Area of a Sector: Definition, Formulae and Solved Examples

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The part of the circle that is enclosed between two radii and the adjoining arc is known as the area of a sector. The area of a sector is the area enclosed between two radii and the arc. The red shaded part OAB is a sector. The AB is the corresponding arc.

Table of Content

Area of a Sector
Formulae
Things to Remember
Sample Questions

Sectors can be major or minor. The larger sector is called the major sector and the smaller sector is called the minor sector. OAPB is the minor sector and OAQB is the major sector. Remember that there is a difference between a sector and a segment. A segment is the part of the circle enclosed between a chord and the adjoining arc.

Important Terms

Radius: The distance from the center of the circle to any point on the circle is the radius of the circle. Radius is denoted by the symbol r.

Diameter: The length of a line from one point on the circle to another point that passes through the center of the circle is the diameter of the circle. Diameter is denoted by the symbol d.

Circumference: The perimeter of the circle is called the circumference of the circle.Circumference is denoted by the symbol c.

Arc: A part of the circumference of the circle is known as the arc of a circle. The AC in the figure below is the arc of a circle.

Also Read:

Angle Formula	Conditional Probability Formula
Differentiation and Integration Formula	Calculus Formula

Formulae

If we know the radius and the angle of the sector, we can calculate the area of the sector. The r is radius and the angle of the sector is θ.

So, we know that the circle with center O is forming an angle of 360°. By applying the unitary method, we can arrive at the formula for the area of a sector.

When the angle at center O i.e. the sector is 360°, the area of the sector is πr².

When the angle forms a degree measure of 1 at the center, the area of the sector is πr2360 .

When the angle of the sector is θ, the area of the sector will be θ360 ×πr²

So, the area of a sector can be calculated by using the following formula:

Area of the sector of angle θ = θ360×πr²

If we know the length of the arc and the radius, we can calculate the area of the sector. The r is radius and the length of the arc is l.

The area of the sector can be calculated using the following formula:

Area of the sector = lr²

For using any of the above formulae, it is important to know the radius of the circle.

One can also calculate the length of the arc, if the angle of the sector and the radius is known, by using the following formula:

Length of the arc = θ360×2πr

Also Read:

Quadrilateral Angle Sum Property	Determinant of a Matrix
Sin 30 Degrees	Tan 0 Degrees

Things to Remember:

The portion of the circle that is enclosed between two radii and the adjoining arc is known as the area of a sector. Sectors can be major or minor. The larger sector is the major sector and the smaller sector is the minor sector.
Area of the sector of angle θ = θ360 ×πr²

Area of the sector = lr²

Length of the arc = θ360×2πr

A part of the circumference of the circle is known as the arc of a circle.

Sample Questions

Question: Find the area of a sector of a circle with radius 6 cm if the angle of the sector is 60°

Solution: If the radius of the circle is 6 cm and the angle of the sector is 60°, the area of the sector can be calculated using the formula θ360×πr²

So, area of the sector = θ360 ×πr²

= 60360×227×(6×6)

= 18.85 cm²

The area of the sector is 18.85cm².

Question: Find the area of a sector of a circle with radius 12 cm if the angle of the sector is 35°

Answer: if the radius is 12 cm and the angle of the sector is 35°, the area of the sector can be calculated using the following formula θ360 ×πr²

So, area of the sector = θ360 ×πr²

= 35360×227×(12×12)

= 35360×227×144

= 44cm²

The area of the sector is 44cm².

Question: Find the area of the sector of a circle with radius 8 cm and the angle of the sector is 42°. Also find the area of the corresponding major sector.

Solution: If the angle of the sector is 42° and the radius of the circle is 8 cm, then the area of the sector:

Area of the minor sector = θ360 ×πr²

=42360×227×(8×8)

=42360×227×64

= 23.46 cm²

Area of the major sector = area of circle – area of minor sector

= 227×64 - 23.46

= 201.64 cm²

Question: Find the area of the sector if the radius of the circle is 8 cm and the length of the arc is 5 cm.

