Radians, arc length, sector and segment area for HSC Maths Advanced

Syllabus dot point

“Use radian measure to find arc length, the area of a sector, and the area of a segment of a circle”

HSCMaths AdvancedYear 12: Trigonometric Functions14 min readUpdated 20 Jun 2026

Quick answer

In radians, arc length is $\ell = r \theta$ , sector area is $\frac{1}{2} r^2 \theta$ , and the segment cut off by a chord has area $\frac{1}{2} r^2 (\theta - \sin \theta)$ .

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What this dot point is asking
The answer
Exam-style questions
Practice questions

What this dot point is asking

NESA wants you to use the radian as the natural unit of angle, convert between radians and degrees, and apply the formulas for arc length and sector area, including segments cut off by a chord. Radians are the unit assumed by all calculus involving trig in Maths Advanced.

Note

Think about cutting a slice from a round pizza. How big your slice is depends on the angle you open at the centre, and the crust along the edge of your slice is the "arc". A radian is just a natural way of measuring that opening angle: instead of counting in $360$ degrees, you measure the angle by how far you walk along the crust compared with the distance from the tip to the edge. Open the angle so the crust is exactly as long as one straight side of the slice, and you have opened by one radian. Once you measure angles this way, working out the length of crust and the amount of topping on your slice becomes a couple of tidy multiplications rather than a mess of conversion factors.

Why bother with a new unit at all? Because the radian is not an arbitrary choice like the degree (why $360$ ?), it is the angle measure that makes the geometry come out clean. Defining the angle as "arc length per radius" means the arc-length and sector-area formulas have no stray conversion factor, and, crucially for later, it is the only unit in which $\frac{d}{dx}(\sin x) = \cos x$ . Almost every error on this dot point traces back to one habit: substituting a degree value into a formula that is built for radians. Fix that habit and the topic is short.

The answer

One radian is the angle subtended at the centre of a circle by an arc of length equal to the radius. Equivalently, the radian measure of an angle is the ratio of arc length to radius:

\theta = \frac{\ell}{r}.

A full revolution is $2 \pi$ radians, because the full circumference $2 \pi r$ divided by the radius $r$ is $2 \pi$ .

From radius to sector area, stage by stage

The whole topic grows from one circle in four steps. Each formula is just the previous picture with one more piece added.

Stage 1, the radius. Start with a circle of radius $r$ and centre $O$ . Everything that follows is measured against this one length $r$ ; that is the point of radians.

Stage 2, define one radian. Swing a second radius round until the arc between the two radii is itself of length $r$ . The angle at the centre is then exactly $1$ radian (about $57.3^\circ$ ). This is the definition: the radian is the angle for which arc length equals radius.

Stage 3, arc length for any angle. Open the angle to a general size $\theta$ (in radians). Because $1$ radian spans an arc of $r$ , an angle of $\theta$ radians spans an arc of $\theta$ lots of $r$ :
$\ell = r\theta.$
This is just the definition $\theta = \frac{\ell}{r}$ rearranged, and it only works with $\theta$ in radians.

Stage 4, sector and segment area. Shade the sector. Its area is the fraction $\frac{\theta}{2\pi}$ of the whole circle $\pi r^2$ , which simplifies to $\frac{1}{2}r^2\theta$ . Join the two arc ends with a chord, and the slice between the chord and the arc is the segment, whose area is the sector minus the triangle, $\frac{1}{2}r^2(\theta - \sin\theta)$ .

Conversion

180^\circ = \pi \text{ radians}, \qquad 1^\circ = \frac{\pi}{180} \text{ rad}, \qquad 1 \text{ rad} = \frac{180^\circ}{\pi} \approx 57.30^\circ.

To convert: multiply degrees by $\frac{\pi}{180}$ to get radians; multiply radians by $\frac{180}{\pi}$ to get degrees.

