National 5 Maths
Arcs and Sectors

The angle at the centre is \(41^\circ\) so this sector is \(\frac{41}{360}\) of the matching circle.

The radius \(r=5,\) so the diameter \(d=2r=10\small.\)

$$ \begin{eqnarray} \small \textsf{(a) Arc length} \normalsize \ &=&\ \frac{41}{360}\times\ \small \textsf{circumference} \normalsize \\[8pt] &=&\ \frac{41}{360}\times \pi d\\[8pt] &=&\ \frac{41}{360}\times \pi\times 10\\[8pt] &\approx&\ 3.58\ \small\textsf{cm}\\[12pt] \small \textsf{(b) Sector area} \normalsize \ &=&\ \frac{41}{360}\times\ \small \textsf{area of circle} \normalsize \\[8pt] &=&\ \frac{41}{360}\times \pi r^2\\[8pt] &=&\ \frac{41}{360}\times\pi\times 5^2\\[8pt] &\approx&\ 8.94\ \small\textsf{cm²} \end{eqnarray} $$

Video solution by Clelland Maths 

The angle at the centre is \(120^\circ\) so this sector is \(\frac{120}{360}\) of the matching circle.

For non-calculator questions like this, you should always simplify the calculation as far as possible before multiplying by 3.14. The number that you have to multiply by 3.14 will only ever have one significant figure, so it will be a short multiplication, possibly followed by multiplying by a power of 10.

$$ \begin{eqnarray} \small \textsf{(a) Arc length} \normalsize \ &=&\ \frac{120}{360}\times\ \small \textsf{circumference} \normalsize \\[8pt] &=&\ \frac{120}{360}\times \pi d\\[8pt] &\approx&\ \frac{1}{3}\times 3.14\times 6\\[8pt] &=&\ 2\times 3.14\\[8pt] &=&\ 6.28\ \small\textsf{cm}\\[12pt] \small \textsf{(b) Sector area} \normalsize \ &=&\ \frac{120}{360}\times\ \small \textsf{area of circle} \normalsize \\[8pt] &=&\ \frac{120}{360}\times \pi r^2\\[8pt] &\approx&\ \frac{1}{3}\times 3.14\times 3^2\\[8pt] &=&\ 3\times 3.14\\[8pt] &=&\ 9.42\ \small\textsf{cm²} \end{eqnarray} $$

Video solution by Clelland Maths 

The angle at the centre is \(72^\circ\) so this sector is \(\frac{72}{360}\) of the matching circle.

For non-calculator questions like this, you should always simplify the calculation as far as possible before multiplying by 3.14. The number that you have to multiply by 3.14 will only ever have one significant figure, so it will be a short multiplication, possibly followed by multiplying by a power of 10, as in this example.

$$ \begin{eqnarray} \small \textsf{(a) Arc length} \normalsize \ &=&\ \frac{72}{360}\times\ \small \textsf{circumference} \normalsize \\[8pt] &=&\ \frac{72}{360}\times \pi d\\[8pt] &\approx&\ \frac{1}{5}\times 3.14\times 100\\[8pt] &=&\ 20\times 3.14\\[8pt] &=&\ 10\times 2\times 3.14\\[8pt] &=&\ 10\times 6.28\\[8pt] &=&\ 62.8\ \small\textsf{cm}\\[12pt] \small \textsf{(b) Sector area} \normalsize \ &=&\ \frac{72}{360}\times\ \small \textsf{area of circle} \normalsize \\[8pt] &=&\ \frac{72}{360}\times \pi r^2\\[8pt] &\approx&\ \frac{1}{5}\times 3.14\times 50^2\\[8pt] &=&\ \frac{1}{5}\times 3.14\times 2500\\[8pt] &=&\ 500\times 3.14\\[8pt] &=&\ 100\times 5\times 3.14\\[8pt] &=&\ 1570\ \small\textsf{cm²} \end{eqnarray} $$

Video solution by Clelland Maths 

In this question, we are given the arc length and have to work backwards to find the angle. Note that we use the same formula as before. The working involves multiplying both sides by 360 to get rid of the fraction.

