Areas Related to Circles Class 10 Notes Maths Chapter 12

Last Updated : 27 Jun, 2023

CBSE Class 10 Maths Notes Chapter 12 Area related to Circles is an excellent resource for learning the concepts in a fast and friendly manner. At GfG we have created these comprehensive notes to help students to better understand the complex topic of “Areas related to Circles”.

Chapter 12 of the NCERT Class 10 Maths textbook delves into the world of Areas related to circles, the chapter covers the concepts, of finding areas of circles, areas of the segment, areas of a sector, and areas in combined figures. These notes are designed to give students a comprehensive summary of the entire chapter and include all the essential topics, formulae, and concepts needed to succeed in their exams.

Need of Studying Area Related to Circles

Circles are the most general figures, in real-world scenarios, and hence calculating their area would be in turn useful. For example, if we want to calculate, the area of an athlete’s track, the area of a circular table, the area of a segment of a wheel outside the car, etc. The scope of the chapter is limited, to calculating the area of the combined figures, using circles, and basic 2-D figures.

Circumference of a Circle

Circumference is the distance around the circle. Archimedes was a mathematician and a physicist, who first calculated the value of π. He defined π(pie), as the ratio of the circumference of its circle, and the diameter.

$\bold{ π\,=\,\frac{C}{D}}$

Where,

C is the Circumference of the circle and
D is the diameter of the circle (where D=2r).

⇒ C = 2D

⇒ C 2πr

Length of Arc of a Circle

Arc is the sub-length of the circumference of the circle. Given, a circle, whose centre is O, we need to find the length of the arc AB, ACB. We have two types of arcs in a circle i.e. Major arc, and Minor arc. The length of the major arc is more than the length of the minor arc. Here, $\stackrel \frown {AB}$ is the minor arc, and ACB, is the major arc.

Length of Minor Arc,

∠AOB(Interior side) = θ°

Length of arc in 360° = 2πr

Length of arc in 1° is,

$\Rightarrow\,\,\frac{1}{360}2πr$

Length of arc in θ° is,

$\Rightarrow\,\frac{θ}{360}2πr$

Length of Major Arc,

∠AOB(Exterior side) = (360° – θ°)

Length of arc in (360-θ)° is,

$\Rightarrow\,\frac{(360-θ)}{360}2πr$

Example: Given a circle with centre O and a radius of 3 cm. Find the length of the arc, such that, the angle formed between them is 60°.

Solution:

We know that,

Length of minor arc is,

$\Rightarrow\,\frac{θ}{360}2πr$

where, r = 3cm, θ = 60°,

Hence, putting the values in the function,

$\Rightarrow\,\frac{60}{360}2.π.3$

On solving the above equation, the length of the arc comes out to be 3.14cm.

Area of a Circle

The proof of the area of the circle is outside the scope of the current syllabus. Archimedes, found that area of the circle, by the method of exhaustion.

The area of the circle = A = πr²

where,

π = 22/7 or 3.14 (pie constant)

r = radius of the circle

Learn more about Area of Circle

Area of Sector of a Circle

A sector is an area enclosed between 2 radii and the arc of a circle. Given, a circle, whose centre is O, we need to find the area of the sector OAB, and OACB. We have two types of sectors in a circle i.e. Major sector, and Minor sector. The area of the major sector is more than the area of the minor sector. Here, OAB is the minor sector, and OACB is the major sector.

Area of Minor Sector

Area of Minor Sector,

∠AOB(Interior side) = θ°

Area of sector in 360° = πr²

Area of sector in 1° is,

$\Rightarrow\frac{1}{360} πr^2$

Area of sector in θ° is,

$\Rightarrow\frac{θ}{360} πr^2$

Area of Major Sector

Area of Major Sector,

∠AOB(Exterior side) = (360° – θ°)

area of sector in (360-θ)° is,

$\frac{360-θ}{360} πr^2$

Example: Find the area of the sector of the circle, whose radius is 10cm, and the angle formed between the 2 radii of the circle is 36°.

Solution:

We know that,

Area of a sector of a circle is,

$\Rightarrow\frac{θ}{360} πr^2$

where, r = 10cm, and θ = 36°,

$\Rightarrow\frac{36}{360} π10^2$

On solving, the area of the sector of the circle comes out to be 10π cm².

