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.
Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km.
(v) A fraction becomes\(\frac{ 9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

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2.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

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3.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

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4.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

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5.
Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, . . . .
(ii) \(2, \frac{5}{2},3,\frac{7}{2}\), . . . .
(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . .
(iv) – 10, – 6, – 2, 2, . . .
(v) 3, \(3 + \sqrt{2} , 3 + 3\sqrt{2} , 3 + 3 \sqrt{2}\) . . . .
(vi) 0.2, 0.22, 0.222, 0.2222, . . . .
(vii) 0, – 4, – 8, –12, . . . .
(viii) \(\frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}\), . . . .
(ix) 1, 3, 9, 27, . . . .
(x) a, 2a, 3a, 4a, . . . .
(xi) a, \(a^2, a^3, a^4,\) . . . .
(xii) \(\sqrt{2}, \sqrt{8} , \sqrt{18} , \sqrt {32}\) . . . .
(xiii) \(\sqrt {3}, \sqrt {6}, \sqrt {9} , \sqrt {12}\) . . . . .
(xiv) \(1^2 , 3^2 , 5^2 , 7^2\), . . . .
(xv) \(1^2 , 5^2, 7^2, 7^3\), . . . .

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6.
Find the sums given below :
\(7 + 10\frac 12+ 14 + ....... + 84\)
\(34 + 32 + 30 + ....... + 10\)
\(–5 + (–8) + (–11) + ....... + (–230)\)

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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

2.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

3.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

4.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

6.
Find the sums given below :
\(7 + 10\frac 12+ 14 + ....... + 84\)
\(34 + 32 + 30 + ....... + 10\)
\(–5 + (–8) + (–11) + ....... + (–230)\)

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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

2. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

3. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

6. Find the sums given below :\(7 + 10\frac 12+ 14 + ....... + 84\)\(34 + 32 + 30 + ....... + 10\)\(–5 + (–8) + (–11) + ....... + (–230)\)

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

3.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

4.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

6.
Find the sums given below :
\(7 + 10\frac 12+ 14 + ....... + 84\)
\(34 + 32 + 30 + ....... + 10\)
\(–5 + (–8) + (–11) + ....... + (–230)\)