Arc Length and Sector Area: Radians Explained with Real Examples

In 30 seconds

A radian is the angle that cuts off an arc equal in length to the radius. A full turn is $ 2\pi $ radians, so $ 180^\circ = \pi $ radians.
With $ \theta $ in radians, arc length and sector area are simple: $ s = r\theta $ and $ A = \tfrac{1}{2}r^2\theta $.
A segment (the piece between a chord and its arc) is a sector minus a triangle: $ \tfrac{1}{2}r^2(\theta-\sin\theta) $.
Real uses: the area a windscreen wiper clears, the stagger on a running track, the crust on a slice of pizza.
Exam point: put your calculator in radian mode, or convert degrees first. Using $ 110 $ instead of $ 1.92 $ in $ \tfrac{1}{2}r^2\theta $ is the classic disaster.

The case: how much glass does a wiper clear?

A car designer is comparing wiper set-ups (the dimensions here are illustrative). The arm pivots at the bottom of the windscreen. The rubber blade runs along the arm from 15 cm to 60 cm from the pivot, and on each pass the arm swings through $ 110^\circ $. One pass takes 0.6 seconds.

The designer asks three things: what area of glass does one pass clear, how far does the tip of the blade travel, and how fast is the tip moving? Figure 1 shows the shape. It is not a full sector, because the rubber does not reach the pivot. It is the band between two arcs, a shape sometimes called an annular sector.

The concept: measuring angles with the radius

Degrees are a human invention: someone long ago decided a full turn should be 360 units. Radians come from the circle itself. Take the radius of a circle, bend it round the edge, and the angle it covers at the centre is 1 radian, about $ 57.3^\circ $.

The circumference of a circle is $ 2\pi r $, which is $ 2\pi $ radius-lengths. So a full turn is $ 2\pi $ radians, half a turn is $ \pi $ radians, and a right angle is $ \tfrac{\pi}{2} $. To convert, multiply degrees by $ \tfrac{\pi}{180} $; to go back, multiply radians by $ \tfrac{180}{\pi} $^[1].

The big advantage is that an angle in radians tells you directly how many radius-lengths of arc it cuts off. An angle of 2 radians on a circle of radius 5 cm cuts off an arc of $ 2 \times 5 = 10 $ cm. In degrees the same calculation needs a fraction of $ 360 $ and a factor of $ \pi $ every time.

Key termRadian: the angle at the centre of a circle subtended by an arc whose length equals the radius. $ \pi $ radians $ = 180^\circ $.

The formula

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

\[ s = r\theta \qquad A = \tfrac{1}{2}r^2\theta \]

Here $ s $ is the arc length (same units as $ r $) and $ A $ is the sector area (units squared). Both come from proportion: the sector is the fraction $ \dfrac{\theta}{2\pi} $ of the whole circle, so $ s = \dfrac{\theta}{2\pi} \times 2\pi r $ and $ A = \dfrac{\theta}{2\pi} \times \pi r^2 $. Two more results follow at once:

\[ A_{\text{band}} = \tfrac{1}{2}\theta\,(R^2-r^2) \]

\[ A_{\text{segment}} = \tfrac{1}{2}r^2(\theta-\sin\theta) \]

The band between radii $ r $ and $ R $ is the big sector minus the small one. The segment is the sector minus the triangle formed by the two radii and the chord, whose area is $ \tfrac{1}{2}r^2\sin\theta $^[2].

Chapter 7 of Mathematics Made Simple builds radians, arcs, sectors and trigonometric graphs from the ground up.Get the textbook · $12

Why this formula fits the case

A wiper blade sweeps through an angle, and every point on it moves along a circular arc centred on the pivot. That is exactly the situation $ s = r\theta $ describes. The area it clears is a sector with a smaller sector cut out of the middle, so the band formula applies directly.

The formulas are not right if the blade does not move in a circle. Some buses use a pantograph wiper that keeps the blade upright as it moves; the cleared shape is then different. And the formulas only work with $ \theta $ in radians. If you prefer degrees, you must use the longer forms $ s = \dfrac{\theta^\circ}{360} \times 2\pi r $ and $ A = \dfrac{\theta^\circ}{360} \times \pi r^2 $.

