CBSE Class 10 Maths Revision Notes Chapter 12

Class 10 Mathematics Areas Related to Circles Notes for Circles of Chapter 12.

With time, exams are getting more competitive; the difficulty level of the curriculum is also increasing. With the pressure of performing well in the board exams and a vast course to cover, it is essential for students of Class 10 to continuously revise and retain all the concepts. Thus, to make it easy for students, Extramarks has provided a comprehensive listing of all class 10 Chapter 12 Mathematics Notes. These notes are framed keeping in mind the latest CBSE syllabus and NCERT books. 

Chapter 12 Mathematics Class 10 Notes that are available on this website are the best way to prepare before the exams. Apart from revision notes, students can also access CBSE Sample Papers, CBSE Previous Year Questions, important questions, and CBSE Extra Questions. These notes combine all of CBSE’s important questions and guidelines following a simpler approach. A thorough revision of these notes will help students in gaining thorough clarity of all the concepts as well as formulas. 

Class 10 Mathematics Revision Notes for Areas Related to Circles of Chapter 12

Access Class 10 Mathematics Chapter 12 – Areas Related to Circles Notes in 30 Minutes

CBSE Class 10 Mathematics Notes Chapter 12 Areas Related to Circles

Introduction of Area Related to Circles Important Topics

One of the most crucial lessons for Class 10 students is the introduction to the chapter on areas related to circles. Few topics are designed in higher classes based on the properties and applications of circles. So, let’s take a look at the key concepts of the circles covered in this chapter.

Introduction:

1. A circle is defined as a collection of points separated by a fixed distance, known as the radius, from a fixed point, known as the centre.
2. When a line and a circle are both in the same plane, they do not intersect. The line may come close to touching the circle at some point. The tangent to the circle is a type of line like this. The line is the circle’s secant because it intersects the circle at two points.

Tangent to a circle:

• A line that only encounters a circle once is said to be tangent to it. The tangent is said to touch the circle at the common point, which is known as the point of contact between the tangent and the circle.
• At each circle point, there is a single tangent.
• A circle can have a maximum of two parallel tangents.
• It intersects the tangent at a point perpendicular to the radius of the circle.
• Tangent has a unique equation since it is a line.
• When the two endpoints of the corresponding chord coincide, the tangent to a circle is a particular instance of the secant.

Conditions of Tangency:

• A tangent is simply called a line if it touches the circle at more than one point.
• We can therefore express the tangent criteria as follows, depending on the point of tangency and where it lies relative to a circle:
  • whenever the point is contained within the circle.
  • when the circle’s centre is at the point.
  • whenever the point is not inside the circle.

Circumference of a Circle:

• The perimeter of a circle has a special name which is called the circumference.
• The distance covered by travelling once around a circle is its perimeter, usually called its circumference.
• The circumference of a circle bears a constant ratio to its diameter.
• Circumference diameter = π

              Circumference of a circle = 2πr.

              also

              circumference = π × diameter = π × 2r (where r is the radius of the circle) = 2πr.

• The circumference of a circle is normally expressed in units like centimetres or metres.

Solved example on the circumference of a circle.

The cost of fencing a circular field at the rate of 24 per meter is 5280. The field is to be ploughed at the rate of 0.50 per m2. Find the cost of ploughing the field (Take π = 22/7 ).

Solution: Length of the fence (in metres) = Total cost / Rate

 = 5280 / 24 

So, the circumference of the field = 220 m.

Therefore, if r meters is the radius of the field, then

2πr = 220 or, 

2 × 22 / 7 × r = 220 or,

r = 220 × 7 / 2 × 22 = 35 i.e., radius of the field is 35 m.

Therefore, area of the field = πr2  = 22 / 7 x 35 × 35 m2.

Now, the cost of ploughing 1 m2 of the field = rupees 0.50.

So, the total cost of ploughing the field = rupees 22 × 5 × 35 × 0.50 

= rupees 1925.

Area of a Circle:

• The area of a circle is πr2, where π=22/7 or ≈ 3.14.
• A circle’s area is just the area that it takes up in a 2D plane.
• The circle’s area is equal to 1 / 2 × 2πr × r = πr2.
• The ratio of a circle’s circumference to its diameter is called the radius.

Solved example on the area of a circle:

Example: Find the area of a circle with a radius of 7 cm.

Solution: Given, radius of circle = 7cm

By the formula we know:

Area of circle = πr2

= π(7)2

= (22/7) (7)2

= 154 sq.cm.

The perimeter of a Semi Circle:

A semicircle’s perimeter is just the product of its diameter and half of its circumference. We are aware that a circle’s circumference is 2r or d. As a result, a semicircle’s perimeter is 1 / 2 (d) + d or πr + 2r, where r is its radius.

The sector of a Circle:

• The sector of a circle is defined as the region of a circle enclosed by an arc and two radii. 
• The smaller area is called the minor sector and the larger area is called the major sector.
• When we write ‘segment’ and ‘sector’ we will mean the ‘minor segment’ and the ‘minor sector’ respectively unless stated otherwise.
• When the degree measure of the angle at the centre is 360, the area of the sector = πr2.
• Area of the sector of angle θ = θ / 360 x πr2.
• Length of an arc of a sector of angle θ = θ / 360 x 2π.

Solved example on the Sector of a Circle: 

Find the area of the sector of a circle with a radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector (Use π = 3.14).

Solution: Area of the sector = θ / 360 x πr2

= 30 / 360 x 3.14 x 4 x 4 cm2

= 12.56 / 3 cm2

= 4.19 cm2

Area of the corresponding major sector.

=Πr2 – the area of the minor sector.

= ( 3.14 x 16 – 4.19 ) cm2

= 46.05 cm2 

= 46.1 cm2 (approx.)

Benefits of studying Extramarks revision notes:

When it comes to Mathematics, both students and parents are pretty apprehensive concerning exam preparation and its result. Extramarks understands this concern and tries to break down complex topics into simpler forms. The platform Extramarks help the students in the following ways by providing notes: 

• These notes help in predicting the type of questions that can be asked in the exams.
• These notes are error-free and have been formulated with the best precision.
• Notes come with solved practice questions important from the exam’s point of view.
• The content provided in the notes explains topics in the simplest engaging ways.
• These are useful as one-hour revision notes before the exam.

Tips on How to Prepare for Exams

• Students should focus more on learning the material rather than being concerned only about marks.
• All the important fundamental formulas must be memorized for their use while solving questions.
• Extramarks notes have been curated keeping in mind the exam-related important topics only, hence should not waste time on other irrelevant topics.
• Using these notes students should solve question papers keeping a note of time and speed.
• Students must first revise these notes before sitting to solve practice papers which would build a stronger base of all the given concepts.

Conclusion:

Extramarks Class 10 mathematics chapter 12 notes are very important for students to go through for securing colourful marks in the exams. These notes are formed by special subject matter experts with expertise in their respective fields. Mathematics is no doubt a difficult paper to face if not prepared well. By providing these notes to the students for preparing before exams, Extramarks help them face the fear of Mathematics and excel with higher grades.