11.1 Areas of Sector and Segment of a Circle

You have already come across the terms sector and segment of a circle in your earlier classes. 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. 11.1, shaded region OAPB is a sector of the circle with centre $\mathrm{O} . \angle \mathrm{AOB}$ 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 $\mathrm{OAQB}$ is called the major sector. You can also see that angle of the major sector is $360^{\circ}-\angle \mathrm{AOB}$.

Fig. 11.1

Now, look at Fig. 11.2 in which AB is a chord of the circle with centre $\mathrm{O}$. So, shaded region APB is a segment of the circle. You can also note that unshaded region $\mathrm{AQB}$ 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. 11.2

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 $\mathrm{O}$ and radius $r$ (see Fig. 11.3). Let the degree measure of $\angle \mathrm{AOB}$ be $\theta$.

Fig. 11.3

You know that area of a circle (in fact of a circular region or disc) is $\pi r^{2}$.

In a way, we can consider this circular region to be a sector forming an angle of $360^{\circ}$ (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 $=\pi r^{2}$

So, when the degree measure of the angle at the centre is 1 , area of the sector $=\frac{\pi r^{2}}{360}$.

Therefore, when the degree measure of the angle at the centre is $\theta$, area of the sector $=\frac{\pi r^{2}}{360} \times \theta=\frac{\theta}{360} \times \pi r^{2}$.

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

$$ \text { Area of the sector of angle } \theta=\frac{\theta}{360} \times \pi r^{2} \text {, } $$

where $r$ is the radius of the circle and $\theta$ 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 $360^{\circ}$ ) as $2 \pi r$, we can obtain the required length of the arc APB as $\frac{\theta}{360} \times 2 \pi r$.

So, length of an arc of a sector of angle $\theta=\frac{\theta}{360} \times 2 \pi r$.

Fig. 11.4

Now let us take the case of the area of the segment APB of a circle with centre $\mathrm{O}$ and radius $r$ (see Fig. 11.4). You can see that :

Area of the segment $\mathrm{APB}=$ Area of the sector $\mathrm{OAPB}-$ Area of $\triangle \mathrm{OAB}$

$$ =\frac{\theta}{360} \times \pi r^{2}-\text { area of } \Delta \mathrm{OAB} $$

Note : From Fig. 11.3 and Fig. 11.4 respectively, you can observe that:

Area of the major sector $\mathrm{OAQB}=\pi r^{2}-$ Area of the minor sector $\mathrm{OAPB}$

and

Area of major segment $\mathrm{AQB}=\pi r^{2}$ - Area of the minor segment APB

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

Example 1 : Find the area of the sector of a circle with radius $4 \mathrm{~cm}$ and of angle $30^{\circ}$. Also, find the area of the corresponding major sector (Use $\pi=3.14$ ).

Solution : Given sector is OAPB (see Fig. 11.5).

Fig. 11.5

$$ \begin{aligned} \text { Area of the sector } & =\frac{\theta}{360} \times \pi r^{2} \\ & =\frac{30}{360} \times 3.14 \times 4 \times 4 \mathrm{~cm}^{2} \\ & =\frac{12.56}{3} \mathrm{~cm}^{2}=4.19 \mathrm{~cm}^{2} \text { (approx.) } \end{aligned} $$

Area of the corresponding major sector

$$ \begin{aligned} & =\pi r^{2}-\text { area of sector OAPB } \\ & =(3.14 \times 16-4.19) \mathrm{cm}^{2} \\ & =46.05 \mathrm{~cm}^{2}=46.1 \mathrm{~cm}^{2} \text { (approx.) } \end{aligned} $$

Alternatively, area of the major sector $=\frac{(360-\theta)}{360} \times \pi r^{2}$

$$ \begin{aligned} & =\left(\frac{360-30}{360}\right) \times 3.14 \times 16 \mathrm{~cm}^{2} \\ & =\frac{330}{360} \times 3.14 \times 16 \mathrm{~cm}^{2}=46.05 \mathrm{~cm}^{2} \\ & =46.1 \mathrm{~cm}^{2} \text { (approx.) } \end{aligned} $$

