Voronoi 1 · Arc length & sector area

Module 20 of 47Lesson 1 of 7 · Voronoi 1 · Arc length & sector area

Arcs, sectors, perpendicular bisectors & Voronoi diagrams · Essential understanding 1

Arc length & sector area

IB AI SL 3.4with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: An arc is a piece of the edge of a circle and a sector is a slice of it. Both are just a fraction of the whole circle, and that fraction is always

\dfrac{\theta}{360}

By the end of this lesson you can:

find the length of an arc, given the radius and the angle at the centre
find the area of a sector, given the radius and the angle at the centre
work backwards to find an unknown angle or radius
use arcs and sectors in real situations (fences, fields, slices)

🧑🏾 Imagine this…

A round pizza is cut into 8 equal slices. Each slice has an angle of

\dfrac{360^\circ}{8}

= 45° at the tip. If the crust all the way round the pizza is 96 cm long, the crust of one slice is

\dfrac{96}{8}

= 12 cm. If the whole pizza has an area of 640 cm², one slice covers

\dfrac{640}{8}

= 80 cm². That is all this lesson is: find what fraction of the circle you have, then take that fraction of the whole.

Circumference

the distance all the way round a circle:

C = 2\pi r

Radius r

the distance from the centre O to the edge.

Arc

a part of the circumference (a curved piece of the edge). Its length is written s.

Sector

a slice of a circle, bounded by two radii and an arc (like a slice of pizza).

Angle at the centre θ

the angle between the two radii that bound the sector, measured in degrees. A full turn is 360°.

A sector with angle θ. The thick orange curve is the arc (length s). The shaded slice is the sector (area A).

The angle tells you the fraction of a full turn: 90° is a quarter, 180° is a half, 60° is a sixth. In general it is

\dfrac{\theta}{360}

★ Arc length and sector area (both in the formula booklet)

Length of an arc: $s = \dfrac{\theta}{360} \times 2\pi r$
Area of a sector: $A = \dfrac{\theta}{360} \times \pi r^2$
r is the radius and θ is the angle at the centre in degrees. The fraction $\dfrac{\theta}{360}$ is the same in both formulas.
The whole circle is $C = 2\pi r$ and $A = \pi r^2$ . A sector is just $\dfrac{\theta}{360}$ of these.

⚠️ Two formulas, two different measurements

The arc is a length, so it uses

2\pi r

(r to the power 1) and the answer is in cm, m or km. The sector is an area, so it uses

\pi r^2

(r squared) and the answer is in cm², m² or km². If your area answer has the unit cm, something has gone wrong.

⚠️ Radius, not diameter

If the question gives the diameter, halve it first. A circular pool of diameter 10 m has radius 5 m. Using 10 in the formulas gives an arc twice too long and an area four times too big.

💡 Fraction method (easy to remember)

Say it in words: arc = fraction × circumference and sector = fraction × circle area. If you can find the fraction

\dfrac{\theta}{360}

, you cannot mix up the formulas.

🧪 Try it yourselfArc length grows steadily with the angle

The line shows the arc length s of a circle of radius r (the horizontal axis is the angle θ in degrees). Set r = 6 and drag the marker to 80: the height is the answer to Example 1. Double the angle and the arc doubles, which is why the graph is a straight line.

Worked examples

Example 1.1Length of an arc

A sector of a circle has radius 6 cm and an angle of 80° at the centre. Find the length of its arc.

🧠 Think it through

The fraction of the circle is $\dfrac{\theta}{360} = \dfrac{80}{360}$ .
The whole circumference is 2πr = 2 × π × 6.
Arc = fraction × circumference. Round only at the very end (3 s.f.).

✍️ By hand

s = \dfrac{80}{360} \times 2\pi \times 6 = \dfrac{2}{9} \times 12\pi = \dfrac{8\pi}{3}

s = 8.38

cm (3 s.f.)

🔢 On your TI-Nspire CX II

Press ⌂ on💻 click ⌂ on keypad and choose 1: Add Calculator (or use a Calculator page you already have).
Type 80/360*2*π*6 and press ctrl💻 Ctrl enter ↵💻 Enter for a decimal. (Plain enter ↵💻 Enter may show the exact answer 8· $\dfrac{\pi}{3}$ .)

8.37758 → 8.38 cm (3 s.f.)

