Area of Segment of a Circle: Types, Formulas, Theorems, Examples

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Area of segment of a circle is obtained through subtracting the area of the triangle from the area of the sector. A line segment that joins two points on a circle or curve with both the end points lying on the arc is called a Chord. The region of the circle cut off from the rest of the circle by this chord is called as segment of a circle. This segment is a part of a circle in two dimensional space bounded by a chord whose end points lie on the arc. In other words, the segments can be defined as the parts that are divided by the circle's arc and connected with the chord through its end points.

Keyterms: Circle, Traingle, Line segment, Chord, Major segment, Minor segment, Radians, Degree

Also Read: Surface Areas and Volumes


Types of Segments in a Circle

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The segments of a circle can be classified into two types that are Major segment and Minor segment. When the chord divides the regions of a circle into segments, the segment that has larger area is known as a major segment and the segment that has smaller area is known as Minor segment.

The major segment consists of a major arc and the minor segment consists of a minor arc. Here it can be observed that ADC is the major segment and ABC is the minor segment. Besides, the area of these segments of a circle can also be calculated using formulas.

Types of Segments in a Circle

Types of Segments in a Circle

Also, read: Areas Related to Circles Revision Notes


Area of Segment of a Circle: Formula

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The formula that can be used to calculate the area of segments of a circle is as follows. It can be calculated either in terms of degree or radian.

  • Area of Segment in Radians: A= (½) × r^2 (θ – Sin θ)
  • Area of Segment in Degree: A= (½) × r^ 2 × [(π/180) θ – sin θ]

Derivation

Let us consider a circle which has a triangle AOB circumscribed within. 

Let  ∠AOB = θ° 

And area of triangle AOB is AΔAOB

Then the Area of sector AOBC = θ/360° × πr2 (Formula).

So, the area of Segment of Circle can be calculated as

Area of Segment = Area of Sector – Area of Triangle

Sector of a Circle

Sector of a Circle

A = θ/360° × πr2 – A ΔAOB

To find the area of triangle AOB we need to calculate the sides.

Using Pythagoras theorem  

OP =\({\sqrt{[r^2 -(\frac{AB}{2}) 2]}}\) in case the length of AB is given.

Or OP= rcos(θ/2) in case the angle θ is given.

Area of triangle= ½ × Base × Height

= ½ × AB × OP.

In this way the area of Segment of a circle can be calculated by substituting the given values. 

Area of Segment in Degree: A= (½) × r^ 2 × [(π/180) θ – sin θ]

Sector vs Segment of a Circle

Sector vs Segment of a Circle

Also Read:


Theorems on Segment of Circle

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There are two theorems on segments of a circle. 

Alternate Segment Theorem

The theorem states that in a circle, the angle which lies between the chord and tangent passing through the end points is equal to the angle in the alternate segment. 

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Here the angle lying between the tangent DE and chord BC is ∠BCE and the angle in alternate segment is ∠BAC.

According to the theorem, ∠BCE = ∠BAC.

Another angle lying between tangent DE and chord AC is ∠ACD and the angle in alternate segment is ∠ABC.

According to the theorem, ∠ACD = ∠ABC.

Angle in the same Segment Theorem

The theorem states that the angles formed in a segment are all equal i.e., angles in the same segment of a circle are equal.

According to the theorem, ∠PAQ = ∠PBQ.

From the inscribed angle Theorem, it can be inferred that an angle formed by an arc at the center of a circle is double the angle divided by the same arc at any other point in the circle.

Therefore, ∠POQ = 2∠PAQ

∠POQ = 2∠PBQ.

From these two we can conclude that 

2∠PAQ= 2∠PBQ

Hence, ∠PAQ = ∠PBQ

Read more Important Questions on Circles


Things to Remember

  • The region of the circle cut off from the rest of the circle by this chord is called a Segment of a circle.
  • It is a part of a circle in two dimensional space bounded by a chord that has its end points lying on the arc.
  • There are two types of Segments: Major segment and Minor segment.
  • The segment that has a larger area is called the Major segment and the segment that has smaller area is called the minor segment.
  • The formula that can be used to calculate Area of Segment in Degree: A= (½) × r^ 2 × [(π/180) θ – sin θ]
  • There are two theorems on Segment of Circle that are Alternate segment theorem and Angle in the same segment theorem.
  • Alternate Segment Theorem states that in a circle, the angle which lies between the chord and tangent passing through the end points is equal to the angle in the alternate segment. 
  • Angle in the same segment Theorem states that the angles formed in a segment are all equal i.e., angles in the same segment of a circle are equal.

Also, read: Circles Revision Notes


Sample Questions

Ques. Calculate the area of the major segment of a circle if the area of its minor segment is 54sq. units and the radius is 7units. Use π=22/7. (2 Marks)

Ans. From the formula

area of the major segment = area of the circle − the area of the minor segment

=πr^2−54

=22/7×7×7−62

=92sq.units

Ques. PQ is a chord of a circle that subtends an angle of 90° at the center of a circle, and the diameter of the circle is 14cm. Calculate the area of the minor sector of this circle? (2 Marks)

Ans. The diameter of the circle =14 cm.