Solution: If the radius of the circle is 8 cm and the length of the arc is 5 cm, the area of the sector can be calculated using the formula lr²

So, area of the sector =lr²

= 5×82

= 20 cm²

So, the area of the sector is 20cm².

Question: If the area of the sector is 80 cm² and the length of the arc is 8 cm, find the radius of the circle.

Solution: The area of the circle is 80 cm² and the length of the arc is 8 cm radius of the circle can be calculated using the following formula.

Area of the sector = lr²

80 cm²= 8 ×r²

r = 80 × 28

r = 20 cm

So, the radius of the circle is 20 cm.

CBSE X Related Questions

1.
Find length and breadth of a rectangular park whose perimeter is $100 \, \text{m}$ and area is $600 \, \text{m}^2$.
View Solution
2.
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$
View Solution
3.
$\alpha, \beta$ are zeroes of the polynomial $3x^2 - 8x + k$. Find the value of $k$, if $\alpha^2 + \beta^2 = \dfrac{40}{9}$
View Solution
4.
If the zeroes of the polynomial $ax^2 + bx + \dfrac{2a}{b}$ are reciprocal of each other, then the value of $b$ is
- $\dfrac{1}{2}$
- 2
- -2
- $-\dfrac{1}{2}$
View Solution
5.
Directions: In Question Numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R).
Choose the correct option from the following:
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
Assertion (A): For any two prime numbers $p$ and $q$, their HCF is 1 and LCM is $p + q$.
Reason (R): For any two natural numbers, HCF × LCM = product of numbers.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
View Solution
6.
AB and CD are diameters of a circle with centre O and radius 7 cm. If $\angle BOD = 30^\circ$, then find the area and perimeter of the shaded region.
View Solution

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Area of a Sector: Definition, Formulae and Solved Examples

Important Terms

Formulae

Things to Remember:

Sample Questions

CBSE X Related Questions

1.
Find length and breadth of a rectangular park whose perimeter is \(100 \, \text{m}\) and area is \(600 \, \text{m}^2\).

2.
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$

3.
\(\alpha, \beta\) are zeroes of the polynomial \(3x^2 - 8x + k\). Find the value of \(k\), if \(\alpha^2 + \beta^2 = \dfrac{40}{9}\)

4.
If the zeroes of the polynomial $ax^2 + bx + \dfrac{2a}{b}$ are reciprocal of each other, then the value of $b$ is

6.
AB and CD are diameters of a circle with centre O and radius 7 cm. If \(\angle BOD = 30^\circ\), then find the area and perimeter of the shaded region.

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Area of a Sector: Definition, Formulae and Solved Examples

Important Terms

Formulae

Things to Remember:

Sample Questions

CBSE X Related Questions

1.Find length and breadth of a rectangular park whose perimeter is \(100 \, \text{m}\) and area is \(600 \, \text{m}^2\).

2.From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$

3.\(\alpha, \beta\) are zeroes of the polynomial \(3x^2 - 8x + k\). Find the value of \(k\), if \(\alpha^2 + \beta^2 = \dfrac{40}{9}\)

4.If the zeroes of the polynomial $ax^2 + bx + \dfrac{2a}{b}$ are reciprocal of each other, then the value of $b$ is

6.AB and CD are diameters of a circle with centre O and radius 7 cm. If \(\angle BOD = 30^\circ\), then find the area and perimeter of the shaded region.

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Find length and breadth of a rectangular park whose perimeter is \(100 \, \text{m}\) and area is \(600 \, \text{m}^2\).

2.
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$

3.
\(\alpha, \beta\) are zeroes of the polynomial \(3x^2 - 8x + k\). Find the value of \(k\), if \(\alpha^2 + \beta^2 = \dfrac{40}{9}\)

4.
If the zeroes of the polynomial $ax^2 + bx + \dfrac{2a}{b}$ are reciprocal of each other, then the value of $b$ is

6.
AB and CD are diameters of a circle with centre O and radius 7 cm. If \(\angle BOD = 30^\circ\), then find the area and perimeter of the shaded region.