Standard exact values:

Degrees	$0$	$30$	$45$	$60$	$90$	$180$	$270$	$360$
Radians	$0$	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$	$\pi$	$\frac{3 \pi}{2}$	$2 \pi$

Arc length

For a sector of radius $r$ with central angle $\theta$ in radians, the arc length is

\ell = r \theta.

This is the formula behind the definition. Use radians, not degrees.

Sector area

The area of a sector of radius $r$ with central angle $\theta$ in radians is

A_{\text{sector}} = \frac{1}{2} r^2 \theta.

Derivation: the area is the fraction $\frac{\theta}{2 \pi}$ of the full circle area $\pi r^2$ , giving $\frac{\theta}{2 \pi} \cdot \pi r^2 = \frac{1}{2} r^2 \theta$ .

Triangle and segment

The triangle formed by the two radii and the chord has area

A_{\text{triangle}} = \frac{1}{2} r^2 \sin \theta.

The minor segment is the region between the chord and the arc. Its area is

A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} = \frac{1}{2} r^2 (\theta - \sin \theta).

The major segment (the larger region on the other side of the chord) has area $\pi r^2 - A_{\text{segment}}$ .

Key fact

For a circle of radius $r$ and central angle $\theta$ measured in radians:

\text{arc length } \ell = r\theta, \qquad \text{sector area } A = \tfrac{1}{2} r^2 \theta, \qquad \text{segment area} = \tfrac{1}{2} r^2 (\theta - \sin\theta).

The arc length and sector formulas use

\theta

alone; the segment subtracts the triangle

\tfrac{1}{2} r^2 \sin\theta

, so it mixes

\theta

(radians) with

\sin\theta

. Note the shared structure: both

\ell = r\theta

and

A = \tfrac{1}{2} r^2\theta

are the same fraction

\frac{\theta}{2\pi}

of the whole circle's circumference

2\pi r

and area

\pi r^2

respectively.

Chord length

By the cosine rule (or by splitting the isosceles triangle), the chord opposite the central angle $\theta$ has length

c = 2 r \sin\frac{\theta}{2}.

How exam questions ask about radians and circular measure

The wording tells you which formula to reach for:

"A sector subtends an angle of $\frac{\pi}{3}$ at the centre ... find the arc length / perimeter." Arc length is $\ell = r\theta$ . If they ask for the perimeter of the sector, add the two radii: $P = r\theta + 2r$ .
"... find the area of the sector." $A = \frac{1}{2} r^2 \theta$ , with $\theta$ in radians.
"Find the area of the segment / the area cut off by the chord / the shaded region." Sector minus triangle: $\frac{1}{2} r^2 (\theta - \sin\theta)$ . The word segment (or a shaded region between a chord and the arc) is the cue to subtract the triangle.
"A chord subtends an angle of $\theta$ at the centre." The angle named is the central angle; use it directly. The chord length, if needed, is $2r\sin\frac{\theta}{2}$ .
"The angle is $60^\circ$ ..." but the formula needs radians. Convert first: $60^\circ = \frac{\pi}{3}$ . A question can mix the two; the formula always wants radians.
"Find the angle, given the arc length / area." Rearrange: $\theta = \frac{\ell}{r}$ from arc length, or $\theta = \frac{2A}{r^2}$ from sector area.
"Express in radians / in degrees." A straight conversion: $\times \frac{\pi}{180}$ for degrees to radians, $\times \frac{180}{\pi}$ the other way.

Exam technique

The first move on every circular-measure question is to get $\theta$ into radians and write it down; substituting a degree value is the error that loses whole questions. Quote the formula you are using ( $\ell = r\theta$ , $A = \frac{1}{2}r^2\theta$ , segment $= \frac{1}{2}r^2(\theta - \sin\theta)$ ) before substituting, give an exact answer in terms of $\pi$ first and then a decimal if asked, and always attach units ( $\text{cm}$ for length, $\text{cm}^2$ for area). For a segment, set the working out in three labelled lines, sector, then triangle, then subtract, so a marker can follow it and award method marks even if the arithmetic slips. Keep the calculator in radian mode whenever a $\sin$ or $\cos$ of a radian value appears.