Area of a Segment of a circle

The area enclosed, in 2 radii and an arc is called the area of the segment of a circle. In a circular region with centre O, two distinct segments can be identified: the minor segment, defined by the arc ABC, and the major segment, defined by the arc ACB. These segments are differentiated by their respective areas, with the major segment encompassing a larger area than the minor segment.

Area of Major Segment

Area of Minor Segment,

∠AOB(Interior side) = θ°

Area of segment = Area of minor sector – Area of Triangle

Area of Minor Segment

Area of Major segment,

∠AOB(Exterior side) = (360° – θ°)

Area of major segment = Area of major sector + Area of Triangle

Example: Given a circle with centre O, with radius = 8cm.

$\bold{OC = 4\sqrt{3}}$ ,

AB = 8cm,

∠AOB = 60°.

Find the area of the major segment of the circle.

Solved examples of Area Related To Circle

Answer:

We know that,

Area of major segment = area of major sector + area of triangle,

Area of major sector is,

$\Rightarrow \frac{360-θ}{360}πr^2$

$\Rightarrow\frac{300}{360}π8^2$

=> 167.467cm²

Area of triangle,

$\Rightarrow\frac{1}{2}.AB.OC$

$\Rightarrow \frac{1}{2}.8.4\sqrt{3}$

=> 27.68cm²

Hence, the area of the major segment = 167.467 + 27.68 = 195.16cm².

Area of Combined Figures

The most important part of this chapter is to find the area of the combined figures. In such problems, one figure is embedded into another figure, and we need to calculate the area of the remaining portion by calculating, by removing the extra area i.e.,

Area of Combined Figure = Total Area – Extra Area

Let’s consider an example for better understanding.

Example: In the following figure, if 3 circles are embedded in a larger triangle then what is the area of the region shaded in color yellow?

Answer:

⇒ Area of Shaded portion = Area of triangle – Area of circles

⇒ Area of Shaded portion = Area of triangle – 3 . Area of circle

⇒ Area of Shaded portion = $\frac{1}{2}l^2\,-\,3πr^2$

Learn to solve NCERT Questions with the help of NCERT Solutions for Chapter 12 Areas Related to Circles

Practice Problems on Area related to Circles

Problem 1. Find the area of the sector of a circle with a radius of 2 cm and of angle 60°. Also, find the area of the corresponding major sector (Use π = 3.14).

Answer:

Given,

Radius = r = 2cm
θ = 60°.

Apply, the formula of the major sector and the minor sector, to find the answer.

Area of Minor Sector = $\frac{θ}{360} πr^2$

⇒ Area of Minor Sector = $\frac{60}{360} π2^2$

Thus, the area of Minor Sector is 2.09cm².

Area of Major Sector = $\frac{(360-θ)}{360} πr^2$

⇒ Area of Major Sector = $\frac{300}{360} π2^2$

Thus, the area of major sector is 10.467cm².

Problem 2. A square of dimensions 32 × 32 cm. Two circles, each of diameter 16 cm, are embedded in the square, as shown in the figure below. Find the area of the shaded region.

Answer:

Given,

l = 32cm, D = diameter = 16cm, radius = D/2 = 8cm.

Apply, the formula for combined figures, and find the area of the shaded region.

Required area = Total Area – Extra Area

where, Total area = Area of the square

Extra area = Area of the circles

Area of the Square = l²

Area of the circles = 2. Area of a circle

Area of a circle = πr²

Area of the cricles = 2.πr²

Area of shaded region = l² – 2πr²

Area of shaded region = 32² – 2π8²

Area of shaded region = 1024 – 401.92

Area of shaded region = 622.08cm².

Problem 3. Find the area of a regular hexagon, whose each side is 2cm.

Answer:

A regular hexagon, with center O. Divide the hexagon into 6 parts, joining OA, OB, OC, OD, OE, and OF.

As, the hexagon is a regular polygon, hence, area of all triangles will be equal,

Area of hexagon = 6 . Area of △AOB,

∠AOB = 60°,

Also, OA = OB, so, ∠OAB = ∠OBA,

Also, ∠AOB + ∠OAB + ∠OBA = 180°,

∠OAB = ∠OBA = 60°.

