Area of a segment of a circle – Solved Examples

Area of a segment of a circle

Segment of a circle

A chord divides a circle into two regions known as segments. The measure of such a region is known as the area of a segment of the circle. It is bounded by a chord and an arc. 

Based on the area these regions are classified into two types

1. Major segment

The segment formed by a chord and an arc larger than the semi-circle is known as the major segment.

This region contains the centre of the circle.

2. Minor segment

The segment formed by a chord and an arc smaller than the semi-circle is known as minor segment.

This region doesn’t contain the centre of the circle.

In the figure given above

1. O – Centre of the circle
2. PQ – Chord 
3. PSQ – Major segment
4. PRQ – Minor segment

△OPQ is a triangle formed by the chord PQ and the lines joining the centre of the circle to the endpoints i.e., OP and OQ.

OP = OQ = radius of circle = r

Chord PR subtends an angle ϴ (in degree) at the centre of the circle.

For finding the area of the major segment and minor segment we need to find the area of the sectors and the area of △OPQ.

Area of △OPQ 

Given:

In △OPQ 

1. OP = OQ = radius of circle = r
2. ∠POQ = ϴ (in degree)

Construction:

1. OT is a perpendicular line drawn from O on PQ 
2. ∠OTP = ∠OTQ = 90°

In △OTP and △OTQ

OP = OQ (given)

∠OTP = ∠OTQ = 90° (by construction)

OT = OT (common side)

Hence △OTP≅△OTQ (RHS)

⇒ ∠POT = ∠QOT = ϴ/2 (Congruent parts of congruent triangles)

⇒ PT = QT (Congruent parts of congruent triangles)

In △POT

O T=O P \cos \frac{\theta}{2}=r \cos \frac{\theta}{2}

P T=O P \sin \frac{\theta}{2}=r \sin \frac{\theta}{2}

In △OPQ 

\text { Height }=O T=r \cos \frac{\theta}{2}

Base =\mathrm{PQ}=\mathrm{PT}+\mathrm{QT}=2 \mathrm{r} \sin \frac{\theta}{2}

\text { Area of } \triangle \mathrm{OPQ}

=\frac{1}{2} \times \text { base } \times \text { height }

=\frac{1}{2} \times P Q \times O T

=\frac{1}{2} \times 2 r \sin \frac{\theta}{2} \times r \cos \frac{\theta}{2}

=\frac{r^{2}}{2} \times 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}

=\frac{r^{2}}{2} \times \sin \theta

Hence the area of △OPQ is equal to \frac{r^2}{2} \times \sin \theta, where r is the radius of the circle and ϴ is the angle made by the chord at the centre of the circle.

Area of minor segment

= Area of sector POQR – Area of △OPQ

= \left(\frac{\theta}{360^{\circ}} \times \pi r^{2}\right)-\left(\frac{r^{2}}{2} \times \sin \theta\right)

= r^{2}\left(\frac{\theta}{360^{\circ}} \pi-\frac{\sin \theta}{2}\right)

Area of major segment

= Area of sector POQS + Area of △OPQ

= \left(\frac{360^{\circ}-\theta}{360^{\circ}} \times \pi r^{2}\right)+\left(\frac{r^{2}}{2} \times \sin \theta\right)

= r^{2}\left(\frac{360^{\circ}-\theta}{360^{\circ}} \pi+\frac{\sin \theta}{2}\right)

Area of a major segment (Alternative Formula)

= Area of the circle – Area of the minor segment 

= \pi r^{2}-r^{2}\left(\frac{\theta}{360^{\circ}} \pi-\frac{\sin \theta}{2}\right)

= r^{2}\left(\pi-\frac{\theta}{360^{\circ}} \pi+\frac{\sin \theta}{2}\right)

= r^{2}\left(\frac{360^{\circ}-\theta}{360^{\circ}} \pi+\frac{\sin \theta}{2}\right)

Solved Examples

1. A chord PQ subtends an angle of 120° at the centre of the circle having a radius of 8 cm. Find the area of the major segment formed by the chord PQ. Also, calculate the area of the minor segment.

Solution:

Radius (r) = 8 cm

Angle subtended by the chord at the centre = ϴ = 120°

\text { Area of the major segment }=r^{2}\left(\frac{360^{\circ}-\theta}{360^{\circ}} \pi+\frac{\sin \theta}{2}\right) \qquad

                                                        =8^{2}\left(\frac{240}{360^{\circ}} \pi+\frac{\sin 120^{\circ}}{2}\right)

                                                       =64\left(\frac{2}{3} \pi+\frac{\sqrt{3}}{4}\right)

                                                       =161.75 \mathrm{~cm}^{2}

Area of the minor segment 

= Area of the circle – Area of the major segment 

= \pi r^{2}-161.75

= 64 \pi-161.75

= 39.31 \mathrm{~cm}^{2}

2. A chord MN subtends an angle of π/3 radians at the centre of the circle having a radius of 6 m. Find the area of the minor segment formed by the chord MN.

Solution:

Radius ( r) = 6 m

Angle subtended by the chord at the centre = ϴ = π/3 radians = 180°/3 = 60°

\text { Area of the minor segment }

=r^{2}\left(\frac{\theta}{360^{\circ}} \pi-\frac{\sin \theta}{2}\right)

=6^{2}\left(\frac{60^{\circ}}{360^{0}} \pi-\frac{\sin 60^{\circ}}{2}\right)

=36\left(\frac{1}{6} \pi-\frac{\sqrt{3}}{4}\right)

=3.26 \mathrm{~m}^{2}