Arc, Sector & Radian Measure

The question	Where it is answered	The tool
How long is the curved edge of the slice?	10.1 Arc length	arc = (`x`⁄360) × 2`πr`
How far is it all the way round the slice?	10.1 Perimeter of a sector	arc + `r` + `r` — the two straight edges count
How much area is in the slice?	10.2 Area of a sector	sector = (`x`⁄360) × `πr`²
How much is left if you cut straight across, along a chord?	10.2 Area of a segment	sector − triangle, the classic two-formula question
What is 2.35 rad in degrees? What is 252° in radians?	10.3 Radian measure	`π` rad = 180°
The same arc and area questions, but the angle is given in radians	10.4 Arc & sector in radians	`s` = `rθ` and `A` = ½`r`²`θ`

The five words — chord, arc, segment, sector, and “minor” ⓘ Recap

Every question in this chapter is about a piece of a circle, so it starts with naming the piece. Two of these five words look alike and mean different things.

A chord is a straight line joining two points on the circle. An arc is part of the circle itself — part of the curve. A chord cuts the disc into two segments; two radii cut it into two sectors. A segment is bounded by a chord and an arc; a sector is bounded by two radii and an arc.

Fig. 10.1

The same circle and the same chord PQ three times. (a) The chord splits the circle into a shorter minor arc PRQ and a longer major arc PSQ. (b) It splits the disc into a minor segment and a major segment. (c) Draw the radii OP and OQ instead and you get a minor sector and a major sector. The word minor always means the smaller of the pair.

Three letters for an arc, four for a sector. “Arc PQ” names two different arcs, so a middle letter picks one: arc PRQ runs through R, arc PSQ through S. A sector adds the centre: sector OPRQ.

More detail

That middle letter is worth reading before any calculation, because it tells you which side of the chord the question means — and so whether the angle at the centre is the one marked on the diagram or its reflex. The same applies to a four-letter sector: OPRQ and OPSQ are the two halves of the disc, and only one of them is the one asked for.

Arc length — a fraction of the way round

The angle at the centre of a full circle is 360°. A sector of x° is therefore x⁄360 of the whole circle, and its arc is that same fraction of the whole circumference.

For an arc subtending an angle of x° at the centre of a circle of radius r,

arc length = \frac{x °}{360 °} \times circumference = \frac{x °}{360 °} \times 2 π r

Fig. 10.2

A quarter of the turn cuts a quarter of the circumference; a third of the turn cuts a third of it. The arc length is directly proportional to the angle at the centre, which is exactly why one fraction does all the work.

The formula runs in all three directions. It holds the arc, the angle and the radius, so any one of them can be the unknown: multiply (Walkthrough 1), or solve for r (Walkthrough 3) or for x (Walkthrough 4).

Write the formula, then substitute. That first line carries a mark of its own, whichever of the three quantities the question has hidden.

Where this comes from

Nothing more sophisticated is going on than one proportion, applied to a length. Arc length is directly proportional to the angle at the centre: double the angle and you double the arc, because you walk twice as far round the same circle. That is why the single fraction x⁄360 does all the work here — and why the same fraction will do it again for area in 10.2.

The perimeter of a sector — the arc plus two radii

A sector is a closed shape with three edges, and only one of them is curved. Walk round the boundary and you travel out along a radius, round the arc, and back along the other radius.

Fig. 10.3

Perimeter of sector OAB = arc AB + OA + OB = arc + 2r. The blue edges are as much a part of the boundary as the red one.

Perimeter is not arc. Perimeter is the distance all the way round the region named, so a sector's is arc + 2r. Backwards: given the perimeter and the radius, start with arc = perimeter − 2r (Walkthrough 4).

A region that is not a sector works the same way. For a shape bounded by an arc and two straight cuts, list the edges first, then add them (Walkthrough 5).

Walkthrough 1 — arc length, and the perimeter of a major sector Basic

In the diagram, O is the centre of a circle of radius 15 cm and ∠AOB = 108°. Find (i) the length of the minor arc AXB, (ii) the perimeter of the major sector OAYB, giving each answer correct to 3 significant figures.

Length of minor arc AXB = $\frac{x °}{360 °} \times 2 π r$

The minor arc is the one through X, and it is the arc the 108° angle cuts off. Quoting the formula before any numbers go in is the first line of working, and it carries a mark of its own.
= $\frac{108}{360} \times 2 π \times 15$

Substitute x = 108 and r = 15. The 15 is the radius, not the diameter — if a question ever gives you a diameter, halve it here and nowhere else.
= $\frac{3}{10} \times 30 π$ = 9π cm

108⁄360 cancels to 3⁄10, and the circumference 2π × 15 is 30π. Holding the answer as 9π keeps it exact: nothing has been rounded yet, so nothing can drift.
Reflex ∠AOB = 360° − 108° = 252° (∠s at a point)

The major sector OAYB is the one through Y, on the other side, so its arc is subtended by the reflex angle. 108° is the wrong angle for part (ii); the whole turn minus 108° is the right one.
Length of major arc AYB = $\frac{252}{360} \times 2 π \times 15$ = $\frac{7}{10} \times 30 π$ = 21π cm

Same formula, same radius, new angle. Notice the check that comes free: 9π + 21π = 30π, which is the whole circumference — the two arcs make it up between them.
Perimeter of major sector OAYB = arc AYB + OA + OB = 21π + 15 + 15 = (21π + 30) cm

A sector is closed by its two radii as well as its arc, so both radii, 15 cm each, go in. Leaving the line as 21π + 30 means only one rounding happens, and it happens next.
(i) 9π = 28.3 cm (3 s.f.) (ii) 21π + 30 = 96.0 cm (3 s.f.)

Round once, at the end, to the accuracy the question asked for. Keep the 0 in 96.0: 96 on its own is only two significant figures and is not the answer that was asked for.

Walkthrough 2 — the perimeter of a sector Basic

The diagram shows a sector OAB of a circle with centre O and radius 9 cm, in which ∠AOB = 150°. Find the perimeter of the sector, correct to 3 significant figures.

Arc AB = $\frac{x °}{360 °} \times 2 π r$ = $\frac{150}{360} \times 2 π \times 9$

The curved edge comes first, because it is the only part of the boundary that needs a formula. 150° is less than 180°, so this is the minor arc — the one drawn.
= $\frac{5}{12} \times 18 π$ = 7.5π cm

150⁄360 cancels to 5⁄12 and the circumference is 18π. Keeping it as 7.5π rather than 23.56… means the rounding is still ahead of you, not behind you.
Perimeter of sector OAB = arc AB + OA + OB

This line is the whole question. A sector has three edges: the arc and two radii. Writing the three names down before any numbers go in is what stops the answer being just the arc.
= 7.5π + 9 + 9 = (7.5π + 18) cm

Both radii are 9 cm, because every radius of a circle is. The answer is exact at this point: 7.5π + 18.
Perimeter = 23.561… + 18 = 41.6 cm (3 s.f.)

One press of the calculator, at the end. Rounding 7.5π to 23.6 first happens to give 41.6 here too — the habit is what protects the questions where it does not.

Walkthrough 3 — running the formula backwards to find the radius Intermediate

The minor arc AB of a circle is 37 cm long and subtends an angle of 132° at the centre. Find the radius of the circle, correct to 3 significant figures.

Arc AB = $\frac{x °}{360 °} \times 2 π r$ , so $37 = \frac{132}{360} \times 2 π r$

Exactly the same formula as before — only the unknown has moved. Put in everything the question gives you and let r stay a letter; there is no separate “find the radius” formula to remember.
$37 = \frac{11}{30} \times 2 π r = \frac{11 π}{15} r$

132⁄360 cancels to 11⁄30, and 11⁄30 × 2 is 11⁄15. Tidying the coefficient of r into one fraction before dividing keeps the next line short and keeps π exact.
$r = \frac{37 \times 15}{11 π}$

Make r the subject: multiply both sides by 15 and divide by 11π. Everything is still exact, so the only rounding left is the one the question asked for.
r = 16.060… = 16.1 cm (3 s.f.)

A radius is a length, so write the unit: cm. Sanity check: a radius near 16 cm gives a circumference of about 101 cm, and 132° is a bit over a third of a turn — about 37 cm. It fits.

Walkthrough 4 — a wire bent into a sector Intermediate

A piece of wire 38 cm long is bent to form a sector of a circle of radius 8 cm. Find the angle subtended by the wire at the centre of the circle, correct to the nearest degree.

