Geometry and measures

Arcs and sectors

Q: How do you find the arc length of a sector?

Multiply the full circumference 2\pi r by the fraction \frac{\theta}{360}, where \theta is the angle at the centre.

11 past-paper questions, easiest first. Try each one, then reveal the answer.

Foundation & Higher3 of 11 from Foundation papers

Revision guide by The Full Marks

Arcs and sectors: what you need to know

An arc is part of a circle's edge and a sector is a slice of the circle. Both are a fraction of the full circle, set by the angle at the centre. At Higher they appear in multi-step problems with segments, cones and shaded regions.

Both tiers need arc lengths and sector areas; segments and multi-step shaded-region problems are mostly asked on Higher papers.

Key facts

Fraction of the circle $= \frac{\theta}{360}$ , where $\theta$ is the angle at the centre.
Arc length $= \frac{\theta}{360} \times 2\pi r$ .
Sector area $= \frac{\theta}{360} \times \pi r^2$ .
Perimeter of a sector $=$ arc length $+ 2r$ .
A semicircle is $\frac{180}{360} = \frac{1}{2}$ of a circle; a quarter circle is $\frac{90}{360}$ .

Arc length or sector area

Write the fraction $\frac{\theta}{360}$ .
Multiply by the full circumference (for arc) or full area (for sector).

Working backwards (angle or radius)

If you are given the perimeter, subtract the two radii first to get the arc length.
Put what you know into the arc or sector formula, leaving $\theta$ (or $r$ ) unknown.
For $\theta$ : divide both sides by $2\pi r$ (or $\pi r^2$ ), then multiply by 360. For $r$ : rearrange and solve, taking a square root for a sector area.

Worked examples

Example 1 Foundation · grade 5

A sector has radius $8$ cm and angle $45^\circ$ . Find its area and its perimeter, to 1 decimal place.

Show the solution

Area $= \frac{45}{360} \times \pi \times 8^2 = 8\pi = 25.13...$
Area $= 25.1$ cm $^2$
Arc $= \frac{45}{360} \times 2 \times \pi \times 8 = 2\pi = 6.28...$ cm
Perimeter $= 2\pi + 8 + 8 = 22.28...$
Perimeter $= 22.3$ cm

Example 2 Higher · grade 7

A sector has radius $12$ cm and arc length $5\pi$ cm. Find the angle of the sector, then its exact area.

Show the solution

$\frac{\theta}{360} \times 2\pi \times 12 = 5\pi$
$\frac{\theta}{360} \times 24\pi = 5\pi$
$\theta = \frac{5}{24} \times 360 = 75^\circ$
Area $= \frac{75}{360} \times \pi \times 12^2 = 30\pi$ cm $^2$

Common mistakes

Forgetting the two radii when finding the perimeter of a sector (or forgetting to take them off when you are given the perimeter).
Using the area formula for an arc length, or the reverse.
Using the diameter as the radius.
Leaving a reflex angle as the small angle: check whether the sector is more than half the circle.

Exam tips

For a segment, find the sector area and subtract the triangle. If the angle is $90^\circ$ the triangle is $\frac{1}{2} \times r \times r$ ; for other angles (Higher) use $\frac{1}{2}ab \sin C$ .
Give exact answers with $\pi$ when asked; round only at the very end otherwise.

Questions people ask

How do you find the arc length of a sector?

Multiply the full circumference $2\pi r$ by the fraction $\frac{\theta}{360}$ , where $\theta$ is the angle at the centre.

How do you find the perimeter of a sector?

Add the arc length to the two straight edges, which are both radii.

Exam questions: Arcs and sectors (11, easiest first)

Arcs and sectors4 marksGrade 5CalculatorOn both tiers

November 2021 · Paper 3 Foundation · Q27

Find the area of a segment cut off a quarter circle by a chord.

Show answer

Answer	18.3 cm²	4 marks

Worked solution Checked against the mark scheme

Quarter circle: $\frac{1}{4} \times \pi \times 8^2 = 50.265...$ cm²

Triangle $OPR$ : $\frac{1}{2} \times 8 \times 8 = 32$ cm²

Segment $= 50.265... - 32 = 18.265...$

So the area is $18.3$ cm² (3 s.f.).

