NSWMaths Standard 2Syllabus dot point

How do you calculate the area of a circle, a sector and shapes built from circular and straight-sided parts?

Calculate the area of circles, sectors and composite figures, including the annulus, by adding and subtracting the areas of simpler shapes

A focused answer to the HSC Maths Standard 2 dot point on area. The area of a circle, the fraction method for a sector, the annulus (ring) as a difference of two circles, and composite shapes built by adding or subtracting simpler parts, with worked Australian examples and clear rounding.

Generated by Claude Opus 4.815 min answerUpdated 2026-06-21

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA wants you to find areas that go beyond rectangles and triangles. You need the area of a full circle, the area of a sector (a "pizza slice" - a fraction of a circle set by its central angle), the area of an annulus (a flat ring, like a washer), and the area of a composite shape: a figure built from simpler pieces that you add together or subtract from one another. The formula for each piece is short. The marks are won and lost on three decisions: using the radius and not the diameter, taking the right fraction of a circle for a sector, and choosing whether to add or subtract when the shape is made of parts. Get those right and round sensibly at the end, and the topic is routine.

The answer

Every area on this page is built from one formula, the area of a circle:

A = \pi r^2,

where $r$ is the radius (the distance from the centre to the edge). A sector is a fraction of that circle, an annulus is one circle minus another, and a composite shape is just simple pieces added or subtracted. Throughout, keep the full value of $\pi$ in your calculator and round only the final answer, stating how many decimal places you used.

The area of a circle

The radius is the key length. If a question gives the diameter $d$ (the full width across the centre), halve it first: $r = \dfrac{d}{2}$ . Then square the radius and multiply by $\pi$ . For a circle of radius $7$ cm,

A = \pi \times 7^2 = \pi \times 49 = 153.9380\ldots \approx 153.94 \text{ cm}^2.

The single most common mistake in the whole topic is squaring the diameter by accident, so always write the radius down before you substitute.

The area of a sector

A sector is the region between two radii and the arc joining them, like a slice of pizza. Its area is simply a fraction of the whole circle, and that fraction is the central angle $\theta$ measured against the full turn of $360^\circ$ :

A = \frac{\theta}{360} \times \pi r^2.

The diagram shows a sector with its central angle $\theta$ marked. A half circle is $\frac{180}{360} = \frac{1}{2}$ of the circle, a quarter circle is $\frac{90}{360} = \frac{1}{4}$ , and a slice of $120^\circ$ is $\frac{120}{360} = \frac{1}{3}$ .

So a sector of radius $10$ cm with a $72^\circ$ angle has area

A = \frac{72}{360} \times \pi \times 10^2 = \frac{1}{5} \times \pi \times 100 = 62.8318\ldots \approx 62.83 \text{ cm}^2.

The annulus: a ring as a difference of circles

An annulus is the flat ring between two circles that share a centre - think of a washer, a CD, or a paved border round a pond. Its area is the big circle minus the small circle. With outer radius $R$ and inner radius $r$ ,

A = \pi R^2 - \pi r^2 = \pi \left(R^2 - r^2\right).

The safe way is to find $R^2 - r^2$ first, then multiply by $\pi$ . For $R = 8$ mm and $r = 5$ mm,

A = \pi \left(8^2 - 5^2\right) = \pi \left(64 - 25\right) = \pi \times 39 = 122.5221\ldots \approx 122.52 \text{ mm}^2.

Note that $R^2 - r^2$ is not the same as $(R - r)^2$ : here $64 - 25 = 39$ , while $(8-5)^2 = 9$ , a very different ring.

Composite shapes: add the parts or subtract a cut-out

A composite shape is a figure made from simpler pieces. The method is always the same:

Decompose the figure into shapes you know (rectangles, triangles, circles, semicircles, sectors).
Find each part's area separately.
Add the parts that are joined together, or subtract a part that has been cut out (a hole).

The diagram below shows a common composite shape, a rectangle with a semicircle on top, broken into its two named parts. Because the semicircle sits on the top edge, its diameter equals the width of the rectangle, so its radius is half that width.

For that rectangle ( $8 \times 12$ ) with a semicircle of radius $4$ on top, the total is

A = \underbrace{8 \times 12}_{\text{rectangle}} + \underbrace{\tfrac{1}{2}\pi \times 4^2}_{\text{semicircle}} = 96 + 25.1327\ldots = 121.1327\ldots \approx 121.13 \text{ units}^2.

When the circular piece is a hole instead, you subtract it. A $20$ cm by $12$ cm metal plate with a circular hole of radius $4$ cm has area $20 \times 12 - \pi \times 4^2 = 240 - 50.2654\ldots = 189.73$ cm $^2$ .

