You might have read about different ‘Shapes and Solid Figures‘ in Mathematics. Can you answer then what the shape of a cricket ball is? Isn’t your answer sphere shape? Objects like a football, a cricket ball, etc., are said to have the shape of a sphere.
A sphere is a real object in many fields of Mathematics. From a bubble of soap to the shape of the Earth, you can easily identify various spherical and hemispherical objects in nature. Because of its smooth-rolling property, most of the balls used in sports are given spherical shapes.
Mensuration, the branch of Mathematics, is an important topic for Class 10 students. You can easily spot a weightage of 6 marks from this unit in the CBSE Class 10 Mathematics exam. This unit consists of immediate problems related to the ‘Volume and Surface Area chapter.
This article will help you quickly revise important topics of the Class 10 Mathematics and calculate the area of a sphere.
Syllabus Of Class 10 Mathematics Exam
Before moving to revision or preparation of any subject for exams, it is a must to get an update about the exam syllabus. It is the most basic and important step for achieving an exceptional outcome.
The recently updated Class 10 Mathematics syllabus includes the following units and their weightage:
Name Of The Unit | Weightage |
Algebra(Cont.) | 10 |
Geometry(Cont.) | 9 |
Trigonometry(Cont.) | 7 |
Mensuration(Cont.) | 6 |
Statistics & Probability(Cont.) | 8 |
Total | 40 |
Internal Assessment | 10 |
Grand Total | 50 |
Exam Pattern Of Class 10 Mathematics
COVID-19 brings a lot of changes in the life of students too. CBSE has changed its exam pattern to help students with safety and precaution measures. Now, students of Class 10 and 12 have to face Term 1 and 2 board exams in different patterns with different time slots, two times a year.
The recently updated exam pattern of Class 10 Mathematics is as follows:
Section | Types Of Questions | Number Of Questions | Weightage |
Section-A | Very Short Answer Type Questions | 6 | 12 |
Section-B | Short Answer Type Questions | 4 | 12 |
Section-C | Long Answer Type Questions | 4 | 16 |
Total | 14 | 40 |
Know About Sphere
A sphere is a geometrical object with a three-dimensional analogue to a two-dimensional circle. It is the centre of a point that moves in space so that its distance from a fixed point remains constant. This fixed point is called the sphere’s centre, and the constant distance is called the sphere’s radius.
From a visual perspective, a sphere has a three-dimensional structure that forms by rotating a circular disc with one of the diagonals. The term sphere was the earliest mentioned by ancient Greek mathematicians. The famous Greek Mathematician Archimedes was the first person to determine the volume and area of the sphere.
Generally, the sphere’s radius is denoted by ‘r’ and its centre by ‘O’. If a radius is extended through the centre to the opposite side of the sphere, it creates a diameter. And it is denoted as ‘d’.
Surface Area Of A Sphere
The area covered by the sphere’s outer surface is known as the surface area. A sphere is a three-dimensional form of a circle.
Because a sphere is three-dimensional, it does not have an area. However, when depicted in two dimensions, a sphere is circular. Thus, to best understand a sphere in three dimensions, it is often helpful to start by understanding the area of a circle.
The area of a circle is found using the following equation:
Area of a circle = r².
Therefore, for a sphere,
Area of a sphere = 4r²
The great Mathematician Archimedes, that if the radius of a cylinder and sphere is ‘r’, the surface area of a sphere is equal to the lateral surface area of the cylinder.
Hence, the relation between the surface area of a sphere and the lateral surface area of a cylinder is given as
Surface Area of Sphere = Lateral Surface Area of Cylinder
As the lateral surface area of the cylinder = 2rh
Where r = radius of the circular side and h = height of the cylinder
But, in the case of a sphere,
Height of a sphere = Radius of a sphere,
Therefore,
The surface area of a sphere = 4r²
How To Calculate the Surface Area Of A Sphere?
The surface area of a sphere is the space occupied by its surface. The surface area of the sphere can be calculated using the formula of the surface area of the sphere. The steps to calculate the surface area are given below:
Step 1: Write down all the given data in the question.
Step 2: Write down the formula needed according to the question (whether CSA or TSA).
Step 3: Put the values and calculate the solution step-by-step.
Let’s take an example for a better understanding.
Example: Find the surface area of a spherical ball with a radius of 9 inches.
Solution: Given, Radius of a spherical ball, r = 9 inches,
We know that
CSA = 4r².
Now,
On putting values, we get
CSA = 4 x 3.14 x 9 x 9 square inches.
CSA = 1017.36 square inches.
Curved Surface Area Of A Sphere (CSA)
The curved surface area is the area of all the curved regions of the solid. In the case of a sphere, since there is no flat surface in the sphere, the curved surface area of a sphere is equal to the total surface area of the sphere. Therefore, the formula for the curved surface area of a sphere is expressed as
The curved surface area of a sphere = The total surface area of a sphere.
Therefore,
The curved surface area of a sphere = 4r²,
Where r = radius of the sphere
Lateral Surface Area Of A Sphere (LSA)
The lateral surface area is the area of all the regions except the bases (i.e., top and bottom). As in the case of a sphere, it has no flat surface, and it is not possible to separate the area of its base from other sides.
Therefore, for all spheres,
The lateral surface area of a sphere = The total surface area of a sphere = 4r²
Total Surface Area Of A Sphere (TSA)
The total surface area is the area of all the sides, top, and bottom of the solid object.
In the case of a sphere, it has no flat surface. Hence, the total surface area of the sphere is 4r².
