Areas Related To Circles Formula: Definition, Sector, Segment and Sample Questions

The region of a circle that is enclosed by an arc and two radii is a circle sector or sector of a circle, where the smaller area is the minor sector and the larger area is called the major sector. It is one of the important topics from which a significant number of questions are asked in the board exam. This article breaks down the chapter in a simple way for students to understand the different concepts discussed in it along with some practice and important questions from this topic.

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Area of a Circle

Circle and some of its basic concepts like Area of a circle and circumference or perimeter of a circle are not new to Class 10 students. Such concepts are introduced to students in lower classes.

Area of a Circle = πr²

where π=227 or 3.14 and ‘r’ stands for the radius of the circle in question here.

Area of a Semicircle

Semicircle, like the name suggests, is the half of a circle. Thus,

Area of a Semicircle = 12πr²

Area of a Quadrant

Quadrant forms the one-fourth of a circle. Thus,

Area of a Quadrant = 14πr²

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Circumference of a Circle

Circumference or perimeter of a circle is in simple words, the total distance covered by going around the boundary of the circle in question. Thus, the circumference will be π times the diameter of the circle.

Thus,

Circumference of a Circle = 2πr or πd

Circumference of a Semicircle

A semicircle forms half of a circle. Thus,

Circumference of a Semicircle = πr+2r

Solved Example 

Ques: It costs Rs. 5280 in order to fence a circular shaped field at the fixed rate of Rs. 24 per meter. The field will be ploughed at the rate of Rs. 0.50 per m2. How much will it cost to plough the field?

Ans: Length of the fence = Total costRate= 528024=220

Thus, the circumference of the field = 220

Considering ‘r’ as the radius of the field, the circumference of the field becomes,

2πr=220

2×227×r=220

r= 220×722×2=35

Thus, the radius of the field is 35.

So, area of the field,

πr2= 227 ×35 ×35

=269507=3850

We know that cost of ploughing 1 m² of the field = Rs. 0.50

Now,

Total cost of ploughing the field = Area of the field × Cost of ploughing 1 m² = 3850 × 0.50

= Rs. 1925

Thus, the total cost of ploughing the field = Rs. 1925

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Sector of a Circle

The sector of a circle can be defined as the area or region of a circle that is enclosed by two radii and an arc. Like in the segment, the smaller area formed is called the minor area and the other bigger area forms the major arc.

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Angle of a Sector

The angle of a sector is the angle in which two radii form the sector with an arc.

Length of the Arc of a Sector

One can find the length of an arc that forms a sector by multiplying the circumference of the circle with the angle of the sector. Thus,

Length of the arc of a sector = θ360 ×2πr

Here θ is the angle of the sector and ‘r’, the radius of the circle.

Perimeter of the Sector of a Circle

Perimeter of a sector is the total distance covered by going around the boundary of the sector in question. We can find the perimeter by adding the length of the arc with the length of the radii. Thus,

Perimeter of the sector of a circle = θ360 ×2πr+2r

Area of the Sector of a Circle

One can find the area of the sector of a circle by multiplying the area of the circle with the angle of the sector. Thus,

Area of a Sector = θ360 × πr²

Here θ is the angle of the sector and ‘r’, the radius of the circle.

Note,

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

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

Solved Example 

Ques. Find the area of the sector of a circle having a radius, 4 cm. The angle of the sector is 30°. Find the area of the corresponding major sector of the circle as well.

Ans: Let us consider the two radii of the circle as OA and OB. Then the sector in question here is OAPB.

We know that,

Area of a sector = θ360 × πr²

Here, θ = 30°, π = 3.14, r = 4cm.

Equating these values into the above equation, the area of the sector becomes

= 30360 ×3.14 ×4 ×4

12.563=4.19 cm²

Thus, the area of the minor sector OAPB = 4.19 cm²

Now to find the Area of the major segment, the below equation can be used.

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

= πr²- θ360 × πr²

Here, π = 3.14, r = 4cm, Area the sector OAPB (minor sector) = 4.19 cm²

Equating these values in the equation, we get,

=3.14 ×4 ×4 -4.19

=46.05 cm²

Thus, the area of the minor sector OAPB = 46.05 cm²

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Segment of a Circle

The segment of a circle is a region cut off from a circle by a secant or a chord. The smaller area formed by a secant cutting a circle forms the minor segment while the other bigger area is called the major segment.

