The Segment of a Circle with Example and FAQs

What is the Segment of a Circle?

A segment of a circle is a region that is enclosed by a chord and an arc. 

Where:

1. A chord is a line segment joining any two points on a circle’s circumference.
2. An arc is a fraction or a part of the circle’s circumference.

A Chord divides the circle into two regions, where one is larger than the other forming two types of segments, that are:

• Major segment: The segment which encloses a larger area. It includes the centre of the circle.
• Minor segment: The segment which encloses a smaller area.

Segment of a Circle

Area of a Segment of a Circle

Area of a Segment

In the circle above A is the centre of the circle, CD is a chord that forms a minor segment which is shown as the shaded region. Let ‘r’ be the radius of the circle and the angle subtended at the centre by the arc CD be θ.

Now to calculate the area of the segment, first, we need to find the area of the ∆ACD and area of the sector ACD. The area of the segment is the difference between the area of the sector ACD and the area of the ∆ACD. 

From trigonometry, we know,

The area of ∆ACD =\frac{1}{2} r^{2} \sin \theta

Now, we also know the formula for the area of a sector is,

Area of sector =\left(\frac{\theta}{360^{\circ}}\right) \times \pi r^{2}, if θ is in degrees.

And, Area of sector =\frac{1}{2} \times r^{2} \theta if θ is in radians.

Thus, Area of the segment ACD (when θ is in degrees)

 = Area of sector ACD – Area of ∆ACD

= \left(\left(\frac{\theta}{360^{\circ}}\right) \times \pi r^{2}\right)-\left(\frac{1}{2} r^{2} \sin \theta\right)

Area of the segment ACD (when θ is in radians)

= Area of sector ACD – Area of ∆ACD

= \left(\frac{1}{2} \times r^{2} \theta\right)-\left(\frac{1}{2} r^{2} \sin \theta\right)

This formula is used to calculate the area of a minor segment when the value of the angle subtended at the centre and the radius of the circle is provided.

Unless specified segment is usually referred to as a minor segment only. But if asked to determine the area of the major segment, then the difference between the area of the circle and the area of the minor segment is calculated.

Perimeter of a Segment of a Circle

The perimeter of the segment = length of the chord + length of the arc.

Now, we know the formula for the length of an arc is, which is:

Length of Arc = rθ, if θ is in radians,

                         = \pi r \frac{\theta}{180^{\circ}} if θ is in degrees.

Length of the Chord =2 r \sin \left(\frac{8}{2}\right)

Therefore, the perimeter of a segment can be easily calculated.

Theorems

There are two important theorems based on the segment of a circle, which are:

1. Angles in the same segment theorem

Statement: The angles subtended in the segment of a circle are always equal.

There are two angles subtended in the segment AB. According to the theorem:

∠AOB = ∠AQB

2. Alternate segment theorem

Statement: The angle formed by the tangent and the chord at the point of contact is equal to the angle formed in the alternate segment on the circumference of the circle through the endpoints of the chord.

According to the theorem ∠PRS = ∠PQR and also ∠RPQ = ∠QRT.

Example

1. Find the area and perimeter of the segment of a circle that extends an angle of measurement and the radius of the circle is 5 units. Take π = 3.14.

Solution:

The radius of the given circle is 5 units and θ = 60°.

Area of the segment OCD (when θ is in degrees)

= Area of sector OCD – Area of ∆OCD

= \left(\left(\frac{\theta}{360^{\circ}}\right) \times \pi r^{2}\right)-\left(\frac{1}{2} r^{2} \sin \theta\right)

= \left(\left(\frac{60^{\circ}}{360^{\circ}}\right) \times 3.14 \times 5^{2}\right)-\left(\frac{1}{2} \times 5^{2} \times \sin 60^{\circ}\right) \text { sq units }

= \left(\frac{1}{6} \times 3.14 \times 25\right)-\left(\frac{1}{2} \times 25 \times \frac{\sqrt{3}}{2}\right) \text { sq units },\sin 60^{\circ}=\frac{\sqrt{3}}{2}

= 13.08 – 10.83 sq units = 2.25 sq units

 ∴ Area of the segment = 2.25 sq units.

The perimeter of segment OCD

= length of chord CD + length of arc CD

=2 r \sin \left(\frac{\theta}{2}\right)+\pi r \frac{\theta}{180^{\circ}}

=\left(2 \times 5 \times \sin \left(\frac{60^{\circ}}{2}\right)\right)+\left(3.14 \times 5 \times \frac{60^{\circ}}{180^{\circ}}\right) \text { units }

=\left(2 \times 5 \times \sin 30^{\circ}\right)+\left(3.14 \times 5 \times \frac{1}{3}\right) \text { units }

=\left(2 \times 5 \times \frac{1}{2}\right)+(5.23) \text { units }

=5+5.23 \text { units }=10.23 \text { units }

∴ The perimeter of the segment of the circle is 10.23 units.