Example 2 : Find the area of the segment AYB shown in Fig. 11.6, if radius of the circle is $21 \mathrm{~cm}$ and $\angle \mathrm{AOB}=120^{\circ}$. (Use $\pi=\frac{22}{7}$ )

Fig. 11.6

Solution : Area of the segment AYB

$$ =\text { Area of sector OAYB }- \text { Area of } \Delta \mathrm{OAB} \tag{1} $$

$$ \text{ Now, area of the sector OAYB } =\frac{120}{360} \times \frac{22}{7} \times 21 \times 21 \mathrm{~cm}^{2}=462 \mathrm{~cm}^{2} \tag{2}$$

For finding the area of $\Delta \mathrm{OAB}$, draw $\mathrm{OM} \perp \mathrm{AB}$ as shown in Fig. 11.7.

Fig. 11.7

Note that $\mathrm{OA}=\mathrm{OB}$. Therefore, by RHS congruence, $\Delta \mathrm{AMO} \cong \Delta \mathrm{BMO}$.

So, $\mathrm{M}$ is the mid-point of $\mathrm{AB}$ and $\angle \mathrm{AOM}=\angle \mathrm{BOM}=\frac{1}{2} \times 120^{\circ}=60^{\circ}$.

Let

$$ \mathrm{OM}=x \mathrm{~cm} $$

So, from $\Delta$ OMA,

$$ \frac{\mathrm{OM}}{\mathrm{OA}}=\cos 60^{\circ} $$

or,

$$ \frac{x}{21}=\frac{1}{2} \quad\left(\cos 60^{\circ}=\frac{1}{2}\right) $$

or,

$$ x=\frac{21}{2} $$

So,

$$ \mathrm{OM}=\frac{21}{2} \mathrm{~cm} $$

Also,

$$ \frac{\mathrm{AM}}{\mathrm{OA}}=\sin 60^{\circ}=\frac{\sqrt{3}}{2} $$

So,

$$ \mathrm{AM}=\frac{21 \sqrt{3}}{2} \mathrm{~cm} $$

Therefore,

$$ \mathrm{AB}=2 \mathrm{AM}=\frac{2 \times 21 \sqrt{3}}{2} \mathrm{~cm}=21 \sqrt{3} \mathrm{~cm} $$

So,

$$ \text { area of } \begin{aligned} \Delta \mathrm{OAB} & =\frac{1}{2} \mathrm{AB} \times \mathrm{OM}=\frac{1}{2} \times 21 \sqrt{3} \times \frac{21}{2} \mathrm{~cm}^{2} \end{aligned} $$

$$ =\frac{441}{4} \sqrt{3} \mathrm{~cm}^{2}\tag{3}$$

Therefore, area of the segment AYB $=\left(462-\frac{441}{4} \sqrt{3}\right) \mathrm{cm}^{2}$ [From (1), (2) and (3)]

$$ =\frac{21}{4}(88-21 \sqrt{3}) \mathrm{cm}^{2} $$

EXERCISE 11.1

Unless stated otherwise, use $\pi=\frac{22}{7}$.

1. Find the area of a sector of a circle with radius $6 \mathrm{~cm}$ if angle of the sector is $60^{\circ}$.

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Solution

Let $OACB$ be a sector of the circle making $60^{\circ}$ angle at centre $O$ of the circle.

Area of sector of angle $\theta=\frac{\theta}{360^{\circ}} \times \pi r^{2}$

Area of sector $OACB=\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(6)^{2}$

$=\frac{1}{6} \times \frac{22}{7} \times 6 \times 6=\frac{132}{7} cm^{2}$

2. Find the area of a quadrant of a circle whose circumference is $22 \mathrm{~cm}$.

3. The length of the minute hand of a clock is $14 \mathrm{~cm}$. Find the area swept by the minute hand in 5 minutes.

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Solution

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates $360^{\circ}$.

