64 cm^{3} are joined end to end. Find the surface area of the resulting cuboid.
Ans
$\begin{array}{l}\text{Given that,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}volume of each cube}={\text{64 cm}}^{3}\\ \text{or}{\left(\text{edge}\right)}^{3}=\text{64}\\ \text{or edge\hspace{0.17em}of each cube}=4\text{cm}\\ \text{The two cubes are joined end to end to form a cuboid of}\\ \text{dimensions 4 cm, 4 cm and 8 cm.}\\ \therefore \text{surface area of the cuboid}=2(\mathrm{lb}+\mathrm{bh}+\mathrm{lh})\\ \text{}=2(4\times 4+4\times 8+4\times 8)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=160{\text{cm}}^{2}\end{array}$
Q.2 A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Ans
$\begin{array}{l}\text{Height of cylindrical part}=\mathrm{h}=137=6\text{cm}\\ \text{Inner surface area of the vessel}\\ \text{}=\mathrm{Curved}\text{}\mathrm{surface}\text{}\mathrm{area}\text{}\mathrm{of}\text{cylindrical part}\\ \text{}\mathrm{Curved}\text{surface area of hemispherical part}\\ \text{}=2\mathrm{\pi rh}+2{\mathrm{\pi r}}^{2}=2\times \frac{22}{7}\times 7(6+7)=572{\text{cm}}^{2}\end{array}$
Q.3 A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Ans
$\begin{array}{l}\text{Radius of cone}=\text{Radius of hemisphere}=\mathrm{r}=3.5\text{cm}\\ \text{Height of hemisphere}=\text{Radius of hemisphere}=3.5\text{cm}\\ \text{Height of cone}=\mathrm{h}=15.53.5=12\\ \text{Slant height of cone}=\mathrm{l}=\sqrt{{\mathrm{h}}^{2}+{\mathrm{r}}^{2}}=\sqrt{{12}^{2}+{(3.5)}^{2}}=\frac{25}{2}\text{cm}\\ \text{Total surface area of toy}=\text{CSA of conical part}\\ \text{}+\text{CSA of hemispherical part}\\ \text{}=\mathrm{\pi rl}+2{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{22}{7}\times 3.5\times \frac{25}{2}+2\times \frac{22}{7}\times \text{\hspace{0.17em}}{(3.5)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=214.5{\text{\hspace{0.17em}cm}}^{2}\end{array}$
Q.4 A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Ans
$\begin{array}{l}\text{From figure it is obvious that greatest diameter of the}\\ \text{hemisphere is 7 cm.}\\ \therefore \text{Radius of hemisphere}=\mathrm{r}=\frac{7}{2}\text{cm}\\ \text{Total surface area}=\text{Surface area of cubical part}\\ \text{}+\text{CSA of hemispherical part}\\ \text{}\text{area of base of hemispherical part}\\ \text{}=6\times {\left(7\right)}^{2}+2{\mathrm{\pi r}}^{2}{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6\times {\left(7\right)}^{2}\text{\hspace{0.17em}\hspace{0.17em}}+\frac{22}{7}\times {\left(\frac{7}{2}\right)}^{2}\\ \text{}=332.5{\text{cm}}^{2}\end{array}$
Q.5 A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Ans
$\begin{array}{l}\text{Diameter of hemisphere}=\text{edge of cube}=\mathrm{l}\\ \therefore \text{Radius of hemisphere}=\frac{\mathrm{l}}{2}\text{}\\ \text{Total surface area of solid}\\ \text{}=\text{Surface area of cubical part}\\ \text{}+\text{CSA of hemispherical part}\\ \text{}\text{area of base of hemispherical part}\\ \text{}=6\times {\mathrm{l}}^{2}+2{\mathrm{\pi r}}^{2}{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6{\mathrm{l}}^{2}+\mathrm{\pi}{\frac{\mathrm{l}}{4}}^{2}\text{}\\ \text{}=\frac{1}{4}{\mathrm{l}}^{2}(24+\mathrm{\pi}){\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}unit}}^{2}\end{array}$
Q.6 A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see the following figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Ans
$\begin{array}{l}\text{Surface area of capsule}\\ \text{}=2\times \text{CSA of hemispherical part}\\ \text{}+\text{CSA of cylindrical part}\\ \text{}=4\mathrm{\pi}{\left(\frac{5}{2}\right)}^{2}+2\mathrm{\pi}\frac{5}{2}\times 9\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=220{\text{mm}}^{2}\end{array}$
Q.7 From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm^{2}.