Worked example

Five calculations covering angle conversion, arc length, sector area, segment area, and chord length, each requiring $\theta$ to be in radians before substituting into a formula.

Convert between degrees and radians

To convert degrees to radians, multiply by $\frac{\pi}{180}$ ; to go the other way, multiply by $\frac{180}{\pi}$ .

120^\circ = 120 \times \frac{\pi}{180} = \frac{2\pi}{3} \text{ radians.}

\frac{5\pi}{4} \text{ rad} = \frac{5\pi}{4} \times \frac{180}{\pi} = 225^\circ.

Apply the arc length formula

A bicycle wheel of radius $35$ cm rotates through $4.5$ radians. The angle is already in radians, so substitute directly into $\ell = r\theta$ :

\ell = 35 \times 4.5 = 157.5 \text{ cm.}

The formula $\ell = r\theta$ only works with $\theta$ in radians because it comes from the definition of the radian as arc length divided by radius.

Convert the angle then compute sector area

A sector has radius $8$ cm and central angle $45^\circ$ . Convert the angle first, because the sector area formula requires radians:

45^\circ = \frac{\pi}{4} \text{ rad.}

A = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 64 \times \frac{\pi}{4} = 8\pi \approx 25.13 \text{ cm}^2.

Find segment area by subtracting the triangle from the sector

For a circle of radius $5$ cm with central angle $\frac{2\pi}{3}$ , compute sector and triangle areas separately, then subtract.

Sector area:

A_{\text{sector}} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 25 \times \frac{2\pi}{3} = \frac{25\pi}{3} \approx 26.18 \text{ cm}^2.

Triangle area (the isosceles triangle formed by the two radii):

A_{\text{triangle}} = \frac{1}{2} r^2 \sin\theta = \frac{1}{2} \times 25 \times \sin\frac{2\pi}{3} = \frac{25}{2} \times \frac{\sqrt{3}}{2} = \frac{25\sqrt{3}}{4} \approx 10.83 \text{ cm}^2.

Segment area:

A_{\text{segment}} = \frac{25\pi}{3} - \frac{25\sqrt{3}}{4} \approx 26.18 - 10.83 \approx 15.35 \text{ cm}^2.

Compute the chord length

For a circle with $r = 7$ and $\theta = \frac{\pi}{3}$ , the chord formula is $c = 2r\sin\frac{\theta}{2}$ . Halve the central angle first:

c = 2 \times 7 \times \sin\frac{\pi}{6} = 14 \times \frac{1}{2} = 7 \text{ cm.}

This result makes sense geometrically: a central angle of $\frac{\pi}{3}$ with equal radii of $7$ forms an equilateral triangle, so the chord equals the radius.

Final answers: $120^\circ = \frac{2\pi}{3}$ rad, $\frac{5\pi}{4}$ rad $= 225^\circ$ ; arc length $157.5$ cm; sector area $8\pi \approx 25.13$ cm $^2$ ; segment area $\approx 15.35$ cm $^2$ ; chord $= 7$ cm.

Common traps

Using degrees in radian formulas: $\ell = r \theta$ and $A = \frac{1}{2} r^2 \theta$ require $\theta$ in radians. Substituting $90$ instead of $\frac{\pi}{2}$ gives a wildly wrong answer.
Forgetting to subtract the triangle: A segment is not the sector; subtract the triangle.
Wrong formula for the triangle: The triangle area is $\frac{1}{2} r^2 \sin \theta$ (with $\sin \theta$ , not $\theta$ ). Confusing this with the sector formula gives a wrong segment.
Calculator in the wrong mode: Always check that your calculator is in radian mode when working from $\frac{\pi}{6}$ etc.
Confusing the minor and major segment: "Minor" is the smaller piece, between the chord and the shorter arc. Make sure $\theta$ refers to the central angle of that smaller piece (less than $\pi$ ).

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

HSC-style3 marks

A circle has radius

12

cm. A sector subtends an angle of

\frac{\pi}{3}

radians at the centre. Find the arc length and the area of the sector.