The radius of circle =14/2=7 cm

Using the formula of the area of a sector of the circle

The area of minor sector =(θ/360?)×πr^2=(90°/360°)×(22/7)×7^2=11×7/2=38.5sq.cm.

Therefore, the area of the minor sector is 38.5sq⋅cm.

Ques. If the area of a sector is 80 sq.cm and the area of the enclosed triangle is 68sq.cm, calculate the area of the segment? (2 Marks)

Ans. From the formula

Area of the segment = area of the sector – area of the triangle

=(80−68)sq.cm

=12sq.cm

Thus, the area of the segment is 12sq.cm.

Ques. What is the difference between Sector and Segment of a circle? (3 Marks)

Ans. The sector and segment are both parts of a circle. A segment is bound by the circumference of the circle and chord. The sector is bound by the circumference of a circle and two radii. Radius of the circle is important and considered in the sector. But the radius is not much considered for the segment. 

If the chord is double the radius of a circle, the segments of a circle can be equal and referred to as semicircles. If the angle between two radii is 180° the sectors can be referred to as semicircles. 

Ques. Calculate the area of a segment of a circle with a central angle of 165 degrees and a radius of 4. Express answer to nearest integer. (3 Marks)

Ans. From the formula

Area of the segment = area of the sector – area of the triangle

Area of sector = n/360 × πr^2

= 165/360 × π4^2

= 23.02 sq.cm

Area of triangle = ½ × B × H

= ½ × 2\({\sqrt{}}\)3 × 2

= 3.46 sq.cm

Area of Segment = 23.02 – 3.46

= 20 sq.cm

Ques. Explain the formula of area of Segment of Circle with derivation. (5 Marks)

Ans. The formula used to calculate the area of Segment of a circle is 

Area of Segment in Radians: A= (½) × r^2 (θ – Sin θ)

Derivation:

Let us consider a circle which has a triangle AOB circumscribed within. 

Let  ∠AOB = θ° 

And area of triangle AOB is AΔAOB

Then the Area of sector AOBC = θ/360° × πr2 (Formula).

So, the area of Segment of Circle can be calculated as

Area of Segment = Area of Sector – Area of Triangle

A = θ/360° × πr2 – AΔAOB

To find the area of triangle AOB we need to calculate the sides.

Using Pythagoras theorem

OP = \({\sqrt{[r^2 -(\frac{AB}{2}) 2]}}\) incase the length of AB is given.

Or OP= rcos(θ/2) incase the angle θ is given.

Area of triangle= ½ × Base × Height

= ½ × AB × OP.

In this way the area of Segment of a circle can be calculated by substituting the given values. 

Area of Segment in Degree: A= (½) × r^ 2 × [(π/180) θ – sin θ]


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CBSE X Related Questions

1.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

      2.

      A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

          3.

          Solve the following pair of linear equations by the substitution method. 
          (i) x + y = 14 
              x – y = 4   

          (ii) s – t = 3 
              \(\frac{s}{3} + \frac{t}{2}\) =6 

          (iii) 3x – y = 3 
                9x – 3y = 9

          (iv) 0.2x + 0.3y = 1.3 
               0.4x + 0.5y = 2.3 

          (v)\(\sqrt2x\) + \(\sqrt3y\)=0
              \(\sqrt3x\) - \(\sqrt8y\) = 0

          (vi) \(\frac{3x}{2} - \frac{5y}{3}\) =-2,
              \(\frac{ x}{3} + \frac{y}{2}\) = \(\frac{ 13}{6}\)

              4.
              Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
              (i) 2, 4, 8, 16, . . . .
              (ii) \(2, \frac{5}{2},3,\frac{7}{2}\), . . . .
              (iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . .
              (iv) – 10, – 6, – 2, 2, . . .
              (v) 3, \(3 + \sqrt{2} , 3 + 3\sqrt{2} , 3 + 3 \sqrt{2}\) . . . .
              (vi) 0.2, 0.22, 0.222, 0.2222, . . . .
              (vii) 0, – 4, – 8, –12, . . . .
              (viii) \(\frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}\), . . . .
              (ix) 1, 3, 9, 27, . . . .
              (x) a, 2a, 3a, 4a, . . . .
              (xi) a, \(a^2, a^3, a^4,\)  . . . .
              (xii) \(\sqrt{2}, \sqrt{8} , \sqrt{18} , \sqrt {32}\) . . . .
              (xiii) \(\sqrt {3}, \sqrt {6}, \sqrt {9} , \sqrt {12}\) . . . . .
              (xiv) \(1^2 , 3^2 , 5^2 , 7^2\), . . . .
              (xv) \(1^2 , 5^2, 7^2, 7^3\), . . . .

                  5.

                  Form the pair of linear equations for the following problems and find their solution by substitution method.

                  (i) The difference between two numbers is 26 and one number is three times the other. Find them.

                  (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

                  (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

                  (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km.

                  (v) A fraction becomes\(\frac{ 9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.

                  (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

                      6.

                      Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

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