How To Find The Area Of A Sector Of A Circle

If the arc subtends an angle of θ at the centre, then its arc length is
$$\frac{\text{ }\!\!\theta\!\!\text{ }}{\text{180}}\text{ }\!\!\times\!\!\text{ }\!\!\pi\!\!\text{ r}$$
Hence, the arc length ‘l’ of a sector of angle θ in a
circle of radius r is given by
$$l=\frac{\text{ }\!\!\theta\!\!\text{ }}{\text{180}}\text{ }\!\!\times\!\!\text{ }\!\!\pi\!\!\text{ r }…….\text{ (i)}$$
If the arc subtends an angle θ, then area of the corresponding sector is
$$\frac{\pi {{r}^{2}}\theta }{360}$$
Thus, the area A of a sector of angle θ in a circle of radius r is given by
$$A=~\frac{\theta }{360}\text{ }\times \text{ }\!\!\pi\!\!\text{ }{{r}^{2}}$$
$$=\frac{\theta }{360}\text{ }\times \text{ }\left( \text{Area of the circle} \right)\text{ }……..\text{ (ii)}$$
= × (Area of the circle) ….(ii)
Some useful results to remember:
(i) Angle described by minute hand in 60 minutes = 360º
Angle described by minute hand in one minute
$$={{\left( \frac{360}{60} \right)}^{0}}=\text{ }6{}^\text{o}$$
Thus, minute hand rotates through an angle of 6º in one minute.
(ii) Angle described by hour hand in 12 hours = 360º
Angle described by hour hand in one hour
$$=\left( \frac{360}{12} \right)_{{}}^{0}=30{}^\text{o}$$

Area Of A Sector Of A Circle With Examples

Example 1:    A sector is cut from a circle of radius 21 cm. The angle of the sector is 150º. Find the length of its arc and area.
Sol.    The arc length l and area A of a sector of angle θ in a circle of radius r are given by
$$l=\frac{\theta }{360}\times 2\pi r\text{ and A=}\frac{\theta }{360}\times \pi {{r}^{2}}\text{ respectively}\text{.}$$
Here, r = 21 cm and q = 150
$$\text{ }l=\left\{ \frac{150}{360}\times 2\times \frac{22}{7}\times 21 \right\}\text{cm}=\text{55 cm}$$
$$\text{and A}=\left\{ \frac{150}{360}\times \frac{22}{7}\times {{(21)}^{2}} \right\}c{{m}^{2}}=\frac{1155}{2}c{{m}^{2}}$$
= 577.5 cm2

Example 2:    Find the area of the sector of a circle whose radius is 14 cm and angle of sector is 45º.
Sol.    We know that the area A of a sector of angle θ in a circle of radius r is given by
$$A=\frac{\theta }{360}\text{ }\times \text{ }\pi {{r}^{2}}$$
Here, r = 14 cm and θ = 45
$$A=\left\{ \frac{45}{360}\times \frac{22}{7}\times {{(14)}^{2}} \right\}c{{m}^{2}}$$
$$=\left\{ \frac{1}{8}\times \frac{22}{7}\times 14\times 14 \right\}c{{m}^{2}}$$
= 77 cm2

Example 3:    In Fig. there are shown sectors of two concentric circles of radii 7 cm and 3.5 cm. Find the area of the shaded region. (Use π = $$\frac { 22 }{ 7 }$$).
Sol.    Let A1 and A2 be the areas of sectors OAB and OCD respectively. Then, A1 = Area of a sector of angle 30º in a circle of radius 7 cm

$${{A}_{1}}=\left\{ \frac{30}{360}\times \frac{22}{7}\times {{7}^{2}} \right\}c{{m}^{2}}\text{ }\left[ U\sin g:A=\frac{\theta }{360}\times \pi {{r}^{2}}\, \right]$$
⇒ A1 = $$\frac { 77 }{ 6 }$$ cm2
A2 = Area of a sector of angle 30º in a circle of radius 3.5 cm.
∴  Area of the shaded region
$${{A}_{2}}=\left\{ \frac{30}{360}\times \frac{22}{7}\times {{(3.5)}^{2}} \right\}c{{m}^{2}}$$
$${{A}_{2}}=\left\{ \frac{1}{12}\times \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2} \right\}c{{m}^{2}}=\frac{77}{24}c{{m}^{2}}$$
$$={{A}_{1}}-{{A}_{2}}=\left( \frac{77}{6}-\frac{77}{24} \right)$$
= $$\frac { 77 }{ 24 }$$ × (4 – 1) cm2 = $$\frac { 77 }{ 8 }$$ cm2 = 9.625 cm2