How exam questions ask about area

The wording changes but always points to one of the four methods:

"Find the area of the circle / circular ..." is the plain $A = \pi r^2$ ; watch for a diameter that must be halved first.
"... of the sector / the shaded slice / the pizza piece" with an angle given is the fraction method $\dfrac{\theta}{360} \times \pi r^2$ .
"... of the ring / the washer / the path around / the border / the annulus" is the difference of two circles, $\pi(R^2 - r^2)$ .
"... of the shaded region / the figure / the shape" with a picture is a composite shape: decompose it, then add or subtract.
"... the remaining area / the area left over / with a hole cut out" signals subtraction: whole shape minus the cut-out.
"How much will it cost / how much turf / paint / paving is needed" means find the area first, then multiply by the rate per square metre, rounding the dollars only at the end.

Four areas across Australian contexts

One of each kind: a sprinkler sector, an annulus washer, a rectangle-plus-semicircle garden, and a plate with a hole cut out. Each follows the same plan: write the formula, substitute the radius, evaluate with full $\pi$ , round at the end.

A garden sprinkler sector

A garden sprinkler sweeps an angle of $120^\circ$ and throws water $9$ m. Find the area of lawn it waters, correct to two decimal places.

Recognise a sector and write the fraction. The watered region is a sector of radius $9$ m with central angle $120^\circ$ :

A = \frac{\theta}{360} \times \pi r^2 = \frac{120}{360} \times \pi \times 9^2.

Simplify the fraction and substitute. Since $\frac{120}{360} = \frac{1}{3}$ and $9^2 = 81$ ,

A = \frac{1}{3} \times \pi \times 81 = \frac{1}{3} \times 254.469\ldots = 84.8230\ldots

Round. So the sprinkler waters $84.82$ m $^2$ of lawn (to two decimal places).

A washer (annulus)

A steel washer has an outer radius of $10$ mm and an inner radius of $6$ mm. Find the area of its flat metal face, correct to two decimal places.

Set up the annulus as a difference of circles.

A = \pi\left(R^2 - r^2\right) = \pi\left(10^2 - 6^2\right).

Do the bracket first, then multiply.

A = \pi\left(100 - 36\right) = \pi \times 64 = 201.0619\ldots

Round. The metal face has area $201.06$ mm $^2$ . (Subtract the squares, $100 - 36 = 64$ ; do not square the difference.)

A rectangle-plus-semicircle garden bed

A garden bed is a rectangle $8$ m wide and $12$ m long with a semicircle joined across the full $8$ m width at one end. Find its total area, correct to two decimal places.

Decompose into a rectangle and a semicircle. The semicircle's diameter is the $8$ m width, so its radius is $4$ m.

Find each part.

A_{\text{rectangle}} = 8 \times 12 = 96 \text{ m}^2, \qquad A_{\text{semicircle}} = \frac{1}{2} \times \pi \times 4^2 = 25.1327\ldots \text{ m}^2.

Add them.

A = 96 + 25.1327\ldots = 121.1327\ldots \approx 121.13 \text{ m}^2.

So the garden bed has area $121.13$ m $^2$ .

A plate with a circular hole

A rectangular steel plate measures $20$ cm by $12$ cm. A circular hole of radius $4$ cm is drilled through it. Find the area of metal remaining, correct to two decimal places.

Recognise a subtraction. The remaining metal is the rectangle minus the circular hole:

A = 20 \times 12 - \pi \times 4^2.

Find each part.

A_{\text{rectangle}} = 240 \text{ cm}^2, \qquad A_{\text{hole}} = \pi \times 16 = 50.2654\ldots \text{ cm}^2.

Subtract.

A = 240 - 50.2654\ldots = 189.7345\ldots \approx 189.73 \text{ cm}^2.

So $189.73$ cm $^2$ of metal remains.

Final answers: the sprinkler waters $84.82$ m $^2$ ; the washer face is $201.06$ mm $^2$ ; the garden bed is $121.13$ m $^2$ ; and the plate has $189.73$ cm $^2$ of metal left.

Common traps

Squaring the diameter instead of the radius: The formula uses the radius. If a question gives the diameter, halve it first: a diameter of $10$ gives $r = 5$ and $A = \pi \times 25$ , not $\pi \times 100$ .
Forgetting the $\frac{\theta}{360}$ fraction for a sector: A sector is only part of a circle. Multiply $\pi r^2$ by $\frac{\theta}{360}$ ; using the full circle's area is the most common sector error.
Squaring the difference in an annulus: Use $\pi(R^2 - r^2)$ , not $\pi(R - r)^2$ . For $R = 8$ , $r = 5$ these give $\pi \times 39$ and $\pi \times 9$ - completely different rings.
Adding when you should subtract (or the reverse): A piece joined on is added; a hole or cut-out is subtracted. Read the picture: a notch or a hole means subtract.
Rounding too early: Keep the full value of $\pi$ (and any part-areas) in the calculator. Rounding mid-way, then multiplying by a cost rate, can shift the final dollar figure.
Dropping the square units: An area is always in square units ( $\text{cm}^2$ , $\text{m}^2$ ); leaving the unit off, or writing a plain unit, can cost the final mark.

