One to one maths interventions built for KS4 success

Weekly online one to one GCSE maths revision lessons now available

In order to access this I need to be confident with:

Parts of a circle Area of a sectorArea of a triangle

Arc length Pythagoras’ theorem Cosine rule Area of a triangle (1/2 abSinC) RoundingThis topic is relevant for:

Here we will learn about the segment of a circle including how to identify the segment of a circle and how to find the area of a segment given the different parts of a circle.

There are also parts of a circle worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

A **segment **of a circle is the area enclosed by an arc of a circle and a chord.

There are two main types of segment:

- The
**major segment**is the segment where the arc length is greater than half the circumference of the circle - The
**minor segment**is

An **arc **is a fraction of the circumference of a circle.

A **chord **is a line segment that connects two points of a circle.

A segment where the chord passes through the centre of the circle is called a **semicircle**.

In order to solve problems involving a segment of a circle:

__Questions involving area__

**Find the length of the radius.****Find the size of the angle creating the sector.****Find the area of the sector.****Find the area of the triangle created by the radii and the chord of a circle.****Subtract the area of the triangle from the area of the sector.****Clearly state your answer.**

__Questions involving perimeter__

**Find the length of the radius.****Find the size of the angle creating the sector.****Find the length of the arc of the segment of a circle.****Find the length of the chord of the segment.****Add the length of the arc and the chord.****Clearly state your answer.**

Get your free segment of a circle worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free segment of a circle worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREECalculate the area of the segment shown below. Give your answer to 3 significant figures

**Find the length of the radius.**

Length of the radius of the circle: 7cm

2**Find the size of the angle creating the sector.**

The size of the angle creating the sector (made by the two radii) is 90^{\circ} .

3**Find the area of the sector.**

\[\text{Area of sector }=\frac{\theta}{360} \times \pi r^{2}\\
=\frac{90}{360} \times \pi \times 7^{2} \\
=\frac{49}{4} \pi\]

\frac{49}{4} \pi \mathrm{cm}^{2}
It is important to** not round** the answer at this stage of the question

4**Find the area of the triangle created by the radii and the chord.**

\[\text{Area of triangle} = \frac{1}{2} a b \sin C \\
=\frac{1}{2} \times 7 \times 7 \times \sin(90)\\
=24.5\]

Area of triangle: 24.5cm^{2}

5**Subtract the area of the triangle from the area of the sector.**

\[\text{Area of sector – Area of triangle} = \frac{49}{4} \pi-24.5\\
=13.9845…\]

6**Clearly state your answer.**

The question asked you to round your answer to 3 significant figures

Area of the segment of the circle= 13.9845...cm^{2}=14.0cm^{2} ( 3 .s.f.)

Calculate the perimeter of the segment shown below. Give your answer to 3 decimal places

**Find the length of the radius.**

Length of radius: 7cm

**Find the size of the angle creating the sector.**

The size of the angle creating the sector (made by the two radii) is 90^{\circ}

**Find the length of the arc of the segment.**

\[\text{Arc length}=\frac{\theta}{360} \times 2 \pi r \\
=\frac{90}{360} \times 2 \pi \times 7 \\
=\frac{7}{2} \pi\]

Length of the arc: \frac{7}{2} \pi

**Find the length of the chord of the segment.**

In this question you can see the two radii and the chord form a right angled triangle. This means you can use Pythagoras’ theorem to find the length of the chord, which is the hypotenuse of the triangle.

\[a^{2}+b^{2}=c^{2} \\
7^{2}+7^{2}=c^{2} \\
49+49=c^{2} \\
98=c^{2} \\
7 \sqrt{2}=c \quad \quad \quad \text{square root both sides of the equation}\]

Therefore the length of the chord is 7 \sqrt{2} \mathrm{cm}

**Add the length of the arc and the chord.**

\[\text{Length of the arc }+\text{ Length of the chord } = \frac{7}{2} \pi+7 \sqrt{2}\\
=20.895069…\]

**Clearly state your answer.**

The question asked you to round your answer to 3 decimal places

Perimeter of segment: 20.895069…cm= 20.895cm (3.d.p)

Calculate the area of the segment shown below. Give your answer to 3 decimal places

**Find the length of the radius.**

Length of radius: 5cm

**Find the size of the angle creating the sector.**

The size of the angle creating the sector (made by the two radii) is 100^{\circ} . This is given in the question.