✅ Answer s = 8.38 cm

Does it make sense? 80° is a bit less than a quarter turn. A quarter of the circumference 2π × 6 = 37.7 cm is 9.42 cm, and 8.38 cm is just under that. ✓

Example 1.2Area of a sector

A circular lawn has radius 12 m. A sprinkler waters a sector of the lawn with an angle of 135° at the centre.

(a) What fraction of the lawn is watered? Give it in its simplest form.
(b) Find the area of the whole lawn.
(c) Find the area of the sector that is watered.

🧠 Think it through

Fraction = $\dfrac{135}{360}$ . Divide top and bottom by 45.
Whole lawn: $\pi r^2$ with r = 12.
Sector = fraction × whole lawn. The unit is m² because it is an area.

✍️ By hand

(a)

\dfrac{135}{360} = \dfrac{3}{8}

(b)

\pi \times 12^2 = 144\pi = 452

m²

(c)

A = \dfrac{3}{8} \times 144\pi = 54\pi = 170

m²

🔢 On your TI-Nspire CX II

Type 135/360 and press enter ↵💻 Enter. The calculator gives the simplest fraction $\dfrac{3}{8}$ .
Type π*12^2 and press ctrl💻 Ctrl enter ↵💻 Enter.
Type 135/360*π*12^2 and press ctrl💻 Ctrl enter ↵💻 Enter.

Answer (c): 169.646 → 170 m² (3 s.f.)

✅ Answer (a)

\dfrac{3}{8}

(b) 452 m² (c) 170 m²

Example 1.3Finding the angle

A circular garden bed has radius 10 m. A curved path along the edge of the bed (an arc) is 15 m long. Find the angle that this arc makes at the centre of the bed.

🧠 Think it through

Here the arc length is known and the angle is the unknown.
Write the formula with the numbers in: $15 = \dfrac{\theta}{360} \times 2\pi \times 10$ .
Undo the multiplication: multiply both sides by 360, then divide by 2π × 10.

✍️ By hand

15 = \dfrac{\theta}{360} \times 20\pi

\theta = \dfrac{15 \times 360}{20\pi} = \dfrac{270}{\pi}

\theta = 85.9°

(3 s.f.)

🔢 On your TI-Nspire CX II

Type 15*360/(2*π*10) and press ctrl💻 Ctrl enter ↵💻 Enter.
Or let the calculator solve it: use menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve and type nSolve(15=x/360*2*π*10,x).

Both routes give 85.9437°

✅ Answer θ = 85.9°

An angle at the centre of a sector must be between 0° and 360°. If your answer is outside that range, check your working.

Example 1.4Finding the radius from an area

A sector has an angle of 70° at the centre and an area of 40 cm².

(a) Find the radius of the circle.
(b) Find the length of the arc of the sector.

🧠 Think it through

Put the numbers in: $40 = \dfrac{70}{360} \times \pi r^2$ .
Get r² on its own, then take the square root. Keep the full calculator value for part (b).
For part (b) use your unrounded radius in the arc formula.

✍️ By hand

r^2 = \dfrac{40 \times 360}{70\pi}

r = \sqrt{\dfrac{14400}{70\pi}}

(a)

r = 8.09

(b)

s = \dfrac{70}{360} \times 2\pi \times 8.09203 = 9.89

🔢 On your TI-Nspire CX II

Type √(40*360/(70*π)) and press ctrl💻 Ctrl enter ↵💻 Enter. Square-root is ctrl💻 Ctrl x²💻 type ^2, then →.
📍 Where to find it on the TI-Nspire
⌨️ square root: ctrl then x² on the keypad, or type sqrt(
Press ctrl💻 Ctrl (-) (the answer key) so the next line uses that value, or store it with ctrl💻 Ctrl var💻 Ctrl+L ▸ r.
Type 70/360*2*π*ans and press ctrl💻 Ctrl enter ↵💻 Enter.

r = 8.09203 cm, arc = 9.88628 cm

✅ Answer (a) r = 8.09 cm (b) s = 9.89 cm

Never use a rounded radius in part (b). Rounding 8.09 first and then calculating can change the third significant figure.

Example 1.5A centre-pivot irrigation field

A farm near Francistown uses a centre-pivot sprinkler. The arm is 300 m long and turns about a fixed centre, watering a sector of angle 250° because a road cuts across the rest of the circle. A fence is to be built along the curved outer edge of the watered field.

(a) Find the length of fence needed.
(b) Find the area watered, in m² and in hectares (1 hectare = 10 000 m²).
(c) Fencing costs P38 per metre. Find the cost of the fence.

