The shape of the lesson
By the end of the lesson, learners can
- Arc length, s = rθ: A sector with angle \(\theta\) is the fraction \(\dfrac{\theta}{2\pi}\) of the whole circle, so its arc is \(\dfrac{\theta}{2\pi} \times 2\pi r = r\theta\).
- Sector area: The same fraction gives the area: \(\dfrac{\theta}{2\pi} \times \pi r^2 = \tfrac{1}{2}r^2\theta\).
- Segments and compound shapes: A segment is the region between a chord and its arc.
- Angle in radians: convert first if it is given in degrees.
- Arc: \(s = r\theta\). Sector: \(A = \tfrac{1}{2}r^2\theta\).
- Triangle: \(\tfrac{1}{2}r^2\sin\theta\); chord: \(2r\sin\dfrac{\theta}{2}\).
- Tangents: right angle with the radius, then right-angled trigonometry.
- Compound regions: add and subtract the simple pieces; perimeters use only the edges of the region.
One picture every idea returns to
For a sector of radius \(r\) and angle \(\theta\) radians: arc length \(s = r\theta\) and area \(A = \tfrac{1}{2}r^2\theta\). A segment is a sector minus a triangle, so its area is \(\tfrac{1}{2}r^2\theta - \tfrac{1}{2}r^2\sin\theta\), and its chord is \(2r\sin\dfrac{\theta}{2}\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Two formulae, endless shapes.. With the angle in radians, the arc of a sector is just \(r\theta\) and its area is \(\tfrac{1}{2}r^2\theta\). Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Arc length, s = rθ | A sector with angle \(\theta\) is the fraction \(\dfrac{\theta}{2\pi}\) of the whole circle, so its arc is \(\dfrac{\theta}{2\pi} \times 2\pi r = r\theta\). Model the worked example Arc and perimeter on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Sector area | The same fraction gives the area: \(\dfrac{\theta}{2\pi} \times \pi r^2 = \tfrac{1}{2}r^2\theta\). Model the worked example Exact sector area on the board, one line at a time. Your turn: A sector has radius 10 cm and angle \(\dfrac{\pi}{5}\). Find the exact arc length and area. | Pairs |
| 21 to 30 min | Teach and check: Segments and compound shapes | A segment is the region between a chord and its arc. Model the worked example Segment area and perimeter on the board, one line at a time. Your turn: Show that the sector of radius \(r\) whose perimeter is \(4r\) has area \(r^2\). | Rally Coach |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 25 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Segments and compound shapes, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| \(s = r\theta\) only works with \(\theta\) in radians. | If the angle is given as \(60^\circ\), convert it to \(\dfrac{\pi}{3}\) first; \(9 \times 60 = 540\) cm is nonsense. |
| Forgetting the \(\tfrac{1}{2}\) doubles the answer. | A useful check: the sector must be smaller than the whole circle, \(\pi r^2\). |
| The triangle in a segment has area \(\tfrac{1}{2}r^2\sin\theta\), using the angle at the centre, not \(\tfrac{1}{2} \times\) chord \(\times r\). | And in a perimeter, use the straight chord, not another arc. |
Arc length and sector area: worksheet
6 questions, 25 marks. Written in the style of the papers; no past paper question is reproduced.
A sector has radius 9 cm and angle \(\dfrac{2\pi}{9}\) radians. Find the exact arc length and the exact area.
A sector of a circle of radius 12 cm has a perimeter of 40 cm. Find the angle of the sector and its area.
A chord \(PQ\) of a circle with centre \(O\) and radius 10 cm subtends an angle of 2.2 radians at \(O\). Find the area and the perimeter of the minor segment cut off by \(PQ\).
A sector \(OAB\) has radius \(r\) cm and a perimeter of \(4r\) cm.
(a) Show that the angle \(AOB\) is 2 radians.
(b) Show that the area of the minor segment cut off by the chord \(AB\) is \(\tfrac{1}{2}r^2(2 - \sin 2)\), and find its value when \(r = 5\).
Sector \(OAB\) has centre \(O\), radius 10 cm and angle \(AOB = \dfrac{\pi}{6}\). The tangent to the arc at \(A\) meets the line \(OB\) extended at \(C\).
(a) Find the exact lengths of \(AC\) and \(OC\).
(b) Find the exact perimeter of the region bounded by the arc \(AB\) and the lines \(BC\) and \(CA\).
(c) Find the exact area of this region.
A windscreen wiper arm \(OP\) of length 50 cm rotates about \(O\) through an angle of 2.2 radians. The rubber blade lies along the arm from 15 cm to 50 cm from \(O\).
(a) Find the area of windscreen wiped by the blade.
(b) Find the perimeter of the wiped region.
(c) Find the straight-line distance between the positions of the tip \(P\) at the two ends of its sweep.