**Find the area of the sector.**

\[\text{Area of sector}=\frac{\theta}{360} \times \pi r^{2}\\
=\frac{100}{360} \times \pi \times 5^{2} \\
=\frac{125}{18} \pi\]

Area of sector: \frac{125}{18} \pi \mathrm{cm}^{2}

**Find the area of the triangle created by the radii and the chord.**

\[\text{Area of triangle }=\frac{1}{2} a b \sin C \\
=\frac{1}{2} \times 5\times 5 \times \sin(100)\\
=12.3100…\]

Area of triangle: 12.3100…cm^{2}

**Subtract the area of the triangle from the area of the sector.**

\[\text{Area of sector} – \text{Area of triangle } =\frac{125}{18} \pi\\
=9.50661565…\]

**Clearly state your answer.**

The question asked you to round your answer to 3 decimal places

Area of segment: 9.50661565...cm^{2}= 9.507cm^{2} (3.d.p)

Calculate the perimeter of the segment shown below. Give your answer to 3 decimal places.

**Find the length of the radius.**

Length of radius: 9cm

**Find the size of the angle creating the sector. **

Size of angle: 80^{\circ}

**Find the length of the arc of the segment of a circle.**

\[\text{Arc length}=\frac{\theta}{360} \times 2 \pi r \\
=\frac{80}{360} \times 2 \pi \times 9 \\
=4 \pi\]

Length of the arc: 4 \pi

**Find the length of the chord of the segment.**

In this question, the triangle is not a right angled triangle so we cannot use Pythagoras’ Theorem to find the missing length. Instead we can use the cosine rule.

Length of chord:

\[a^{2}=b^{2} + c^{2} -2bc \cos(A)\\
a^{2}=9^{2} + 9^{2} – 2 \times 9 \times 9 \times \cos(80)\\
a^{2}=162-162 \cos(80)\\
a^{2}=133.8689952\\
a=11.57017697\]

**Add the length of the arc and the chord.**

\[\text{Length of the arc }+\text{ Length of the chord } = 4 \pi+11.57017697\\
=24.13654759\]

**Clearly state your answer.**

The question asked you to round the answer to 3 decimal places.

Perimeter = 24.137cm

Calculate the area of the segment shown below. Give your answer to 1 decimal place

**Find the length of the radius.**

Length of radius: 20cm

**Find the size of the angle creating the sector. **

The angle of the sector, created by the two radii, is not given to you in this question. Here you can use the triangle created by the two radii and the chord to find the angle

You need to apply the cosine rule to find the size of the angle.

A is the angle you are trying to find. You can therefore use the rearranged cosine rule to find the angle.

\[\cos{A}=\frac{b^{2}+c^{2}-a^{2}}{2 b c} \\
\cos{A}=\frac{20^{2}+20^{2}-28^{2}}{2 \times 20 \times 20} \\
\cos{A}=\frac{1}{50} \\
A=\cos^{-1}\left(\frac{1}{50}\right) \\
A=88.8540008^{\circ}\]

The size of the angle creating the sector (made by the two radii) is 88.854^{\circ} .

**Find the area of the sector.**

31.0159..+28=59.0159cm\[\text{Area of sector }=\frac{\theta}{360}\times \pi r^{2}\\
=\frac{88.854}{360} \times \pi \times 20^{2} \\
=310.15897\]

Area of sector: 310.15897cm^{2}

**Find the area of the triangle created by the radii and the chord.**

\[\text{Area of triangle}=\frac{1}{2} a b \sin C \\
=\frac{1}{2} \times 20\times 20 \times\sin(88.854)\\
=199.95999…\]

Area of triangle: 199.95999…cm^{2}

**Subtract the area of the triangle from the area of the sector.**

\[\text{Area of sector} – \text{Area of triangle } =310.15897-199.959..\\
=110.19897…\]

**Clearly state your answer.**

The question asked you to round your answer to 1 decimal place

Area of segment: 110.19897cm^{2}= 110.2cm^{2} (1.d.p)

Calculate the perimeter of the segment shown below. Give your answer to 1 decimal place

**Find the length of the radius.**

Length of radius: 20cm (given to you in the question)

**Find the size of the angle creating the sector. **

The angle of the sector, created by the two radii, is not given to you in this question. Here you can use the triangle created by the two radii and the chord to find the angle

You need to apply the cosine rule to find the size of the angle.

