# 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

## 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 Level

## 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 Level

## 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

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

A Level

A Level

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