Perimeter of the sector = arc + r + r, so arc = 38 − 8 − 8 = 22 cm

The wire is bent into the whole boundary, not just the curved bit — two straight 8 cm pieces of it are the radii. Taking them off first leaves 22 cm of wire as the arc.
$22 = \frac{x °}{360 °} \times 2 π \times 8$

Now the arc is a known 22 cm and the angle is the unknown. The radius is still 8, because bending the wire did not change it.
$x = \frac{22 \times 360}{16 π}$

Multiply both sides by 360 and divide by 2π × 8 = 16π. Doing it in one move, with the numbers still exact, avoids two roundings.
x = 157.56…, so the angle is 158° (nearest degree)

The question named its own accuracy — nearest degree — so that beats the paper's default of 1 decimal place. And 158° is a sensible answer: the arc, 22 cm, is a little under half the circumference 16π ≈ 50 cm.

Walkthrough 5 — the perimeter of a region that is not a sector Advanced

In the figure, O is the centre of a circle of radius 10 cm. The points A and C lie on the circle, OA is perpendicular to AB and OCB is a straight line. Given that ∠AOB = 64°, find the perimeter of the shaded region ABC, correct to 3 significant figures.

Fig. 10.4

Perimeter of ABC = arc AC + AB + BC

Before any calculation, walk round the shaded region and name its three edges: one arc and two straight pieces. Each of the three now becomes its own small problem, and none of them can be forgotten.
Arc AC = $\frac{64}{360} \times 2 π \times 10$ = 11.170 cm (5 s.f.)

The 64° angle at O is subtended by arc AC, because C is where OB crosses the circle. Carrying 5 significant figures through the middle of the working is what keeps the third figure of the final answer correct.
In △OAB, ∠OAB = 90°, so tan 64° = $\frac{A B}{10}$ and AB = 10 tan 64° = 20.503 cm (5 s.f.)

“OA is perpendicular to AB” is there to hand you a right-angled triangle. Relative to the 64° angle, AB is opposite and OA is adjacent — that pair is tangent.
cos 64° = $\frac{10}{O B}$ , so OB = $\frac{10}{cos 64 °}$ = 22.812 cm (5 s.f.)

In the same triangle, OB is the hypotenuse and OA is adjacent — that pair is cosine. Note that OB is not one of the three edges; it is a stepping stone to BC.
BC = OB − OC = 22.812 − 10 = 12.812 cm

OCB being a straight line is what lets you subtract, and C being on the circle is what makes OC a radius, so OC = 10. Two given facts, both used in one line.
Perimeter = 11.170 + 20.503 + 12.812 = 44.5 cm (3 s.f.)

Add the three edges named in step 1 and round once, here. An answer of about 44 cm is believable: the region is roughly a triangle with a 20 cm side and a 13 cm side.

More detail

“Find the perimeter” and “find the arc” ask for different things, and on a sector the difference is the two straight edges — which is where more of this chapter’s marks sit than anywhere else. One line of writing settles it — name the edges before adding them — and the same line goes on working for a shape with four edges, or with one curved edge and two straight cuts, as Walkthroughs 4 and 5 show.

Check yourself — 10.1

Find the length of the minor arc of a circle of radius 21 cm that subtends 80° at the centre, leaving your answer in terms of π and also correct to 3 significant figures.
Answer
Arc = $\frac{80}{360} \times 2 π \times 21$ = $\frac{2}{9} \times 42 π$ = $\frac{28 π}{3}$ = 28π⁄3 cm = 29.3215… = 29.3 cm (3 s.f.).
A sector of a circle of radius 6 cm has an angle of 210° at the centre. Find its perimeter, correct to 3 significant figures.
Answer
Arc = $\frac{210}{360} \times 2 π \times 6$ = 7π = 21.991… cm.
Perimeter = 7π + 6 + 6 = 33.991… = 34.0 cm (3 s.f.). The angle is reflex, so this is a major sector — but the formula does not care, and the two radii still have to be added.
The arc of a sector of radius 14 cm is 33 cm long. Find the angle at the centre, correct to the nearest degree.
Answer
$33 = \frac{x °}{360 °} \times 2 π \times 14$ , so x = $\frac{33 \times 360}{28 π}$ = 135.05… = 135° (nearest degree).
A sector has a perimeter of 50 cm and a radius of 11 cm. Find the angle at the centre, correct to the nearest degree.
Answer
Arc = 50 − 11 − 11 = 28 cm.
x = $\frac{28 \times 360}{22 π}$ = 145.84… = 146° (nearest degree). Taking the two radii off first is the whole question.
A semicircular sector has radius r. A student says its perimeter is πr. What has gone wrong, and what is the perimeter?
Answer
πr is the length of the arc alone — half of the circumference 2πr. A semicircular sector is closed by its diameter, which is two radii laid end to end, so the perimeter is πr + 2r. Arc is not perimeter, even when the two straight edges happen to line up.

Area of a sector — the same fraction, applied to area

Everything said about the arc in 10.1 is true again here, with one word changed. A sector whose angle is x° is x⁄360 of the whole circle, so its area is that fraction of the whole circle's area — just as its arc was that fraction of the circumference.

For a sector subtending an angle of x° at the centre of a circle of radius r,

area of sector = \frac{x °}{360 °} \times area of circle = \frac{x °}{360 °} \times π r^{2}

Two routes to a major sector. Use the reflex angle, or work out the minor sector and take it off the whole circle. Both are one line; the second gives a check for free, since the two sectors add to πr².

An area answer takes square units — cm², m² — and a mark goes for writing them.

Walkthrough 6 — minor and major sector, exactly Basic

In the diagram, O is the centre of a circle of radius 12 cm and ∠AOB = 105°. Find the area of (i) the minor sector, (ii) the major sector, leaving your answers in terms of π.

Fig. 10.5

Area of minor sector = $\frac{x °}{360 °} \times π r^{2}$ = $\frac{105}{360} \times π \times 12^{2}$

The minor sector is the one the 105° angle opens onto, shaded in the diagram. Quote the formula, then substitute x = 105 and r = 12 — and square the radius, not the whole product.
= $\frac{7}{24} \times 144 π$ = 42π cm²

105⁄360 cancels to 7⁄24, and π × 12² is 144π. “In terms of π” means stop here — this line is the answer, and pressing the calculator now would lose the exactness the question asked for.
Reflex ∠AOB = 360° − 105° = 255° (∠s at a point)

The major sector is everything else, so its angle is the reflex one. The reason “∠s at a point” beside it is three words, and it is the line the marker looks for.
Area of major sector = $\frac{255}{360} \times π \times 12^{2}$ = $\frac{17}{24} \times 144 π$ = 102π cm²

Same formula, same radius, the reflex angle. The two fractions 7⁄24 and 17⁄24 add to 1, which is the arithmetic version of “the two sectors make the whole circle”.
(i) 42π cm² (ii) 102π cm²

Check: 42π + 102π = 144π, and the whole circle is π × 12² = 144π cm². Both answers are exact, so neither is rounded — and both carry cm², because they are areas.

Walkthrough 7 — from the area back to the radius Intermediate

A sector of a circle has an area of 132 cm² and an angle of 100° at the centre. Find (i) the radius, (ii) the perimeter of the sector, each correct to 3 significant figures.

$\frac{100}{360} \times π r^{2} = 132$

The area formula again, with the area known and the radius unknown. Everything the question gives goes in at once; nothing needs rearranging until the numbers are all in place.
$r^{2} = \frac{132 \times 360}{100 π}$ = 151.26 (5 s.f.)

Multiply by 360 and divide by 100π in one move. This is r², not r — a very easy line to stop one step too early on.
r = $\sqrt{151.26}$ = 12.299 cm (5 s.f.)

Take the square root. Only the positive root is kept: a radius is a length, and the negative root has no meaning here.
Arc = $\frac{100}{360} \times 2 π \times 12.299$ = 21.466 cm (5 s.f.)

The perimeter needs the arc, and the arc needs the radius just found. Use the 5-figure 12.299, not the 12.3 that is about to be written as the answer: a rounded number fed back carries its error forward.
Perimeter = arc + r + r = 21.466 + 12.299 + 12.299 = 46.064 cm

Three edges again: the arc and both radii. This is where a “find the perimeter” part quietly reuses the answer to part (i).
(i) r = 12.3 cm (ii) perimeter = 46.1 cm (3 s.f.)

Both answers are rounded only now, and both to 3 significant figures because that is what was asked. A length gets cm, not cm² — only the 132 in the question was an area.