Show worked answer →

Arc length: $\ell = r \theta = 12 \cdot \frac{\pi}{3} = 4 \pi \approx 12.57$ cm.

Sector area: $A = \frac{1}{2} r^2 \theta = \frac{1}{2} \cdot 144 \cdot \frac{\pi}{3} = 24 \pi \approx 75.40$ cm $^2$ .

Markers reward the correct formulas, the correct substitution with $\theta$ in radians, and exact then approximate answers with units.

HSC-style3 marks

A chord of a circle of radius

10

cm subtends an angle of

\frac{\pi}{2}

at the centre. Find the area of the minor segment cut off by the chord.

Show worked answer →

Segment area = sector area minus triangle area.

Sector: $\frac{1}{2} r^2 \theta = \frac{1}{2} \cdot 100 \cdot \frac{\pi}{2} = 25 \pi$ .

Triangle: $\frac{1}{2} r^2 \sin \theta = \frac{1}{2} \cdot 100 \cdot \sin\frac{\pi}{2} = 50$ .

Segment: $25 \pi - 50 \approx 78.54 - 50 = 28.54$ cm $^2$ .

Markers expect the segment formula, the substitution into both terms, and a numerical answer with units.

Practice questions

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

foundation2 marks

Convert

210^\circ

to radians, giving your answer as an exact multiple of

\pi

Show worked solution →

Multiply by $\frac{\pi}{180}$ to go from degrees to radians.

210^\circ = 210 \times \frac{\pi}{180} = \frac{210\pi}{180} = \frac{7\pi}{6} \text{ radians.}

Marker's note: one mark for using the factor $\frac{\pi}{180}$ , one for the simplified exact answer $\frac{7\pi}{6}$ . A decimal such as $3.67$ instead of the exact multiple of $\pi$ would not earn the second mark when an exact form is asked for.

foundation2 marks

A sector of a circle has radius

10

cm and a central angle of

1.2

radians. Find the area of the sector.

Show worked solution →

The angle is already in radians, so substitute straight into $A = \frac{1}{2} r^2 \theta$ .

A = \frac{1}{2} \times 10^2 \times 1.2 = \frac{1}{2} \times 100 \times 1.2 = 60 \text{ cm}^2.

Marker's note: one mark for the correct formula with substitution, one for $60$ cm $^2$ with units. There is no conversion to do here, so a student who "converts" $1.2$ as if it were degrees has misread the question.

foundation3 marks

A sector of a circle of radius

6

cm has an arc length of

8.4

cm. (a) Find the central angle in radians. (b) Find the perimeter of the sector.

Show worked solution →

Part (a): rearrange $\ell = r\theta$ to make $\theta$ the subject.

\theta = \frac{\ell}{r} = \frac{8.4}{6} = 1.4 \text{ radians.}

Part (b): the perimeter of a sector is the arc plus the two bounding radii.

P = \ell + 2r = 8.4 + 2 \times 6 = 8.4 + 12 = 20.4 \text{ cm.}

Marker's note: one mark for $\theta = 1.4$ from $\theta = \frac{\ell}{r}$ , one for adding the two radii to the arc, one for $20.4$ cm with units. Quoting only the arc length as the "perimeter" is the common slip.

core3 marks

A chord of a circle of radius

6

cm subtends an angle of

\frac{\pi}{2}

at the centre. Find the exact area of the minor segment cut off by the chord, then give it correct to two decimal places.

Show worked solution →

Segment area is the sector minus the triangle. For radius $r = 6$ and $\theta = \frac{\pi}{2}$ ,

A_{\text{sector}} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 36 \times \frac{\pi}{2} = 9\pi.

Triangle area uses $\frac{1}{2} r^2 \sin\theta$ :

A_{\text{triangle}} = \frac{1}{2} \times 36 \times \sin\frac{\pi}{2} = 18 \times 1 = 18.

Subtract to get the segment.

A_{\text{segment}} = 9\pi - 18 \approx 28.27 - 18 = 10.27 \text{ cm}^2.

