Sector of the circle :The portion (or part) of the circular region enclosed by two radii and the
corresponding arc is called a sector of the circle .
Segment of the circle.:The portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle.
= $(θ/360)$ ×π $r^2$
= π $r^2$ - area of minor sector
= Area of minor sector (AOBP) -Area of ∆ AOB
= Area of circle -Area of minor segment (ABP)
Thus, in Fig., 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.
Remark : When we write ‘segment’ and ‘sector’
we will mean the ‘minor segment’ and the ‘minor sector’ respectively, unless stated otherwise.
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.
Since the area of a circle is $πr^2$.
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 = ${πr^2}/360$
Therefore, when the degree measure of the angle at the centre is θ, area of the sector = ${πr^2}/360$ × θ
Area of the sector of angle θ =$θ/360 × πr^2$
where $r$ is the radius of the circle and θ the angle of the sector in degrees.
Similarily:
length of an arc of a sector of angle θ = $θ/360 × 2πr$
Area of the segment APB = Area of the sector OAPB – Area of ∆ OAB
Area of the segment APB =$θ/360 × πr^2$ - Area of ∆ OAB Area of the major sector OAQB = $πr^2$ – Area of the minor sector OAPBArea of major segment AQB = $πr^2$ – Area of the minor segment APB
Example: 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).
Solution : Given sector is OAPB (see Fig. 12.8).
Area of the sector of angle θ =$θ/360 × πr^2$
= $30/360 × 3.14 × 4 × 4$
=$4.19$ $cm^2$
Area of the corresponding major sector:
= $πr^2$ – area of sector OAPB
= (3.14 × 4 × 4 – 4.19) $cm^2$
= 46.05 $cm^2$ = 46.1 $cm^2$ (approx.)
Alternatively, area of the major sector = ${360 – θ }/360 × πr^2$
=${360 – 30 }/360$ × 3.14 × 4 × 4
=46.05 $cm^2$
Example: Find the area of the segment AYB shown in Fig., if radius of the circle is 21 cm
and ∠ AOB = 120°. (Use π = $22/7$ )
Solution : Area of the segment AYB = Area of sector OAYB – Area of ∆ OAB (1)
Now, area of the sector OAYB = $120/360$ × $22/7$ × 21 × 21 $cm^2$ = 462 $cm^2$ (2)
For finding the area of ∆ OAB, draw OM ⊥AB as shown in Fig. 12.10. Note that OA = OB.
Therefore, by RHS congruence, ∆AMO ≅ ∆ BMO.
So, M is the mid-point of AB and ∠ AOM = ∠BOM = $1/2$ ×120° = 60°.
Let OM = $x$ $cm$
So, from ∆OMA, ${OM}/{OA}$ = cos 60°
$ x/21$ = $1/2$ (Since cos 60°=$1/2$ )
$x$ = $21/2$
So, OM = $21/2$ cm
Also, ${AM}/{OA}$= sin 60° = $√3/2$
${AM}$ = ${21√3}/2$ $cm$
Therefore, AB = 2 AM = ${21√3}$ $cm$
So, area of ∆ OAB = $1/2$ AB × OM = $1/2$ × ${21√3}$ × $21/2$ $cm^2$
=${441√3}/4$ $cm^2$
Therefore, area of the segment AYB = ${441√3}/4$ $cm^2$ [From (1), (2) and (3)]
= $21/4$ (88 – 21√3) $cm^2$