Watch out: Using $2\pi r$ (circumference) instead of $\pi r^2$ .

Arcs and sectors4 marksGrade 5No calculatorOn both tiers

November 2018 · Paper 1 Foundation · Q26

Show that a shaded region between a quarter circle and a semicircle is π/8 of a square.

Show answer

Answer	$\frac{100\pi - 50\pi}{400} = \frac{\pi}{8}$	4 marks

Worked solution Checked against the mark scheme

Quarter circle, radius 20: $\frac{1}{4}\pi \times 20^2 = 100\pi$

Semicircle, radius 10: $\frac{1}{2}\pi \times 10^2 = 50\pi$

Shaded $= 100\pi - 50\pi = 50\pi$

Square $= 20 \times 20 = 400$

$\frac{50\pi}{400} = \frac{\pi}{8}$

Watch out: Using 20 as the radius of the semicircle.

Arcs and sectors5 marksGrade 5CalculatorFoundation

June 2018 · Paper 2 Foundation · Q19

Find the cost of fencing a semicircular field, charged per metre plus a daily fee.

Show answer

Answer	$£4378.20$	5 marks

Worked solution Checked against the mark scheme

Curved edge: $\frac{1}{2} \times \pi \times 50 = 78.54$ m

Perimeter: $78.54 + 50 = 128.54$ m

Fence: $128.54 \times 29.86 = £3838.20$

Labour: $3 \times 180 = £540$

Total: $3838.20 + 540 = £4378.20$

Watch out: Forgetting the straight edge (the diameter).

Arcs and sectors3 marksGrade 6CalculatorHigher

November 2024 · Paper 3 Higher · Q11

Use the perimeter of a sector to find its reflex angle.

Show answer

Answer	$304^\circ$	3 marks

Worked solution Checked against the mark scheme

Arc length $= 34.3 - 2 \times 4.7 = 24.9$ m

Arc $= \frac{\theta}{360} \times 2\pi r$ , so

$\theta = \frac{24.9}{2\pi \times 4.7} \times 360 = 303.5... \approx 304^\circ$

Watch out: Forgetting to subtract the two radii from the perimeter.

Arcs and sectors2 marksGrade 6CalculatorHigher

November 2023 · Paper 3 Higher · Q12

Find the angle of a reflex sector given its area and the radius of the circle.

Show answer

Answer	$246°$ (246 to 247)	2 marks

Worked solution Checked against the mark scheme

$\dfrac{x}{360} \times \pi \times 6.2^2 = 82.6$

$x = \dfrac{82.6 \times 360}{\pi \times 6.2^2} = \dfrac{29\,736}{120.76...} = 246.2...$

$x = 246°$ (3 s.f.)

Watch out: Using the formula for arc length, or using the diameter instead of the radius.

Arcs and sectors3 marksGrade 6No calculatorHigher

November 2022 · Paper 1 Higher · Q15

Find the angle of a sector given its radius and its arc length in terms of π.

Show answer

Answer	$40$	3 marks

Worked solution Checked against the mark scheme

Full circumference $= 2\pi \times 18 = 36\pi$

Fraction of the circle: $\frac{4\pi}{36\pi} = \frac{1}{9}$

$x = \frac{1}{9} \times 360 = 40$

Watch out: Using the area formula $\pi r^2$ instead of the circumference.

Arcs and sectors4 marksGrade 7No calculatorHigher

June 2024 · Paper 1 Higher · Q15

Given the radius and arc length of a sector, find its area in terms of pi.

Show answer

Answer	$15\pi\text{ cm}^2$	4 marks

Worked solution Checked against the mark scheme

Full circumference $= 2\pi \times 6 = 12\pi$

Fraction of the circle: $\frac{5\pi}{12\pi} = \frac{5}{12}$ (angle $= 150^\circ$ )

Full area $= \pi \times 6^2 = 36\pi$

Sector area $= \frac{5}{12} \times 36\pi = 15\pi\text{ cm}^2$

Watch out: Using $\pi r^2$ with the arc length instead of finding the fraction of the circle first.