Answer key and marking notes
- Arc \(= 9 \times \dfrac{2\pi}{9} = 2\pi\) cm
- Area \(= \tfrac{1}{2} \times 81 \times \dfrac{2\pi}{9} = 9\pi\ \text{cm}^2\)
- Arc \(= 40 - 24 = 16\), so \(\theta = \dfrac{16}{12} = \dfrac{4}{3}\) radians.
- Area \(= \tfrac{1}{2} \times 144 \times \dfrac{4}{3} = 96\ \text{cm}^2\)
- Area \(= \tfrac{1}{2} \times 100 \times (2.2 - \sin 2.2) = 50(2.2 - 0.80850) = 69.6\ \text{cm}^2\)
- Arc \(= 10 \times 2.2 = 22\); chord \(= 2 \times 10\sin 1.1 = 17.82\)
- Perimeter \(= 22 + 17.82 = 39.8\) cm
- (a) \(2r + r\theta = 4r\), so \(r\theta = 2r\) and \(\theta = 2\).
- (b) Segment = sector \(-\) triangle \(= \tfrac{1}{2}r^2 \times 2 - \tfrac{1}{2}r^2\sin 2 = \tfrac{1}{2}r^2(2 - \sin 2)\).
- When \(r = 5\): \(12.5 \times (2 - 0.90930) = 13.6\ \text{cm}^2\)
- (a) Angle \(OAC = \dfrac{\pi}{2}\). \(AC = 10\tan\dfrac{\pi}{6} = \dfrac{10}{\sqrt{3}} = \dfrac{10\sqrt{3}}{3}\); \(OC = \dfrac{10}{\cos(\pi/6)} = \dfrac{20}{\sqrt{3}} = \dfrac{20\sqrt{3}}{3}\).
- (b) Arc \(= 10 \times \dfrac{\pi}{6} = \dfrac{5\pi}{3}\); \(BC = \dfrac{20\sqrt{3}}{3} - 10\).
- Perimeter \(= \dfrac{5\pi}{3} + \dfrac{20\sqrt{3}}{3} - 10 + \dfrac{10\sqrt{3}}{3} = \dfrac{5\pi}{3} + 10\sqrt{3} - 10\) cm
- (c) Triangle \(OAC = \tfrac{1}{2} \times 10 \times \dfrac{10\sqrt{3}}{3} = \dfrac{50\sqrt{3}}{3}\); sector \(= \tfrac{1}{2} \times 100 \times \dfrac{\pi}{6} = \dfrac{25\pi}{3}\).
- Area \(= \dfrac{50\sqrt{3}}{3} - \dfrac{25\pi}{3}\ \text{cm}^2\)
- (a) The wiped region is a large sector minus a small sector: \(\tfrac{1}{2} \times 2.2 \times (50^2 - 15^2) = 1.1 \times 2275 = 2502.5\), so \(2500\ \text{cm}^2\) (3 s.f.).
- (b) Outer arc \(= 50 \times 2.2 = 110\); inner arc \(= 15 \times 2.2 = 33\); two straight edges of \(50 - 15 = 35\).
- Perimeter \(= 110 + 33 + 35 + 35 = 213\) cm
- (c) The chord is \(2 \times 50\sin 1.1 = 100 \times 0.89121 = 89.1\) cm
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
A sector of a circle has radius 7.5 cm and angle 1.2 radians. Find the length of its arc.
- A9 cm
- B33.75 cm
- C0.157 cm
- D6.25 cm
A sector has radius 9 cm and angle \(\dfrac{2\pi}{9}\) radians. Find the exact area of the sector.
- A\(9\pi\ \text{cm}^2\)
- B\(18\pi\ \text{cm}^2\)
- C\(2\pi\ \text{cm}^2\)
- D\(\pi\ \text{cm}^2\)
A sector of a circle of radius 5 cm has an arc of length 14 cm. Find the area of the sector.
- A\(35\ \text{cm}^2\)
- B\(70\ \text{cm}^2\)
- C\(4.46\ \text{cm}^2\)
- D\(24\ \text{cm}^2\)
A sector of a circle of radius 6 cm has a perimeter of 21 cm. Find the angle of the sector in radians.
- A1.5
- B3.5
- C2.5
- D0.667
A chord of a circle of radius 8 cm subtends an angle of 1.8 radians at the centre. Find the area of the minor segment cut off by the chord.
- A\(26.4\ \text{cm}^2\)
- B\(56.6\ \text{cm}^2\)
- C\(52.9\ \text{cm}^2\)
- D\(64.9\ \text{cm}^2\)
A chord \(AB\) subtends an angle of \(\dfrac{2\pi}{3}\) at the centre of a circle of radius 10 cm. Find the exact perimeter of the minor segment.
- A\(\dfrac{20\pi}{3} + 10\sqrt{3}\) cm
- B\(\dfrac{20\pi}{3} + 20\) cm
- C\(\dfrac{20\pi}{3} + 5\sqrt{3}\) cm
- D\(\dfrac{10\pi}{3} + 10\sqrt{3}\) cm
A sector has area \(27\ \text{cm}^2\) and perimeter 21 cm. Find the possible values of the radius \(r\).