Why this works

Cut the disc at the centre into 360 equal wedges and a sector of x° is exactly x of them. So it holds x⁄360 of the circle's area for the same reason its arc holds x⁄360 of the circumference: both are shares of the same 360 wedges. Nothing about area needs a new idea — only the quantity being shared out changes, from the circumference 2πr to the area πr².

Past-paper questions

G3 N2018 · Paper 2 · Q7(b)(ii) 3 marks

AB is a diameter of the large circle, centre O.
CD is a diameter of the small circle, centre O.
AC and BD are tangents to the small circle.
The radius of the large circle is 7 cm and angle OAC = 30°.

Calculate the shaded area. [3]

Worked solution

A tangent is perpendicular to the radius at the point of contact, so angle OCA = 90°. In right-angled triangle OAC the hypotenuse is OA = 7 cm, so

$O C = 7 \sin 30° = 3.5$ cm and $A C = 7 \cos 30° = 6.0621…$ cm, giving angle AOC = 180° − 90° − 30° = 60° (∠ sum of △).

Each unshaded bite out of the ring is triangle OAC with the 60° sector of the small circle removed:

$\frac{1}{2} \times 3.5 \times 6.0621… - \frac{60 °}{360 °} \times π \times {3.5}^{2} = 10.6088… - 6.4140… = 4.1947…$ cm²

Area of the ring $= π (7^{2} - {3.5}^{2}) = 36.75 π = 115.4535…$ cm²

Shaded area $= 115.4535… - 2 \times 4.1947… = 107.06…$ = 107 cm² (3 s.f.).

Why this works. Three ideas: the tangent fixes the small radius at 3.5 cm, the angle sum of triangle OAC gives 60°, and the ring is a difference of two circles. Shaded = ring less two bites at C and D, each a triangle minus a 60° sector. Square first: $π (7^{2} - {3.5}^{2})$ , not $π \times {(7 - 3.5)}^{2}$ .

G3 N2024 · Paper 1 · Q9 2 marks

In the diagram, the shaded area represents the area cleaned by a windscreen wiper.
All lengths are in centimetres.

Calculate the shaded area. [2]

Worked solution

The blade sweeps out a 96° sector of radius 90 cm and leaves behind a 96° sector of radius 24 cm, so the cleaned region is the difference of two sectors that share a centre and an angle:

$shaded = \frac{96 °}{360 °} \times π \times 90^{2} - \frac{96 °}{360 °} \times π \times 24^{2}$

$= \frac{96 °}{360 °} \times π \times (8100 - 576) = 2006.4 π = 6303.29…$ = 6300 cm² (3 s.f.).

Why this works. Two marks, one idea: the blade misses the pivot, so the swept region is a sector with a smaller sector removed. Same angle in both, so $\frac{96 °}{360 °} π$ factors out and only $90^{2} - 24^{2}$ is left — one keying sequence. Not $\frac{96 °}{360 °} π \times {(90 - 24)}^{2}$ , a 96° sector of radius 66.

Area of a segment — the two-formula question

A chord cuts a sector into two pieces: a triangle, whose two equal sides are radii, and the segment that is left over between the chord and the arc. There is no formula for a segment on any formula sheet, and there does not need to be — it is a subtraction.

area of minor segment = area of minor sector - area of triangle

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

\frac{1}{2} a b sin C = \frac{1}{2} r^{2} sin θ

Fig. 10.6

Sector OAB minus triangle OAB leaves the minor segment. The triangle's two equal sides are both radii, so its area is ½r² sin θ — the standard ½ab sin C with a = b = r.

Minor segment = sector − triangle. The triangle sits inside the minor sector, so taking it away leaves the piece between the chord and the arc.

Major segment = major sector + triangle. Here the triangle lies outside the major sector — the two radii separate them — but on the same side of the chord as the major arc, so it is added.

In ½ab sin C, the angle C sits between the sides a and b — here the angle at the centre, between the two radii.

Walkthrough 8 — the area of a minor segment Intermediate

The diagram shows a sector OAB of a circle with centre O and radius 20 cm. M is the midpoint of the chord AB and OM = 16 cm. (i) Show that ∠AOB = 73.7°, correct to 1 decimal place. (ii) Find the area of the shaded minor segment, correct to 3 significant figures.

OM ⊥ AB, so in △OMA, cos ∠AOM = $\frac{O M}{O A} = \frac{16}{20}$

The line from the centre to the midpoint of a chord is perpendicular to it, so OMA is right-angled at M. There, OM is adjacent to the angle at O and the radius OA is the hypotenuse — that pair is cosine.
∠AOM = cos⁻¹(0.8) = 36.8699° (4 d.p.)

16⁄20 = 0.8 exactly, and the inverse cosine turns it back into an angle. Keep four decimal places: this angle is about to be doubled and then used twice more.
∠AOB = 2 × 36.8699° = 73.7398° = 73.7° (1 d.p.) (shown)

By symmetry OM bisects the angle at O as well as the chord, so the full angle is twice ∠AOM. In a “show that” part, finish by writing the printed value and adding (shown).
Area of sector OAB = $\frac{73.7398}{360} \times π \times 20^{2}$ = 257.40 cm² (5 s.f.)

Use the unrounded 73.7398°, not the 73.7° you just wrote down. The printed value is what the question wanted to see; the accurate value is what the next lines need.
Area of △OAB = $\frac{1}{2} \times 20 \times 20 \times sin 73.7398 °$ = 192.00 cm² (5 s.f.)

½ab sin C with both sides equal to the radius 20 and C the angle between them. The calculator is in degree mode here — 73.7398 is degrees, not radians.
Area of minor segment = area of sector − area of triangle = 257.40 − 192.00

The chord splits the sector into exactly two pieces, the triangle and the segment, so taking one from the whole leaves the other. Write the words before the numbers — that is the line that shows the method.
= 65.4 cm² (3 s.f.)

Round once, here, and write cm². This is exactly the question where early rounding bites: feeding 73.7° into the sector formula instead of 73.7398° gives 65.3, and the third significant figure — the one being asked for — is wrong.

Walkthrough 9 — a region between two sectors Advanced

The figure shows two sectors with the same centre O and the same angle of 120°, with OAD and OBC straight lines. Given that OA = OB = 7 cm and OD = OC = 16 cm, find (i) the area of the shaded region ABCD, leaving your answer in terms of π; (ii) its perimeter, correct to 3 significant figures.

Fig. 10.7

Area of ABCD = area of sector OCD − area of sector OAB

The shaded region is the big slice with the small slice punched out of it. Naming that before calculating anything turns an unfamiliar shape into two sectors you already know how to handle.
= $\frac{120}{360} \times π \times 16^{2} - \frac{120}{360} \times π \times 7^{2}$

Both sectors have the same angle — that is what “OAD and OBC are straight lines” guarantees — so the only difference between the two terms is the radius.
= $\frac{1}{3} π (256 - 49)$ = $\frac{1}{3} \times 207 π$ = 69π cm²

120⁄360 = 1⁄3, so it factors out and the two radii collapse into one difference of squares. “In terms of π” means this line is the answer to (i).
Arc DC = $\frac{120}{360} \times 2 π \times 16$ = $\frac{32 π}{3}$ = 33.510 cm (5 s.f.)

The outer edge of the region is an arc of the large circle, so it uses r = 16. Part (ii) needs a decimal in the end, so five figures are carried from here.
Arc AB = $\frac{120}{360} \times 2 π \times 7$ = $\frac{14 π}{3}$ = 14.661 cm (5 s.f.)

The inner edge is an arc of the small circle, radius 7. It is part of the boundary too — the region is a ring segment, not a sector, and it is curved on both sides.
AD = BC = 16 − 7 = 9 cm

The two straight edges run along the radii, from the small circle out to the large one, so each is the difference of the two radii.
Perimeter = arc DC + arc AB + AD + BC = 33.510 + 14.661 + 9 + 9

Four edges, not three: both arcs and both straight pieces. Listing them by name before adding is what stops one of them going missing.
(i) 69π cm² (ii) 66.2 cm (3 s.f.)

Part (i) was asked for exactly, so it is not rounded; part (ii) was asked to 3 significant figures, so it is rounded once, here. Units: cm² for the area, cm for the perimeter.

More detail

½ab sin C is one of the formulas printed on the paper, so the triangle half of a segment question is handed to you; the degree sector formula and the subtraction itself are not. A segment has no formula of its own because it never needed one — two shapes you already know, one taken from the other.