Hence, △AOB is an equilateral triangle,

We know that,

Area of an equilateral triangle is,

$\Rightarrow\frac{\sqrt{3}}{4}a^2$

$\Rightarrow\frac{\sqrt{3}}{4}2^2$

$\Rightarrow\sqrt{3}cm^2$

Area of hexagon is,

$\Rightarrow 6\cdot \sqrt{3}\ \text{cm}^2$

Problem 4: Given a circle, with centre O, of radius 6cm, and the side of the triangle 12cm. Find the percentage of the portion of the triangle which is outside the circle. (Percentage has to be found, w.r.t. the triangle). (∠DOE = 60°, OB = 12cm).

Answer:

Area of portion, outside the circle is ADCEB,

Area of ADCEB = area of triangle – area of ODCE

where,

ODCE is the area of the sector of the circle,

We know that,

Area of sector $=\frac{θ}{360}πr^2$

⇒ Area of sector = $\frac{60}{360}π6^2$

⇒ Area of ODCE = 18.84cm²

Area of traingle OAB is,

As two sides of traingle OAB are same, and one angle is 60°, hence, it’s an equilateral triangle.

Area of equilateral triangle $= \frac{\sqrt{3}}{4}a^2$

⇒ Area of equilateral triangle = $\frac{\sqrt{3}}{4}\times12^2$

⇒ Area of equilateral triangle = 62.28cm².

⇒ Area of ADCEB = 62.28 – 18.84 = 43.44cm².

Percentage of area outside the circle $=\frac{43.44}{62.28}\times 100$

⇒ Percentage of area outside the circle = 69.74%.

FAQs on Areas related to Circles

Q1: What is a Circle?

Answer:

A circle is the collection of points, such that the distance between the set of points, and a given point is constant. A circle can also be defined as a 2-D shape, which has no vertices, and edges.

Q2: What is the Area of a Right-Angled Triangle?

Answer:

The area of a right angled triangle is,

$\bold{\text{Area of triangle} = \frac{1}{2}\cdot Base \cdot Height}$

Q3: What is the Area of a Triangle Between 2 Radii and a Segment?

Answer:

The area of the triangle is,

$\bold{\text{Area of triangle} = \frac{1}{2} \cdot r^2 \cdot θ}$

Where,

r is the radius of the circle,
θ is the angle formed between radii, and center of the circle

Q4. Give Some Real-Life Examples of finding the Area of the Circle, of Combined Figures.

Answer:

Some of the real-life examples, of the area of combined figures are

Flower Beds
Drain Covers,
Window Designs,
Designs on Table Covers

Q5. How to Calculate the Area of the Sector of a Circle?

Answer:

The area of the sector of a circle is given as,

$\bold{\text{Area of Sector} = \frac{θ}{360}\cdot πr^2 }$

Where,

θ is the angle formed between the 2 radii and the arc of the circle, and
r is the radius of the circle.

Q6. How to Calculate the Length of the Arc of a Circle?

Answer:

The length of the arc of a circle is,ircle is given as,

$\bold{\text{Perimeter of Sector} = \frac{θ}{360}\cdot 2πr }$

Where,

θ is the angle formed between the 2 radii and the arc of the circle, and
r is the radius of the circle.

Suggest improvement

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles

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Areas Related to Circles Class 10 Notes Maths Chapter 12

Need of Studying Area Related to Circles

Circumference of a Circle

Length of Arc of a Circle

Area of a Circle

Area of Sector of a Circle

Area of Minor Sector

Area of Major Sector

Area of a Segment of a circle

Area of Major Segment

Area of Minor Segment

Area of Combined Figures

Practice Problems on Area related to Circles

FAQs on Areas related to Circles

Q1: What is a Circle?

Q2: What is the Area of a Right-Angled Triangle?

Q3: What is the Area of a Triangle Between 2 Radii and a Segment?

Q4. Give Some Real-Life Examples of finding the Area of the Circle, of Combined Figures.

Q5. How to Calculate the Area of the Sector of a Circle?

Q6. How to Calculate the Length of the Arc of a Circle?

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