1. The longest side is the ‘hypotenuse’. Students could be presented with triangles in a variety of orientations and discuss how they can identify the hypotenuse. They could try identifying right-angled triangles within other common figures, such as diagonals in rectangles and heights in isosceles and other triangles.

2. The formula given is c² = a² + b². These letters are used throughout the programme. This could be compared with versions of the formula given in classroom texts. Students could be challenged to express the theorem algebraically for triangles labelled in a different order, or with a different set of letters, or using vertices to name the sides.

3. The height of the tower is 15m, and the distance of the anchor point from the tower is 25m. The length of rope needed was found to be 30m (to the nearest 10m). Pupils could work through this calculation and then generate more examples in the same context. For example, what if the tower was 12m high? What if the field was bigger or smaller?

4. Students could find the symbol on their calculators and establish how to use the function.

5. The Babylonians, the Chinese and the Egyptians were mentioned in the programme. Pythagoras’ school produced the earliest known proof of the theorem. The Babylonians knew many Pythagorean triples. The Egyptians knew the (3,4,5) triple and used it as shown in the programme. The Chinese produced the first diagrammatic proof of the theorem. The Hindus also knew the theorem and discovered many triples. According to the archaeologist Alexander Thom, megalithic man used properties of right-angled triangles to construct stone circles. The theorem entered the Guinness Book of Records in 1995 as the most proved theorem of all time.

6. 12 knots can be counted (actually 13 are needed to produce the necessary 12 spaces, but two are coincident here). Pupils could replicate the experiment using string. Equal spaces may be constructed more easily if the string is marked rather than knotted.

7. (6,8,10) and (5,12,13) are illustrated in the programme. Some more are: (8,15,17), (7,24,25), (20,21,29), (12,35,37), (9,40,41), (28,45,53), (11,60,61), (16,63,65), (33,56,65), (48,55,73). These are all ‘primitive’ triples: that is, they are not multiples of a smaller triple. For more information on triples see the ‘Further Ideas’ and ‘Background’ sections.

8. Yes, it is Ben in the boat and Katie at the top of the cliff.

9. The units in the question are metres. As Ben explains, he clearly should have spotted his mistake since 10,600 metres is inappropriate for a length of rope. You could discuss ways of checking answers and units for measuring lengths.

10. The guy ropes are 10m long. The mast was 9m high. The formula was rearranged to give b² = c² — a². Students could look at this example in detail and answer the final question asked in the programme: How far should the engineers fix the guy ropes from the base of the mast?

Worksheet 2: Tick or Trash

Question 1

Student B is correct. Student A has forgotten to halve the base before applying Pythagoras’ Theorem. Student B’s use of a clear diagram was helpful.

Question 2

Student B is correct. Student A has assumed that the side to be found is the hypotenuse. Student A’s stated formula a² = b² + c² does not correspond with the a given in the question. The orientation of the triangle could also have contributed to the confusion. Students could compare the rounding of answers in the two solutions, and their appropriateness.

Worksheet 3: Exam Practise Questions (Edexcel)

This is a selection of past exam questions from Edexcel, targeted at the middle ability range. You can use the sequence as it stands or select individual questions to suit your needs.

Given below is the mark scheme for each question, which will help you to see how marks are allocated and what the examiner is especially looking out for. The mark schemes use the following abbreviations:
oe
or equivalent
cao
correct answer only
ft
follow-through marks
dep
dependent
indep
independent
M
method marks
A
accuracy marks
B
benefit-of-doubt mark
SC
special case

Question 1

Diagonals cross at centre

20² + 20² or 800

= 28.28...

= 28.3 cm

root(20² + 20²)

Notes:

B1 (can be implied).M1. M1 (dep on 1^st M1). A1 cao for 28.3 or better.

Question 2

(a) isosceles

(b) ¹/₂ x 28.6 x (19.7 — 11.3) = 120.12 m²

root(275.05)

= 16.6 m

(d) tan x = ^8.4/_14.3 = 0.5874 = 30.4º

cos x= ^14.3/_16.6 = 0.8614 = 30.4º

sin x = ^8.4/_16.6 = 0.5060 = 30.4º

Notes:

(a) B1 cao (recognisable).

(b) M1 for ¹/₂ x 28.6 x (19.7 — 11.3) or ¹/₂ x 28.6 x 8.4. A1 cao for 120.1, 120.12; A0 for 120.

(c) M1 squaring and adding ‘height’ and ‘base’. M1 (dep on M1) finding the square root. A1 cao anything rounding to 16.6.

(d) M1 tan = ^‘8.4’/_‘14.3’ (oe). M1 (dep) finding tan^—1. A1 cao anything rounding to 30.4.

Question 3

root(18² — 12²) = 13.4

Question 4

(a) 5.3² + ‘0.9’² = 5.38 (C)

(b) tan angle = ^‘0.9’/_5.3; angle = 9.6º (B)

Question 5

root(15²+ 12²) = 19.2cm