Arc Length & Area of a Sector

Arc Length & Area of a Sector

A LevelAQAEdexcelOCRAQA 2022OCR 2022

Arc Length & Area of a Sector

We’ll now look at how to calculate arc lengths and sector areas.

Make sure you’re up to snuff on your radians!

A Level AQA Edexcel OCR

Arc Length

Let’s say you’ve got a section of a circle, and you want to find the length of the curved edge.

For a circle with radius \textcolor{red}{r} and angle \textcolor{blue}{\theta}, we have the arc length \textcolor{purple}{l} = \textcolor{red}{r}\textcolor{blue}{\theta}.

As mentioned, it’s important that you’re using radians for your value of \textcolor{blue}{\theta}.

We can actually use this formula to derive the circumference of a circle.

Set \textcolor{blue}{\theta} = 2\pi. Then we have \textcolor{purple}{l} = 2\textcolor{red}{r}\pi = d\pi, where d is the diameter.

A LevelAQAEdexcelOCR

Area of a Sector

Say we have the same section of a circle, but we now wish to calculate the area of the sector.

For a circle with radius \textcolor{red}{r} and angle \textcolor{blue}{\theta}, we have the sector area \textcolor{orange}{A} = \dfrac{1}{2}\textcolor{red}{r}^2\textcolor{blue}{\theta}.

Again, using \textcolor{blue}{\theta = 2\pi} gives us the equation for the area of a circle:

\textcolor{orange}{A} = \dfrac{1}{2}\textcolor{red}{r}^2 \textcolor{blue}{2\pi} = \pi \textcolor{red}{r}^2.

A LevelAQAEdexcelOCR

Example Questions

Firstly, we have

\theta = 45° = \dfrac{\pi}{4}

Then the length of the arc is

l = r\theta = \dfrac{9\pi}{4}

Giving the total perimeter

\dfrac{9\pi}{4} + 9 + 9 = 9(2 + \dfrac{\pi}{4})\text{ mm}

First of all, the angle \theta = 60° = \dfrac{\pi}{3}.

The area of one sector is

\dfrac{1}{2}r^2 \theta = \dfrac{1}{2} \times 4^2 \times \dfrac{\pi}{3} =  \dfrac{8\pi}{3}\text{ cm}^2

 

so the area of all three identical sectors is 8\pi\text{ cm}^2.

Total area of garden:

5 \times 3.5 = 17.5

Area of patio:

(1 \times 3.5) + (\dfrac{1}{2} \times 1^2 \times \pi) = 3.5 + \dfrac{\pi}{2}

Area of pool:

\dfrac{1}{2} \times 1.25^2 \times \dfrac{\pi}{2} = \dfrac{25\pi}{64}

Then, the total area covered by grass is

17.5 - (3.5 + \dfrac{\pi}{2} + \dfrac{25\pi}{64}) = 14 - \dfrac{57}{64}\pi

Additional Resources

MME

Exam Tips Cheat Sheet

A Level
MME

Formula Booklet

A Level

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