Some Important Questions Related To The Sphere For The Class 10 Mathematics Exam
Some important questions related to the sphere topic from the Mensuration unit are as follows:
- The diameter of a copper sphere is 6 cm. The sphere is melted and is drawn into a long wire of a uniform circular cross-section. If the length of the wire is 36 cm, find its radius.
- A sphere diameter of 12 cm is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 3×5/9 cm. Find the diameter of the cylindrical vessel.
- There were 504 cones, each 3.5 cm diameter and 3 cm height, melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area.
- Two spheres of the same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the new sphere.
- Calculate the cost required to paint a football in the shape of a sphere with a radius of 7 cm. If the painting cost is INR 2.5/square cm. (Take π = 22/7)
- Calculate the curved surface area of a sphere having a radius equal to 3.5 cm(Take π= 22/7)
- A vessel full of water is in the form of an inverted cone of a height of 8 cm and the radius of its top, which is open, is 5 cm. 100 spherical lead balls are dropped into the vessel. One-fourth of the water flows out of the vessel. Find the radius of a spherical ball.
Important Topics For The Class 10 Mathematics Exam Preparations
Some important topics for the Class 10 Mathematics exam preparations are given below:
Chapter 4 – Quadratic Equations:
- Solutions of quadratic equations (only real roots) by factorisation and using quadratic formulas.
- Applications of quadratic equations.
Chapter 5 – Arithmetic Progressions (AP):
- Sum to n terms of an AP.
- Simple day-to-day problems using AP.
Chapter 10 – Circles:
- Tangent based problems.
- Theorem based problems.
Chapter 11 – Constructions:
- Construction of tangents.
- Constructions of incircle and circumcircle of a triangle.
- Construction of figures.
Chapter 9 – Some Applications of Trigonometry:
- Simple problems on height and distance.
- The angle of elevation/depression-based problems.
Chapter 13 – Surface Areas and Volumes:
- Problems based on finding volume and surface areas.
- Problems based on day-to-day activities.
Chapter 14 – Statistics:
- Mean, Median, and Mode based problems.
- Problems based on cumulative frequency tables.
Some Important Tips For Revising Mathematics Syllabus For The Class 10 Exam
Here are some tips that can be beneficial for you with the revision of Mathematics for the Class 10 exam. These are as follows:
- Update yourself with the CBSE Class 10 Mathematics exam syllabus.
- Prepare a notes diary and write all the formulas in it.
- Revise these formulas daily.
- Practice solving problems after revising formulas.
- Solve all the examples and back exercises of the NCERT book and NCERT exemplar.
- Try to solve numerical step-by-step. Do not use shortcuts with calculations.
- Practice case-study based problems.
- Solve sample papers, Class 10 previous years’ papers, and latest mock papers according to board exam duration.
- Put your focus more on the important topics from each chapter.
- Try not to overwrite and make cuttings, and write clearly on the answer sheet.
Conclusion
Although Mathematics is a tricky and formula-based subject, you can easily score 90+ in the Class 10 exam if you get over it. Mensuration is one of the simplest but tricky units. Simplest in the form that you can easily identify the formulas by just looking at the problems. And it is tricky because you need first to identify the rough diagram draft from the problem to get a clear idea about the question.
In the above article, we tried to explain to students about the area of sphere terms and some preparation tips that can help them with the revision of Mathematics. We hope it may be helpful to you.
FAQs
1. What type of questions from the Mensuration unit have a high chance to occur in the Class 10 Mathematics exam?
From the years of experience and records of the Mensuration units of CBSE Class 10 exam patterns, the following problems have been recorded:
- Problems in finding volumes and surface areas of a right circular cone, right circular cylinder, hemisphere, and sphere.
- Problems involving converting one type of metallic solid into another and other mixed problems.
2. What are the important chapters for the Class 10 Mathematics exam?
Below is the list of important chapters for the Class 10 Mathematics exam:
Name Of The Unit | Important Chapter |
Algebra | Quadratic Equations
Arithmetic Progressions |
Geometry | Circles
Constructions |
Trigonometry | Some Applications of Trigonometry |
Mensuration | Surface Areas and Volumes |
Statistics and Probability | Statistics |
3. How can I identify any shape as a circle or a sphere?
A figure is called a sphere if a three-dimensional shape must meet the following requirements:
- Be symmetrical along all dividing axes
- All points on the surface are equidistant from the centre
- The sphere has no edges
- There are no corners
- There is a single surface, which is rounded and thus not a face (which is flat)
- The curvature of the surface remains the same along the entire surface.
The cross-section of a sphere is circular. A two-dimensional representation of a sphere is a circle when only two axes (X, Y) are available.
4. What are the best reference books that I can use to revise the Class 10 Mathematics exam?
The following are the best reference books to revise the Class 10 Mathematics exam:
- Mathematics Exemplar Problems for Class X, NCERT Publication
- Secondary School Mathematics for Class 10 – RS Aggarwal
- All in one mathematics – Arihant
- Mathematics for Class 10 – R D Sharma
- NCERT Solutions for Class 10 Mathematics
- Together With For Mathematics for Class X – Rachna Publication
5. What are the important formulas related to sphere shape?
The important formulas related to sphere share are as follows:
- Surface Area:
The surface area of a sphere is the total area covered by the surface of a sphere in three-dimension space. The formula for surface area is as given below:
SA = 4πr² square units
- Volume:
The volume of a sphere is the space occupied by a sphere in three-dimension space. The formula is as given below:
V = (4/3)πr³ cubic units