A is the angle you are trying to find. You can therefore use the rearranged cosine rule to find the angle.

\[\cos{A}=\frac{b^{2}+c^{2}-a^{2}}{2 b c} \\
\cos{A}=\frac{20^{2}+20^{2}-28^{2}}{2 \times 20 \times 20} \\
\cos{A}=\frac{1}{50} \\
\cos^{-1}\left(\frac{1}{50}\right) \\
A=88.8540008^{\circ}\]

The size of the angle creating the sector (made by the two radii) is 88.854^{\circ} .

**Find the length of the arc of the segment.**

\[\text{Arc length }=\frac{\theta}{360} \times 2 \pi r \\
=\frac{88.854}{360} \times 2 \pi \times 20 \\
=31.0159…\]

Area of sector: 31.0159cm^{2}

**Find the length of the chord of the segment.**

Length of chord: 28cm

**Add the length of the arc and the chord.**

Length of the arc + Length of chord = 31.0159.. + 28 = 59.0159cm

**Clearly state your answer.**

The question asked you to round your answer to 1 decimal place

Perimeter of segment: 59.0159...cm = 59.0 cm (1.d.p)

**Confusing the segment/sector**

A segment is the area enclosed by the arc of a circle and a chord. A sector is the area enclosed by the arc of a circle and two radii.

**Confusing the chord/diameter**

A chord can cross a circle at any point, but a diameter but go through the centre of the circle.

**Incorrect use of the cosine rule**

Many mistakes are made when applying other rules within a segment question e.g the cosine rule. Take your time with these parts and regularly check that your answer makes sense within the context of the question.

**Not including the chord length when finding the perimeter of a segment**

Remember the perimeter of a shape is the sum of the lengths of each of the sides. Therefore the perimeter of a segment is made up the arc and the chord

**Degrees and radians**

Angles can be measured in both degrees and radians, however at GCSE we only use degrees.

Segment of a circle is part of our series of lessons to support revision on circles, sectors and arcs. You may find it helpful to start with the main circles, sectors and arcs lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

__Circles, sectors and arcs__(coming soon)- Circumference of a circle
- Area of a sector
- Equation of a circle
- Arc length
- Parts of a circle
- Perimeter of a sector

1. What is the area of the segment in the diagram? Give your answer to 1 decimal place

36\pi \mathrm{cm}^{2}

72 cm^{2}

41.1 cm^{2}

41.097 cm^{2}

\text{Area of sector }=\frac{\theta}{360} \pi r^{2}\\
=\frac{90}{360}\times \pi \times 12^{2}\\
=36 \pi

\text{Area of triangle }=\frac{1}{2}ab \sin C\\ =\frac{1}{2}\times 12 \times 12 \sin(90)\\ =72

\text{Area of segment }=36 \pi – 72\\ =41.09733553\\ =41.1\mathrm{cm}^{2}

2. What is the perimeter of the segment in the diagram? Give your answer to 1 decimal place

12\pi

40.97 cm

20.7cm

35.8 cm

\text{Arc length}=\frac{\theta}{360} \times 2 \pi r \\
=\frac{90}{360} \times 2 \pi \times 12 \\
=6 \pi

Length of cord:

a^{2}+b^{2}=c^{2} \\ 12^{2}+12^{2}=c^{2} \\ 144+144=c^{2} \\ 288=c^{2} \\ 12 \sqrt{2}=c\text{Length of the arc }+\text{ Length of the chord } = 6\pi + 12 \sqrt{2}\\ =35.82011867\\ =35.8\mathrm{cm}

3. Calculate the area of the segment shown. Give your answer to 1dp .

22.3 m^{2}

19.6m^{2}

25.1m^{2}

15.9m^{2}

\text{Area of sector }=\frac{\theta}{360} \pi r^{2}\\
=\frac{160}{360}\times \pi \times 4^{2}\\
=\frac{64}{9} \pi

\text{Area of triangle }=\frac{1}{2}ab \sin C\\ =\frac{1}{2}\times 4 \times 4 \sin(160)\\ =2.73616…

\text{Area of segment }=\frac{64}{9} \pi – 2.73616\\ =19.60405\\ =19.6\mathrm{m}^{2}

4. Calculate the perimeter of the segment shown. Give your answer to 3 significant figures.

1.92 m

3.767 m

1.85 m

3.77 m

\text{Arc length}=\frac{\theta}{360} \times 2 \pi r \\
=\frac{55}{360} \times 2 \pi \times 2 \\
=\frac{11}{18} \pi

Length of cord:

a^{2}=b^{2} + c^{2} -2bc \cos(A)\\ a^{2}=2^{2} + 2^{2} – 2 \times 2 \times 2 \times \cos(55)\\ a^{2}=8-8 \cos(55)\\ a^{2}=3.411388509\\ a=1.846994453\text{Length of the arc }+\text{ Length of the chord } =\frac{11}{18} \pi + 1.846994453\\ =3.76685663

5. Calculate the area of the segment shown. Give your answer to 2 decimal places.

14.21cm^{2}

25.40cm^2

11.19cm^{2}

18.66cm^{2}

First we need to find the angle using the cosine rule.