- A\(r = 4.5\) or \(r = 6\)
- B\(r = 6\) only
- C\(r = 1.5\) or \(r = 9\)
- D\(r = 4.5\) only
In the diagram, \(OAB\) is a sector of a circle with centre \(O\), radius 6 cm and angle 0.9 radians. The tangent to the arc at \(A\) meets \(OB\) extended at \(C\). Find the area of the shaded region bounded by the arc \(AB\), \(BC\) and \(CA\).
- A\(6.48\ \text{cm}^2\)
- B\(29.2\ \text{cm}^2\)
- C\(12.8\ \text{cm}^2\)
- D\(22.7\ \text{cm}^2\)
Quiz key
\(s = r\theta = 7.5 \times 1.2 = 9\) cm. The tempting 33.75 is \(\tfrac{1}{2}r^2\theta\), the sector's area, not its arc. 0.157 cm uses the degree formula \(\dfrac{\theta}{360} \times 2\pi r\) with an angle that is already in radians, and 6.25 cm divides \(r\) by \(\theta\).
\(A = \tfrac{1}{2} \times 9^2 \times \dfrac{2\pi}{9} = \tfrac{1}{2} \times 81 \times \dfrac{2\pi}{9} = 9\pi\ \text{cm}^2\). The tempting \(18\pi\) leaves out the \(\tfrac{1}{2}\). \(2\pi\) is the arc length \(r\theta\), and \(\pi\) uses \(\tfrac{1}{2}r\theta\) with \(r\) instead of \(r^2\).
\(\theta = \dfrac{s}{r} = \dfrac{14}{5} = 2.8\) radians, so \(A = \tfrac{1}{2} \times 5^2 \times 2.8 = 35\ \text{cm}^2\) (equivalently \(A = \tfrac{1}{2}rs\)). The tempting \(70\) leaves out the \(\tfrac{1}{2}\). \(4.46\) uses \(\theta = \dfrac{r}{s} = \dfrac{5}{14}\), the ratio upside down, and 24 is the perimeter \(14 + 5 + 5\) in cm.
\(2r + r\theta = 21\) gives \(12 + 6\theta = 21\), so \(\theta = \dfrac{9}{6} = 1.5\) radians. The tempting 3.5 is \(\dfrac{21}{6}\), treating the whole perimeter as the arc. 2.5 takes off only one radius, and 0.667 is \(\dfrac{6}{9}\), the division upside down.
Area \(= \tfrac{1}{2}r^2\theta - \tfrac{1}{2}r^2\sin\theta = 32(1.8 - \sin 1.8) = 32(1.8 - 0.9738\ldots) = 26.4\ \text{cm}^2\). The tempting 56.6 takes \(\sin 1.8\) in degree mode, so almost nothing is subtracted. 52.9 forgets the \(\tfrac{1}{2}\) on both parts, and 64.9 uses \(\cos 1.8\) in place of \(\sin 1.8\).
Arc \(= 10 \times \dfrac{2\pi}{3} = \dfrac{20\pi}{3}\). Chord \(= 2 \times 10\sin\dfrac{\pi}{3} = 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3}\). Perimeter \(= \dfrac{20\pi}{3} + 10\sqrt{3}\) cm. The tempting \(\dfrac{20\pi}{3} + 20\) adds two radii, which is the perimeter of the sector, not the segment. \(5\sqrt{3}\) is only half the chord, and \(\dfrac{10\pi}{3}\) halves the arc.
\(r\theta = 21 - 2r\), so the area is \(\tfrac{1}{2}r(21 - 2r) = 27\), giving \(2r^2 - 21r + 54 = 0\), \((2r - 9)(r - 6) = 0\), so \(r = 4.5\) or \(r = 6\). Both work: \(\theta = \dfrac{12}{4.5} = 2.67\) or \(\theta = \dfrac{9}{6} = 1.5\), each less than \(2\pi\). The tempting "6 only" stops at one root. \(1.5\) or \(9\) comes from forgetting the \(\tfrac{1}{2}\): \(r(21 - 2r) = 27\).
In right-angled triangle \(OAC\), \(AC = 6\tan 0.9 = 7.561\ldots\), so its area is \(\tfrac{1}{2} \times 6 \times 6\tan 0.9 = 22.68\ldots\). The sector is \(\tfrac{1}{2} \times 36 \times 0.9 = 16.2\). Shaded area \(= 22.68 - 16.2 = 6.48\ \text{cm}^2\). The tempting 29.2 forgets the \(\tfrac{1}{2}\) in the triangle. 12.8 uses \(OC = \dfrac{6}{\cos 0.9}\) as the height instead of \(AC\), and 22.7 is the triangle alone, without taking the sector away.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.