Arcs and sectors4 marksGrade 7CalculatorHigher

June 2019 · Paper 2 Higher · Q12

Find the perimeter of a sector given its radius and area, to 3 significant figures.

Show answer

Answer	$25.4$ cm	4 marks

Worked solution Checked against the mark scheme

Sector angle: $\frac{\theta}{360} \times \pi \times 7^2 = 40$ , so $\theta = \frac{40 \times 360}{49\pi} = 93.54...^\circ$

Arc $= \frac{93.54...}{360} \times 2\pi \times 7 = 11.43$ cm

Perimeter $= 7 + 7 + 11.43 = 25.43 = 25.4$ cm (3 s.f.)

Watch out: Forgetting to add the two radii to the arc length.

Arcs and sectors5 marksGrade 8No calculatorHigher

November 2020 · Paper 1 Higher · Q22

Set the area of a 45° sector ring equal to the area of a circle, form a quadratic and solve for x.

Show answer

Answer	$x = 0.5$	5 marks

Worked solution Checked against the mark scheme

Outer radius $= (3x - 1) + 2x = 5x - 1$

Area A $= \frac{45}{360}\pi[(5x - 1)^2 - (3x - 1)^2] = \frac{\pi}{8}(16x^2 - 4x)$

Area B $= \pi(1 - x)^2 = \pi(x^2 - 2x + 1)$

Equate and cancel $\pi$ : $2x^2 - \frac{x}{2} = x^2 - 2x + 1$ , so $2x^2 + 3x - 2 = 0$

$(2x - 1)(x + 2) = 0$ , so $x = 0.5$ or $x = -2$

$x = -2$ gives negative lengths, so $x = 0.5$

Watch out: Using the diameter $(2 - 2x)$ as the radius of circle B.

Arcs and sectors5 marksGrade 9No calculatorHigher

June 2022 · Paper 1 Higher · Q21

Find the exact area of the two shaded regions formed inside the middle of three overlapping equal circles.

Show answer

Answer	$16\sqrt{3} - \frac{16\pi}{3}$ cm²	5 marks

Worked solution Checked against the mark scheme

Join A and B to a crossing point of their circles: all three sides are 4 cm, so the triangle is equilateral and the overlap of circles A and B uses a $120^\circ$ angle at A.

One segment $= \frac{120}{360}\pi(4^2) - \frac{1}{2}(4)(4)\sin 120^\circ = \frac{16\pi}{3} - 4\sqrt{3}$

Each overlap (lens) is 2 segments, and circle B has 2 lenses, so the lenses total $4\left(\frac{16\pi}{3} - 4\sqrt{3}\right)$

Shaded = circle B minus the two lenses $= 16\pi - \frac{64\pi}{3} + 16\sqrt{3} = 16\sqrt{3} - \frac{16\pi}{3}\text{ cm}^2$

Watch out: Using a 60° sector for the segment of the lens instead of 120°.

Arcs and sectors5 marksGrade 9CalculatorHigher

November 2019 · Paper 3 Higher · Q23

Find the reflex angle of a sector that folds into a cone of given volume and height.

Show answer

Answer	$264°$	5 marks

Worked solution Checked against the mark scheme

$56.8 = \frac{1}{3}\pi r^2 \times 3.6$ , so $r^2 = \dfrac{56.8}{1.2\pi} = 15.07$ and $r = 3.882$ cm

Slant height $l = \sqrt{3.882^2 + 3.6^2} = 5.294$ cm (the sector's radius)

Arc length = base circumference: $\frac{\theta}{360} \times 2\pi l = 2\pi r$

$\theta = 360 \times \dfrac{r}{l} = 360 \times \dfrac{3.882}{5.294} = 264°$

Watch out: Using the height 3.6 cm as the sector's radius instead of the slant height.

Also tested in these questions

Their main topic is different, but part of each one tests this topic.

June 2023 · Paper 3 Higher · Q14: Use the cosine rule to find a radius, then work out the area of a sector.
November 2022 · Paper 2 Foundation · Q21: Find the area of a garden made of two rectangles, a triangle and a quarter circle, then the cost of enough bags of seed.
November 2018 · Paper 3 Higher · Q16: Find the perimeter of a shape made of a triangle and a major sector, using the sine rule for a missing side.