Sketch the chord and shade the piece the question wants before writing a minus sign. If a minor segment comes out negative, the two terms are the wrong way round: the sector is always the larger of the two, because the triangle sits inside it.

Check yourself — 10.2

A sector of a circle of radius 9 cm has an angle of 40° at the centre. Find its area, exactly and correct to 3 significant figures.
Answer
Area = $\frac{40}{360} \times π \times 9^{2}$ = $\frac{1}{9} \times 81 π$ = 9π cm² = 28.274… = 28.3 cm² (3 s.f.).
A minor sector has an area of 55 cm² and a radius of 8 cm. Find its angle at the centre, correct to the nearest degree.
Answer
$\frac{x °}{360 °} \times π \times 8^{2} = 55$ , so x = $\frac{55 \times 360}{64 π}$ = 98.47… = 98° (nearest degree). Under 180°, so “minor” is consistent.
A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the minor segment it cuts off, correct to 3 significant figures.
Answer
Sector = $\frac{90}{360} \times π \times 10^{2}$ = 25π = 78.5398 cm².
Triangle = ½ × 10 × 10 × sin 90° = 50 cm².
Segment = 78.5398 − 50 = 28.539… = 28.5 cm² (3 s.f.).
For the same circle and chord, find the area of the major segment, correct to 3 significant figures, and check your two answers against each other.
Answer
Major sector = $\frac{270}{360} \times 100 π$ = 75π = 235.619 cm². The triangle is on the same side of the chord as the major arc's centre, so here it is added: 235.619 + 50 = 285.619… = 286 cm² (3 s.f.).
Check: 28.539 + 285.619 = 314.16 = π × 10², the whole circle.
A student writes “area of segment = area of triangle − area of sector” and gets −28.5 cm² for question 3. What is the error, and how could the minus sign have warned them?
Answer
The subtraction is the wrong way round. For a minor segment the triangle lies inside the sector, so the sector is always the larger of the two and the correct order is sector − triangle. A negative area is impossible, so the minus sign in the answer is itself the signal to swap the two terms.

Past-paper questions

G3 N2016 · Paper 1 · Q24 5 marks

ABCD is a square, centre O.
BC = 2r.
E, F, G and H are the midpoints of the sides of the square.
EFG is a semi-circular arc, centre O.
The other arc EG has centre H.
HG = $\sqrt{2}$ r.
What fraction of the square ABCD is not shaded? [5]

Worked solution

Area of the square $= {(2 r)}^{2} = 4 r^{2}$ .

EFG is a semicircle of radius OG = r, so its area is $\frac{1}{2} π r^{2}$ .

Angle GHE = 90°, so GHE is a quadrant of a circle of radius $\sqrt{2} r$ :

$sector G H E = \frac{1}{4} \times π \times {(\sqrt{2} r)}^{2} = \frac{1}{2} π r^{2}$ , $triangle G H E = \frac{1}{2} \times \sqrt{2} r \times \sqrt{2} r = r^{2}$

The piece of the semicircle lying between the chord GE and the arc drawn from H is that sector less that triangle: $\frac{1}{2} π r^{2} - r^{2}$ .

Shaded area $= \frac{1}{2} π r^{2} - (\frac{1}{2} π r^{2} - r^{2}) = r^{2}$

Unshaded area $= 4 r^{2} - r^{2} = 3 r^{2}$ , so the fraction not shaded is $\frac{3 r^{2}}{4 r^{2}}$ = $\frac{3}{4}$ .

Why this works. The algebra stays exact: no radius is given. Quadrant H has radius $\sqrt{2} r$ , area $\frac{1}{4} π \times 2 r^{2} = \frac{1}{2} π r^{2}$ — the semicircle's. Every π cancels; the crescent, semicircle less sector-minus-triangle, is $r^{2}$ . Square $\sqrt{2} r$ to $2 r^{2}$ , not $\sqrt{2} r^{2}$ , and answer the fraction not shaded, not $\frac{1}{4}$ .

G3 N2017 · Paper 2 · Q8 11 marks

The diagram shows a semicircle, centre O, radius 30 cm.
M is the midpoint of the chord AB.
OM = 20 cm.

Show that angle AOB = 96.4°, correct to 3 significant figures. [2]
Calculate the shaded area. [4]
The semicircle is the cross-section of a water trough of length 1.5 m, standing on level ground.
The shaded area represents the water in the trough.
1. Calculate the volume of water, in cm³, in the trough. [2]
2. Calculate the number of litres of water that must be added to fill the trough. Give your answer correct to the nearest 10 litres. [3]

Worked solution

(a) M is the midpoint of the chord AB, and the line from the centre to the midpoint of a chord is perpendicular to it, so angle OMA = 90°. In right-angled triangle OMA the hypotenuse is OA = 30 cm and OM = 20 cm is adjacent to angle AOM:

$\cos ∠ A O M = \frac{20}{30} \Rightarrow ∠ A O M = 48.1896…°$

$∠ A O B = 2 \times 48.1896…° = 96.3793…°$ = 96.4° (3 s.f.) (shown)

(b) The shaded region is the segment cut off by the chord AB — sector AOB less triangle AOB. Keep the angle unrounded:

$sector = \frac{96.3793… °}{360 °} \times π \times 30^{2} = 756.96…$ cm²

$triangle = \frac{1}{2} \times 30 \times 30 \times \sin 96.3793…° = 447.21…$ cm²

$shaded = 756.96… - 447.21… = 309.748…$ = 310 cm² (3 s.f.).

(c)(i) The trough is a prism whose cross-section is the shaded region, and 1.5 m = 150 cm:

$volume = 309.748… \times 150 = 46 462.2…$ = 46 500 cm³ (3 s.f.).

(c)(ii) A full trough is a half-cylinder of radius 30 cm and length 150 cm:

$capacity = \frac{1}{2} \times π \times 30^{2} \times 150 = 212 057.5…$ cm³

$water to add = 212 057.5… - 46 462.2… = 165 595.2…$ cm³ $= 165.595…$ litres

= 170 litres (to the nearest 10 litres).

Why this works. Eleven marks. (a) is a show that: write the reason (OM ⊥ AB because M is the midpoint), since $\cos x = \frac{20}{30}$ is only half the answer and the doubling is the rest. (b) is the segment recipe fed with 96.3793°, not 96.4°; (c) turns on units — 1.5 m = 150 cm, 1 litre = 1000 cm³, water added, nearest 10 litres.

G3 N2024 · Paper 2 · Q8(b) 5 marks

The diagram shows two circles, centres P and Q.
S is a point of intersection of the circles and PQR is a straight line.
Angle RPS = 43° and angle PRS = 13°.
The radius of the larger circle is 7 cm and the radius of the smaller circle is 4.5 cm.

Calculate the shaded area. [5]

Worked solution

QS and QR are both radii of the larger circle, so triangle QSR is isosceles and angle QSR = angle QRS = 13° (base ∠s of isos. △). Hence

$∠ S Q P = 13 ° + 13 ° = 26 °$ (exterior ∠ of △QSR).

The circles meet again at the mirror image of S in the line PQR, so the two centres each see the common chord through twice the angle marked:

angle at P = 2 × 43° = 86°, angle at Q = 2 × 26° = 52°.

The shaded lens is two segments back to back — one in each circle:

$smaller: \frac{86 °}{360 °} \times π \times {4.5}^{2} - \frac{1}{2} \times 4.5 \times 4.5 \times \sin 86° = 15.197… - 10.100… = 5.0971…$ cm²

$larger: \frac{52 °}{360 °} \times π \times 7^{2} - \frac{1}{2} \times 7 \times 7 \times \sin 52° = 22.235… - 19.306… = 2.9292…$ cm²

Shaded area $= 5.0971… + 2.9292… = 8.026…$ = 8.03 cm² (3 s.f.).

Why this works. The overlap of two circles is two segments joined along the common chord, so the segment recipe runs twice. The two central angles come from the rest: 43° at P doubles to 86° by symmetry in the line of centres, and 13° at R becomes 26° at Q (QS and QR are radii), then 52°. Both doublings carry marks; 8 cm² against 63.6 cm² is believable.

Why 360? — and a measure that does not need a number

There is no mathematical reason for a full turn to be 360 degrees. So here is a way of measuring an angle that involves no arbitrary number at all: measure it by the arc it cuts, in units of the radius.

