The shape of the lesson
By the end of the lesson, learners can
- Where the formula comes from: In the lab, fix two sides and swing the angle between them.
- Working backwards: If the area is given, the formula becomes an equation.
- Polygons, segments and three sides: Many area problems split into triangles.
- Formula: Area \(= \tfrac{1}{2}ab\sin C\), with \(C\) between \(a\) and \(b\).
- Obtuse angles: work exactly the same way; \(\sin C\) is still positive.
- Backwards: \(\sin C = \dfrac{2 \times \text{Area}}{ab}\); consider \(180^\circ - C\) as well.
- Three sides: cosine rule for an angle first.
- Segment: sector area minus triangle area, both using the angle at the centre.
One picture every idea returns to
Area \(= \tfrac{1}{2}ab\sin C\), where \(C\) is the angle BETWEEN the sides \(a\) and \(b\). It works because the perpendicular height onto side \(a\) is \(b\sin C\).
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, Half base times height, without the height.. Extended. 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: Where the formula comes from | In the lab, fix two sides and swing the angle between them. Project tab 3 of the lab (Sine and cosine rules lab, area half ab sin C) and let a student drive it while the class predicts. Model the worked example Finding the area on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Working backwards | If the area is given, the formula becomes an equation. Model the worked example Finding the angle on the board, one line at a time. Your turn: A triangle has sides 6 cm and 10 cm and an area of 18 cm 2. Find the two possible values of the angle between those sides. | Pairs |
| 21 to 30 min | Teach and check: Polygons, segments and three sides | Many area problems split into triangles. Model the worked example A regular hexagon on the board, one line at a time. | Round Table |
| 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, 16 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 Polygons, segments and three sides, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 |
|---|---|
| Using an angle that is not between the two sides. | If you know sides 9 cm and 12 cm and an angle opposite one of them, first find the included angle (with the sine rule), then use the formula. |
| Forgetting the \(\tfrac{1}{2}\) when working backwards. | With area 30 cm 2 and sides 8 and 10, \(\sin C = \dfrac{30}{80}\) gives \(22.0^\circ\), which is wrong. The half becomes a 2 when you rearrange: \(\sin C = \dfrac{2 \times \text{Area}}{ab}\). |
| In a segment problem, using the chord length as a side in \(\tfrac{1}{2}ab\sin C\). | The two sides of the triangle are the RADII, and \(C\) is the angle at the centre. |
Area of a triangle: half ab sin C: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
Calculate the area of triangle \(ABC\) in which \(a = 7.4\) cm, \(b = 5.9\) cm and \(C = 63^\circ\).
Two sides of a triangle are 12 cm and 15 cm, and the angle between them is \(145^\circ\). Calculate the area.
A triangle has area 45 cm2. Two of its sides meet at an angle of \(28^\circ\). One of these sides is 14 cm. Calculate the length of the other.
A triangle has sides 6 cm and 10 cm and an area of 18 cm2. Find the two possible sizes of the angle between these two sides.
Without a calculator, show that the area of an equilateral triangle of side \(x\) cm is \(\dfrac{\sqrt{3}}{4}x^2\) cm2.
A circular table top has radius 50 cm and centre \(O\). A straight cut \(AB\) removes a segment, where angle \(AOB = 80^\circ\).
(a) Calculate the area of triangle \(AOB\).
(b) Calculate the area of the segment removed.
Answer key and marking notes
- Area \(= \tfrac{1}{2} \times 7.4 \times 5.9 \times \sin 63^\circ\)
- \(= 19.5\) cm2 (3 s.f.)
- Area \(= \tfrac{1}{2} \times 12 \times 15 \times \sin 145^\circ = 90\sin 145^\circ\)
- \(= 51.6\) cm2 (3 s.f.)
- \(45 = \tfrac{1}{2} \times 14 \times x \times \sin 28^\circ = 7x\sin 28^\circ\)
- \(x = \dfrac{45}{7\sin 28^\circ} = 13.7\) cm (3 s.f.)
- \(18 = \tfrac{1}{2} \times 6 \times 10 \times \sin C = 30\sin C\), so \(\sin C = 0.6\).
- \(C = \sin^{-1}(0.6) = 36.9^\circ\)
- or \(C = 180^\circ - 36.9^\circ = 143.1^\circ\)
- Every angle of an equilateral triangle is \(60^\circ\), and \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\).
- Area \(= \tfrac{1}{2} \times x \times x \times \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{4}x^2\).