Example 4:    A pendulum swings through an angle of 30º and describes an arc 8.8 cm in length. Find the length of the pendulum.
Sol.      Here, q = 30º, l = arc = 8.8 cm
$$l=\frac{\theta }{360}\times 2\pi r$$
$$8.8=~\frac{30}{360}\times 2\times \frac{22}{7}\times r~$$
$$r=\frac{8.8\times 6\times 7}{22}=16.8\text{ }cm$$

Example 5:    The length of minute hand of a clock is 14 cm. Find the area swept by the minute hand in one minute. (Use π = 22/7)
Sol.     Clearly, minute hand of a clock describes a circle of radius equal to its length i.e., 14 cm.
Since the minute hand rotates through 6º in one minute. Therefore, area swept by the minute hand in one minute is the area of a sector of angle 6º in a circle of radius 14 cm. Hence, required area A is given by
$$A=\frac{\theta }{360}\times \pi {{r}^{2}}~$$
$$A=\left\{ \frac{6}{360}\times \frac{22}{7}\times {{(14)}^{2}} \right\}$$
$$A=\left\{ \frac{1}{60}\times \frac{22}{7}\times 14\times 14 \right\}=\frac{154}{15}$$
= 10.26 cm2

Example 6:    The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm. Find the area of the sector.
Sol.    Let OAB be the given sector. Then,
Perimeter of sector OAB = 16.4 cm

⇒ OA + OB + arc AB = 16.4 cm
⇒ 5.2 + 5.2 + arc AB = 16.4
⇒ arc AB = 6 cm
l = 6 cm
∴ Area of  sector OAB = $$\frac { 1 }{ 2 }$$ lr
= $$\frac { 1 }{ 2 }$$ × 6 × 5.2 cm2 = 15.6 cm2

Example 7:    The minute hand of a clock is 10cm long. Find the area of the face of the clock described by the minute hand between 9 A.M. and 9.35 A.M.
Sol.    We have,
Angle described by the minute hand in one minute = 6º
∴ Angle described by the minute hand in
35 minutes = (6 × 35)º = 210º
∴ Area swept by the minute hand in 35 minutes
= Area of a sector of angle 210º in a circle of radius 10 cm
$$=\left\{ \frac{210}{360}\times \frac{22}{7}\times {{(10)}^{2}} \right\}c{{m}^{2~}}=\text{ }183.3\text{ }c{{m}^{2}}~\text{ }\left[ \text{Using }A=\frac{\theta }{360{}^\text{o}}\times \pi {{r}^{2}} \right]$$

Example 8:    The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of distances travelled by their tips in 2 days.
(Take π = 22/7)
Sol.    In 2 days, the short hand will complete 4 rounds.
∴ Distance moved by its tip = 4 (Circumference of a circle of radius 4 cm)
$$=4\times \left( 2\times \frac{22}{7}\times 4 \right)=\frac{704}{7}\text{ cm}$$
In 2 days, the long hand will complete 48 rounds.
∴ Distance moved by its tip
= 48 (Circumference of a circle of radius 6 cm)
$$=48\times \left( 2\times \frac{22}{7}\times 4 \right)=\frac{12672}{7}\text{ cm}$$
Hence,
Sum of the distance moved by the tips of two hands of the clock
$$=\left( \frac{704}{7}+\frac{12672}{7} \right)=\text{ }1910.57\text{ }cm$$

Example 9:    An elastic belt is placed round the rim of a pulley of radius 5 cm. One point on the belt is pulled directly away from the centre O of the pulley until it is at P, 10 cm from O. Find the length of the belt that is in contact with the rim of the pulley. Also, find the shaded area.
Sol.    In the adjacent figure, let ∠AOP = ∠BOP = θ. Clearly, portion AB of the belt is not in contact with the rim of the pulley. In right triangle OAP, we have