θ = \frac{s}{r}

where θ is the angle, in radians, subtended at the centre of a circle of radius r by an arc of length s. In particular, an arc as long as the radius subtends one radian.

Fig. 10.8

Bend a piece of string of length r round the circle from A to B. The angle it opens up at the centre is 1 radian — a little under 60°, whatever the size of the circle.

A radian has no units. θ is a length divided by a length, so the centimetres cancel and a pure number is left — one that already is the fraction of a turn, which is why the 10.4 formulas have no 360 in them.

rad is written only to say the number is not degrees. “θ = 2.25” and “θ = 2.25 rad” mean the same thing; “θ = 2.25°” is a different angle.

Where this comes from

360 is a choice inherited from Babylonian astronomy; 100 would have worked just as well. Radian measure sidesteps the choice completely.

Take a circle of radius r and an arc of length s on it. The ratio s ÷ r does not depend on how big the circle is — double the circle and both lengths double — so it is a property of the angle alone. That ratio is the angle, measured in radians.

`π` rad = 180° — the one line to memorise

A full turn is 2π radians, and it is also 360°. Halve both, and one line converts an angle either way.

π rad = 180 °

so to go from radians to degrees, multiply by

\frac{180 °}{π}

, and to go from degrees to radians, multiply by

\frac{π}{180 °}

.

Fig. 10.9

The four worth knowing on sight. Each is 180° divided by something, so each is π divided by the same thing: 180 ÷ 6 = 30 and π ÷ 6 = π⁄6. A full turn is 2π rad and a straight line is π rad.

Which way round? Sanity-check the size. One radian is about 57°, so a radian is a big unit: an angle in radians is a much smaller number than the same angle in degrees. 43° comes out under 1; 2.35 rad comes out over 100°.

The calculator has a mode. sin 1.4 in degree mode is 0.0244; in radian mode it is 0.985, and only a small D or R at the top of the screen says which. Set it to Rad when a question gives radians, and back afterwards.

Walkthrough 10 — radians into degrees Basic

Convert (a) 5π⁄12 rad and (b) 2.35 rad to degrees, giving a non-exact answer correct to 1 decimal place.

π rad = 180°, so 1 rad = $\frac{180 °}{π}$

Everything in this topic comes out of this one line, so write it down first. Half a turn is π radians and it is also 180°; dividing both sides by π gives what one radian is worth.
(a) $\frac{5 π}{12}$ rad = $\frac{5 π}{12} \times \frac{180 °}{π}$ = 5 × 15° = 75°

A multiple of π lets the π cancel, so no calculator is needed: 180 ÷ 12 = 15, and five of those is 75. That cancellation is why exam angles are so often written as fractions of π.
(b) 2.35 rad = 2.35 × $\frac{180 °}{π}$ = 134.645°

Same conversion, but with no π to cancel the answer is not exact. A quick check on the size: 2.35 radians is a bit over two lots of 57°, so something near 135° is right.
(a) 75° (b) 134.6° (1 d.p.)

Only (b) gets rounded. (a) came out exact, so writing “75.0°” would suggest an accuracy it does not need — the instruction said “a non-exact answer”.

Walkthrough 11 — degrees into radians Basic

Convert (a) 252° to radians, leaving your answer in terms of π, and (b) 43° to radians, correct to 3 significant figures.

180° = π rad, so 1° = $\frac{π}{180}$ rad

The same fact as before, turned round. Going into radians means multiplying by π⁄180, which makes the number smaller — as expected, since a radian is a bigger unit than a degree.
(a) 252° = 252 × $\frac{π}{180}$ = $\frac{252}{180} π = \frac{7}{5} π$ rad

Cancel 252⁄180 by 36 to get 7⁄5. Leaving the π standing is what “in terms of π” asks for, and it is also the form the next question is most likely to hand back to you.
(b) 43° = 43 × $\frac{π}{180}$ = 0.750491… rad

43 does not share a factor with 180 worth cancelling, so this one goes to the calculator. Under 1 radian, as expected: 43° is well under 57°.
(a) 7π⁄5 rad (b) 0.750 rad (3 s.f.)

Keep the final 0 in 0.750 — without it the answer reads as two significant figures. And (a) is exact, so it is not rounded at all.

Walkthrough 12 — an angle straight from the arc and the radius Intermediate

An arc of length 27 cm subtends an angle θ at the centre of a circle of radius 12 cm. Find θ (i) in radians, (ii) in degrees, correct to 1 decimal place.

$θ = \frac{s}{r}$

This is the definition of a radian, not a formula to be looked up: the angle in radians is arc divided by radius. Nothing needs converting first.
$θ = \frac{27}{12}$ = 2.25 rad

Both lengths are in centimetres, so the units cancel and what is left is a pure number — that number is the angle in radians. 27⁄12 is exactly 2.25, so nothing is rounded.
2.25 rad = 2.25 × $\frac{180 °}{π}$ = 128.916°

Only now is a conversion needed, because part (ii) asked for degrees. 2.25 is a little over two radians, so a little over 114° — 129° fits.
(i) θ = 2.25 rad (ii) θ = 128.9° (1 d.p.)

Part (i) is exact and stays exact; only the degree answer is rounded, and to 1 decimal place because that is the paper's rule for an angle in degrees.

Walkthrough 13 — trigonometry with the calculator in radian mode Intermediate

In triangle PQR, ∠PQR = π⁄2 rad, QR = 6.5 cm and ∠QPR = 0.83 rad. Find (i) PQ, (ii) PR, each correct to 3 significant figures.

π⁄2 rad = 90°, so △PQR is right-angled at Q

Half of π rad is half of 180°. Writing the right angle as π⁄2 keeps the whole question in radians; spotting it unlocks the ordinary trigonometric ratios.
tan 0.83 = $\frac{Q R}{P Q} = \frac{6.5}{P Q}$

Looking from the 0.83 rad angle at P: QR is opposite it and PQ is adjacent to it — that pair is tangent. Set the calculator to Rad before touching the tan key; 0.83 is radians, not degrees.
PQ = $\frac{6.5}{tan 0.83}$ = 5.9446 cm (5 s.f.)

The unknown is on the bottom, so it swaps with the 6.5. Carrying five figures leaves room for the final rounding to be honest.
sin 0.83 = $\frac{6.5}{P R}$ , so PR = $\frac{6.5}{sin 0.83}$ = 8.8084 cm (5 s.f.)

PR is the hypotenuse, opposite the right angle, and QR is opposite the 0.83 rad angle — that pair is sine. Going back to the given 6.5 rather than the calculated PQ keeps this answer independent of the last one.
(i) PQ = 5.94 cm (ii) PR = 8.81 cm (3 s.f.)

Round both once, at the end. Check with Pythagoras' Theorem: 5.9446² + 6.5² = 77.588, and √77.588 = 8.8084 — the two answers agree.

Where this comes from

Go all the way round: the arc is the whole circumference, s = 2πr, so θ = 2πr ÷ r = 2π. A full turn is therefore 2π radians as well as 360°, and halving both gives the line worth memorising, π rad = 180°.

The size check catches the commoner of the two slips. Converting 43° and getting 2463 means the multiplication went the wrong way round: 43° is well under one radian.

The accuracy convention for this chapter, the same as the front of the paper: a non-exact angle in degrees goes to 1 decimal place, and a non-exact angle in radians to 3 significant figures.

Check yourself — 10.3

Convert 5π⁄6 rad to degrees.
Answer
$\frac{5 π}{6} \times \frac{180 °}{π}$ = 5 × 30° = 150°. The π cancels, so this is exact and needs no calculator.
Convert 108° to radians, leaving your answer in terms of π.
Answer
108 × $\frac{π}{180}$ = $\frac{108}{180} π$ = 3π⁄5 rad (cancelling by 36). As a decimal that is about 1.88 — sensibly less than half of 2π.
Convert 1.7 rad to degrees, correct to 1 decimal place.
Answer
1.7 × $\frac{180 °}{π}$ = 97.402… = 97.4° (1 d.p.). Just over π⁄2, so just over 90° — it fits.
An arc of length 9.6 cm subtends an angle of 0.75 rad at the centre of a circle. Find the radius.
Answer
$θ = \frac{s}{r}$ , so r = $\frac{s}{θ} = \frac{9.6}{0.75}$ = 12.8 cm (exact). No π appears anywhere — that is the radian definition doing the work.
Without using a calculator, decide whether 2 rad is bigger or smaller than 120°, and explain how you know.
Answer
120° is a third of 360°, so it is a third of 2π rad, i.e. 2π⁄3 rad. Since π is a little more than 3, 2π⁄3 is a little more than 2. So 2 rad is smaller than 120°. (Equivalently: 1 rad ≈ 57°, so 2 rad ≈ 115°.)

