Course Content
Chapter 01 – Operations on Sets
The set operations are performed on two or more sets to obtain a combination of elements as per the operation performed on them. In a set theory, there are three major types of operations performed on sets, such as: Union of sets (∪) The intersection of sets (∩) Difference of sets ( – ) In this lesson we will discuss these operations along with their Venn diagram and will learn to verify the following laws: Commutative, Associative, Distributive, and De-Morgans' law.
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Chapter 02 – Real Numbers
All real numbers follow three main rules: they can be measured, valued, and manipulated. Learn about various types of real numbers, like whole numbers, rational numbers, and irrational numbers, and explore their properties. In this chapter, we will learn about Squares and cubes of real numbers and find their roots.
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Chapter 03 – Number System
The number system or the numeral system is the system of naming or representing numbers. There are different types of number systems in Mathematics like decimal number system, binary number system, octal number system, and hexadecimal number system. In this chapter, we will learn different types and conversion procedures with many number systems.
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Chapter 04 – Financial Arithmetic
Financial mathematics describes the application of mathematics and mathematical modeling to solve financial problems. In this chapter, we will learn about partnership, banking, conversion of currencies, profit/markup, percentage, and income tax.
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Chapter 05 – Polynomials
In algebra, a polynomial equation contains coefficients, exponents, and variables. Learn about forming polynomial equations. In this chapter, we will study the definition and the three restrictions of polynomials, we'll tackle polynomial equations and learn to perform operations on polynomials, and learn to avoid common mistakes.
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Chapter 06 – Factorization, Simultaneous Equations
In algebra, factoring is a technique to simplify an expression by reversing the multiplication process. Simultaneous Equations are a set of two or more algebraic equations that share variables and are solved simultaneously. In this chapter, we will learn about factoring by grouping, review the three steps, explore splitting the middle term, and work examples to practice verification and what simultaneous equations are with examples. Find out how to solve the equations using the methods of elimination, graphing, and substitution.
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Chapter 07 – Fundamentals of Geometry
Geometry is the study of different types of shapes, figures, and sizes. It is important to know and understand some basic concepts. We will learn about some of the most fundamental concepts in geometry, including lines, polygons, and circles.
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Chapter 08 – Practical Geometry
Geometric construction offers the ability to create accurate drawings and models without the use of numbers. In this chapter, we will discover the methods and tools that will aid in solving math problems as well as constructing quadrilaterals and right-angled triangles.
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Chapter 09 – Areas and Volumes
The volume and surface area of a sphere can be calculated when the sphere's radius is given. In this chapter, we will learn about the shape sphere and its radius, and understand how to calculate the volume and surface area of a sphere through some practice problems. Also, we will learn to use and apply Pythagoras' theorem and Herons' formula.
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Chapter 10 – Demonstrative Geometry
Demonstrative geometry is a branch of mathematics that is used to demonstrate the truth of mathematical statements concerning geometric figures. In this chapter, we will learn about theorems on geometry that are proved through logical reasoning.
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Chapter 11 – Trigonometry
Sine and cosine are basic trigonometric functions used to solve the angles and sides of triangles. In this chapter, we will review trigonometry concepts and learn about the mnemonic used for sine, cosine, and tangent functions.
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Chapter 12 – Information Handling
Frequency distribution, in statistics, is a graph or data set organized to show the frequency of occurrence of each possible outcome of a repeatable event observed many times. Measures of central tendency describe how data sets are clustered in a central value. In this chapter, we will learn to construct the frequency distribution table, and learn more about three measures of central tendency, its importance, and various examples.
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Mathematics – VIII
About Lesson
How much wrapping paper is needed to completely cover the outside of a gift like one of these, assuming no gaps or overlaps?

The answer to this question requires an understanding of the surface area. This is an example that can be used to answer the question of “what is surface area?

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What is Surface Area?

The surface area measures the area of the outside of a three-dimensional shape. It can be described as the space covering the outside of a solid figure. Surface area and area are both measured in square units, but they are not the same.

For example:

  • The area of a flat, two-dimensional picture is the space covered by the paper. This painting is on the page of a magazine. The page is flat, with just two dimensions: length and width.
  • Contrast that with the surface area of a painted box. The box is a three-dimensional object and has three dimensions: length, width, and depth.

There are formulas for calculating the area of a two-dimensional shape, there are formulas for calculating the surface area of a three-dimensional solid. There are also formulas for calculating the surface area of different three-dimensional objects. Different solid shapes have different formulas.

Sphere

sphere is a three-dimensional object that is round in shape. The sphere is defined in three axes, i.e., x-axis, y-axis and z-axis. This is the main difference between a circle ana d sphere. A sphere does not have any edges or vertices, like other 3D shapes. 

The points on the surface of the sphere are equidistant from the center. Hence, the distance between the center and the surface of the sphere are equal at any point. This distance is called the radius of the sphere. Examples of spheres are a ball, a globe, the planets, etc

Sphere

The Greek mathematician Archimedes discovered that the surface area of a sphere is the same as the lateral surface area of a cylinder having the same radius as the sphere and the height the length of the diameter of the sphere.

How to calculate the surface area of a sphere - Quora

Surface Area of a Sphere

The surface area of a sphere is defined as the region covered by its outer surface in three-dimensional space.  A Sphere is a three-dimensional solid having a round shape, just like a circle. The formula for the total surface area of a sphere in terms of pi (π) is given by:

Surface Area  = 4 π rsquare units

The difference between a sphere and a circle is that a circle is a two-dimensional figure or a flat shape, whereas, a sphere is a three-dimensional shape. Therefore, the area of circle is different from  the area of the sphere. 

