Circles, Arcs and Sectors

Learn how to work with circles, arcs and sectors for IGCSE Maths, including how to find the arc length and sector area of any circle. This page covers the circumference and area of a circle, the perimeter and area of a semicircle, and how to take a fraction of the whole circle for arcs and sectors. Each method is shown with clear worked examples, including how to give answers in terms of π, followed by auto-marked practice rooms so you can build confidence.

Prior Knowledge You should know the parts of a circle (radius, diameter, circumference) and be comfortable substituting into a formula and rearranging. A calculator with a π key is useful, though many answers can be left exact in terms of π.

⇩ Jump to Practice Questions ⇩

The four key formulae

Everything on this page comes from two circle formulae, plus the idea that an arc or a sector is just a fraction of the whole circle. The radius is \(r\) and the diameter is \(d = 2r\).

Circumference \(C = \pi d\) or \(2\pi r\) The distance all the way around the circle.

Area of a circle \(A = \pi r^2\) The space enclosed inside the circle.

Arc length \(\dfrac{\theta}{360}\times 2\pi r\) A fraction of the circumference.

Sector area \(\dfrac{\theta}{360}\times \pi r^2\) A fraction of the circle's area.

💡 Never mix up the two circle formulae

Area measures a 2D space, so it is in square units, and the formula has the square in it: \(A = \pi r^2\). Circumference is a length (just once around), so there is no square: \(C = 2\pi r\).

Quick check: if your answer is an area it should end in cm² and the working must contain \(r^2\). If you did not square the radius, you found a length, not an area.

Radius \(r\): centre to edge.

Diameter \(d = 2r\): edge to edge through the centre.

Circumference and area of a circle

If you are given the diameter, use \(C = \pi d\). If you are given the radius, use \(C = 2\pi r\). For the area you always need the radius, so halve the diameter first if necessary.

Circumference and area \[\begin{array}{rcl} C &=& \pi d \\ &=& 2\pi r \\ A &=& \pi r^2 \end{array}\]

You can give an answer two ways. Leaving it in terms of \(\pi\) (for example \(12\pi\)) is exact. Multiplying out gives a decimal, which should usually be rounded to three significant figures.

Semicircles and quadrants

A semicircle is half a circle and a quadrant is a quarter. The area is the matching fraction of the circle's area; the perimeter is the fraction of the circumference plus the straight edges (the diameter for a semicircle, the two radii for a quadrant).

A semicircle: the curved half plus the straight diameter.

Semicircle \[ \text{Area} = \tfrac{1}{2}\pi r^2 \] \[ \text{Perimeter} = \tfrac{1}{2}\pi d + d \]

A quadrant: a quarter arc plus the two radii.

Quadrant \[ \text{Area} = \tfrac{1}{4}\pi r^2 \] \[ \text{Perimeter} = \tfrac{1}{2}\pi r + 2r \]

Arcs and sectors

A sector is a "pizza slice" of a circle, bounded by two radii and an arc. The angle at the centre is \(\theta\). Because a full turn is \(360^\circ\), a sector with angle \(\theta\) is the fraction \(\dfrac{\theta}{360}\) of the whole circle. The same fraction gives the arc length from the circumference and the sector area from the circle area.

Arc length and sector area \[ \text{Arc length} = \frac{\theta}{360}\times 2\pi r \] \[ \text{Sector area} = \frac{\theta}{360}\times \pi r^2 \]

A sector: two radii, an arc, and the centre angle \(\theta\).

The perimeter of a sector is the arc length plus the two straight radii:

Perimeter of a sector \[ \text{Perimeter} = \frac{\theta}{360}\times 2\pi r + 2r \]

Foundation worked examples

💡 Example 1: circumference

A circle has diameter \(9\) cm. Find its circumference, to 3 significant figures.

\[\begin{array}{rcl} C &=& \pi d \\ &=& \pi \times 9 \\ &=& 28.27\ldots \\ &=& 28.3 \text{ cm} \end{array}\]

What's happening?