$$\cos \theta =\frac{OA}{OP}=\frac{5}{10}=\frac{1}{2}\text{ }\Rightarrow \text{ }\theta =\text{ }60{}^\text{o}$$
⇒ ∠AOB = 2θ = 120º
$$\text{Arc AB}=\frac{120{}^\text{o}\times 2\times \pi \times 5}{360}=\frac{10\pi }{3}cm$$
$$\left[ \text{Using }l=\frac{\theta }{360}\times 2\pi r \right]$$
Hence, Length of the belt that is in contact with the rim of the pulley
= Circumference of the rim – Length of arc AB
$$=\text{ }2\pi \times 5-\frac{10\pi }{3}=\frac{20\pi }{3}~cm$$
Now,
Area of sector OAQB = $$\frac { 1 }{ 2 }$$ × π × 52 cm2
$$=\frac{25\pi }{3}\text{ c}{{\text{m}}^{2}}\text{ }\left[ \text{Using}Area=\frac{\theta }{360}\times \pi {{r}^{2}} \right]$$
Area of quadrilateral OAPB = 2 (Area of ∆OAP)
= 2 × (1/2 × OA × AP)
= 5 × 5√3 cm2
[∵ OP2 = OA2 + AP2   ⇒  AP = $$\sqrt{100-25}$$ = 5√3 ]
= 25√3 cm2
Hence,
OAPB – Area of sector OAQB.
$$=\left( 25\sqrt{3}-\frac{25\pi }{3} \right)=\frac{25}{3}(3\sqrt{3}-\pi )\text{ c}{{\text{m}}^{\text{2}}}$$

Example 10:    An arc of a circle is of length 5π cm and the sector it bounds has an area of 20 π cm². Find the radius of the circle.
Sol.    Let the radius of the circle be r cm and the arc AB of length 5π cm subtends angle θ at the centre O of the circle. Then,
Arc AB = 5π cm and
Area of sector OAB = 20π cm2

$$\Rightarrow \frac{\theta }{360}\times 2\pi r=5\pi \text{ and }\frac{\theta }{360}\times \pi {{r}^{2}}=20\pi$$
$$\Rightarrow \frac{\frac{\theta }{360}\times \pi {{r}^{2}}}{\frac{\theta }{360}\times 2\pi r}=\frac{20\pi }{5\pi }$$
⇒ r/2 = 4  ⇒   r = 8 cm
ALTER: We have, Area = 1/2 r ⇒20π
= 1/2 × 5π × r = 8 cm

Example 11:    An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm. Find the area between the two consecutive ribs of the umbrella.
Sol.    Since ribs are equally spaced. Therefore,
Angle made by two consecutive ribs at the centre = $$\frac { 360 }{ 8 }$$ = 45º

Thus,
Area between two consecutive ribs
= Area of a sector of a circle of radius 45 cm and sector angle 45º
$$=\left\{ \frac{45}{360}\times \frac{22}{7}\times 45\times 45 \right\}\text{c}{{\text{m}}^{2}}\text{ }\left[ \text{Using }Area=\frac{\theta }{360}\times \pi {{r}^{2}} \right]$$
=  $$\frac { 1 }{ 8 }$$ × $$\frac { 22 }{ 7 }$$ × 45× 45 cm2 = 795.53 cm2

Example 12:    A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. Find:
(i) the total length of the silver wire required
(ii) the area of each sector of the brooch.
Sol.    (i) We have,

Total length of the silver wire = Circumference of the circle of radius
35/2 mm + Length of five diameters
= 2π × $$\frac { 35 }{ 2 }$$ + 5 × 35 mm
$$=\left( 2\times \frac{22}{7}\times \frac{35}{2}+175 \right)\text{ mm}$$
= 285 mm
(ii)  The circle is divided into 10 equal sectors, Therefore, Area of each sector of the brooch
= 1/10 (Area of the circle)
= 1/10 × π  × (35/2)2 cm2
$$\text{=}\frac{1}{10}\times \frac{22}{7}\times \frac{35}{2}\times \frac{35}{2}\text{ m}{{\text{m}}^{2}}$$
$$=\frac{385}{4}\text{ m}{{\text{m}}^{2}}$$

What are the Parts of a Circle

So far, we have discussed about the triangle and quadrilateral that have linear boundaries. Circle is a closed figure that has a curvilinear boundary.
When we think of circles, the very first thing that comes to our mind is its round shape, for example, bangles, coins, rings, plates, chapattis, pizzas, CDs etc. Wheels of a car, bus, cycle, truck, train, and aeroplane are also round in shape. If we take a stone, tie it to one end of a string and swing it in the air by holding the other end of the string, the path traced by the stone will be a circular path and it will make a circle.

1. Circle: A circle is a collection of all those points in a plane that are at a given constant distance from a given fixed point in the plane.
2. Centre: Circle is a closed figure made up of points in a plane that are at the same distance from a fixed point, called the centre of the circle. In the figure O is the centre.