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Area of the Segment of a Circle

Area of the highlighted segment APB = Area of the OAPB – Area of the triangle AOB

We all know that,

Area of a Triangle = 12 × base ×height

So here,

Area of the segment APB = θ360 × πr²- 12 ×AB ×OM

One can use trigonometric ratios to find the base (AB) and height (OM), in order to find the area of the triangle.

If the angle of the sector is available, you can use the below formula directly to find the Area of a segment.

Area of Segment = θ360 × πr²- r2 ×sin θ2 ×cos θ2

Note

Area of the major segment = Area of the circle – Area of the minor segment

Area of the minor segment = Area of the circle – Area of the major segment

If the chord forms a right angle at the centre, then,

Area of the corresponding segment = π4- 12 r²

If the chord at the centre, subtends an angle of 60°,

Area of the corresponding segment = π³- 32 r²

If the chord forms an angle of 120° at the centre,

Area of the corresponding segment = π³- 34 r²

Solved Example 

Ques. Find the area of a segment, if the radius of the corresponding circle is 21 cm and the angle of the sector formed by two radii is 120°.

Ans: Let us consider the segment whose area is to be found as PQR.

It is given that the radius of the circle is, OQ = OP = 21 cm

The given angle of the sector °

Here,

Area of the segment PRQ = Area of the sector OPRQ – Area of the triangle POQ

Area of the sector OPRQ = θ360 × πr²

Where, θ = 120°, r = 21 cm and π = 227

Equating the values in the above-given equation, we get,

Area of the sector OPRQ = 120360 × 227 ×21 ×21 = 462 cm²

In order to find the area of the triangle POQ, draw a line OS ⊥ PQ

We know that OP = OQ, thus ΔOSQ ≅ ΔOSP

This means that S, is the mid-point of PQ and 12 ×120=60°

Let, OS = x cm

Thus, from ΔOSQ,

OSOQ = cos 60°

x21= 12

(since,cos 60°= 12)

2x=21

x=OM= 212 cm

Also,

QSOQ=sin 60°= 32

QS= 213 2 cm

We know,

PQ = 2QS

=2 × 213 2=213 cm

So,

Area of ΔPOQ= 12 ×PQ ×OS

= 12 × 213 × 212

= 44143 cm²

Thus,

Area of segment PRQ = 462- 44143

= 21488- 213 cm²

Thus, the area of the segment PRQ = 21488- 213 cm²

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Areas of Combination of Plain Figures

Students are familiar with finding the areas of individual plain figures. This chapter’s last subheading, Areas of combination of Plain Figures, tries to familiarize the students on how to find the area of something which has a combination of plain figures of different shapes as well.

A student must be familiar with the areas of some of the common plain figures in order to attend questions from this section.

• Area of a square with a side ‘l’ = l2
• Area of a rectangle with length ‘l’ and breadth ‘b’ = l × b
• Area of a parallelogram with base ‘b’ and ‘h’ = b × h
• Area of a trapezium with ‘a’ and ‘b’ as the length of the parallel sides and height ‘h’ = [(a + b) × h]/2?

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Solved Example 

Ques. Find the area of the shaded region in the below figure where the length of the side of the square ABCD is 14 cm.

Ans. We know that,

Area of a square = l²

Here l = 14 cm. Therefore,

Area of the square ABCD = 14 × 14 = 196 cm²

Diameter of each circle = 142 = 7 cm

So,

Radius of each circle = 72 cm

Thus, we can calculate the area of each circle.

Area of a circle = πr²

where, π = 227 , r = 72 cm. So,

Area of each circle in the image = 227 × 72 × 72 = 772 cm² 

Therefore,

Area of all the 4 circles = 4 × 772 = 154 cm²

To find the area of the shaded region,

Area of the shaded region = Area of the square ABCD – Area of the four circles

= (196 – 154) cm²

= 42 cm²

Thus, the area of the shaded region = 42 cm²

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Things to Remember

• The region of a circle that is enclosed by an arc and two radii is a circle sector or sector of a circle.
• The sector of a circle can be defined as the area or region of a circle that is enclosed by two radii and an arc.
• The segment of a circle is a region cut off from a circle by a secant or a chord.
• The smaller area formed by a secant cutting a circle forms the minor segment while the other bigger area is called the major segment.
• Area of the major segment = Area of the circle – Area of the minor segment
• Area of the minor segment = Area of the circle – Area of the major segment

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