Setting it up

Angle: $ 110^\circ = 110 \times \dfrac{\pi}{180} = \dfrac{11\pi}{18} \approx 1.920 $ radians.
Radii: inner $ r = 15 $ cm, outer $ R = 60 $ cm.
Time per pass: 0.6 s.
Units: arc lengths in cm, area in cm², speed in cm/s (convert to m/s at the end if it reads better).

Solving it

Worked example

Find the area cleared in one pass, the distance the tip travels and the tip’s speed.

Step 1Convert: $ \theta = \dfrac{11\pi}{18} \approx 1.9199 $ rad.

Step 2Outer sector: $ \tfrac{1}{2} \times 60^2 \times \theta = 1800\theta \approx 3455.8 $ cm².

Step 3Inner sector (not wiped): $ \tfrac{1}{2} \times 15^2 \times \theta = 112.5\theta \approx 216.0 $ cm².

Step 4Wiped band: $ 1800\theta-112.5\theta = 1687.5\theta \approx 3239.8 $ cm². Check with the band formula: $ \tfrac{1}{2}\theta(3600-225) = 1687.5\theta $.

Step 5Tip arc: $ s = 60\theta \approx 115.2 $ cm.

Step 6Tip speed: $ \dfrac{115.2}{0.6} \approx 192 $ cm/s, about 1.92 m/s.

One pass clears about 3240 cm² of glass (roughly a third of a square metre), while the tip travels about 1.15 m.

Arc length also explains a puzzle at every athletics meeting: why the runners in outside lanes start ahead of those inside.

Worked example

A running track has two semicircular bends. In lane 1 the running line on the bends has radius 36.8 m, and each lane is 1.22 m wide (figures close to a standard outdoor track, but treat them as illustrative). In a 400 m race everyone runs both bends in their own lane. How much further forward must lane 2 start than lane 1? And lane 8?

Step 1Two semicircles turn through $ \pi + \pi = 2\pi $ radians in total.

Step 2Lane 1 bends: $ s_1 = 36.8 \times 2\pi \approx 231.2 $ m.

Step 3Lane 2 bends: radius $ 36.8 + 1.22 = 38.02 $ m, so $ s_2 = 38.02 \times 2\pi \approx 238.9 $ m.

Step 4Difference: $ s_2-s_1 = 1.22 \times 2\pi \approx 7.67 $ m.

Step 5Lane 8 is $ 7 \times 1.22 = 8.54 $ m further out, so its extra distance is $ 8.54 \times 2\pi \approx 53.7 $ m.

Lane 2 starts about 7.67 m ahead of lane 1, and lane 8 about 53.7 m ahead. Notice Step 4: the extra distance is $ (\text{extra radius}) \times \theta $, whatever the radius of lane 1. That is $ s = r\theta $ doing the work.

Exam questions also like the segment: the region between a chord and its arc.

Worked example

A round table of radius 60 cm has a drop leaf. The fold line is a chord that subtends $ \tfrac{2\pi}{3} $ at the centre. Find the area of the leaf and its perimeter, exactly and to 1 decimal place.

Step 1Sector: $ \tfrac{1}{2} \times 60^2 \times \tfrac{2\pi}{3} = 1200\pi $ cm².

Step 2Triangle: $ \tfrac{1}{2} \times 60^2 \times \sin\tfrac{2\pi}{3} = 1800 \times \tfrac{\sqrt3}{2} = 900\sqrt3 $ cm².

Step 3Leaf (segment): $ 1200\pi-900\sqrt3 $, which is $ 3769.9-1558.8 \approx 2211.1 $ cm².

Step 4Arc: $ 60 \times \tfrac{2\pi}{3} = 40\pi $ cm. Chord: $ 2 \times 60\sin\tfrac{\pi}{3} = 60\sqrt3 $ cm.

Step 5Perimeter: $ 40\pi + 60\sqrt3 \approx 125.7 + 103.9 = 229.6 $ cm.

The leaf is about 2211 cm², about a fifth of the whole tabletop ($ 3600\pi \approx 11\,310 $ cm²).