Exam technique

Write the formula with the radius substituted before you reach for the calculator - that line is where the method mark sits even if the arithmetic slips. For a sector, state the fraction $\frac{\theta}{360}$ explicitly; for an annulus, write $\pi(R^2 - r^2)$ and compute the bracket first; for a composite shape, sketch the decomposition and label each part so the marker can follow your "add" or "subtract". Carry the unrounded value through to the very end, then round to the number of decimal places the question asks for (two is the usual default) and write the square unit. If a cost is wanted, find the area first and multiply by the rate last. A quick sanity check - a sector should be a believable fraction of $\pi r^2$ , a ring smaller than its outer circle - catches a wrong method before it costs marks.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC-style2 marksA sector of a circular garden has a radius of

6

m and a central angle of

100^\circ

. Find the area of the sector, correct to two decimal places.

Show worked answer →

Award one mark for setting up the fraction-of-a-circle method, $A = \frac{100}{360} \times \pi \times 6^2$ , and one mark for the correct evaluated answer $31.42$ m $^2$ (from $\frac{100}{360} \times \pi \times 36 = 31.4159\ldots$ ). A marker accepts the equivalent fraction $\frac{5}{18}$ . The method mark is lost if the student uses $\frac{100}{360}$ of the circumference instead of the area, or forgets to square the radius; the accuracy mark is lost for premature rounding or for omitting the m $^2$ unit.

2021 HSC-style3 marksA circular fountain of radius

4

m is surrounded by a paved ring (an annulus) extending out to an outer radius of

9

m. Find the area of the paved ring, correct to two decimal places.

Show worked answer →

Award one mark for recognising the annulus method (subtract the inner circle from the outer circle), $A = \pi(9^2 - 4^2)$ ; one mark for evaluating the bracket correctly as $81 - 16 = 65$ ; and one mark for the final area $204.20$ m $^2$ (from $\pi \times 65 = 204.2035\ldots$ ). The most penalised error is computing $\pi(9-4)^2 = \pi \times 25$ , which a marker treats as the wrong method and awards zero of the last two marks. A correct method with one arithmetic slip can still earn the method mark.

2020 HSC-style4 marksA stained-glass window is a rectangle

0.8

m wide and

1.5

m tall with a semicircle on top. (a) Find the total area of glass, correct to two decimal places. (b) The glass costs $120 per square metre. Find the cost, correct to the nearest dollar.

Show worked answer →

Part (a): award one mark for decomposing into a rectangle plus a semicircle and identifying the semicircle radius as $0.4$ m (half the $0.8$ m width), and one mark for the total area $1.45$ m $^2$ (from $0.8 \times 1.5 + \frac{1}{2}\pi \times 0.4^2 = 1.2 + 0.2513\ldots$ ). Part (b): award one mark for multiplying area by the rate and one mark for the rounded cost $174 (from $1.4513\ldots \times 120 = 174.16\ldots$ ). Markers reward carrying the unrounded area into the cost; rounding the area to $1.45$ first still rounds to $174 here, but premature rounding is a habit that loses marks elsewhere. A bare answer with no decomposition shown caps part (a) at one mark.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksFind the area of a circle with radius

7

cm. Give your answer correct to two decimal places.

Show worked solution →

Use the circle area formula. The area of a circle is $A = \pi r^2$ with $r = 7$ :

A = \pi \times 7^2 = \pi \times 49

Evaluate and round. Using the calculator value of $\pi$ ,

A = 153.9380\ldots \approx 153.94 \text{ cm}^2.

So the area is $153.94$ cm $^2$ (to two decimal places). Keep the full value of $\pi$ in the calculator and round only at the very end.

foundation2 marksA circular tabletop has a diameter of

10

cm in a scale drawing. Find its area, correct to two decimal places.

Show worked solution →

Halve the diameter to get the radius. The formula needs the radius, not the diameter, so

r = \frac{10}{2} = 5 \text{ cm}.

Apply $A = \pi r^2$ .

A = \pi \times 5^2 = \pi \times 25 = 78.5398\ldots \approx 78.54 \text{ cm}^2.

So the area is $78.54$ cm $^2$ . (The most common slip here is using $10$ as the radius; always halve the diameter first.)

foundation2 marksFind the area of a semicircle with radius

6

m, correct to two decimal places.