$$ \begin{eqnarray} \small \textsf{Arc length} \normalsize \ &=&\ \frac{a}{360}\times\ \small \textsf{circumference} \normalsize \\[8pt] 12\ &=&\ \frac{a}{360}\times \pi\times 14\\[8pt] 12 \times 360 &=&\ a \times \pi\times 14\\[8pt] 4320 &=&\ 14\pi a \\[8pt] a &=&\ \frac{4320}{14\pi} \\[8pt] a &\approx&\ 98.2^\circ \end{eqnarray} $$

Video solution by Clelland Maths 

This is another working backwards question, but this time we are given the sector area.

$$ \begin{eqnarray} \small \textsf{Sector area} \normalsize \ &=&\ \frac{a}{360}\times\ \small \textsf{area of circle} \normalsize \\[8pt] 115\ &=&\ \frac{a}{360}\times \pi \times 8^2\\[8pt] 115\ &=&\ \frac{64\pi a}{360}\\[8pt] 115 \times 360 &=&\ 64\pi a\\[8pt] 41400 &=&\ 64\pi a \\[8pt] a &=&\ \frac{41400}{64\pi} \\[8pt] a &\approx&\ 205.9^\circ \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} \small \textsf{Arc length} \normalsize \ &=&\ \frac{72}{360}\times\ \small \textsf{circumference} \normalsize \\[8pt] 7.79\ &=&\ \frac{72}{360}\times \pi d\\[8pt] 7.79\ &=&\ \frac{72\pi d}{360}\\[8pt] 7.79 \times 360 &=&\ 72\pi d\\[8pt] 2804.4 &=&\ 72\pi d\\[8pt] d &=&\ \frac{2804.4}{72\pi} \\[8pt] d &\approx&\ 12.398\\[8pt] \small\textsf{So}\normalsize \ r &\approx&\ \frac{12.398}{2} \approx 6.20\ \small\textsf{cm} \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} \small \textsf{Sector area} \normalsize \ &=&\ \frac{153}{360}\times\ \small \textsf{area of circle} \normalsize \\[8pt] 17.3\ &=&\ \frac{153}{360}\times \pi r^2\\[8pt] 17.3 \times 360 &=&\ 153\pi r^2\\[8pt] 6228 &=&\ 153\pi r^2\\[8pt] r^2 &=&\ \frac{6228}{153\pi} \\[8pt] r &=&\ \sqrt{\frac{6228}{153\pi}} \\[8pt] r &\approx&\ 3.6\ \small\textsf{cm} \end{eqnarray} $$

Video solution by Clelland Maths 

A non-calculator question requiring backwards working has never appeared on an exam paper, but this example is worth studying.

The answer below requires us to be able to divide 6.28 by 3.14 without a calculator. This isn't difficult if you realise that 2 \(\times\) 3.14 = 6.28, but we can imagine some students not noticing that and getting stuck as a result.

Because we cannot use a calculator, trying to multiply both sides by 360 would be a huge mistake! This answer needs us to be careful to keep the numbers as easy as possible.

$$ \begin{eqnarray} \small \textsf{Arc length} \normalsize \ &=&\ \small\frac{\textsf{angle}}{360}\normalsize\times\ \small \textsf{circumference} \normalsize \\[8pt] \small \textsf{Arc length} \normalsize \ &=&\ \small\frac{a}{360}\normalsize\times\ \pi d \\[8pt] 6.28\ &=&\ \small\frac{a}{360}\normalsize\times 3.14\times 10\\[8pt] 6.28\ &=&\ \small\frac{10a}{360}\normalsize\times 3.14\\[8pt] \small\frac{6.28}{3.14}\normalsize &=&\ \small\frac{10a}{360}\normalsize\\[8pt] 2 &=&\ \small\frac{a}{36}\normalsize\\[8pt] a &=&\ 2\times 36 \\[8pt] a &=&\ 72^\circ \end{eqnarray} $$

Note: Would a question like this be fair? That's debatable. We ran a Twitter poll of maths teachers. 75% of them thought that this would be fair, but 25% disagreed. So we included it on the website, just in case!