Master arcs, sectors and segments with 1,565 practice questions, every one with a fully worked answer.Get 1,565 questions · $9

What the answer tells us

The wiper clears about 3240 cm² per pass. That sounds like a lot, but compare it with the full sector of radius 60 cm, about 3456 cm²: the unwiped core near the pivot is only about 6% of it, so a 15 cm gap costs little. The tip moves at about 1.9 m/s while the inner end of the rubber, at 15 cm, travels only $ 15\theta \approx 28.8 $ cm per pass, or 48 cm/s. Every point turns through the same angle in the same time (about 3.2 radians per second), so speed is proportional to distance from the pivot: $ v = r\omega $. That is why wiper rubber wears fastest at the tip.

Reasonableness checks. A $ 110^\circ $ sweep is a little under a third of a full turn ($ \tfrac{110}{360} \approx 0.31 $), and the full circle of radius 60 cm has area about 11 310 cm², so the outer sector should be roughly 3500 cm². It is. The units also check: $ r^2\theta $ is cm² because $ \theta $ has no units.

Sensitivity. Area grows with the square of the radius. Lengthening the arm to 65 cm (inner end still at 15 cm) raises the cleared area to $ \tfrac{1}{2}\theta(65^2-15^2) = 2000\theta \approx 3840 $ cm², about 18.5% more for a blade only 11% longer. Increasing the angle, by contrast, raises the area in direct proportion.

Area swept by a windscreen wiper: an annular sector

arc traced by the tip: s = 60θinner end of the rubber: radius 15 cm

Figure 1. The rubber runs from 15 cm to 60 cm along the arm, so it wipes the band between two arcs. Area = ½θ(60² − 15²) with θ in radians. Illustrative wiper dimensions, drawn to scale.

Common mistakePutting degrees into $ s = r\theta $ or $ A = \tfrac{1}{2}r^2\theta $. With $ \theta = 110 $ the “area” would be $ 1687.5 \times 110 = 185\,625 $ cm², bigger than a car. If an answer is absurdly large, check the angle units first.

Common mistakeForgetting the straight edges in a perimeter. A sector’s perimeter is $ r\theta + 2r $ (arc plus two radii), and a segment’s perimeter is the arc plus the chord. Giving only the arc length is a frequent one-mark loss.

Exam tipCambridge 9709 tests arcs, sectors and segments in Paper 1 section 1.4, Circular measure, usually in a composite shape with a “show that $ \theta = \dots $” start^[3]. In IB Mathematics: analysis and approaches it is SL topic 3.4^[4]. AP Calculus does not ask for sector areas directly, but it uses radians throughout: results such as $ \dfrac{d}{dx}\sin x = \cos x $ are only true when $ x $ is in radians. Give exact answers in terms of $ \pi $ and $ \sqrt3 $ when the angle is a “nice” fraction of $ \pi $.

To see how radians lead into trigonometric graphs, identities and later calculus, the Mathematics Made Simple textbook covers them across Chapters 7 and 8.

Get both Mathematics Made Simple books for $16 and save $5: trigonometry to calculus, explained and drilled.Both books · $16

Practice questions

Radian mode on. For a much larger set at every level, the Mathematics Made Simple practice book has 1,565 questions with worked solutions and two mock papers.

EasyQ1. A pizza of diameter 30 cm is cut into 8 equal slices. Find the area of one slice and the length of its crust (the arc). Show worked answer

Radius $ r = 15 $ cm and angle $ \theta = \dfrac{2\pi}{8} = \dfrac{\pi}{4} $.
Area: $ \tfrac{1}{2} \times 15^2 \times \tfrac{\pi}{4} = \dfrac{225\pi}{8} \approx 88.4 $ cm².
Crust: $ s = 15 \times \tfrac{\pi}{4} = \dfrac{15\pi}{4} \approx 11.8 $ cm.

EasyQ2. Convert $ 135^\circ $ to radians exactly, and 2.5 radians to degrees. Show worked answer

$ 135 \times \dfrac{\pi}{180} = \dfrac{3\pi}{4} $.
$ 2.5 \times \dfrac{180}{\pi} \approx 143.2^\circ $.

