The shape of the lesson
By the end of the lesson, learners can
- The square on the hypotenuse: In the lab, drag the corners of the right-angled triangle and watch the three squares.
- Pythagoras everywhere: Many shapes hide a right-angled triangle: the diagonal of a rectangle, the height of an isosceles triangle, the distance between two points on a grid.
- Exact answers and multi-step problems: Without a calculator, leave the answer as a surd in its simplest form.
- Sketch: draw the right-angled triangle and mark the right angle.
- Label: the hypotenuse \(c\) is opposite the right angle.
- Hypotenuse missing: \(c^2 = a^2 + b^2\), then square root.
- Shorter side missing: \(a^2 = c^2 - b^2\), then square root.
- Check: the hypotenuse must be the longest side; round to 3 s.f. at the end.
One picture every idea returns to
In a right-angled triangle with hypotenuse \(c\) (the side opposite the right angle) and shorter sides \(a\) and \(b\): \(a^2 + b^2 = c^2\). To find the hypotenuse, add the squares; to find a shorter side, subtract.
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 squares make the third.. In every right-angled triangle the square on the longest side has exactly the same area as the squares on the other two sides put together. 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: The square on the hypotenuse | In the lab, drag the corners of the right-angled triangle and watch the three squares. Project tab 1 of the lab (Pythagoras and trigonometry lab, Pythagoras' theorem) and let a student drive it while the class predicts. Model the worked example Finding the hypotenuse on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Pythagoras everywhere | Many shapes hide a right-angled triangle: the diagonal of a rectangle, the height of an isosceles triangle, the distance between two points on a grid. Model the worked example Height and area of an isosceles triangle on the board, one line at a time. Your turn: A rectangle measures 9 cm by 4 cm. Find the length of its diagonal. | Pairs |
| 21 to 30 min | Teach and check: Exact answers and multi-step problems | Without a calculator, leave the answer as a surd in its simplest form. Model the worked example Exact answer, without a calculator on the board, one line at a time. | 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, 13 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 Exact answers and multi-step problems, 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 |
|---|---|
| Adding the squares when you need a shorter side. | If the hypotenuse is given, you must SUBTRACT. Using \(6.5^2 + 1.8^2\) for the ladder gives 6.74 m, longer than the ladder itself, which is impossible. |
| For the isosceles triangle, use HALF the base. | Using the full 10 cm gives \(\sqrt{169 - 100} = 8.31\) cm, which is wrong. |
| Rounding too early. | If a first length is \(\sqrt{97} = 9.848\ldots\), keep it in the calculator for the next step; using 9.8 can change the final 3 s.f. answer. |
Pythagoras' theorem: worksheet
6 questions, 13 marks. Written in the style of the papers; no past paper question is reproduced.
A right-angled triangle has shorter sides 8.5 cm and 6.2 cm. Calculate the length of the hypotenuse.
The hypotenuse of a right-angled triangle is 14.2 cm and one of the shorter sides is 11.3 cm. Calculate the length of the third side.
Without a calculator, find the length of the hypotenuse of a right-angled triangle whose shorter sides are 3 cm and 6 cm. Give your answer in the form \(k\sqrt{5}\).
A triangle has sides 20 cm, 21 cm and 29 cm. Show that the triangle is right-angled.
Calculate the length of the line joining \(P(1, -3)\) and \(Q(6, 9)\).
A ship sails 12 km due north, then 7 km due east, then a further 5 km due north.
Calculate the straight-line distance of the ship from its starting point.
Answer key and marking notes
- \(c^2 = 8.5^2 + 6.2^2 = 72.25 + 38.44 = 110.69\)
- \(c = \sqrt{110.69} = 10.5\) cm (3 s.f.)
- \(a^2 = 14.2^2 - 11.3^2 = 201.64 - 127.69 = 73.95\)
- \(a = \sqrt{73.95} = 8.60\) cm (3 s.f.)
- \(c^2 = 3^2 + 6^2 = 9 + 36 = 45\)
- \(c = \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}\) cm
- The longest side is 29 cm: \(29^2 = 841\).
- \(20^2 + 21^2 = 400 + 441 = 841\).
- Since \(20^2 + 21^2 = 29^2\), by the converse of Pythagoras' theorem the triangle is right-angled (the right angle is opposite the 29 cm side).
