Circle is one of the easy-grasping topics of geometry , you may observe circular shapes around you. Having an understanding of circular geometry can be of greater importance as we sometimes deal with the problems related to circular objects in our daily lives.
In this article we have brought you combined information that includes fundamentals of Circle, perimeter, area , Area of a sector of a circle and length of a arc of a circle. so let's first begin with the basics of Circle next the topics we have listed above.
What is Circle
Circle is the locus of points (a group of points) that are at equal distance from a fixed point and that fixed point is called the centre of the circle.
Diameter : Diameter is a straight line on the circle that connects any two points on the circle and passes through the centre of Circle.It is generally denoted by (D) or (d).
Here the length \( AB \) in the Diagram called the diameter of the circle
Radius : Radius is the length that is exactly half of the diameter of a circle Or It is defined as the fixed distance From the centre to any point on the locus of Circle and it is denoted by (R) or (r).
Radius = \( D \over 2 \)
Also Read :
Perimeter and Area of the Circle
Perimeter/ Circumference : The total length that is required to form the boundary of a circle is called Perimeter
The perimeter of a circle is calculated by the following formula:
\( Perimeter = 2\pi R \)
Semi - Perimeter : Half of the perimeter of Circle.
The semi- perimeter of a circle is
= \( \pi R \)
Area of the circle
Area : The area of a circle is the space that is occupied by it .
Area of the circle= \( \pi R^2 \)
Area of the Semi- Circle is given by:
= \( {\pi R^2}\over 2\)
Sector and Segment of a Circle
Sector : A Sector is the area bounded between two radii and an arc of the circle.
Minor Sector : Area bounded between two radii and a minor arc is called minor Sector.
Major Sector : Area bound between two radii and a major arc is called major Sector.
Area of the sector = \( {\theta\over 360°}\times \pi R^2\)
Where \(\theta\) is the angle subtended by the arc at the centre.
Length of an Arc = \( {\theta\over 360°}\times 2\pi R\)
Perimeter of a Sector : Arc length+ 2R
Segment : A Segment is the area bounded between a chord and an arc of a circle.
Minor Segment: Area bounded between a chord and a minor arc of the circle.
Major Segment: Area bounded between a chord and a major arc of the circle.
Area of the segment = Area of the sector - area of the triangle
Case 1 : If the chord subtens an angle of 60° then the area of the segment is given by
Area of the Segment = Area of the sector - area of equilateral triangle
Case 2 : If the chord subtens an angle which is not 60° then the area of the segment is given by
Area of the Segment = Area of the sector - \( {1\over 2} r^2 sin\theta\)
Solved Examples for Sector and Segment of a Circle
Example 1 : An arc subtends an angle of 30° at the centre of a circle of radius 7 cm. Find the area of the minor and major sector.
Ans : Radius of the circle = 7 cm
Angle subtended by arc \(\theta\) = 30°
Area of the the minor Sector = \( {\theta\over 360°}\pi R^2\)
= \( {30°\over 360°}\times {22 \over 7 }\times 7^2\)
= 12.83 \( cm^2 \)
Area of the Major sector = Area of the circle- Area of the minor Sector
Area of the circle = \(\pi R^2\)
= \({22\over 7}\times 7^2\)
= \( 22 \times 7\)
=154 \(cm^2\)
Area of the Major sector = \(154-12.83 \)
= 141.16 \(cm^2\)
Example 2 : An arc subtends an angle of 60° at the centre of a circle of radius 42 cm. Find the length of minor and major arc
Ans : length of the arc = \( {\theta\over 360°}2\pi R\)
Angle subtended by arc = 60°
Radius of circle = 42 cm
= \( {60\over 360°}\times 2 \times{22\over 7}\times 42\)
= 44 \(cm\)
Angle subtended by arc = 60°
Radius of circle = 42 cm of the minor arc = 44 cm.
Length of Major Arc = \(2\pi R \) - Length of minor arc
= \( 2 \times {22\over 7}\times 42\)- 44
= 264 - 44
=220 cm
Example 3: In a circle of radius 14 cm an arc subtens an angle of 120° at the centre then find the perimeter of the minor and major sector
Ans :
Angle subtended by arc = 120°
Radius of circle = 14 cm
length of the arc = \( {\theta\over 360°}\times 2\pi R\)
= \( {120°\over 360°}\times 2\times{22\over 7 }\times 14\)
= \(88\over 7\)
Perimeter of the minor Sector =\( 88\over 7\)+ \(2R\)
= \(88\over 7\)+\(2\times 14\)
= \(284 \over 7 \)
Length of the major arc = \(2\pi R \) - Length of the minor arc
= \( 2 \times {22\over 7}\times 14 \) -\( 284 \over 7\)
= \( 88 - {284\over 7 } \)
= \( 332 \over 7\)
Perimeter of the major arc = length of the major arc + 2R
= \( 332 \over 7\) + \( 2 \times 14\)
= \( 528 \over 7\)
Example 4: In a circle of radius 21 cm, a chord subtends an angle of 60° at the centre.Find the area of major segment and minor segment
Ans: Since the chord make an angle of 60 at the centre hence the triangle form by the two radii and chord must be an equilateral triangle.
