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### Areas of Sector and Segment of a Circle

♦ Areas of Sector and Segment of a Circle

You have already come across the terms sector and segment of a circle..

Recall that the portion (or part) of the circular region enclosed by two radii and the corresponding arc is called a sector of the circle and the portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle.

Thus, in Fig. 12.4, 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`.

Now, look at Fig. 12.5 in 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.

`text ( Remark : )` When we write ‘segment’ and ‘sector’ we will mean the ‘minor segment’ and the ‘minor sector’ respectively, unless stated otherwise.

Now with this knowledge, let us try to find some relations (or formulae) to calculate their areas.

Let `OAPB` be a sector of a circle with centre `O` and radius `r` (see Fig. 12.6). Let the degree measure of `∠ AOB` be `θ`.

You know that area of a circle (in fact of a circular region or disc) is `πr^2`.

In a way, 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 degree measure of the angle at the centre is `360`, area of the sector `= πr^2`

So, when the degree measure of the angle at the centre is 1, area of the sector ` = (pi r^2)/(360)`

Therefore, when the degree measure of the angle at the centre is `θ`, area of the sector ` = (pi r^2)/(360) xx θ = θ/(360) xx pi r^2` .

Thus, we obtain the following relation (or formula) for area of a sector of a circle:

`text ( Area of the sector of angle ) θ = θ/(360) xx pi r^2` ,

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

Now, a natural question arises : Can we find the length of the arc `APB` corresponding to this sector? Yes. Again, by applying the Unitary Method and taking the whole length of the circle (of angle` 360°`) as `2πr`, we can obtain the required

length of the arc `APB` as ` θ/(360) xx 2 pi r` .

So,`text ( length of an arc of a sector of angle ) θ = θ/(360) xx 2 pi r` ,

Now let us take the case of the area of the segment `APB` of a circle with centre `O` and radius `r` (see Fig. 12.7). You can see that :

Area of the segment `APB = text (Area of the sector) OAPB – text (Area of) Delta OAB`

` = θ/(360) xx pi r^2 - text (area of) Delta OAB`

Note : From Fig. 12.6 and Fig. 12.7 respectively, you can observe that :

` text (Area of the major sector ) OAQB = πr^2 – text (Area of the minor sector) OAPB`

and ` text ( Area of major segment ) AQB = πr^2 – text (Area of the minor segment ) APB`

Let us now take some examples to understand these concepts (or results).

Recall that the portion (or part) of the circular region enclosed by two radii and the corresponding arc is called a sector of the circle and the portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle.

Thus, in Fig. 12.4, 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`.

Now, look at Fig. 12.5 in 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.

`text ( Remark : )` When we write ‘segment’ and ‘sector’ we will mean the ‘minor segment’ and the ‘minor sector’ respectively, unless stated otherwise.

Now with this knowledge, let us try to find some relations (or formulae) to calculate their areas.

Let `OAPB` be a sector of a circle with centre `O` and radius `r` (see Fig. 12.6). Let the degree measure of `∠ AOB` be `θ`.

You know that area of a circle (in fact of a circular region or disc) is `πr^2`.

In a way, 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 degree measure of the angle at the centre is `360`, area of the sector `= πr^2`

So, when the degree measure of the angle at the centre is 1, area of the sector ` = (pi r^2)/(360)`

Therefore, when the degree measure of the angle at the centre is `θ`, area of the sector ` = (pi r^2)/(360) xx θ = θ/(360) xx pi r^2` .

Thus, we obtain the following relation (or formula) for area of a sector of a circle:

`text ( Area of the sector of angle ) θ = θ/(360) xx pi r^2` ,

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

Now, a natural question arises : Can we find the length of the arc `APB` corresponding to this sector? Yes. Again, by applying the Unitary Method and taking the whole length of the circle (of angle` 360°`) as `2πr`, we can obtain the required

length of the arc `APB` as ` θ/(360) xx 2 pi r` .

So,`text ( length of an arc of a sector of angle ) θ = θ/(360) xx 2 pi r` ,

Now let us take the case of the area of the segment `APB` of a circle with centre `O` and radius `r` (see Fig. 12.7). You can see that :

Area of the segment `APB = text (Area of the sector) OAPB – text (Area of) Delta OAB`

` = θ/(360) xx pi r^2 - text (area of) Delta OAB`

Note : From Fig. 12.6 and Fig. 12.7 respectively, you can observe that :

` text (Area of the major sector ) OAQB = πr^2 – text (Area of the minor sector) OAPB`

and ` text ( Area of major segment ) AQB = πr^2 – text (Area of the minor segment ) APB`

Let us now take some examples to understand these concepts (or results).

Q 3250401314

Find the area of the sector of a circle with radius `4` cm and of angle `30°`. Also, find the area

of the corresponding major sector (Use `π = 3.14` ).

Class 10 Chapter 12 Example 2

of the corresponding major sector (Use `π = 3.14` ).

Class 10 Chapter 12 Example 2

Given sector is `OAPB` (see Fig. 12.8).

Area of the sector ` = θ/(360) xx pi r^2`

` = 30/360 xx 3.14 xx 4 xx 4 cm^2`

`= 12.56/3 cm^2 = 4.19 cm^2` (approx )

Area of the corresponding major sector

` = pi r^2 - text ( area of sector ) OAPB`

`(3.14 × 16 – 4.19) cm^2`

`= 46.05 cm^2 = 46.1 cm^2`(approx.)

Alternatively, area of the major sector `= ( 360 - θ)/360 xx pi r^2`

`= ( (360 -30)/360 ) xx 3.14 xx 16 cm^2`

` = 330/360 xx 3.14 xx 16 cm^2 = 46.05 cm^2`

` = 46.1 cm^2` (approx )

Q 3200401318

Find the area of the segment `AYB` shown in Fig. 12.9, if radius of the circle is `21 cm` and

` ∠ AOB = 120^o ` (Use` π = 22/7` )

Class 10 Chapter 12 Example 3

` ∠ AOB = 120^o ` (Use` π = 22/7` )

Class 10 Chapter 12 Example 3

Area of the segment `AYB`

`=` Area of sector `OAYB –` Area of `Delta OAB`............ (1)

Now, area of the sector `OAYB = 120/360 xx 22/7 xx 21 xx 21 cm^2 = 462 cm^2` ............... (2)

For finding the area of `Delta OAB`, draw `OM ⊥ AB` as shown in Fig. 12.10.

Note that `OA = OB`. Therefore, by `RHS` congruence, `Delta AMO ≅ Delta BMO`.

So, M is the mid-point of `AB` and `∠ AOM = ∠ BOM = 1/2 xx 120^o = 60^o` .

Let `OM = x` cm

So, from `Delta OMA, (OM)/(OA) = cos 60^o`

or, ` x/21 = 1/2 (cos 60^o = 1/2 )`

or, `x = 21/2 cm`

Also, ` (AM)/(OA) = sin 60^o = (sqrt 3)/2`

So, `AM = (2 sqrt 3)/2 cm`

Therefore, `AB =2 AM = ( 2 xx 21 sqrt 3)/2 cm = 21 sqrt 3 cm`

So, area of `Delta OAB = 1/2 AB xx OM = 1/2 xx 21 sqrt 3 xx 21/2 cm^2`

` = 441/4 sqrt 3 cm^2` .......(3)

Therefore, area of the segment `AYB = ( 462 - 441/4 sqrt 3) cm^2 ` [ From (1), (2) and (3) ]

`= 21/4 ( 88 -21 sqrt 3) cm^2`