In 5 minutes, minute hand will rotate $=\frac{360^{\circ}}{60} \times 5=30^{\circ}$

Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of $30^{\circ}$ in a circle of $14 cm$ radius.

Area of sector of angle $\theta=\frac{\theta}{360^{\circ}} \times \pi r^{2}$

Area of sector of $30^{\circ}=\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14$

$=\frac{22}{12} \times 2 \times 14$

$=\frac{11 \times 14}{3}$

$=\frac{154}{3} cm^{2}$

Therefore, the area swept by the minute hand in 5 minutes is $\frac{154}{3} cm^{2}$.

4. A chord of a circle of radius $10 \mathrm{~cm}$ subtends a right angle at the centre. Find the area of the corresponding:

(i) minor segment

(ii) major sector. (Use $\pi=3.14$ )

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Solution

Let $A B$ be the chord of the circle subtending $90^{\circ}$ angle at centre $O$ of the circle.

Area of major sector $OADB=(\frac{360^{\circ}-90^{\circ}}{360^{\circ}}) \times \pi r^{2}=(\frac{270^{\circ}}{360^{\circ}}) \pi r^{2}$ $=\frac{3}{4} \times 3.14 \times 10 \times 10$

$=235.5 cm^{2}$

Area of minor sector $OACB=\frac{90^{\circ}}{360^{\circ}} \times \pi r^{2}$

$=\frac{1}{4} \times 3.14 \times 10 \times 10$

$=78.5 cm^{2}$

Area of $\triangle OAB=\frac{1}{2} \times OA \times OB=\frac{1}{2} \times 10 \times 10$

$=50 cm^{2}$

Area of minor segment $A C B=$ Area of minor sector $OACB$ -

Area of $\triangle OAB=78.5-50=28.5 cm^{2}$

5. In a circle of radius $21 \mathrm{~cm}$, an arc subtends an angle of $60^{\circ}$ 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

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Solution

Radius $(r)$ of circle $=21 cm$

Angle subtended by the given arc $=60^{\circ}$

Length of an arc of a sector of angle $\theta=\frac{\theta}{360^{\circ}} \times 2 \pi r$

Length of $arc A C B=\frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21$

$=\frac{1}{6} \times 2 \times 22 \times 3$

$=22 cm$

Area of sector $OACB=\frac{60^{\circ}}{360^{\circ}} \times \pi r^{2}$

$=\frac{1}{6} \times \frac{22}{7} \times 21 \times 21$

$=231 cm^{2}$

In $\triangle OAB$,

$\angle OAB=\angle OBA(As OA=OB)$

$\angle OAB+\angle AOB+\angle OBA=180^{\circ}$

$2 \angle OAB+60^{\circ}=180^{\circ}$

$\angle OAB=60^{\circ}$

Therefore, $\triangle OAB$ is an equilateral triangle.

Area of $\triangle OAB=\frac{\sqrt{3}}{4} \times(\text{ Side })^{2}$

$=\frac{\sqrt{3}}{4} \times(21)^{2}=\frac{441 \sqrt{3}}{4} cm^{2}$

Area of segment $A C B=$ Area of sector $O A C B$ - Area of $\triangle O A B$

$=(231-\frac{441 \sqrt{3}}{4}) cm^{2}$

6. A chord of a circle of radius $15 \mathrm{~cm}$ subtends an angle of $60^{\circ}$ at the centre. Find the areas of the corresponding minor and major segments of the circle.

(Use $\pi=3.14$ and $\sqrt{3}=1.73$ )

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Solution

Radius ( $r$ ) of circle $=15 cm$

Area of sector OPRQ $=\frac{60^{\circ}}{360^{\circ}} \times \pi r^{2}$

$=\frac{1}{6} \times 3.14 \times(15)^{2}$

$=117.75 cm^{2}$

In $\triangle OPQ$,

$\angle OPQ=\angle OQP($ As OP $=OQ)$

$\angle OPQ+\angle OQP+\angle POQ=180^{\circ}$

$2 \angle OPQ=120^{\circ}$

$\angle OPQ=60^{\circ}$

$\triangle OPQ$ is an equilateral triangle.