\cos{A}=\frac{b^{2}+c^{2}-a^{2}}{2 b c} \\ \cos{A}=\frac{5^{2}+5^{2}-8.5^{2}}{2 \times 5 \times 5} \\ \cos{A}=-\frac{89}{200} \\ A=\cos^{-1}\left(-\frac{89}{200}\right) \\ A=116.4233388^{\circ}\text{Area of sector }=\frac{\theta}{360} \pi r^{2}\\ =\frac{116.423}{360}\times \pi \times 5^{2}\\ =25.39955844

\text{Area of triangle }=\frac{1}{2}ab \sin C\\ =\frac{1}{2}\times 5 \times 5 \sin(116.423)\\ =11.194165

\text{Area of segment }=25.39955844 – 11.194165\\ =14.20539344\\ =14.21\mathrm{cm}^{2}

6. Calculate the perimeter of the segment shown. Give your answer to 3 significant figures.

22.0mm

16.5mm

8.52mm

22.5mm

First we need to find the angle using the cosine rule.

\cos{A}=\frac{b^{2}+c^{2}-a^{2}}{2 b c} \\ \cos{A}=\frac{7^{2}+7^{2}-8^{2}}{2 \times 7 \times 7} \\ \cos{A}=\frac{17}{49} \\ A=\cos^{-1}\left(\frac{17}{49}\right) \\ A=69.69980^{\circ}\text{Arc length}=\frac{\theta}{360} \times 2 \pi r \\ =\frac{69.6998}{360} \times 2 \pi \times 7 \\ =8.515436986

Chord length: 8mm

\text{Perimeter }=8.515436986+8\\ =16.515436986\\ =16.5\mathrm{mm}

1. The diagram shows a sector OABC of a circle with centre O .

(a) Calculate the length of the arc ABC of the sector. Give your answer correct to 3 significant figures.

(b) Calculate the area of the shaded segment ABC . Give your answer correct to 3 significant figures.

**(7 marks)**

Show answer

(a)

\frac{120}{360} \times \pi \times 20.8**(1)**

**(1)**

**(1)**

(b)

Either 113.26… or 46.83 seen

**(1)**

**(1)**

**(1)**

**(1)**

2. The diagram shows a sector OPRS of a circle with centre O .

Calculate the area of the shaded segment PRS .

Give your answer correct to 3 significant figures.

**(5 marks)**

Show answer

Area of sector = 22.34

**(1)**

Area of triangle = 20.569

**(1)**

**(1)**

**(1)**

**(1)**

3. The diagram shows a sector OPRS of a circle with centre O .

Calculate the area of the shaded segment RST .

Give your answer correct to 3 significant figures.

**(5 marks)**

Show answer

Area of sector = 1954.76876…

**(1)**

Area of triangle = 1835.444…

**(1)**

**(1)**

**(1)**

**(1)**

4. The diagram below shows a shaded segment.

(a) Find the area of the shaded segment. Give your answer correct to 3 significant figures.

(b) Find the perimeter of the shaded segment. Give your answer correct to 3 significant figures.

**(9 marks)**

Show answer

(a)

Angle at centre = 112.88538 … may be shown on diagram

**(1)**

Area of sector = 141.85595…

**(1)**

Area of triangle = 66.332496…

**(1)**

**(1)**

**(1)**

**(1)**

(b)

Length of arc = 23.6426587…

**(1)**

Length of chord = 20

**(1)**

**(1)**

**(1)**

**(1)**

You have now learned how to:

- Identify and apply circle definitions and properties
- Identify and apply circle definitions and properties, including a segment
- Find the perimeter of a segment
- Calculate areas of segments by applying trigonometry

Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.

Find out more about our GCSE maths revision programme.

x
#### FREE GCSE Maths Practice Papers - 2022 Topics

Download free

Practice paper packs based on the advanced information for the Summer 2022 exam series from Edexcel, AQA and OCR.

Designed to help your GCSE students revise some of the topics that will come up in the Summer exams.