The same two formulas, with the 360 removed

Nothing new happens here: take the degree formulas from 10.1 and 10.2 and change the unit the angle is measured in. A full turn stops being 360° and becomes 2π rad, so every “x out of 360” becomes “θ out of 2π”:

arc = \frac{θ}{2 π} \times 2 π r = r θ

sector area = \frac{θ}{2 π} \times π r^{2} = \frac{1}{2} r^{2} θ

In both, θ is in radians. Write s = rθ and A = ½r²θ.

Fig. 10.10

One sector, two sets of clothes. The picture is identical; only the units on the angle have changed, and with them the formulas. Choosing between them is not a mathematical decision — it is whichever unit the question used.

These two, and only these two, are printed on your paper: “Arc length = rθ” and “Sector area = ½r²θ”, both where θ is in radians. The degree versions are not there, and neither is π rad = 180°.

An angle in degrees gives you a choice. Use the degree formula from memory, or convert the angle to radians first and use the printed one. Either is fine; one of the two needs to be in your head.

rθ needs θ in radians. For a radius of 10 cm and an angle of 60°, the arc is 10 × (π⁄3) = 10.5 cm; putting 60 in gives 600 cm, longer than the whole circumference, 62.8 cm.

Check the size against the whole circle. An arc cannot be longer than 2πr, and a sector cannot hold more area than πr².

Walkthrough 14 — arc length in radians Basic

In the diagram, O is the centre of a circle of radius 24 cm and ∠AOB = 5π⁄8 rad. Find (i) the length of the minor arc APB, (ii) the perimeter of the major sector OAQB, giving each answer correct to 3 significant figures.

Length of minor arc APB = rθ = 24 × $\frac{5 π}{8}$

The angle is already in radians, so the printed formula s = rθ applies directly — no 360, no conversion. Two numbers multiplied, and that is the whole first line.
= 15π cm = 47.1238… cm

24 ÷ 8 = 3 and 3 × 5 = 15, so the arc is exactly 15π cm. The circumference is 48π cm and 5π⁄8 rad is under half a turn, so 15π sits below 24π as it should.
Reflex ∠AOB = 2π − $\frac{5 π}{8}$ = $\frac{11 π}{8}$ rad

A full turn is 2π rad, not 360, once the angle is in radians — this is the line where people forget which unit they are in. 16⁄8 − 5⁄8 = 11⁄8.
Length of major arc AQB = 24 × $\frac{11 π}{8}$ = 33π cm

Same formula, same radius, the reflex angle. Check: 15π + 33π = 48π, the whole circumference.
Perimeter of major sector OAQB = arc AQB + OA + OB = 33π + 24 + 24 = (33π + 48) cm

Radians change the arc formula and nothing else — a sector is still closed by its two radii, so both radii, 24 cm each, go in exactly as they did in degrees.
(i) 15π = 47.1 cm (ii) 33π + 48 = 152 cm (3 s.f.)

Round once, at the end. 151.67… to 3 significant figures is 152 — three figures, so no decimal point is needed here.

Walkthrough 15 — sector area in radians Basic

A and B lie on a circle with centre O and radius 14 cm. Given that ∠AOB = 1.15 rad, find the area of (i) the minor sector, (ii) the major sector, correct to 3 significant figures where necessary.

Area of minor sector = $\frac{1}{2} r^{2} θ$ = $\frac{1}{2} \times 14^{2} \times 1.15$

The second of the two printed formulas, used straight off the sheet. Note it is r that is squared, not rθ — write 14² × 1.15, not (14 × 1.15)².
= $\frac{1}{2} \times 196 \times 1.15$ = 112.7 cm²

This is exact — 98 × 1.15 comes out at 112.7 with nothing left over, so there is nothing to round. “Where necessary” in the question is a warning that not every part needs the calculator's last digit.
Reflex ∠AOB = 2π − 1.15 = 5.1332 rad (5 s.f.)

A full turn in radians is 2π ≈ 6.2832, so the reflex angle is that minus 1.15. Carrying five figures here protects the third figure of the area that follows.
Area of major sector = $\frac{1}{2} \times 196 \times 5.1332$ = 503.05 cm²

Same formula, same radius, the reflex angle. The major sector is far bigger than the minor one, which is what the picture would show.
(i) 112.7 cm² (ii) 503 cm² (3 s.f.)

Check against the whole circle: π × 14² = 615.75, and 112.7 + 503.05 = 615.75. Both answers carry cm²; (i) is exact and keeps its four figures, (ii) is rounded to three.

Walkthrough 16 — from a perimeter to the angle and the area Intermediate

A sector OAB of a circle has radius 12 cm and perimeter 43 cm. Find (i) ∠AOB in radians, (ii) the area of the sector.

Perimeter = arc AB + OA + OB, so arc AB = 43 − 12 − 12 = 19 cm

The two radii come off first. Everything else in this question runs on the arc, and the arc is not what was given — the perimeter was.
s = rθ, so 19 = 12θ

With the arc known and the radius known, the printed formula has only one unknown left in it. No conversion is needed anywhere: the answer is wanted in radians and rθ works in radians.
$θ = \frac{19}{12}$ = 1.5833… = 1.58 rad (3 s.f.)

19⁄12 is the exact angle; 1.58 is what gets written as the answer, to 3 significant figures as the paper requires for a non-exact angle in radians.
Area = $\frac{1}{2} r^{2} θ$ = $\frac{1}{2} \times 12^{2} \times \frac{19}{12}$ = 6 × 19

Use the exact 19⁄12, not the rounded 1.58: the 12 in the denominator cancels beautifully against the 144 and the answer comes out whole. Substituting 1.58 instead would give 113.76 — close, but no longer exact.
(i) ∠AOB = 1.58 rad (3 s.f.) (ii) area = 114 cm²

Two different accuracies in one question, and both are right: (i) is non-exact so it is rounded, (ii) came out exactly 114 so it is not. Area takes cm².

Walkthrough 17 — a segment, in radians Advanced

The diagram shows a sector OAB of a circle with centre O and radius 18 cm, in which ∠AOB = 1.4 rad. Find the area of the minor segment cut off by the chord AB, correct to 3 significant figures.

Area of minor segment = area of sector OAB − area of △OAB

The recipe is the same one as in degrees — the chord splits the sector into a triangle and the segment. Radians change how each piece is calculated, not what the pieces are.
Area of sector OAB = $\frac{1}{2} r^{2} θ$ = $\frac{1}{2} \times 18^{2} \times 1.4$ = 226.8 cm²

Straight off the formula sheet, with the angle already in radians. 162 × 1.4 = 226.8 exactly, so nothing is rounded yet.
Area of △OAB = $\frac{1}{2} \times O A \times O B \times sin θ$ = $\frac{1}{2} \times 18^{2} \times sin 1.4$

½ab sin C with both sides equal to the radius. sin 1.4 means the sine of 1.4 radians, so the calculator is in Rad; in degree mode it returns 0.0244 instead of 0.985 and the answer is far too small.
= 162 × 0.98545 = 159.64 cm² (5 s.f.)

sin 1.4 ≈ 0.985 — close to 1, because 1.4 rad is close to π⁄2, a right angle. Five figures are carried because the next line is a subtraction of two similar numbers.
Area of minor segment = 226.8 − 159.64

Sector minus triangle, in that order. The sector is the larger of the two, as it is for every minor segment — the triangle sits inside it.
= 67.2 cm² (3 s.f.)

Round once, here, and write cm². Roughly 30% of the sector is left over as segment, which is about what the shaded sliver in the diagram looks like.

Walkthrough 18 — a shaded region between a triangle and a sector Advanced

In the figure, OPQ is a sector of a circle with centre O and radius 15 cm. OQR is a straight line with OR = 19 cm and the length of arc PQ is 13.5 cm. Find (i) ∠POQ in radians, (ii) PR, correct to 2 decimal places, (iii) the area of the shaded region PQR, correct to 3 significant figures.

Fig. 10.11

s = rθ, so 15θ = 13.5 and θ = 0.9 rad

The arc and the radius are both given, so the angle falls straight out of s = rθ — and it falls out exactly, at 0.9, which every later line will use.
By the cosine rule, PR² = 15² + 19² − 2(15)(19) cos 0.9

In △OPR you know two sides, OP = 15 and OR = 19, and the angle between them — that is exactly the cosine rule's shape. Calculator in Rad: cos 0.9 is 0.6216, not 0.99988.
= 586 − 570 cos 0.9 = 231.6823 (4 d.p.)