Ans
$\begin{array}{l}\text{Slant height of conical part}=\mathrm{l}=\sqrt{{\mathrm{r}}^{2}+{\mathrm{h}}^{2}}=\sqrt{{(0.7)}^{2}+{(2.4)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2.5\text{cm}\\ \text{Required surface area}\\ \text{}=\text{curved surface area of cylindrical portion}\\ \text{}+\text{curved surface area of conical portion}\\ \text{}+\text{\hspace{0.17em}\hspace{0.17em}area of cylindrical base}\\ \text{}=\text{2}\mathrm{\pi}\text{rh}+\mathrm{\pi rl}+\mathrm{\pi}{\text{r}}^{2}\\ \text{}=2\times \frac{22}{7}\times 0.7\times 2.4+\frac{22}{7}\times 0.7\times 2.5+\frac{22}{7}\times {(0.7)}^{2}\\ \text{}=17.60{\text{cm}}^{2}\approx 18{\text{cm}}^{2}\end{array}$
Q.8 A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the following figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
Ans
$\begin{array}{l}\text{Surface area of the given article}\\ \text{}=\text{curved surface area of cylindrical portion}\\ \text{}+2\times \text{curved surface area of hemispherical portion}\\ \text{}=\text{2}\mathrm{\pi}\text{rh}+2\times 2{\mathrm{\pi r}}^{2}\\ \text{}=2\times \frac{22}{7}\times 3.5\times 10+2\times 2\times \frac{22}{7}\times {(3.5)}^{2}\\ \text{}=374{\text{cm}}^{2}\end{array}$
Q.9 A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
Ans
$\begin{array}{l}\text{Volume of the given solid}=\text{Volume of cone}\\ \text{}+\text{Volume of hemisphere}\\ \text{}=\frac{1}{3}\times {\mathrm{\pi r}}^{2}\mathrm{h}+\frac{2}{3}{\mathrm{\pi r}}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}\times \mathrm{\pi}{\left(1\right)}^{2}\left(1\right)+\frac{2}{3}\mathrm{\pi}{\left(1\right)}^{3}=\mathrm{\pi}{\text{cm}}^{3}\text{}\end{array}$
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FAQs (Frequently Asked Questions)
1. How to cover the maths Class 10 Chapter 13 Exercise 13.1 Mathematics?
The easiest way to start preparing for the Class 10 Chapter 13 Exercise 13.1 is with the CERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1. By using the NCERT Solutions For Class 10 Mathematics materials, the student will save their precious time. The student should also get into the habit of taking notes so that they can be reviewed throughout the preparation. Making notes and references from the CERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1 can help them remember what they have learned. Last but not least, it is advised that students use Exercise 13.1 Class 10 NCERT Solutions as a reference to get the most out of the chapter and to develop a core concept.
2. Is Class 10 Maths Chapter 13 Exercise 13.1 crucial for the Board Examination?
Maths Class 10 Chapter 13 Exercise 13.1 is crucial for the boards. Students might encounter lengthy wordproblemstyle questions in this chapter. All they need to do is remember the key formulas for surface areas and volumes and their applications because the problem is quite intriguing and enjoyable to tackle. The Extramarks study materials can be used to aid in their comprehension of all the concepts. They can refer to the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1.
3. How can CBSE students properly use the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1?
Basic tasks and exercises are included in Exercise 13.1 Class 10 NCERT Solutions to help students better comprehend surface areas and volumes in real space. They must attentively study the entire chapter because each and every line includes important details or information. For students to learn how to approach issues and use the appropriate geometric formulas, they must practise the examples that are given.
4. How many topics are covered in the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1 ?
The topics covered in the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13 are as follows: Explanation of the calculations of the surface area of a cuboid and a cube, surface area of a right circular cylinder and right circular cone, as well as the surface area of a sphere, volume of a cylinder, cuboid, right circular cone and the volume of a sphere.