Area of $\triangle OPQ=\frac{\sqrt{3}}{4} \times(\text{ side })^{2}$

$=\frac{\sqrt{3}}{4} \times(15)^{2}=\frac{225 \sqrt{3}}{4} cm^{2}$

$=56.25 \sqrt{3}$

$=97.3125 cm^{2}$

Area of segment PRQ = Area of sector OPRQ - Area of $\triangle OPQ$

$=117.75$ - 97.3125

$=20.4375 cm^{2}$

Area of major segment PSQ = Area of circle - Area of segment PRQ

$ \begin{aligned} & =\pi(15)^{2}-20.4375 \\ & =3.14 \times 225-20.4375 \\ & =706.5-20.4375 \\ & =\quad 686.0625 cm^{2} \end{aligned} $

7. A chord of a circle of radius $12 \mathrm{~cm}$ subtends an angle of $120^{\circ}$ at the centre. Find the area of the corresponding segment of the circle.

(Use $\pi=3.14$ and $\sqrt{3}=1.73$ )

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Solution

Let us draw a perpendicular OV on chord ST. It will bisect the chord ST.

$SV=VT$

In $\triangle OVS$,

$ \begin{aligned} & \frac{O V}{O S}=\cos 60^{\circ} \\ & \frac{O V}{12}=\frac{1}{2} \\ & O V=6 cm \end{aligned} $

$\frac{SV}{SO}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$

$\frac{SV}{12}=\frac{\sqrt{3}}{2}$

$SV=6 \sqrt{3} cm$

$ST=2 SV=2 \times 6 \sqrt{3}=12 \sqrt{3} cm$

Area of $\triangle OST=\frac{1}{2} \times ST \times OV$

$=\frac{1}{2} \times 12 \sqrt{3} \times 6$

$=36 \sqrt{3}=36 \times 1.73=62.28 cm^{2}$

Area of sector OSUT $=\frac{120^{\circ}}{360^{\circ}} \times \pi(12)^{2}$

$=\frac{1}{3} \times 3.14 \times 144=150.72 cm^{2}$

Area of segment SUT = Area of sector OSUT - Area of $\triangle OST$

$=150.72-62.28$

$=88.44 cm^{2}$

8. A horse is tied to a peg at one corner of a square shaped grass field of side $15 \mathrm{~m}$ by means of a $5 \mathrm{~m}$ long rope (see Fig. 11.8). Find

Fig. 11.8

(i) the area of that part of the field in which the horse can graze.

(ii) the increase in the grazing area if the rope were $10 \mathrm{~m}$ long instead of $5 \mathrm{~m}$. (Use $\pi=3.14$ )

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Solution

From the figure, it can be observed that the horse can graze a sector of $90^{\circ}$ in a circle of $5 m$ radius.

Area that can be grazed by horse $=$ Area of sector $OACB$

$=\frac{90^{\circ}}{360^{\circ}} \pi r^{2}$

$=\frac{1}{4} \times 3.14 \times(5)^{2}$

$=19.625 m^{2}$

Area that can be grazed by the horse when length of rope is $10 m$ long $=\frac{90^{\circ}}{360^{\circ}} \times \pi \times(10)^{2}$

$=\frac{1}{4} \times 3.14 \times 100$

$=78.5 m^{2}$

Increase in grazing area $=(78.5-19.625) m^{2}$

$=58.875 m^{2}$

9. A brooch is made with silver wire in the form of a circle with diameter $35 \mathrm{~mm}$. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 11.9. Find :

(i) the total length of the silver wire required.

(ii) the area of each sector of the brooch.

Fig. 11.9

10. An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius $45 \mathrm{~cm}$, find the area between the two consecutive ribs of the umbrella.

Fig. 11.10

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Solution

There are 8 ribs in an umbrella. The area between two consecutive ribs is subtending $\frac{360^{\circ}}{8}=45^{\circ}$ at the centre of the assumed flat circle.