Marker's note: one mark for the sector $9\pi$ , one for the triangle $18$ (using $\sin\theta$ , not $\theta$ ), one for the segment $9\pi - 18 \approx 10.27$ cm $^2$ . Forgetting to subtract the triangle, or using $\frac{1}{2} r^2 \theta$ for the triangle, loses the final mark.

core4 marks

A pendulum of length

75

cm swings through an angle of

24^\circ

. (a) Convert the swing angle to radians, correct to three decimal places. (b) Find the length of the arc traced by the pendulum bob, correct to the nearest millimetre. (c) Find the area swept out by the pendulum, correct to the nearest square centimetre.

Show worked solution →

Part (a): convert the angle first, because both formulas need radians.

24^\circ = 24 \times \frac{\pi}{180} = \frac{2\pi}{15} \approx 0.419 \text{ radians.}

Part (b): the bob traces an arc of radius $75$ cm.

\ell = r\theta = 75 \times \frac{2\pi}{15} = 10\pi \approx 31.4 \text{ cm} = 314 \text{ mm.}

Part (c): the area swept is the sector of radius $75$ cm.

A = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 75^2 \times \frac{2\pi}{15} = \frac{1}{2} \times 5625 \times \frac{2\pi}{15} = 375\pi \approx 1178 \text{ cm}^2.

Marker's note: one mark for $24^\circ = \frac{2\pi}{15} \approx 0.419$ rad, one for the arc $10\pi \approx 314$ mm, two for the swept area $375\pi \approx 1178$ cm $^2$ (one for the correct substitution, one for the rounded value with units). Using $24$ in place of the radian value is the error that sinks all three parts.

core4 marks

A wooden fan opens out into a single sector, shown in a diagram as sector

OAB

with centre

O

, radius

24

cm and the angle

AOB

marked as

150^\circ

. A ribbon is sewn along the curved edge

AB

, and the fabric fills the whole sector. (a) Find the exact length of ribbon along the arc

AB

. (b) Find the exact area of fabric in the sector. (c) The maker has only

65

cm of ribbon; state whether it is enough, justifying with a calculation.

Show worked solution →

Read the diagram, then convert the angle to radians.

150^\circ = 150 \times \frac{\pi}{180} = \frac{5\pi}{6} \text{ radians.}

Part (a): ribbon length is the arc $AB$ , using $\ell = r\theta$ .

\ell = 24 \times \frac{5\pi}{6} = \frac{120\pi}{6} = 20\pi \text{ cm.}

Part (b): fabric area is the sector, using $A = \frac{1}{2} r^2 \theta$ .

A = \frac{1}{2} \times 24^2 \times \frac{5\pi}{6} = \frac{1}{2} \times 576 \times \frac{5\pi}{6} = 240\pi \text{ cm}^2.

Part (c): compare $20\pi$ with $65$ cm.

20\pi \approx 62.83 \text{ cm} < 65 \text{ cm,}

so the $65$ cm of ribbon is enough (with about $2.2$ cm to spare).

Marker's note: one mark for converting $150^\circ = \frac{5\pi}{6}$ , one for the arc $20\pi$ cm, one for the area $240\pi$ cm $^2$ , one for the decision in (c) backed by $20\pi \approx 62.83 < 65$ . Comparing $65$ with the area $240\pi$ instead of the arc length is the misread to avoid.

exam5 marks

A sector

OAB

is cut from sheet metal. Its perimeter (the arc

AB

plus the two radii

OA

and

OB

) must be exactly

40

cm. Let the radius be

r

cm and the central angle be

\theta

radians. (a) Show that the area of the sector is

A = 20r - r^2

. (b) Find the radius that maximises the area, justifying that it is a maximum. (c) Hence find this maximum area and the corresponding central angle.

Show worked solution →

Part (a): use the fixed perimeter to remove $\theta$ . The perimeter of the sector is the arc plus the two radii:

r\theta + 2r = 40 \quad\Rightarrow\quad r\theta = 40 - 2r \quad\Rightarrow\quad \theta = \frac{40 - 2r}{r}.