The diameter is given, so use \(C = \pi d\) directly.

Round the decimal to 3 significant figures at the end.

💡 Example 2: area in terms of π

A circle has radius \(6\) cm. Find its area, leaving your answer in terms of \(\pi\).

\[\begin{array}{rcl} A &=& \pi r^2 \\ &=& \pi \times 6^2 \\ &=& 36\pi \text{ cm}^2 \end{array}\]

What's happening?

Square the radius first: \(6^2 = 36\).

Leaving the answer as \(36\pi\) is exact, so no rounding is needed.

💡 Example 3: semicircle perimeter

A semicircle has diameter \(10\) cm. Find its perimeter, to 3 significant figures.

\[\begin{array}{rcl} P &=& \tfrac{1}{2}\pi d + d \\ &=& \tfrac{1}{2}\times \pi \times 10 + 10 \\ &=& 15.70\ldots + 10 \\ &=& 25.7 \text{ cm} \end{array}\]

What's happening?

Half the circumference is the curved part.

Add the straight diameter (\(10\) cm) across the bottom. Forgetting this is the most common error.

💡 Example 4: arc length

A sector has radius \(8\) cm and angle \(45^\circ\). Find the arc length, to 3 significant figures.

\[\begin{array}{rcl} \text{Arc} &=& \dfrac{\theta}{360}\times 2\pi r \\ &=& \dfrac{45}{360}\times 2\pi \times 8 \\ &=& \tfrac{1}{8}\times 16\pi \\ &=& 2\pi \\ &=& 6.28 \text{ cm} \end{array}\]

What's happening?

The fraction \(\dfrac{45}{360} = \dfrac{1}{8}\) of the full circle.

Multiply that fraction by the full circumference \(2\pi r = 16\pi\).

💡 Example 5: sector area and perimeter

A sector has radius \(12\) cm and angle \(120^\circ\). Find its area and its perimeter, to 3 significant figures.

\[\begin{array}{rcl} \text{Area} &=& \dfrac{120}{360}\times \pi r^2 \\ &=& \tfrac{1}{3}\times \pi \times 144 \\ &=& 48\pi \\ &=& 151 \text{ cm}^2 \end{array}\] \[\begin{array}{rcl} \text{Arc} &=& \dfrac{120}{360}\times 2\pi \times 12 \\ &=& 8\pi \\ \text{Perimeter} &=& 8\pi + 2\times 12 \\ &=& 49.1 \text{ cm} \end{array}\]

What's happening?

The fraction is \(\dfrac{120}{360} = \dfrac{1}{3}\).

For the area, take a third of \(\pi r^2 = 144\pi\).

For the perimeter, find the arc (\(8\pi\)) then add the two radii (\(2\times 12 = 24\)).

Compound shapes

This is where most exam marks are won. The circle skills above are the building blocks; the questions below show how they are combined.

Exam questions often combine a circle part with straight-sided shapes, such as a rectangle with a semicircle on top (a window shape) or a rectangle with a rounded corner (a quadrant). The method is always the same: break the shape into familiar pieces, work out each piece, then combine.

For area, add the pieces (or subtract, if a piece is removed).
For perimeter, add only the edges on the outside of the whole shape. Where two pieces join, that inner edge is not part of the perimeter.

A compound shape: a rectangle with a semicircle on top. Add the parts; count only the outer edges for perimeter.

💡 Example 6: area of a window shape

A shape is a rectangle \(10\) cm wide and \(6\) cm tall with a semicircle on top. Find its area, to 3 significant figures.

\[\begin{array}{rcl} \text{Area} &=& \text{rectangle} + \text{semicircle} \\ &=& 10 \times 6 + \tfrac{1}{2}\pi (5)^2 \\ &=& 60 + 12.5\pi \\ &=& 60 + 39.27\ldots \\ &=& 99.3 \text{ cm}^2 \end{array}\]

What's happening?

The semicircle sits on the \(10\) cm side, so its diameter is \(10\) and its radius is \(5\).