225 + 361 = 586 and 2(15)(19) = 570. Keeping four decimal places here is what makes the 2-decimal-place answer that follows trustworthy.
PR = $\sqrt{231.6823}$ = 15.22 cm (2 d.p.)

Take the positive root, since PR is a length. The question named 2 decimal places, so that beats the paper's default of 3 significant figures.
Area of △OPR = $\frac{1}{2} \times 15 \times 19 \times sin 0.9$ = 111.62 cm² (5 s.f.)

½ab sin C on the same two sides and the same included angle. This triangle is the whole region OPR, out of which the sector will be taken.
Area of sector POQ = $\frac{1}{2} \times 15^{2} \times 0.9$ = 101.25 cm²

Using the exact 0.9 from step 1. Note that Q lies on OR, so the sector really does sit inside the triangle.
Area of shaded region PQR = area of △OPR − area of sector POQ = 111.62 − 101.25

Here the sector is cut out of the triangle — the opposite of a segment, where the triangle is cut out of the sector. Which way round it goes is read off the diagram, not remembered.
(i) 0.9 rad (ii) 15.22 cm (iii) 10.4 cm² (3 s.f.)

Three parts, three accuracies, all as asked: (i) exact, (ii) 2 d.p., (iii) 3 s.f. The small answer to (iii) is believable — the shaded sliver is a thin crescent between the arc and the straight edge PR.

Why this works

The 2π cancels outright in the first formula and leaves half of it in the second. That is the whole derivation, and it is worth doing once, because then s = rθ and A = ½r²θ stop looking like two more things to memorise: they are the degree formulas of 10.1 and 10.2 with the unit on the angle changed, and nothing else.

Check yourself — 10.4

A sector has radius 25 cm and an angle of 0.64 rad at the centre. Find its arc length and its area.
Answer
Arc = rθ = 25 × 0.64 = 16 cm.
Area = ½r²θ = ½ × 625 × 0.64 = 200 cm². Both are exact — no rounding needed.
A sector of radius 9 cm has an arc of length 12 cm. Find the angle at the centre in radians and the area of the sector.
Answer
12 = 9θ, so θ = 12⁄9 = 4⁄3 = 1.33 rad (3 s.f.).
Area = ½ × 81 × 4⁄3 = 54 cm² (using the exact 4⁄3, not 1.33).
A sector of radius 10 cm has an area of 60 cm². Find the angle at the centre in radians and the perimeter of the sector.
Answer
½ × 100 × θ = 60, so θ = 1.2 rad (exact).
Arc = 10 × 1.2 = 12 cm, so perimeter = 12 + 10 + 10 = 32 cm.
A chord of a circle of radius 8 cm subtends an angle of 2 rad at the centre. Find the area of the minor segment, correct to 3 significant figures.
Answer
Sector = ½ × 64 × 2 = 64 cm².
Triangle = ½ × 64 × sin 2 = 32 × 0.90930 = 29.098 cm² (calculator in Rad).
Segment = 64 − 29.098 = 34.9 cm² (3 s.f.).
For the same circle and chord, find the area of the major segment, correct to 3 significant figures.
Answer
Reflex angle = 2π − 2 = 4.2832 rad, so the major sector = ½ × 64 × 4.2832 = 137.06 cm².
The triangle is added for a major segment: 137.06 + 29.098 = 166 cm² (3 s.f.).
Check: 34.902 + 166.16 = 201.06 = π × 8², the whole circle.

Past-paper questions

G3 N2015 · Paper 2 · Q5(b) 3 marks

The perimeter of a sector of another circle is 14.8 cm.
The angle of the sector is 1.7 radians.
Calculate the area of the sector. [3]

Worked solution

The perimeter of a sector is the arc plus the two straight radii, and with the angle in radians the arc is rθ:

$r \times 1.7 + 2 r = 14.8 \Rightarrow 3.7 r = 14.8 \Rightarrow r = 4$ cm

$area of sector = \frac{1}{2} r^{2} θ = \frac{1}{2} \times 4^{2} \times 1.7$ = 13.6 cm² (exactly).

Why this works. No diagram, so draw one: a sector has three edges, two of them straight, and the first mark is $r θ + 2 r = 14.8$ . Writing 1.7r = 14.8 uses the arc alone. In radians the formulas are the short ones, so a $\frac{1.7}{360}$ in the working means degrees have crept in; the numbers come out exact: r = 4, 13.6.

G3 N2018 · Paper 1 · Q16 4 marks

The diagram shows a circle with radius r cm.
The circle is divided into two sectors.
The angle of the minor sector is θ radians.
The perimeter of the major sector is twice the perimeter of the minor sector.
Find the value of θ.
Give your answer correct to three decimal places. [4]

Worked solution

Both sectors are bounded by the same two radii; only the arcs differ. The major sector turns through 2π − θ, so

$minor perimeter = r θ + 2 r$ , $major perimeter = r (2 π - θ) + 2 r$

$r (2 π - θ) + 2 r = 2 (r θ + 2 r)$

$2 π r - r θ + 2 r = 2 r θ + 4 r$

Collect the θ terms on one side: $2 π r - 2 r = 3 r θ$

Divide every term by r (a radius is never zero): $2 π - 2 = 3 θ$

$θ = \frac{2 π - 2}{3} = 1.42772…$ , so $θ = 1.428$ (3 d.p.).

Why this works. No radius is given: r is meant to cancel. The marks sit on the two perimeter expressions: + 2r belongs in both, since the sectors share radii, and dropping it gives $r (2 π - θ) = 2 r θ$ , so $θ = \frac{2 π}{3} = 2.094$ . The major angle is 2π − θ; three decimal places makes 1.428.

G3 N2019 · Paper 2 · Q6(b) 5 marks

P and Q are points on the circle centre O with radius 4 cm.
R and S are points on the circle centre O with radius 6 cm.
OPR and OQS are straight lines.
The perimeter of the minor sector OPQ is 15.2 cm.

Calculate angle POQ in radians. [2]
Calculate the total shaded area. [3]

Worked solution

(a) The perimeter of sector OPQ is the arc PQ plus the radii OP and OQ:

$4 θ + 2 \times 4 = 15.2 \Rightarrow 4 θ = 7.2 \Rightarrow θ = 1.8$ , so angle POQ = 1.8 radians.

(b) The shaded region is the whole ring with the wedge between OR and OS taken out of it, plus the sector OPQ put back in.

$ring = π (6^{2} - 4^{2}) = 20 π = 62.831…$ cm²

$wedge = \frac{1}{2} \times 6^{2} \times 1.8 - \frac{1}{2} \times 4^{2} \times 1.8 = 32.4 - 14.4 = 18$ cm²

$sector O P Q = \frac{1}{2} \times 4^{2} \times 1.8 = 14.4$ cm²

$shaded = 62.831… - 18 + 14.4 = 59.231…$ = 59.2 cm² (3 s.f.).

Why this works. Part (a) is the perimeter of a sector, and 1.8 comes out exactly. (b) is bookkeeping: name each piece, price it, add or subtract once; leaving out the inner sector gives 44.8, taking the wedge off twice gives 41.2. The lines pass through the centre, so the wedge is two sectors of one angle differenced, ½ × 1.8 × (6² − 4²) = 18.

G3 N2020 · Paper 2 · Q6(b) 6 marks

The diagram shows a circle, centre K, radius 12 cm.
Angle LKM = 1.8 radians.

Calculate the length of the major arc LM. [2]
Calculate the percentage of the circle that is shaded. [4]

Worked solution

(a) The major arc turns through the rest of the full turn:

$reflex ∠ L K M = 2 π - 1.8 = 4.4831…$ rad

$major arc = r θ = 12 \times 4.4831… = 53.798…$ = 53.8 cm (3 s.f.).

(b) The shaded region is the minor segment cut off by the chord LM — sector LKM less triangle LKM:

$sector = \frac{1}{2} \times 12^{2} \times 1.8 = 129.6$ cm²

$triangle = \frac{1}{2} \times 12 \times 12 \times \sin 1.8 = 70.117…$ cm²

$segment = 129.6 - 70.117… = 59.482…$ cm², $circle = π \times 12^{2} = 452.389…$ cm²

$\frac{59.482…}{452.389…} \times 100 = 13.148…$ = 13.1 % (3 s.f.).

