When a pizza chef cuts a slice, they are creating a sector — a wedge-shaped piece of a circle. But how much crust is on the curved edge? And what is the area of the topping? In this lesson, you will learn the elegant formulas that answer both questions, and discover why radians make them beautifully simple.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A pizza has radius 20 cm. A slice is cut with an angle of $45^\circ$ at the centre. Without using any formulas, estimate the arc length of the crust on this slice. Then, consider how you might calculate it exactly if you knew that a full circle's circumference is $2\pi r$.
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: The sine and cosine functions have a period of 360° for all transformations.
Right: The period is affected by horizontal dilation; y = sin(nx) has period 360°/n, not 360°.
📚 Core Content
When an angle $\theta$ is measured in radians, the arc length $l$ swept out by that angle in a circle of radius $r$ is given by:
$$l = r\theta$$
This formula is elegant because it follows directly from the definition of a radian: $\theta = \frac{l}{r}$. In degrees, the equivalent formula is:
$$l = 2\pi r \times \frac{\theta}{360^\circ}$$
Notice how the radian version has no $\pi$ and no $360$ — it is cleaner because radians are the natural unit for circular measure. This is why mathematicians and physicists almost always use radians when working with circular motion, waves, and calculus.
A sector is the "pizza slice" region bounded by two radii and an arc. The area of a sector with central angle $\theta$ (in radians) and radius $r$ is:
$$A = \frac{1}{2}r^2\theta$$
Again, this is much simpler than the degree version:
$$A = \pi r^2 \times \frac{\theta}{360^\circ}$$
The sector area formula comes from taking the same fraction of the full circle's area ($\pi r^2$) as the angle is of a full revolution ($2\pi$ radians). When you simplify $\pi r^2 \times \frac{\theta}{2\pi}$, the $\pi$ cancels and you get $\frac{1}{2}r^2\theta$.
If you know the arc length or sector area, you can rearrange the formulas to find the missing quantity:
🧮 Worked Examples
🧪 Activities
1 $r = 6$ cm, $\theta = \frac{\pi}{3}$ rad. Find $l$.
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2 $r = 10$ cm, $\theta = \frac{3\pi}{4}$ rad. Find $A$.
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3 $r = 12$ cm, $\theta = 45^\circ$. Find $l$ and $A$.
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4 $l = 8\pi$ cm, $r = 4$ cm. Find $\theta$ in radians.
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5 $A = 50\pi$ cm$^2$, $\theta = \frac{\pi}{2}$ rad. Find $r$.
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1 A pendulum swings through an angle of $\frac{\pi}{6}$ radians. If the pendulum is 80 cm long, how far does the tip travel along its arc?
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2 A garden sprinkler waters a sector with radius 5 m and angle $72^\circ$. What area of lawn does it water? Leave your answer in exact form.
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Earlier you were asked: A pizza has radius 20 cm and a slice angle of $45^\circ$. Estimate the arc length of the crust.
A full circle has circumference $2\pi r = 40\pi \approx 125.7$ cm. The slice is $\frac{45}{360} = \frac{1}{8}$ of the full pizza. So the arc length is $\frac{1}{8}$ of the circumference: $\frac{40\pi}{8} = 5\pi \approx 15.7$ cm. In radians, $45^\circ = \frac{\pi}{4}$, so $l = r\theta = 20 \times \frac{\pi}{4} = 5\pi$ cm. The radian formula gives the same answer instantly without needing to calculate the full circumference first.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. A sector has radius 9 cm and central angle $\frac{2\pi}{3}$ radians. (a) Find the exact arc length. (b) Find the exact area of the sector. 3 MARKS
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9. A piece of wire 40 cm long is bent to form the perimeter of a sector. If the radius of the sector is 12 cm, find the angle of the sector in radians. 3 MARKS
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10. Two sectors have the same area. Sector A has radius 6 cm and angle $\frac{\pi}{2}$ radians. Sector B has radius 4 cm. Find the angle of Sector B in radians. Show all working. 3 MARKS
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Answer in your workbook.
1. $l = 6 \times \frac{\pi}{3} = 2\pi$ cm
2. $A = \frac{1}{2}(100) \times \frac{3\pi}{4} = \frac{75\pi}{2}$ cm$^2$
3. $45^\circ = \frac{\pi}{4}$ rad. $l = 12 \times \frac{\pi}{4} = 3\pi$ cm. $A = \frac{1}{2}(144) \times \frac{\pi}{4} = 18\pi$ cm$^2$.
4. $\theta = \frac{8\pi}{4} = 2\pi$ rad (a full circle)
5. $50\pi = \frac{1}{2}r^2 \times \frac{\pi}{2} \Rightarrow r^2 = 200 \Rightarrow r = 10\sqrt{2}$ cm
1. $l = 80 \times \frac{\pi}{6} = \frac{40\pi}{3} \approx 41.9$ cm
2. $72^\circ = \frac{2\pi}{5}$ rad. $A = \frac{1}{2}(25) \times \frac{2\pi}{5} = 5\pi$ m$^2$
1. A — $l = r\theta$.
2. A — $l = 6 \times \frac{\pi}{3} = 2\pi$ cm.
3. A — $A = \frac{1}{2}r^2\theta$.
4. A — $A = \frac{1}{2}(16) \times \frac{\pi}{2} = 4\pi$ cm$^2$.
5. A — $60^\circ = \frac{\pi}{3}$ rad; $l = 10 \times \frac{\pi}{3} = \frac{10\pi}{3}$ cm.
Q8 (3 marks): (a) $l = 9 \times \frac{2\pi}{3} = 6\pi$ cm [1.5]. (b) $A = \frac{1}{2}(81) \times \frac{2\pi}{3} = 27\pi$ cm$^2$ [1.5].
Q9 (3 marks): Perimeter $= 2r + l = 40$ [1]. So $l = 40 - 24 = 16$ cm [0.5]. Then $\theta = \frac{l}{r} = \frac{16}{12} = \frac{4}{3}$ rad [1.5].
Q10 (3 marks): Area of A $= \frac{1}{2}(36) \times \frac{\pi}{2} = 9\pi$ cm$^2$ [1]. Set equal to area of B: $9\pi = \frac{1}{2}(16) \times \theta_B$ [1]. So $\theta_B = \frac{9\pi}{8}$ rad [1].
Sprint through questions on arc length and area of sector calculations. Pool: lessons 1–2.
Tick when you've finished all activities and checked your answers.