Area of circle = π r

The surface area of a sphere formula is given  by,

A = 4 π rsquare units

In terms of diameter, the surface area of a sphere is expressed as,

S = 4π(d/2)2     where d is the diameter of the sphere.

For any three-dimensional shapes, the area of the object can be categorized into three types. They are:

  • Curved Surface Area
  • Lateral Surface Area
  • Total Surface Area

Curved Surface Area: The curved surface area is the area of all the curved regions of the solid.

Lateral Surface Area: The lateral surface area is the area of all the regions except the bases (i.e., top and bottom).

Total Surface Area: 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.

Therefore,

Total surface area of a sphere = Curved surface area of a sphere

Solved Examples

Let us take some examples to learn how to calculate the surface area of a sphere using its formula.

Example: 

Find the surface area of a spherical ball that has a radius of 9 inches.

Solution:

Note the radius of the sphere. Here, the radius of the ball is 9 inches.

As we know, the surface area of sphere = 4πr2, so after substituting the value of r = 9, we get,

surface area of sphere = 4πr2 = 4 × 3.14 × 92 = 4 × 3.14 × 81 = 1017.36

Therefore, the surface area of the sphere is 1017.36 in2

Example: 

Find the surface area of a sphere if its radius is given as 6 units.

Solution: 

Given, the radius ‘r’ = 6 units. So, let us substitute the value of r = 6 units

⇒ The surface area of the sphere = 4πr= 4 × π × 62 = 4 × 3.14 × 36 = 452.16 unit2

∴ The surface area of the sphere is 452.16 unit2

Example:

Calculate the curved surface area of a sphere having a radius equal to 3.5 cm (Take π= 22/7)

Solution:

We know,

Curved surface area = Total surface area = 4 π rsquare units

= 4 × (22/7) × 3.5 × 3.5

Therefore, the curved surface area of a sphere= 154 cm2

Volume

Volume is defined as the space occupied within the boundaries of an object in three-dimensional space. It is also known as the capacity of the object.

Every three-dimensional object occupies some space. This space is measured in terms of its volume. Finding the volume of an object can help us to determine the amount required to fill that object, like the amount of water needed to fill a bottle, an aquarium, or a water tank

The volume of 3-Dimensional Shapes

Since different three-dimensional objects have different shapes, their volumes are also variable. Let us look at some three-dimensional shapes and learn how to calculate their volume(V).

Volume of Sphere

The volume of a sphere is the measurement of the space it can occupy. A sphere is a three-dimensional shape that has no edges or vertices.

Volume of Sphere - Definition, Formula, Derivation, Solved Examples

The volume of Sphere Formula

The volume of sphere formula can be given for a solid as well as a hollow sphere. In the case of a solid sphere, we only have one radius but in the case of a hollow sphere, there are two radii, having two different values of radius one for the outer sphere and one for the inner sphere.

The volume of Solid Sphere

If the radius of the sphere formed is r and the volume of the sphere is V. Then, the volume of the sphere is given by:

Volume of Sphere, V = (4/3) πr3

The volume of Hollow Sphere

If the radius of the outer sphere is R, the radius of the inner sphere is r and the volume of the sphere is V. Then, the volume of the sphere is given by:

The volume of Sphere, V = Volume of Outer Sphere – Volume of Inner Sphere

             V = (4/3) πR3 – (4/3) πr3 = (4/3) π(R– r3)

What is the formula for the volume of a sphere?

How to Calculate Volume of Sphere?

The volume of a sphere is the space occupied inside a sphere. The volume of the sphere can be calculated using the formula of the volume of the sphere. The steps to calculate the volume of a sphere are:

  • Step 1: Check the value of the radius of the sphere.
  • Step 2: Take the cube of the radius.
  • Step 3: Multiply r3 by (4/3)π
  • Step 4: At last, add the units to the final answer.

Let us take an example to learn how to calculate the volume of a sphere using its formula.

Solved Examples

Example:

What is the volume of a sphere with a radius of 12?

Solution:

To solve for the volume of a sphere, you must first know the equation for the volume of a sphere.

In this equation, r is equal to the radius. We can plug the given radius from the question into the equation for r.

V = 4/3(π)(12^3)

Now we simply solve for V.

V = 4/3(π)(1728)

V = (π)(2304) = 2304 π

The volume of the sphere is  2304 π.

Example:

What is the volume of a sphere with a radius of 4? (Round to the nearest tenth)

Solution:

To solve for the volume of a sphere you must first know the equation for the volume of a sphere.

The equation is

Then plug the radius into the equation for r yielding 

V = 4/3(4^3) π

Then cube the radius to get

 V = 4/3(64) π

Multiply the answer by 4/3 and  to yield 85.3 π.

The answer is 85.3 π.

Example:

For a sphere, the volume is given by = (4/3) πr3 , and the surface area is given by = 4 πr2. If the sphere has a surface area of 256 π, what is the volume?

Solution:

Given the surface area, we can solve for the radius and then solve for the volume.

4πr2 = 256 π

4r2 = 256

r2 = 64

r = 8

Now solve the volume equation, substituting for r:

V = (4/3) π (8)3

V = (4/3) π (512)

V = (2048/3) π

V = 683 π

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Exercise Files
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