Circles Sectors and Arcs Worksheets, Questions and Revision

Circles Sectors and Arcs Worksheets, Questions and Revision

GCSE 4 - 5AQAEdexcelOCRWJECFoundationAQA 2022Edexcel 2022WJEC 2022

Circle Sectors and Arcs

It can be useful to calculate the area and arc length of sectors of circles.

Make sure you are happy with the following topics before continuing:

Circles

Level 4-5 GCSE AQA Edexcel OCR WJEC
calculate the area of a sector

Area of a Sector

\textcolor{blue}{\text{Area of a sector}} = \dfrac{\textcolor{limegreen}{x}}{360\degree} \times \textcolor{red}{\text{Area of full circle}}

Where \textcolor{limegreen}{x} is the angle of the sector.

\textcolor{red}{\text{area of the circle}} = \textcolor{red}{\pi r^2}

calculate the area of a sector

Calculating the area of the circle segment shown involves using the angle of the sector \textcolor{limegreen}{82\degree} and the radius \textcolor{red}{12} cm.

We apply these to the equation above:

\dfrac{\textcolor{limegreen}{82\degree}}{360\degree} \times \textcolor{red}{\pi \times 12^2} = 103.04 \text{ cm}^2

Level 4-5 GCSE AQA Edexcel OCR WJEC
calculate arc length

Arc Length

The arc length is the length of the part of the circumference which is part of the circle segment. The equation for this is:

\textcolor{purple}{\text{Arc Length}} = \dfrac{\textcolor{limegreen}{x}}{360\degree} \times \textcolor{red}{\text{circumference of full circle}}

Where \textcolor{limegreen}{x} is the angle of the sector.

\textcolor{red}{\text{Circumference of the full circle}} = \textcolor{red}{\pi D}

calculate arc length

We can calculate the arc length of the circle segment shown by using the angle of the sector \textcolor{limegreen}{118\degree} and the diameter 2 \times\textcolor{red}{9} = \textcolor{red}{18} cm.

Now we apply these to the equation above:

\dfrac{\textcolor{limegreen}{118\degree}}{360\degree} \times \textcolor{red}{\pi \times 18} = 18.54 cm

Level 4-5 GCSE AQA Edexcel OCR WJEC
area and circumference example

Example 1: Area and Circumference

The circle shown has centre C.

Find its area and circumference to 1 decimal place.

 

[2 marks]

area and circumference example

As the line passes through the centre, we know it is a diameter of the circle. So:

Circumference = \pi d = \pi \times 42 = 131.9 \text{cm (1 dp)}

The diameter is twice the length of the radius, so the \text{radius } = 42 \div 2 = 21cm. So:

Area of circle = \pi r^2=\pi\times (21^2) = 1385.4\text{cm}^2\text{ (1 dp)}

Note: questions may ask you to “leave your answer in terms of \pi”, this means that the answer should be in the form of “something \times \pi”.

Level 4-5 GCSE AQA Edexcel OCR WJEC
sector area arc length example

Example 2: Sector Area & Arc Length

The sector of a circle has centre C as shown.

Find the area of the sector and the arc length to 1 decimal place.

[2 marks]

sector area arc length example

The angle is 120 \degree, which means that this sector is \frac{120}{360} as a fraction of the whole circle.

So, we get:

\textcolor{blue}{\text{Sector Area}} = \dfrac{120}{360} \times \pi \times 8^2 

= 67.0\text{ cm}^2 (1 dp)

\textcolor{purple}{\text{Arc Length}} = \dfrac{120}{360} \times \pi \times 2 \times 8

= 16.8\text{ cm} (1 dp)

Level 4-5 GCSE AQA Edexcel OCR WJEC

Example Questions

The formula for the area of a circle is \pi r^2. In this question we are given the diameter rather than the radius.  Given that the diameter is twice the length of the radius:

 

\text{radius} = 5.24 \div 2 = 2.62 \text{cm}

 

Therefore,

\text{area } = \pi r^2 = \pi \times (2.62)^2 = 21.6 \text{cm}^2

The formula for the area of a circle is \pi r^2. The area of the sector is,

 

\text{area} =\dfrac{72\degree}{360\degree} \times \pi (5)^2 = \dfrac{72\degree}{360\degree} \times 25 \pi = 5\pi\text{ m}^2

The formula for the area of a circle is \pi r^2. The area of this sector, 26.15 \text{m}^2, must be equal to \frac{x}{360} of the total area of the circle. So, as an equation, this looks like:

 

26.15 = \dfrac{x\degree}{360\degree} \times \pi \times 9^2

 

26.15=\dfrac{x\degree}{360\degree} \times 81\pi

 

Dividing the left-hand side and right-hand side by 81 \pi gives

 

\dfrac{26.15}{81\pi} = \dfrac{x\degree}{360\degree}

 

Then, multiply both sides by 360 to get,

x=\dfrac{26.15}{81\pi}\times 360\degree

 

x = 37.0\degree (to 1d.p.)

The question asks for the total perimeter of the shape. We know that one side is 14 mm but the other two are missing. Immediately we can identify that the other straight side is also 14 mm. Then, all that remains is to calculate the arc length.

 

The angle in this sector is 165 degrees, meaning that the arc length will be equal to \frac{165}{360} of the total circumference. The formula for the circumference is \pi d, or alternatively (and more helpfully in this case), 2\pi r. So, we get:

Arc length = \dfrac{165}{360} \times 2 \pi \times 14 = 40.3 \text{mm}

Therefore,

\text{total perimeter} = 14 + 14 + 40.3 = 68.3 \text{mm}

The formula for the area of a circle is \pi r^2. The area of this sector, 160 \text{m}^2, must be equal to \frac{x}{360} of the total area of the circle. So, as an equation, this looks like:

 

160 = \dfrac{x\degree}{360\degree} \times \pi \times 9^2

 

160=\dfrac{x\degree}{360\degree} \times 81\pi

Dividing the left-hand side and right-hand side by 81 \pi gives

 

\dfrac{160}{81\pi} = \dfrac{x\degree}{360\degree}

 

Then, multiply both sides by 360 to get,

x=\dfrac{160}{81\pi}\times 360\degree

 

x = 226\degree (to 3s.f.)

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