Area between two consecutive ribs of circle $=\frac{45^{\circ}}{360^{\circ}} \times \pi r^{2}$

$ \begin{aligned} & =\frac{1}{8} \times \frac{22}{7} \times(45)^{2} \\ & =\frac{11}{28} \times 2025=\frac{22275}{28} cm^{2} \end{aligned} $

11. A car has two wipers which do not overlap. Each wiper has a blade of length $25 \mathrm{~cm}$ sweeping through an angle of $115^{\circ}$. Find the total area cleaned at each sweep of the blades.

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Solution

It can be observed from the figure that each blade of wiper will sweep an area of a sector of $115^{\circ}$ in a circle of $25 cm$ radius.

Area of such sector $=\frac{115^{\circ}}{360^{\circ}} \times \pi \times(25)^{2}$

$=\frac{23}{72} \times \frac{22}{7} \times 25 \times 25$

$=\frac{158125}{252} cm^{2}$

Area swept by 2 blades $=2 \times \frac{158125}{252}$

$=\frac{158125}{126} cm^{2}$

12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle $80^{\circ}$ to a distance of $16.5 \mathrm{~km}$. Find the area of the sea over which the ships are warned. (Use $\pi=3.14$ )

13. A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is $28 \mathrm{~cm}$, find the cost of making the designs at the rate of ₹ 0.35 per $\mathrm{cm}^{2}$. (Use $\sqrt{3}=1.7$ )

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Solution

It can be observed that these designs are segments of the circle.

Consider segment APB. Chord AB is a side of the hexagon. Each chord will substitute $\frac{360^{\circ}}{6}=60^{\circ}$ at the centre of the circle.

In $\triangle OAB$,

$\angle OAB=\angle OBA(As OA=OB)$

$\angle A O B=60^{\circ}$

$\angle OAB+\angle OBA+\angle AOB=180^{\circ}$

$2 \angle OAB=180^{\circ}-60^{\circ}=120^{\circ}$

$\angle OAB=60^{\circ}$

Therefore, $\triangle OAB$ is an equilateral triangle.

Area of $\triangle OAB=\frac{\sqrt{3}}{4} \times(\text{ side })^{2}$

$\begin{aligned}=\frac{\sqrt{3}}{4} \times(28)^{2}=196 \sqrt{3} & =196 \times 1.7 \\ & =333.2 cm^{2}\end{aligned}$

Area of sector OAPB $=\frac{60^{\circ}}{360^{\circ}} \times \pi r^{2}$

$=\frac{1}{6} \times \frac{22}{7} \times 28 \times 28$

$=\frac{1232}{3} cm^{2}$

Area of segment APB = Area of sector OAPB - Area of $\triangle O A B$

$=(\frac{1232}{3}-333.2) cm^{2}$

Therefore, area of designs $=6 \times(\frac{1232}{3}-333.2) cm^{2}$

$ \begin{aligned} & =(2464-1999.2) cm^{2} \\ & =464.8 cm^{2} \end{aligned} $

Cost of making $1 cm^{2}$ designs $=$ Rs 0.35

Cost of making $464.76 cm^{2}$ designs $=464.8 \times 0.35=$ Rs 162.68

Therefore, the cost of making such designs is Rs 162.68 .

14. Tick the correct answer in the following :

Area of a sector of angle $p$ (in degrees) of a circle with radius $\mathrm{R}$ is

(A) $\frac{p}{180} \times 2 \pi \mathrm{R}$

(B) $\frac{p}{180} \times \pi \mathrm{R}^{2}$

(C) $\frac{p}{360} \times 2 \pi \mathrm{R}$

(D) $\frac{p}{720} \times 2 \pi R^{2}$

11.2 Summary

In this chapter, you have studied the following points :

1. Length of an arc of a sector of a circle with radius $r$ and angle with degree measure $\theta$ is $\frac{\theta}{360} \times 2 \pi r$.

2. Area of a sector of a circle with radius $r$ and angle with degree measure $\theta$ is $\frac{\theta}{360} \times \pi r^{2}$.

3. Area of segment of a circle

$=$ Area of the corresponding sector - Area of the corresponding triangle.