Show worked solution →

A semicircle is half a circle. Find the full circle's area, then halve it:

A = \frac{1}{2} \times \pi r^2 = \frac{1}{2} \times \pi \times 6^2 = \frac{1}{2} \times \pi \times 36.

Evaluate.

A = \frac{1}{2} \times 113.097\ldots = 56.5486\ldots \approx 56.55 \text{ m}^2.

So the semicircle has area $56.55$ m $^2$ . A semicircle is just the fraction $\frac{180}{360} = \frac{1}{2}$ of a circle.

core3 marksA sector of a circle has radius

10

cm and a central angle of

72^\circ

. Find its area, correct to two decimal places.

Show worked solution →

A sector is a fraction of a circle. The fraction is the central angle over $360^\circ$ :

\text{fraction} = \frac{72}{360} = \frac{1}{5}.

Multiply that fraction by the whole circle's area.

A = \frac{72}{360} \times \pi r^2 = \frac{1}{5} \times \pi \times 10^2 = \frac{1}{5} \times \pi \times 100.

Evaluate and round.

A = \frac{1}{5} \times 314.159\ldots = 62.8318\ldots \approx 62.83 \text{ cm}^2.

So the sector has area $62.83$ cm $^2$ . (Check it makes sense: a fifth of the circle should be roughly a fifth of $314$ , which is about $63$ .)

core3 marksA washer is an annulus (a ring) with an outer radius of

8

mm and an inner radius of

5

mm. Find the area of the metal face, correct to two decimal places.

Show worked solution →

An annulus is a big circle with a small circle removed. Subtract the inner circle's area from the outer circle's area:

A = \pi R^2 - \pi r^2 = \pi \left(R^2 - r^2\right),

where $R = 8$ is the outer radius and $r = 5$ is the inner radius.

Substitute and simplify inside the brackets first.

A = \pi \left(8^2 - 5^2\right) = \pi \left(64 - 25\right) = \pi \times 39.

Evaluate.

A = 122.522\ldots \approx 122.52 \text{ mm}^2.

So the metal face has area $122.52$ mm $^2$ . (Take the difference of the squares before multiplying by $\pi$ , not the square of the difference: $64 - 25 = 39$ , not $(8-5)^2 = 9$ .)

exam4 marksA garden bed is shaped like a rectangle

10

m long and

6

m wide with a semicircle joined to one of the short

6

m ends. (a) Find the radius of the semicircle. (b) Find the total area of the garden bed, correct to two decimal places.

Show worked solution →

Part (a) - the semicircle sits on the $6$ m end. Its diameter equals that side, so

r = \frac{6}{2} = 3 \text{ m}.

Part (b) - split the shape into a rectangle plus a semicircle, then add. The rectangle is $10$ m by $6$ m:

A_{\text{rectangle}} = 10 \times 6 = 60 \text{ m}^2.

The semicircle has radius $3$ m:

A_{\text{semicircle}} = \frac{1}{2} \times \pi \times 3^2 = \frac{1}{2} \times \pi \times 9 = 14.1371\ldots \text{ m}^2.

Add the two parts.

A = 60 + 14.1371\ldots = 74.1371\ldots \approx 74.14 \text{ m}^2.

So the total area is $74.14$ m $^2$ . (Decompose first, label each part, add at the end - and keep the unrounded semicircle value until the final line.)

exam5 marksA square paved courtyard measures

12

m by

12

m. A circular pond of radius

3

m is built into the centre, and the rest is paved. (a) Find the paved area, correct to two decimal places. (b) Paving costs $45 per square metre. Find the total cost of the paving, correct to the nearest dollar.

Show worked solution →

Part (a) - subtract the pond from the square. The paved region is the square with the circle removed:

A_{\text{paved}} = A_{\text{square}} - A_{\text{circle}} = 12 \times 12 - \pi \times 3^2.

Work out each piece.

A_{\text{square}} = 144 \text{ m}^2, \qquad A_{\text{circle}} = \pi \times 9 = 28.2743\ldots \text{ m}^2.

Subtract.

A_{\text{paved}} = 144 - 28.2743\ldots = 115.7256\ldots \approx 115.73 \text{ m}^2.

Part (b) - multiply the area by the rate. Using the unrounded area to keep the cost accurate,

\text{cost} = 115.7256\ldots \times 45 = 5207.65\ldots

so the cost is $5208 to the nearest dollar. (When a question asks for a cost, carry the full area into the multiplication and round the dollars only at the end.)

What this dot point is asking

The answer

The area of a circle

The area of a sector

The annulus: a ring as a difference of circles

Composite shapes: add the parts or subtract a cut-out

How exam questions ask about area

Exam-style practice questions

Practice questions

Related dot points