Video solution by Clelland Maths 

The length OA of the pendulum is the radius r of the sector.

$$ \begin{eqnarray} \small \textsf{Arc length} \normalsize \ &=&\ \frac{\small\textsf{angle}\normalsize}{360}\times\ \small \textsf{circumference} \normalsize \\[8pt] 28.4 &=&\ \frac{65}{360}\times \pi d\\[8pt] 28.4 \times 360 &=&\ 65\pi d\\[8pt] 10\small\,\normalsize 224 &=&\ 65\pi d\\[8pt] d &=&\ \frac{10\small\,\normalsize 224}{65\pi} \\[8pt] d &=&\ 50.067...\\[8pt] \small\textsf{So}\normalsize \ r &=&\ \frac{d}{2} \approx 25.0\ \small\textsf{cm} \end{eqnarray} $$

Video solution by Clelland Maths 

As always with these non-calculator questions, the key is to simplify and leave the 3.14 until the last step.

$$ \begin{eqnarray} \small \textsf{Sector area} \normalsize \ &=&\ \frac{\small\textsf{angle}\normalsize}{360}\times\ \small \textsf{area of circle} \normalsize \\[8pt] &=&\ \frac{45}{360}\times \pi r^2\\[8pt] &\approx&\ \frac{1}{8}\times 3.14\times 20^2\\[8pt] &=&\ \frac{1}{8}\times 3.14\times 400\\[8pt] &=&\ 50\times 3.14\\[8pt] &=&\ 10\times 5\times 3.14\\[8pt] &=&\ 10\times 15.7\\[8pt] &=&\ 157\ \small\textsf{cm²} \end{eqnarray} $$

Video solution by Clelland Maths 

In this question, we are given the arc length and have to work backwards to find the angle a°. Note that we use the same formula as before. The working involves multiplying both sides by 360 to get rid of the fraction.

$$ \begin{eqnarray} \small \textsf{Arc length} \normalsize \ &=&\ \frac{a}{360}\times\ \small \textsf{circumference} \normalsize \\[8pt] 31.5\ &=&\ \frac{a}{360}\times \pi d\\[8pt] 31.5\ &=&\ \frac{a}{360}\times \pi\times (2\times 6.4)\\[8pt] 31.5 \times 360 &=&\ a \times \pi\times 12.8\\[8pt] 11\small\,\normalsize 340 &=&\ 12.8\pi a \\[8pt] a &=&\ \frac{11\small\,\normalsize 340}{12.8\pi} \\[8pt] a &\approx&\ 282.0^\circ \end{eqnarray} $$

Video solution by Clelland Maths 

This is a simple 3-mark question. We just need to substitute the values and evaluate.

$$ \begin{eqnarray} \small \textsf{Arc length} \normalsize \ &=&\ \frac{\small\textsf{angle}\normalsize}{360}\times\ \small \textsf{circumference} \normalsize \\[8pt] &=&\ \frac{320}{360}\times\ \pi d \\[8pt] &=&\ \frac{320}{360}\times \pi \times (2\times 7.4)\\[8pt] &\approx&\ 41.3\ \small\textsf{cm} \end{eqnarray} $$

Video solution by Clelland Maths 

In non-calculator questions like this, always simplify as far as possible before multiplying by 3.14.

$$ \begin{eqnarray} \small \textsf{Arc length} \normalsize \ &=&\ \frac{\small\textsf{angle}\normalsize}{360}\times\ \small\textsf{circumference}\normalsize \\[8pt] &=&\ \frac{240}{360}\times \pi d\\[8pt] &\approx&\ \frac{2}{3}\times 3.14\times (2\times 30)\\[8pt] &=&\ \frac{2}{3}\times 3.14\times 60\\[8pt] &=&\ 40\times 3.14\\[8pt] &=&\ 10\times 4\times 3.14\\[8pt] &=&\ 10\times 12.56\\[8pt] &=&\ 125.6\ \small\textsf{cm} \end{eqnarray} $$

Video solution by Clelland Maths 

This is a straightforward 3-mark question. We just need to substitute and calculate.

$$ \begin{eqnarray} \small \textsf{Arc length} \normalsize \ &=&\ \frac{\small\textsf{angle}\normalsize}{360}\times\ \small \textsf{circumference} \normalsize \\[8pt] &=&\ \frac{106}{360}\times\ \pi d \\[8pt] &=&\ \frac{106}{360}\times \pi \times (2\times 9.15)\\[8pt] &\approx&\ 16.9\ \small\textsf{metres} \end{eqnarray} $$

Video solution by Clelland Maths