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Areas of sector and segment of a circle

Areas of sector and segment of a circle

The segment of a circle

A circular segment is a region of a circle which is “cut off” from the rest of the circle by a secant or a chord.

  • In Fig 1, which AB is a chord of the circle with centre O.
  • So, shaded region APB is a segment of the circle.
  • You can also note that unshaded region AQB is another segment of the circle formed by the chord AB.
  • For obvious reasons, APB is called the minor segment and AQB is called the major segment.

Fig 1

The sector of a circle

A circular sector or circle sector is the portion of a circle enclosed by two radii and an arc.

  • In Fig 2, shaded region OAPB is a sector of the circle with centre O.
  • AOB is called the angle of the sector.
  • Note that in this figure, unshaded region OAQB is also a sector of the circle.
  • For obvious reasons, OAPB is called the minor sector and OAQB is called the major sector. You can also see that angle of the major sector is 360° – AOB.

The angle of a Sector

The angle of a sector is that angle which is enclosed between the two radii of the sector.

Let OAPB be a sector of a circle with centre

O and radius r (see Fig 3). Let the degree

measure of AOB be theta which is known as angle of a sector.

Length of the arc of a sector

The length of the arc of a sector can be found by using the expression for the circumference of a circle and the angle of the sector, using the following formula:

L = × 2πr

where θ is the angle of sector and r is the radius of the circle.

Can we find the length of the arc APB corresponding to this sector?

Yes, by applying the Unitary Method and taking the whole length of the circle (of angle 360°) as 2r, we can obtain the required length of the arc APB as × 2πr

Fig 4

So, length of an arc of a sector of angle = × 2πr.

Area of a Sector of a Circle

we can consider this circular region to be a sector forming an angle of 360° (i.e., of degree measure 360) at the centre O. Now by applying the Unitary Method, we can arrive at the area of the sector OAPB as follows:

When the degree measure of the angle at the centre is 360, area of the sector = —-(1)

So, when the degree measure of the angle at the centre is 1, the area of the sector = Fig 5

Therefore, when the degree measure of the angle at the centre is , area of the sector

Area of the sector of angle 

Now let us take the case of the area of the segment APB of a circle with centre O and radius r see Fig 6.

You can see that: Area of the segment APB = Area of the sector

OAPB – Area of OAB —(1)

And we know the area of sector = 

Put eq (2) in (1)

Fig 6

Area of the segment APB =  – area of  OAB —-(3)

From Eq (3) we can obtain following,

Area of the major sector OAQB = – Area of the minor sector OAPB and

Area of major segment AQB = – Area of the minor segment APB

Lets Solve some Example problems based on sectors and segments:

Q.1. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

        (i) the length of the arc

        (ii) area of the sector formed by the arc

        (iii) area of the segment formed by the corresponding chord

Sol. Here, radius = 21 cm and θ = 60°

https://www.careerlauncher.com/cbse-ncert/class-10/10th-Math-Areas-Related-Nce-UntitOE19.JPG

        (i) Circumference of the circle = 2 πr

                https://www.careerlauncher.com/cbse-ncert/class-10/10th-Math-Areas-Related-Nce-UntitOE20.JPG

        (ii) Area of the sector with sector angle 60°

                https://www.careerlauncher.com/cbse-ncert/class-10/10th-Math-Areas-Related-Nce-UntitOE21.JPG

        (iii) Area of the segment APQ = [Area of the sector AOB] – [Area of ΔAOB]          …(1)

                In ΔAOB, OA = OB = 21 cm

                ∴∠A = ∠B = 60°                                                                                                      [θ∠O = 60°]

                ⇒AOB is an equilateral Δ,

                ∴AB = 21 cm

                Draw OM ⊥ AB such that

                https://www.careerlauncher.com/cbse-ncert/class-10/10th-Math-Areas-Related-Nce-UntitOE22.JPG

Q.2.   A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.

Sol. Here, θ = 120° and r = 12 cm

        https://www.careerlauncher.com/cbse-ncert/class-10/10th-Math-Areas-Related-Nce-UntitOE27.JPG

https://www.careerlauncher.com/cbse-ncert/class-10/10th-Math-Areas-Related-Nce-UntitOE28.JPG

        https://www.careerlauncher.com/cbse-ncert/class-10/10th-Math-Areas-Related-Nce-UntitOE29.JPG

        From (1) and (3)

        Area of the minor segment

        = [Area of minor segment] – [Area of ΔAOB]

        = [150.72 cm2] – [62.28 cm2] = 88.44 cm2

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