🧠 Think it through

The radius is the arm length, 300 m, and θ = 250°.
The fence runs along the arc, so use the arc formula.
The watered area is the sector area. Divide by 10 000 to change m² into hectares.
Cost = length × price. Use the unrounded length.

✍️ By hand

(a)

s = \dfrac{250}{360} \times 2\pi \times 300 = 1309

(b)

A = \dfrac{250}{360} \times \pi \times 300^2 = 196 000

m² = 19.6 ha

= 1308.997 \times 38 =

P49 700

🔢 On your TI-Nspire CX II

Type 250/360*2*π*300 and press ctrl💻 Ctrl enter ↵💻 Enter.
Type 250/360*π*300^2 and press ctrl💻 Ctrl enter ↵💻 Enter, then ans/10000.
Type 250/360*2*π*300*38 and press ctrl💻 Ctrl enter ↵💻 Enter.

✅ Answer (a) 1310 m (b) 196 000 m² = 19.6 ha (c) P49 700

Money answers are usually rounded to the nearest pula or to 3 s.f. The question says which. P49 741.90 can be written as P49 700 (3 s.f.).

Challenge questions

⭐ Challenge 1.6Finding r and A from an arcPaper 1 style

A sector of a circle has an angle of 100° at the centre. The arc of the sector has length 12.5 cm.

(a) Find the radius of the circle.
(b) Find the area of the sector.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Put the numbers in: $12.5 = \dfrac{100}{360} \times 2\pi r$ and solve for r.
Keep the unrounded radius for the area.

✍️ By hand

r = \dfrac{12.5 \times 360}{100 \times 2\pi} = 7.16

A = \dfrac{100}{360} \times \pi \times 7.16197^2 = 44.8

cm²

🔢 On your TI-Nspire CX II

Type 12.5*360/(100*2*π) and press ctrl💻 Ctrl enter ↵💻 Enter, then 100/360*π*ans^2.

✅ Answer (a) r = 7.16 cm (b) A = 44.8 cm²

Shortcut for checking: a sector's area is also ½ × arc × radius = ½ × 12.5 × 7.16 = 44.8 cm². It works because the sector is a thin 'triangle' of base s and height r.

⭐ Challenge 1.7Windscreen wiperPaper 2 style

A windscreen wiper arm pivots at a point O. The rubber blade is fixed to the arm so that it runs from 10 cm to 45 cm away from O. In one sweep the arm turns through an angle of 110°, and the blade cleans a curved band of the windscreen.

(a) Find the distance travelled by the far tip of the blade (45 cm from O) in one sweep.
(b) Find the area of the windscreen cleaned in one sweep.
(c) The wiper makes 40 complete cycles (out and back) in one minute. Find the total distance, in metres, travelled by the tip in one minute.
(d) State one assumption you made in part (b).

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

The tip moves along an arc of radius 45 cm and angle 110°.
The cleaned region is a big sector (radius 45) with a small sector (radius 10) taken away.
One cycle = one sweep out and one sweep back = 2 sweeps. Change cm to m by dividing by 100.
Think about whether the blade really cleans every point in the band.

✍️ By hand

(a)

s = \dfrac{110}{360} \times 2\pi \times 45 = 86.4

(b)

A = \dfrac{110}{360}\pi(45^2 - 10^2) = 1850

cm²

(c)

40 \times 2 \times 86.3938 = 6912

cm = 69.1 m

🔢 On your TI-Nspire CX II

(a) Type 110/360*2*π*45 and press ctrl💻 Ctrl enter ↵💻 Enter.
(b) Type 110/360*π*(45^2-10^2) and press ctrl💻 Ctrl enter ↵💻 Enter. Brackets matter: subtract the squares first.
(c) Type 80*110/360*2*π*45/100 and press ctrl💻 Ctrl enter ↵💻 Enter.

✅ Answer (a) 86.4 cm (b) 1850 cm² (c) 69.1 m (d) For example: the blade is straight, but the region is modelled as a sector band, so the straight blade only matches the curved band approximately; or the blade cleans every point between 10 cm and 45 cm.

Part (b) is the 'subtract two sectors' idea again. It returns in Lesson 2 with rings, paths and segments.

Next: Lesson 2 · Perimeter, segments & composite shapes

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL Voronoi 1 · Arc length & sector area

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL

Arc length & sector area

Worked examples

Challenge questions