- Horizontal change \(6 - 1 = 5\); vertical change \(9 - (-3) = 12\).
- \(PQ = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\) units
- Total distance north: \(12 + 5 = 17\) km. Total distance east: 7 km.
- Distance\(^2 = 17^2 + 7^2 = 289 + 49 = 338\)
- Distance \(= \sqrt{338} = 18.4\) km (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.
Triangle ABC has a right angle at A, with AB = 13 cm and AC = 8 cm. Find the length \(x\) of BC.
- A15.3 cm
- B10.2 cm
- C21 cm
- D233 cm
A rectangle has a diagonal of length 17.5 cm and a width of 10.5 cm. Find the length of the rectangle.
- A14 cm
- B20.4 cm
- C7 cm
- D196 cm
Find the distance between the points \(P(-2, 5)\) and \(Q(4, -3)\).
- A10 units
- B2.83 units
- C14 units
- D100 units
Which of these sets of side lengths makes a right-angled triangle?
- A9 cm, 40 cm, 41 cm
- B6 cm, 8 cm, 11 cm
- C5 cm, 12 cm, 14 cm
- D10 cm, 24 cm, 25 cm
Without a calculator, find the length of the hypotenuse of a right-angled triangle whose shorter sides are 3 cm and 9 cm. Give your answer as a surd in its simplest form.
- A\(3\sqrt{10}\) cm
- B\(9\sqrt{10}\) cm
- C\(6\sqrt{2}\) cm
- D\(2\sqrt{3}\) cm
An isosceles triangle has sides 17 cm, 17 cm and 16 cm. Find its area.
- A120 cm2
- B46.0 cm2
- C240 cm2
- D136 cm2
Quiz key
BC is the hypotenuse, so \(x^2 = 13^2 + 8^2 = 169 + 64 = 233\) and \(x = \sqrt{233} = 15.264\ldots = 15.3\) cm (3 s.f.). The tempting 10.2 cm SUBTRACTS the squares (\(\sqrt{105}\)), which is only right when the hypotenuse is given; it is also shorter than AB, impossible for a hypotenuse. 21 cm adds the lengths, and 233 cm forgets the square root.
The diagonal is the hypotenuse, so length \(= \sqrt{17.5^2 - 10.5^2} = \sqrt{306.25 - 110.25} = \sqrt{196} = 14\) cm. The tempting 20.4 cm adds the squares, giving a side longer than the diagonal. 7 cm subtracts the lengths before squaring, and 196 cm forgets the square root.
Horizontal change \(= 4 - (-2) = 6\), vertical change \(= -3 - 5 = -8\). Distance \(= \sqrt{6^2 + 8^2} = \sqrt{100} = 10\) units. The tempting 2.83 comes from dropping the negative signs (\(4 - 2 = 2\) and \(5 - 3 = 2\), so \(\sqrt{8}\)). 14 adds the changes instead of using Pythagoras, and 100 forgets the square root.
\(9^2 + 40^2 = 81 + 1600 = 1681 = 41^2\), so 9, 40, 41 is right-angled. For 6, 8, 11: \(36 + 64 = 100 \ne 121\). For 5, 12, 14: \(169 \ne 196\). The tempting 10, 24, 25 looks like a doubled 5, 12, 13 triangle, but that would be 10, 24, 26: \(100 + 576 = 676 \ne 625\).
\(c^2 = 9 + 81 = 90\), so \(c = \sqrt{90} = \sqrt{9 \times 10} = 3\sqrt{10}\) cm. The tempting \(9\sqrt{10}\) takes 9 outside the root instead of \(\sqrt{9} = 3\). \(6\sqrt{2} = \sqrt{72}\) subtracts the squares, and \(2\sqrt{3} = \sqrt{12}\) adds the sides without squaring them.
The height meets the base at its midpoint, so it is \(\sqrt{17^2 - 8^2} = \sqrt{225} = 15\) cm. Area \(= \tfrac{1}{2} \times 16 \times 15 = 120\) cm2. The tempting 46.0 cm2 uses the FULL base in Pythagoras (\(\sqrt{17^2 - 16^2} = \sqrt{33}\)). 240 cm2 forgets the half, and 136 cm2 uses the sloping side 17 cm as the height.
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.