# Circle Arcs, Sectors and Segments

## Circle Arcs, Sectors and Segments Revision

**Circle Arcs, Sectors and Segments**

A circle can be split into **arcs**, **segments** and **sectors**.

You will need to be able to find** arc** lengths, **segment** area and perimeter, and **sector** area and perimeter.

**Introduction**

It is important to be able to recognise the parts of a circle, and whether each section is minor and major.

For **sectors**, the **minor sector** is the always the **sector** that has an angle, (x\degree) at the centre smaller than 180\degree, and the remainder of the circle is the **major sector**.

The **minor sector** produces the **minor arc**, as follows:

For **segments**, the **segment** that is smaller than half of the area of the circle is the **minor segment**, enclosed by a **chord** and the **minor arc**, and the remainder of the circle is the **major segment**, as follows:

**Arc Length**

To calculate **arc** length, you will always need to size of the angle at the centre of the **sector**, and the circumference of the full circle.

**Arc** length can be calculated using the following equation:

Length of Arc = \dfrac{x}{360}\times Circumference of full Circle

**Sector **

**Area**

To calculate the area of a **sector**, you will again need the angle in the centre, as well as the area of the full circle.

**Sector area** can be calculated using the following equation:

Area of Sector = \dfrac{x}{360}\times Area of full Circle

**Perimeter**

To find the **perimeter** of a **sector**, you add together the **arc length** and the two radii that outline the **sector**, or:

Perimeter of a **Sector** = Arc Length + 2r

Where r is the radius.

**Segment**

A **segment** can be considered the “end” of a **sector.**

This means, we can use the angle at the centre to help us work out perimeter and area.

**Perimeter**

To calculate the perimeter of a **segment**, you need to add the **arc length** and the length of the chord:

Perimeter of a segment = arc length + chord length

We may also need to work out the chord length if this is not given. We can do this by using the angle size and the length of the radii to use the Cosine rule:

a^2=b^2+c^2-2bc\cos(A)

Where b and c are radii.

As b and c are radii, they are equal to each other, and the equation can be simplified to,

a^2=2b^2-2b^2\cos(A)

Where b is the radius.

**Area**

To calculate the area of a **segment**, you need to find the area of the full **sector**, and subtract the area of the triangle next to the segment, or:

Area of **segment** = Area of sector - Area of triangle

The area of the triangle can be calculated using:

Area of triangle = \dfrac{1}{2} ab\sin(C)

Which can be simplified to,

Area of triangle = \dfrac{1}{2} a^2\sin(C)

Where a is the radius.

**Example 1: Sector Perimeter and Area**

Find the perimeter and area of the following **sector**, which is part of a circle with radius 5\text{ cm}

Give your answers to the nearest whole number.

**[4 marks]**

**Area:**

We will use the equation,

Area of Sector = \dfrac{x}{360}\times Area of full Circle

The area of the full circle is,

\pi r^2 = 25\pi

So the area of the **sector** is,

\dfrac{120}{360}\times25\pi = 26\text{ cm}^2

**Perimeter:**

We need to find the **arc length** to find the perimeter, using:

Length of Arc = \dfrac{x}{360}\times Circumference of full Circle

The circumference of the full circle is,

2\pi r = 10\pi

Therefore, the** arc length** is,

\dfrac{120}{360}\times10\pi=10\text{ cm}

And the perimeter is,

10 + 5 + 5 = 20\text{ cm}

**Example 2: Segment Area**

Calculate the area of the shaded area to 2 decimal places:

**[4 marks]**

The shaded area is a **segment**, so we can use the equation:

Area of segment = Area of sector - Area of triangle

So first, we need to calculate the area of the **sector**,

Area of full circle:

\pi r^2 = 16\pi

Area of **sector**:

\dfrac{60}{360}\times16\pi=\dfrac{8}{3}\pi

And now the area of the triangle,

\dfrac{1}{2} ab \sin(C) = \dfrac{1}{2}\times4\times4\times\sin(60) = 4\sqrt{3}

Finally, we can calculate the area of the **segment**:

Area of **segment** = Area of sector - Area of triangle

Area of **segment** =\dfrac{8}{3}\pi - 4\sqrt{3} = 1.45\text{ cm}

## Circle Arcs, Sectors and Segments Example Questions

**Question 1**: Calculate the length of the arc subtended from a sector with radius 12\text{ cm}, and an angle 40\degree to 1 decimal place.

**[3 marks]**

Using the equation,

Length of Arc = \dfrac{x}{360}\times Circumference of full Circle

Firstly, we need to work out circumference of full circle,

2\pi r = 24\piAnd then the arc length,

\dfrac{40}{360}\times24\pi=8.4\text{ cm}**Question 2**: Work out the area of the major sector in this diagram, to the nearest whole number:

**[4 marks]**

The questions asks for the area of the major sector, so we are doing the larger sector on the circle, this means the angle in the centre is,

360-100=260\degreeWe need to find the full area of the circle first,

\pi r^2 = 100\piAnd now we can work out the area of the major sector,

\dfrac{260}{360}\times 100\pi = 227\text{ cm}^2**Question 3**: Calculate the perimeter of the shaded segment, to 1 decimal place.

**[4 marks]**

To find the perimeter of a segment, we need arc length and chord length.

First, we will work out arc length,

Circumference of circle:

2\pi r = 10\piArc length:

\dfrac{160}{360}\times 10\pi = \dfrac{40}{9}\pi

Now, to work out chord length we will use,

a = \sqrt{b^2+c^2-2bc\cos(A)}\\ \sqrt{5^2+5^2-2(5)(5)\cos(160)}=9.848...\text{cm}

Perimeter:

\dfrac{40}{9}\pi + 9.848... = 23.8\text{ cm}## You May Also Like...

### MME Learning Portal

Online exams, practice questions and revision videos for every GCSE level 9-1 topic! No fees, no trial period, just totally free access to the UK’s best GCSE maths revision platform.