MediumQ3. A sector has radius 8 cm and perimeter 28 cm. Find its angle in radians and its area. Show worked answer

Perimeter $ = 2r + r\theta $, so $ 16 + 8\theta = 28 $.
$ 8\theta = 12 $, so $ \theta = 1.5 $ rad.
Area $ = \tfrac{1}{2} \times 64 \times 1.5 = 48 $ cm².

MediumQ4. A garden sprinkler waters a sector of lawn with angle 1.2 radians and area 30 m². Find the radius it reaches and the length of the curved edge. Show worked answer

$ \tfrac{1}{2}r^2 \times 1.2 = 30 $, so $ 0.6r^2 = 30 $ and $ r^2 = 50 $.
$ r = \sqrt{50} \approx 7.07 $ m.
Curved edge: $ s = r\theta = \sqrt{50} \times 1.2 \approx 8.49 $ m.

ExamQ5. A circle has centre $ O $ and radius 10 cm. A chord $ AB $ has length 12 cm. (a) Show that angle $ AOB $ is 1.287 radians, correct to 3 decimal places. [2] (b) Find the area of the minor segment cut off by $ AB $. [3] (c) Find the perimeter of the minor segment. [2] Show worked answer

(a) The perpendicular from $ O $ bisects $ AB $, giving a right-angled triangle with hypotenuse 10 and opposite side 6 [1]. So $ \sin\tfrac{\theta}{2} = 0.6 $, $ \tfrac{\theta}{2} = 0.6435 $ and $ \theta = 1.287 $ rad [1].
(b) $ \sin\theta = 2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2} = 2 \times 0.6 \times 0.8 = 0.96 $ [1]. Segment $ = \tfrac{1}{2} \times 10^2 \times (\theta-\sin\theta) = 50(1.2870-0.96) $ [1] $ \approx 16.35 $ cm², or 16.4 cm² to 3 s.f. [1].
(c) Arc $ = 10 \times 1.287 = 12.87 $ cm [1]. Perimeter $ = 12.87 + 12 = 24.87 \approx 24.9 $ cm [1].

Frequently asked questions

What are the formulas for arc length and sector area?

With the angle $ \theta $ in radians, arc length is $ s = r\theta $ and sector area is $ A = \tfrac{1}{2}r^2\theta $. In degrees they become $ \dfrac{\theta}{360} \times 2\pi r $ and $ \dfrac{\theta}{360} \times \pi r^2 $.

Why do mathematicians prefer radians?

Because they make formulas simpler. Arc length is just radius times angle, and in calculus the derivative of $ \sin x $ is exactly $ \cos x $ only when $ x $ is in radians. In degrees an awkward factor of $ \tfrac{\pi}{180} $ appears everywhere.

How do I find the area of a segment?

Find the sector area, then subtract the triangle formed by the two radii and the chord: $ \tfrac{1}{2}r^2\theta-\tfrac{1}{2}r^2\sin\theta = \tfrac{1}{2}r^2(\theta-\sin\theta) $.

How many degrees is one radian?

$ \dfrac{180}{\pi} \approx 57.3^\circ $. A useful check: $ \pi \approx 3.14 $ radians is $ 180^\circ $, so 1 radian is a little under a third of that.

References

Abramson, J. et al. (2021) Algebra and Trigonometry, 2nd edn, Section 7.1 Angles. Houston: OpenStax. https://openstax.org/details/books/algebra-and-trigonometry-2e
Stewart, J., Clegg, D. and Watson, S. (2021) Calculus: Early Transcendentals, 9th edn, Appendix D Trigonometry. Boston: Cengage Learning.
Cambridge International. Cambridge International AS & A Level Mathematics 9709: Syllabus for 2026 and 2027. cambridgeinternational.org
International Baccalaureate Organization. Mathematics: analysis and approaches guide (first assessment 2021). ibo.org

Arc Length and Sector Area: Radians Explained with Real Examples

The case: how much glass does a wiper clear?

The concept: measuring angles with the radius

The formula

Why this formula fits the case

Setting it up

Solving it

What the answer tells us

Practice questions

Frequently asked questions

What are the formulas for arc length and sector area?

Why do mathematicians prefer radians?

How do I find the area of a segment?

How many degrees is one radian?

References

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