Class 10 Areas of Sector and Segment of a Circle

Topic Covered

♦ 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).
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
Solution:

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
Solution:

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