- (a) Area of triangle \(= \tfrac{1}{2} \times 50 \times 50 \times \sin 80^\circ = 1231.0\ldots\) so 1230 cm2 (3 s.f.)
- (b) Sector area \(= \dfrac{80}{360} \times \pi \times 50^2 = 1745.3\ldots\) cm2
- Segment \(= 1745.3\ldots - 1231.0\ldots = 514.3\ldots\)
- Segment area \(= 514\) cm2 (3 s.f.)
6 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.
In triangle ABC, BC = 9 cm, AC = 7 cm and angle ACB \(= 52^\circ\). Find the area of the triangle.
- A24.8 cm2
- B49.6 cm2
- C19.4 cm2
- D31.5 cm2
Two sides of a triangle are 10 cm and 13 cm, and the angle between them is \(140^\circ\). Find the area of the triangle.
- A41.8 cm2
- B65.0 cm2
- C83.6 cm2
- D49.8 cm2
A triangle has sides 9 cm and 12 cm and an area of 40 cm2. The angle between these sides is acute. Find it.
- A\(47.8^\circ\)
- B\(21.7^\circ\)
- C\(132.2^\circ\)
- D\(42.2^\circ\)
A triangle has an area of 60 cm2. Two of its sides meet at an angle of \(40^\circ\), and one of these sides is 15 cm. Find the length of the other side.
- A12.4 cm
- B6.22 cm
- C10.4 cm
- D8 cm
A regular pentagon is made of five identical triangles meeting at its centre O. The distance from O to each vertex is 8 cm. Find the area of the pentagon.
- A152 cm2
- B30.4 cm2
- C139 cm2
- D304 cm2
A circle has centre O and radius 9 cm. A chord AB cuts off a minor segment, and angle AOB \(= 70^\circ\). Find the area of the minor segment.
- A11.4 cm2
- B38.1 cm2
- C49.5 cm2
- D87.5 cm2
Quiz key
Area \(= \tfrac{1}{2} \times 9 \times 7 \times \sin 52^\circ = 24.82\ldots = 24.8\) cm2. The tempting 49.6 cm2 forgets the \(\tfrac{1}{2}\). 19.4 cm2 uses cosine instead of sine, and 31.5 cm2 is \(\tfrac{1}{2} \times 9 \times 7\), treating 7 cm as the perpendicular height.
Area \(= \tfrac{1}{2} \times 10 \times 13 \times \sin 140^\circ = 65 \times 0.6428 = 41.8\) cm2. The tempting 65.0 cm2 leaves out \(\sin 140^\circ\), as if the sides were base and height. 83.6 cm2 forgets the half, and 49.8 cm2 uses \(\cos 140^\circ\) (and ignores its minus sign).
\(40 = 54\sin C\), so \(\sin C = \dfrac{40}{54} = 0.7407\) and \(C = 47.8^\circ\). The tempting \(21.7^\circ\) forgets the half: \(\sin C = \dfrac{40}{108}\). \(132.2^\circ\) gives the same area but is obtuse, and \(42.2^\circ\) uses \(\cos^{-1}\) instead of \(\sin^{-1}\).
\(60 = \tfrac{1}{2} \times 15 \times b \times \sin 40^\circ\), so \(b = \dfrac{120}{15\sin 40^\circ} = 12.44\ldots = 12.4\) cm. The tempting 6.22 cm forgets to double 60 when removing the half. 10.4 cm divides by \(\cos 40^\circ\), and 8 cm ignores the angle altogether (\(\dfrac{120}{15}\)).
The angle at the centre of each triangle is \(360^\circ \div 5 = 72^\circ\), so each triangle has area \(\tfrac{1}{2} \times 8 \times 8 \times \sin 72^\circ = 30.43\) cm2 and the pentagon has area \(5 \times 30.43 = 152\) cm2 (3 s.f.). The tempting 30.4 cm2 is just one triangle. 139 cm2 uses \(60^\circ\), the hexagon angle, and 304 cm2 forgets the half.
Sector \(= \dfrac{70}{360} \times \pi \times 9^2 = 49.48\) cm2. Triangle \(= \tfrac{1}{2} \times 9 \times 9 \times \sin 70^\circ = 38.06\) cm2. Segment \(= 49.48 - 38.06 = 11.4\) cm2. The tempting 38.1 cm2 is only the triangle, and 49.5 cm2 only the sector. 87.5 cm2 ADDS them instead of subtracting.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
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.