Arc length and sector area

IB Mathematics Applications and Interpretation SL· Geometry and Trigonometry, SL 3.4 (arc length and area of a sector, angle in degrees)· 12 min read

1. Core Arc Length Formula★★☆☆☆⏱ 10 min

Arc length is the linear distance along the curved edge of a sector. It is a fraction of the full circumference $2 π r$ , where that fraction is the central angle compared to a full turn of $36 0^{\circ}$ .

📘 Definition

Arc Length

$s$

Total distance of the curved segment between two points on a circle's circumference

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📐 Worked Example

Find the arc length of a sector with radius 6 cm and central angle 120°, giving your answer to 2 decimal places.

1
The angle is in degrees, so use the arc length formula with the $\frac{θ}{36 0 ^{\circ}}$ fraction.
2
$s = \frac{120^\circ}{360^\circ} \times 2\pi \times 6$
3
$s = \frac{1}{3} \times 12\pi = 4\pi$
4
Calculate the decimal value: $4 π \approx 12.57$ cm.

2. Sector Area Calculations★★☆☆☆⏱ 10 min

Sector area uses the same proportional logic as arc length, comparing the sector's central angle to the full $36 0^{\circ}$ of a circle, then scaling the total area $π r^{2}$ of the full circle accordingly.

📘 Definition

Sector Area

$A$

Total 2D area enclosed by two radii and their connecting arc

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📐 Worked Example

A sector has radius 8 m and central angle 60°. Calculate its exact area, leaving your answer in terms of $π$ .

1
The angle is in degrees, so use the sector area formula with the $\frac{θ}{36 0 ^{\circ}}$ fraction.
2
$A = \frac{60^\circ}{360^\circ} \times \pi \times 8^2$
3
$A = \frac{1}{6} \times 64\pi = \frac{32\pi}{3} \text{ m}^2$

✓ Quick check

Test your understanding of the formulas below:

What is the arc length of a sector with radius 5 cm and central angle 90°?
- 2.5 cm
- 7.85 cm
- 15.71 cm
- 39.27 cm
Reveal answer
7.85 cm —
Using $s = \frac{90}{360} \times 2 π \times 5 = 2.5 π \approx 7.85$ cm.

3. Reverse Calculations for Radius or Angle★★★☆☆⏱ 12 min

IB AI SL exam questions frequently ask you to rearrange the standard formulas to solve for an unknown radius or central angle, rather than directly calculating arc length or sector area from given values.

🔬 Derivation

Goal:

Rearrange the arc length formula to make the central angle (degrees) the subject

Starting from:

$s = \frac{θ}{36 0 ^{\circ}} \times 2 π r$

1
Multiply both sides by $36 0^{\circ}$ and divide by $2 π r$
2
$\theta = \frac{s \times 360^\circ}{2\pi r}$

Result:

Rearranged formula: $θ = \frac{36 0 ^{\circ} s}{2 π r}$

📐 Worked Example

A sector has arc length 15 cm and radius 5 cm. Find its central angle in degrees, to the nearest degree.

1
Substitute $s = 15$ and $r = 5$ into the rearranged arc length formula.
2
$\theta = \frac{15 \times 360^\circ}{2\pi \times 5} = \frac{5400^\circ}{10\pi}$
3
Evaluate to the nearest degree.
$\theta \approx 172^\circ$

4. Applied Real-World Sector Problems★★★★☆⏱ 15 min

IB AI SL places heavy emphasis on real-world geometry applications, so you will often see arc and sector questions paired with other shapes like triangles, set in contexts such as garden design, clock faces, or road construction.

📐 Worked Example

A garden patio is made of a 120° sector of a circle with radius 4 m, attached to an isosceles triangle with two sides equal to the circle's radius. Calculate the total area of the patio to 1 decimal place.

1
First calculate the area of the sector part of the patio.
2
$A_{sector} = \frac{120^\circ}{360^\circ} \times \pi \times 4^2 = \frac{16\pi}{3} \approx 16.755 \text{ m}^2$
3
Next calculate the area of the isosceles triangle with two sides 4 m and included angle 120° using $A = \frac{1}{2} ab sin θ$ .
4
$A_{triangle} = \frac{1}{2} \times 4 \times 4 \times \sin(120^\circ) \approx 6.928 \text{ m}^2$
5
Add the two areas together for the total patio area.
6
$Total A \approx 16.755 + 6.928 = 23.7 \text{ m}^2$

5. Common Pitfalls

Wrong move:

Multiplying radius by the angle directly (as $r \times θ$ ) with the angle in degrees

Why:

The short $s = r θ$ form only works for radians, which are an AHL topic; at SL you must use the degree formula with the $\frac{θ}{36 0 ^{\circ}}$ fraction

Correct move:

Always use $s = \frac{θ}{36 0 ^{\circ}} \times 2 π r$ with the angle in degrees

Wrong move:

Forgetting to square the radius when calculating sector area

Why:

Students often copy the arc length formula structure and omit the $r^{2}$ term, leading to an area value that is far too small

Correct move:

Write the full formula out before substituting values to confirm you have all required terms

Wrong move:

Dropping the $\frac{θ}{36 0 ^{\circ}}$ fraction and using the full $2 π r$ or $π r^{2}$

Why:

That gives the whole circle, not the sector, so your answer is far too large

Correct move:

Always scale by the fraction of the full turn that the central angle represents

Wrong move:

Using 3.14 as an approximate value for $π$ instead of the calculator's built-in value

Why:

This introduces rounding errors that make your final answer fall outside the acceptable IB mark range

Correct move:

Keep all intermediate values in your calculator memory, only rounding your final answer as specified

Wrong move:

Confusing sector area with segment area

Why:

Segment area is sector area minus the inner triangle, not the full sector area

Correct move:

Draw a quick sketch of the shape described to confirm exactly which region you are asked to calculate

6. Quick Reference Cheatsheet

Quantity	Formula (angle in degrees)
Arc length $s$	$s = \frac{θ}{36 0 ^{\circ}} \times 2 π r$
Sector area $A$	$A = \frac{θ}{36 0 ^{\circ}} \times π r^{2}$
Unknown angle $θ$	$θ = \frac{s \times 36 0 ^{\circ}}{2 π r}$
Unknown radius $r$ (from area)	$r = \frac{36 0 ^{\circ} A}{π θ}$

What's Next

Now that you have mastered core arc length and sector area calculations in degrees, you are ready to tackle more advanced trigonometric applications in IB Math AI SL. You will next learn to solve problems involving segments of circles, which extend sector area concepts by subtracting triangular areas to find the area of curved slices. This skill is frequently combined with sine and cosine rule questions on Paper 2, where you will be asked to find the total perimeter and area of composite 2D shapes. Mastering these fundamentals will also prepare you for 3D geometry problems involving cones and spheres later in the unit, where sector properties are used to calculate curved surface areas. Practice mixed problems to build speed and accuracy under exam pressure.