Why this works. Everything here is in radians: sin 1.8 = 0.9738 in RAD mode, while DEG gives 0.0314, a triangle of 2.26 cm² and a segment bigger than its own sector. Then two familiar moves: “major” means subtract from 2π before multiplying by r, and “percentage of the circle” divides by πr², not by the sector. Keep the segment unrounded; 129.6 is exact.

G3 N2024 · Paper 2 · Q8(a)(ii) 2 marks

A, B, C and D are points on the circle, centre O.
M is the midpoint of AB and N is the midpoint of CD.
AB = CD.

The radius of the circle is 4 cm and angle COD = 1.8 radians.
Calculate the length of the major arc AB. [2]

Worked solution

Equal chords of a circle subtend equal angles at the centre, so

angle AOB = angle COD = 1.8 radians.

$reflex ∠ A O B = 2 π - 1.8 = 4.4831…$ rad

$major arc A B = r θ = 4 \times 4.4831… = 17.932…$ = 17.9 cm (3 s.f.).

Why this works. The angle sits at COD and the arc wanted is AB, so the first mark is the line AB = CD: equal chords subtend equal angles at the centre. The second is for reading major: subtract from 2π, then multiply by r. 4 × 1.8 = 7.2 is the minor arc; the two make the circumference, 8π = 25.1 cm.

Degrees and radians, side by side

There are only two ideas in this chapter and each is written twice — once for each way of measuring the angle. Read the table across, not down: the two columns are the same mathematics.

	Angle in degrees (`x`°)	Angle in radians (`θ`)
A full turn	360°	2`π` rad
Arc length	$\frac{x °}{360 °} \times 2 π r$	$s = r θ$
Sector area	$\frac{x °}{360 °} \times π r^{2}$	$A = \frac{1}{2} r^{2} θ$
Perimeter of a sector	arc + `r` + `r` — the same in both, and the mark most often lost
Area of a triangle made by a chord and two radii	$\frac{1}{2} r^{2} sin θ$ — that is ½`ab` sin `C` with `a` = `b` = `r`
Area of a minor segment	sector − triangle, in whichever units the angle came in
Printed on the paper's MATHEMATICAL FORMULAE page?	NO — both degree formulas must be memorised	YES — both are printed, word for word
Converting between them	`π` rad = 180° — not on the formula page either. × 180⁄`π` to reach degrees, × `π`⁄180 to reach radians

What the paper prints. rθ and ½r²θ (both “where θ is in radians”) and ½ab sin C — and nothing for degrees, nothing for converting, nothing for a segment.

So three things to carry in your head: the two “x out of 360” formulas, the line π rad = 180°, and the sentence segment = sector − triangle.

The segment recipe — four lines, every time

1. Find the angle at the centre, if it is not given. (Often from a right-angled triangle: the line from the centre to the midpoint of the chord is perpendicular to it and bisects the angle, so cos of half the angle = that distance ÷ r.)
2. Area of sector — (x⁄360) × πr² in degrees, or ½r²θ in radians.
3. Area of triangle — ½r² sin(that same angle).
4. Minor segment = sector − triangle. Major segment = major sector + triangle. Check against the whole circle: the two segments must add up to πr².

Which tool for which question

The question asks for…	Reach for	Where the marks go
an arc length, angle in degrees	(`x`⁄360) × 2`πr`	the formula line before the substitution; the exact value before rounding
an arc length, angle in radians	`s` = `rθ`	checking that the angle really is in radians, not degrees
a perimeter of a sector or a shaded region	name every edge first, then add	the two radii — arc alone is the classic lost mark
a radius or an angle, given the arc or the area	the same formula, rearranged	the substitution line; the positive square root only
a major arc or sector	reflex angle = 360° − `x`°, or 2`π` − `θ`	using the reflex angle, and the free check that the two parts make a whole
an area of a segment	sector − triangle (½`r`² sin `θ`)	both areas shown separately; the unrounded angle used in both
a shaded region made of a sector and a triangle	read the diagram: which one is being cut out of which?	the sentence naming the two pieces before the subtraction
a region between two arcs (a ring segment)	big sector − small sector; four edges for the perimeter	both arcs and both straight edges
an angle converted	`π` rad = 180°	leaving a multiple of `π` exact; 1 d.p. for degrees, 3 s.f. for radians
sin, cos or tan of an angle in radians	the calculator, in Rad mode	switching the mode — and switching it back

The habits that earn the marks

1. Write the formula down before substituting into it — that line is worth a mark on its own.
2. Read whether the angle is a degree or a radian before you touch the calculator, and set the mode to match.
3. Perimeter is not arc. Walk round the region and name every edge, including the straight ones.
4. For a major arc or sector, use the reflex angle — 360° − x°, or 2π − θ.
5. Keep π and fractions exact through the middle of the working; if you must go decimal, carry 5 significant figures.
6. Round once, at the end — 3 s.f. unless the question names another accuracy, and 1 d.p. for an angle in degrees.
7. “In terms of π” and “exact” mean do not press the calculator at all.
8. Units: cm for a length, cm² for an area, rad or ° for an angle.
9. Sanity-check the size: an arc cannot beat 2πr, a sector cannot beat πr², and an area can never be negative.

Syllabus at a glance

The two G3 Mathematics (K310) outcomes this chapter covers, quoted from the Secondary Three/Four section of the syllabus.

G3 Mathematics (K310, first examination 2027) — these notes are written to it.

LO	What you must be able to do	Covered in
5.7	arc length, sector area and area of a segment of a circle	10.1 Arc length, 10.1 Perimeter, 10.2 Sector area, 10.2 Segments
5.8	use of radian measure of angle (including conversion between radians and degrees)	10.3 The radian, 10.3 Conversion, 10.4 `rθ` and ½`r`²`θ`

Wording is quoted verbatim from the Secondary Three/Four section of the G3 Mathematics syllabus, and matches the last two bullets of the G5 row of the K310 content table. Two things this chapter leans on are not listed under G5, because they belong elsewhere: the circumference 2πr and area πr² of a whole circle, which are Secondary 1/2 in both bands and are recapped in 10.1; and ½ab sin C, which is outcome G4 4.5 — “use of the formula ½ab sin C for the area of a triangle” — and is what makes the triangle inside a segment computable (10.2, 10.4). The cosine rule used once in Walkthrough 18 is G4 4.6. Nothing in this chapter is outside the G2 course. The G2 Secondary Three/Four list splits the same ground over four outcomes instead of two — 5.7 “arc length as fraction of the circumference and sector area as fraction of the area of a circle”, 5.8 “area of a segment”, 5.9 “use of radian measure of angle (including conversion between radians and degrees)” and 5.10 “problems involving the arc length, sector area of a circle and area of a segment”. G2 5.9 is the same sentence as G3 5.8, so radian measure is not a G3-only topic and nothing on this page is marked as outside the G2 course.

🍕 Arc, Sector & Radian Measure — Revision Notes

Six questions, and where each one is answered

The five words — chord, arc, segment, sector, and “minor” ⓘ Recap

Arc length — a fraction of the way round

The perimeter of a sector — the arc plus two radii

Check yourself — 10.1

Area of a sector — the same fraction, applied to area

Past-paper questions

Area of a segment — the two-formula question

Check yourself — 10.2

Past-paper questions

Why 360? — and a measure that does not need a number

`π` rad = 180° — the one line to memorise

Check yourself — 10.3

The same two formulas, with the 360 removed

Check yourself — 10.4

Past-paper questions

Degrees and radians, side by side

The segment recipe — four lines, every time

Which tool for which question

The habits that earn the marks

Syllabus at a glance

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🍕 Arc, Sector & Radian Measure — Revision Notes

Six questions, and where each one is answered

The five words — chord, arc, segment, sector, and “minor” ⓘ Recap

Arc length — a fraction of the way round

The perimeter of a sector — the arc plus two radii

Check yourself — 10.1

Area of a sector — the same fraction, applied to area

Past-paper questions

Area of a segment — the two-formula question

Check yourself — 10.2

Past-paper questions

Why 360? — and a measure that does not need a number

π rad = 180° — the one line to memorise

Check yourself — 10.3

The same two formulas, with the 360 removed

Check yourself — 10.4

Past-paper questions

Degrees and radians, side by side

The segment recipe — four lines, every time

Which tool for which question

The habits that earn the marks

Syllabus at a glance

`π` rad = 180° — the one line to memorise