Substitute into the sector-area formula, noticing that $\frac{1}{2} r^2 \theta = \frac{1}{2} r \cdot (r\theta)$ :

A = \frac{1}{2} r^2 \theta = \frac{1}{2} r (r\theta) = \frac{1}{2} r (40 - 2r) = 20r - r^2.

Part (b): differentiate and solve $\frac{dA}{dr} = 0$ .

\frac{dA}{dr} = 20 - 2r = 0 \quad\Rightarrow\quad r = 10 \text{ cm.}

Justify the maximum with the second derivative:

\frac{d^2A}{dr^2} = -2 < 0,

so the curve is concave down and $r = 10$ gives the maximum area.

Part (c): evaluate the area and the angle.

A = 20(10) - 10^2 = 200 - 100 = 100 \text{ cm}^2.

\theta = \frac{40 - 2(10)}{10} = \frac{20}{10} = 2 \text{ radians.}

Marker's note: one mark for the perimeter equation $r\theta + 2r = 40$ and making $r\theta$ the subject, one for reaching the shown form $A = 20r - r^2$ , one for solving $\frac{dA}{dr} = 0$ to $r = 10$ , one for the justification via $\frac{d^2A}{dr^2} < 0$ , one for the final $A = 100$ cm $^2$ and $\theta = 2$ radians. Writing the perimeter as $r\theta$ alone (dropping the two radii) is the error that breaks part (a).

exam5 marks

Two circles have the same centre

O

. The inner circle has radius

8

cm and the outer circle has radius

14

cm. A single central angle of

\frac{\pi}{3}

radians cuts an annular region (the region between the two arcs, bounded by two radii of the outer circle). (a) Find the exact area of this annular region. (b) Find the exact total perimeter of the region, which consists of the inner arc, the outer arc, and the two straight segments joining them. (c) Give both the area and the perimeter correct to two decimal places.

Show worked solution →

Part (a): the annular region is the big sector minus the small sector, both with $\theta = \frac{\pi}{3}$ .

A_{\text{outer}} = \frac{1}{2} \times 14^2 \times \frac{\pi}{3} = \frac{1}{2} \times 196 \times \frac{\pi}{3} = \frac{98\pi}{3}.

A_{\text{inner}} = \frac{1}{2} \times 8^2 \times \frac{\pi}{3} = \frac{1}{2} \times 64 \times \frac{\pi}{3} = \frac{32\pi}{3}.

A_{\text{annular}} = \frac{98\pi}{3} - \frac{32\pi}{3} = \frac{66\pi}{3} = 22\pi \text{ cm}^2.

Part (b): add the two arcs and the two straight joining segments. Each straight segment runs from the inner circle to the outer circle along a radius, so its length is $14 - 8 = 6$ cm.

\ell_{\text{outer}} = 14 \times \frac{\pi}{3} = \frac{14\pi}{3}, \qquad \ell_{\text{inner}} = 8 \times \frac{\pi}{3} = \frac{8\pi}{3}.

P = \frac{14\pi}{3} + \frac{8\pi}{3} + 6 + 6 = \frac{22\pi}{3} + 12 \text{ cm.}

Part (c): convert both to decimals.

A = 22\pi \approx 69.12 \text{ cm}^2, \qquad P = \frac{22\pi}{3} + 12 \approx 23.04 + 12 = 35.04 \text{ cm.}

Marker's note: one mark for each sector area and the subtraction giving $22\pi$ in (a); one for the two arc lengths and one for adding the two straight segments of length $6$ to get $\frac{22\pi}{3} + 12$ in (b); one for the two rounded values in (c). The trap in (b) is forgetting the two straight edges, or using $14 + 8$ instead of $14 - 8$ for their length.

Practise this

Sources & how we know this

Mathematics Advanced Stage 6 Syllabus (2017) — NESA
Mathematics Advanced HSC exam papers — NESA