Find each piece, then add. Keep full accuracy until the final line.

💡 Example 7: perimeter of a window shape

Find the perimeter of the same shape (rectangle \(10\) cm by \(6\) cm with a semicircle on top), to 3 significant figures.

\[\begin{array}{rcl} \text{Perimeter} &=& (10 + 6 + 6) + \tfrac{1}{2}\times 2\pi (5) \\ &=& 22 + 5\pi \\ &=& 22 + 15.7\ldots \\ &=& 37.7 \text{ cm} \end{array}\]

What's happening?

Go around the outside: the bottom, the two vertical sides, and the curved arc.

The straight top of the rectangle is not counted: it is inside the shape where the semicircle joins.

💡 Example 8: rectangle with a rounded corner

A shape is a rectangle \(12\) cm by \(8\) cm with a quarter circle of radius \(4\) cm rounding off the top-right corner. Find its area and perimeter, to 3 significant figures.

\[\begin{array}{rcl} \text{Area} &=& \text{rectangle} - \text{corner} + \text{quarter circle} \\ &=& 12 \times 8 - 4^2 + \tfrac{1}{4}\pi (4)^2 \\ &=& 96 - 16 + 4\pi \\ &=& 92.6 \text{ cm}^2 \end{array}\] \[\begin{array}{rcl} \text{Perimeter} &=& 12 + 8 + (12-4) + \tfrac{1}{4}(2\pi \times 4) + (8-4) \\ &=& 32 + 2\pi \\ &=& 38.3 \text{ cm} \end{array}\]

What's happening?

For the area, rounding off a corner cuts away the corner square (\(4^2 = 16\)) and adds back a quarter circle, giving \(96 - 16 + 4\pi\).

For the perimeter, go around the outside: the full bottom (\(12\)), the left side (\(8\)), the shortened top (\(12 - 4 = 8\)), the quarter arc, then the shortened right side (\(8 - 4 = 4\)).

Exact or decimal?

Read the question carefully to see which form is wanted.

The question says...	Give your answer as...	Example
"in terms of \(\pi\)" or "leave your answer as a multiple of \(\pi\)"	An exact value with \(\pi\) in it (do not multiply out)	\(36\pi\) cm²
"to 3 significant figures" or "to 2 decimal places"	A rounded decimal	\(113\) cm²
nothing specific	A decimal rounded sensibly (3 s.f. is safe)	\(28.3\) cm

🔑 Key points

\(C = \pi d\) or \(2\pi r\), and \(A = \pi r^2\).
Area has the square in it; circumference does not.
Area always needs the radius: halve the diameter if needed.
An arc or sector is the fraction \(\dfrac{\theta}{360}\) of the whole circle.
A semicircle is half a circle; a quadrant is a quarter. Both add their straight edges to the perimeter.
Perimeter of a sector adds the two radii to the arc.
For a compound shape, add the parts and count only the outer edges.

⚠️ Common pitfalls

Using the diameter in \(A = \pi r^2\) instead of the radius.
Mixing up \(\pi d\) (circumference) with \(\pi r^2\) (area).
Forgetting to add the diameter to a semicircle's perimeter.
Forgetting to add the two radii to a sector's perimeter.
Counting the inner join as part of a compound shape's perimeter.
Rounding too early; keep full accuracy until the final line.

⇩ Jump to Practice Questions ⇩

Ready to practise? The rooms below cover circumference, area, arc length, sector area and compound shapes, with a mix of exact and decimal answers.

Next: Volume of a Frustum →

Circles, Arcs and Sectors: Practice Room

Practise the circle skills and then put them to work on compound shapes, auto-marked as you type. In Rooms 1 and 2, watch the tag on each question: 3 s.f. wants a rounded decimal (for example 28.3), and in terms of π wants an exact answer such as 36π. For exact answers you can type pi or tap the π button next to the box to insert the symbol; both are accepted. Questions are randomly generated, so refresh for a new set. Difficulty rises left to right.

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