angle subtended by the chord at centre = 60
Radius of the circle = 21 cm (Also one of the sides of the triangle)
\( \therefore \) Area of the Segment = Area of the sector - area of equilateral triangle
area of the sector = \( {\theta\over 360°}\times \pi R^2\)
Length of the major arc = \(2\pi R \) - Length of the minor arc
= \( 2 \times {22\over 7}\times 14 \) -\( 284 \over 7\)
= \( 88 - {284\over 7 } \)
= \( 332 \over 7\)
Perimeter of the major arc = length of the major arc + 2R
= \( 332 \over 7\) + \( 2 \times 14\)
= \( 528 \over 7\)
Example 4: In a circle of radius 21 cm, a chord subtends an angle of 60° at the centre.Find the area of major segment and minor segment
Ans: Since the chord make an angle of 60 at the centre hence the triangle form by the two radii and chord must be an equilateral triangle.
angle subtended by the chord at centre = 60
Radius of the circle = 21 cm (Also one of the sides of the triangle)
\( \therefore \) Area of the Segment = Area of the sector - area of equilateral triangle
area of the sector = \( {\theta\over 360°}\times \pi R^2\)
= \( {60°\over 360°}\times {22 \over 7 }\times 21^2\)
= 231 \( cm^2\)
area of the equilateral triangle = \( {\sqrt(3)\over 4}\times s^2 \)
= \( {\sqrt(3)\over 4}\times 21^2 \)
\( 441\sqrt(3)\over 4\) \( cm^2\)
= 231 \( cm^2\)
area of the equilateral triangle = \( {\sqrt(3)\over 4}\times s^2 \)
= \( {\sqrt(3)\over 4}\times 21^2 \)
\( 441\sqrt(3)\over 4\) \( cm^2\)
Area of the minor Segment = 231 - \( 441\sqrt(3)\over 4\) \( cm^2 \)
Area of the major Segment = area of the circle - area of the minor segment
=\( \pi \times R^2 \) - Area of the minor segment
Area of the circle = \( {22 \over 7 }\times 21^2\)
=1386 \( cm^2\)
Area of the major Segment = 1386 -231 - \( 441\sqrt(3)\over 4\)
Area of the major Segment = area of the circle - area of the minor segment
=\( \pi \times R^2 \) - Area of the minor segment
Area of the circle = \( {22 \over 7 }\times 21^2\)
=1386 \( cm^2\)
Area of the major Segment = 1386 -231 - \( 441\sqrt(3)\over 4\)
= 1155- \( 441\sqrt(3)\over 4\) \( cm^2 \)
Example 5: In a circle of radius 28 cm, a chord subtends an angle of 90° at the centre.Find the area of major segment and minor segment
Ans: Since the chord make an angle of 90 at the centre hence the triangle form by the two radii and chord wouldn't be an equilateral triangle. in this case the area of the trianle will be calculated by
\( {1\over 2} r^2 sin\theta\)
Example 5: In a circle of radius 28 cm, a chord subtends an angle of 90° at the centre.Find the area of major segment and minor segment
Ans: Since the chord make an angle of 90 at the centre hence the triangle form by the two radii and chord wouldn't be an equilateral triangle. in this case the area of the trianle will be calculated by
\( {1\over 2} r^2 sin\theta\)
\( \therefore \) Area of the Segment = Area of the sector - area of triangle
area of the sector = \({\theta\over 360°}\times \pi R^2\)
area of the sector = \({\theta\over 360°}\times \pi R^2\)
= \( {90°\over 360°}\times {22 \over 7 }\times 28^2\)
= 616 \( cm^2\)
Area of triangle = \( {1\over 2} r^2 sin\theta\)
=\( {1\over 2}\times 28^2 \times sin90\)
=392 \( cm^2\)
Area of the minor Segment = 616- 392 = 224 \( cm^2\)
Area of the major Segment = area of the circle - area of the minor segment
Area of the circle = \( {22 \over 7 }\times 28^2\)
= 2464 \( cm^2\)
Area of the major Segment = 2464-224 \( cm^2\)
= 616 \( cm^2\)
Area of triangle = \( {1\over 2} r^2 sin\theta\)
=\( {1\over 2}\times 28^2 \times sin90\)
=392 \( cm^2\)
Area of the minor Segment = 616- 392 = 224 \( cm^2\)
Area of the major Segment = area of the circle - area of the minor segment
Area of the circle = \( {22 \over 7 }\times 28^2\)
= 2464 \( cm^2\)
Area of the major Segment = 2464-224 \( cm^2\)
= 2240 \(cm^2\)
Frequently Asked questions
Q2: What is the relation between radius and tangent?