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**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.**

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°**

** (i) Circumference of the circle = 2 πr**

** **

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

** **

** (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**

** **

**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**

** **

** **

** From (1) and (3)**

** Area of the minor segment**

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

** = [150.72 cm ^{2}] – [62.28 cm^{2}] = 88.44 cm^{2}**

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