1. Radius: The constant distance from its centre is called the radius of the circle. In the figure, OA is radius
2. Chord: A line segment joining two points on a circle is called a chord of the circle. In the figure, AB is a chord of the circle. If a chord passes through centre then it is longest chord. PQ, PR, and ST are chords of the circle. Chord ST passes through the centre, hence it is a diameter.
3. Diameter: A chord passing through the centre of a circle is called the diameter of the circle. A circle has an infinite number of diameters. CD is the diameter of the circle as shown in the figure. If d is the diameter of the circle then d = 2r. where r is the radius. or the longest chord is called diameter.
In the figure, AB is the diameter and the arcs CD and DC are semicircles.
4. Arc: A continuous piece of a circle is called an arc. Let A,B,C,D,E,F be the points on the circle. The circle is divided into different pieces. Then, the pieces AB, BC, CD, DE, EF etc. are all arcs of the circle.
Let P,Q be two points on the circle. These P, Q divide the circle into two parts. Each part is an arc. These arcs are denoted in anti-clockwise direction
5. Circumference of a circle: The perimeter of a circle is called its circumference. The circumference of a circle of radius r is 2πr.
6. Semicircle: The diameter of a circle divides the circle into two equal parts. Each part is called a semi-circle. We can also say that half of a circle is called a semi¬circle. In the figure,  AXB and AYB represents two semi-circles.
7. Segment: Let AB be a chord of the circle. Then, AB divides the region enclosed by the circle (i.e., the circular disc) into two parts. Each of the parts is called a segment of the circle. The segment, containing the minor arc is called minor segment and the segment, containing the major arc, is called the major segment and segment of a circle is the region between an arc and chord of the circle.
8. Central Angles: Consider a circle. The angle subtended by an arc at the centre O is called the central angle. The vertex of the central angle is always at the centre O.
9. Degree measure of an arc: Degree measure of a minor arc is the measure of the central angle subtended by the arc.

The degree measure of the circumference of the circle is always 360°.
10. Interior and Exterior of Circle
A circle divides the plane on which lies into three parts.
(i) Inside the circle. which is called the interior of the circle
(ii) Circle
(iii) Outside the circle, which is called the exterior of the circle.
The circle and its interior make up the circular region.
11. Sector:
A sector is that region of a circular disc which lies between an arc and the two radii joining the extremities of the arc and the centre. OAB is a sector as shown in the figure.
Quadrant: One fourth of a circular disc is called a quadrant.
12. Position of a point:
Point Inside the circle: A point P, such that OP < r, is said to lie inside the circle.
The point inside the circle is also called interior point. (Example : Centre of cirle)
Point outside the circle: A point Q, such that OQ > r, is said to lie outside the circle C (O, r) = {X, OX = r}
The point outside the circle is also called exterior point.
Point on the circle: A point S, such that OS = r is said to lie on the circle C(O, r) = {X ,OX = r}.
Circular Disc: It is defined as a set of interior points and points on the circle. In set notation, it is written as : C(O, r) = {X : P OX ≤ r}
13. Concentric Circles:
Circles having the same centre and different radius are said to be concentric circles.
Remark. The word ‘radius’ is used for a line segment joining the centre to any point on the circle and also for its length.
14. Congruence of Circles & Arcs
Congruent circles: Two circles are said to be congruent if and only if, one of them can be superposed on the other, so as the cover it exactly. It means two circles are congruent if and only if, their radii are equal. i.e., C (O, r) and C (O’ , r) are congruent if only if r = s.
Congruent arcs: Two arcs of a circle are congruent, if either of them can be superposed on the other, so as to cover it exactly. It is only possible, if degree measure of two arcs are the same.

Example 1: Take two points A and B on a plane sheet. Draw a circle with A as a centre, AC as radius and B in its exterior.
Solution: Mark two points A and B on a paper.
A •            • B
As the point B should be in the exterior of the circle, take A as the centre and radius (r) less than AB to draw a circle.

Example 2 :Find the diameter of the circle of radius 6 cm.
Solution: We know,
Diameter = 2 × radius
∴ Diameter =2 × 6 cm =12 cm

What Is Sector Of A Circle

Minor sector:
A sector of a circle is called a minor sector if the minor arc of the circle is a part of its boundary
In Fig. sector OAB is the minor sector.
Major sector:
A sector of a circle is called a major sector if the major arc of the circle is a part of its boundary.
In Fig. sector OACB is the major sector.
Following are some important points to remember:
(i) A minor sector has an angle θ, subtended at
the centre of the circle, whereas a major sector has no angle.
(ii) The sum of the arcs of major and minor sectors of a circle is equal to the circumference of the circle.
(iii) The sum of the areas of major and minor sectors of a circle is equal to the area of the circle.
(iv) The boundary of a sector consists of an arc of
the circle and the two radii.