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24 Oct – 30 Oct 2026
Plenary
Plenary
Simultaneous Equations
Question
In pairs, discuss the following problem. Amira and Ben are solving a pair of simultaneous equations:
3x + 2y = 16
y = x - 3
Amira says: 'We already know what y is in terms of x, so we can substitute y = x - 3 into the first equation. That gives 3x + 2(x - 3) = 16.'
Ben says: 'I think it's easier to rearrange the first equation to make y the subject, then substitute that into the second equation.'
Who has the better method? Work through both approaches together and decide whether they give the same solution. Be ready to explain which method you prefer and why.
In pairs, discuss the following problem. Amira and Ben are solving a pair of simultaneous equations:
3x + 2y = 16
y = x - 3
Amira says: 'We already know what y is in terms of x, so we can substitute y = x - 3 into the first equation. That gives 3x + 2(x - 3) = 16.'
Ben says: 'I think it's easier to rearrange the first equation to make y the subject, then substitute that into the second equation.'
Who has the better method? Work through both approaches together and decide whether they give the same solution. Be ready to explain which method you prefer and why.
Suggested approach / summary
Amira's method: substitute y = x - 3 into 3x + 2y = 16, giving 3x + 2(x - 3) = 16. Expand: 3x + 2x - 6 = 16, so 5x = 22, x = 4.4, and y = 1.4. Ben's method: rearrange 3x + 2y = 16 to y = (16 - 3x)/2, then substitute into y = x - 3: (16 - 3x)/2 = x - 3. Multiply by 2: 16 - 3x = 2x - 6, so 22 = 5x, x = 4.4, y = 1.4. Both give the same solution, (4.4, 1.4). Students should see that Amira's method is more efficient here because y is already the subject of the second equation, avoiding rearrangement. Alternative valid approaches include substituting x = y + 3 into the first equation, or using elimination, but substitution is the focus of this step.
Discussion prompts (from the original resource)
- Does it matter which equation you substitute into, and which variable you substitute? Why?
- What happens if you substitute y = x - 3 into the second equation instead of the first? What do you notice?
- How can you check your solution is correct? What does the solution represent graphically?
- Which method would be more efficient if the equations were 3x + 2y = 16 and 2x - y = 3? Does the 'best' method depend on the equations given?
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Plenary
Tables and Probability
Question
A Year 8 class at a school in Leeds is investigating whether students' preferred method of travelling to school is linked to their year group. They survey 200 students and record the results in the two-way table below.
| Travel method | Year 7 | Year 8 | Year 9 | Total |
|---|---|---|---|---|
| Walk | 28 | 22 | 18 | 68 |
| Bus | 24 | 26 | 30 | 80 |
| Car | 18 | 20 | 14 | 52 |
| Total | 70 | 68 | 62 | 200 |
A student says: "There is no link between year group and travel method because the totals are roughly the same in each column."
Discuss whether you agree with this statement. Use probabilities from the table to support your argument. In particular, calculate and compare:
- P(Walk) overall
- P(Walk | Year 7)
- P(Walk | Year 9)
- P(Bus | Year 7)
- P(Bus | Year 9)
What do these probabilities tell you about whether travel method depends on year group?
A Year 8 class at a school in Leeds is investigating whether students' preferred method of travelling to school is linked to their year group. They survey 200 students and record the results in the two-way table below.
| Travel method | Year 7 | Year 8 | Year 9 | Total |
|---|---|---|---|---|
| Walk | 28 | 22 | 18 | 68 |
| Bus | 24 | 26 | 30 | 80 |
| Car | 18 | 20 | 14 | 52 |
| Total | 70 | 68 | 62 | 200 |
A student says: "There is no link between year group and travel method because the totals are roughly the same in each column."
Discuss whether you agree with this statement. Use probabilities from the table to support your argument. In particular, calculate and compare:
- P(Walk) overall
- P(Walk | Year 7)
- P(Walk | Year 9)
- P(Bus | Year 7)
- P(Bus | Year 9)
What do these probabilities tell you about whether travel method depends on year group?
Suggested approach / summary
First calculate the overall probability: P(Walk) = 68/200 = 0.34. Then calculate the conditional probabilities: P(Walk | Year 7) = 28/70 = 0.4; P(Walk | Year 9) = 18/62 ≈ 0.29. Also P(Bus | Year 7) = 24/70 ≈ 0.34 and P(Bus | Year 9) = 30/62 ≈ 0.48. Since P(Walk | Year 7) ≠ P(Walk | Year 9), the probability of walking depends on the year group, so there is some association. The student's claim is therefore not fully correct: equal column totals do not imply no link; it is the conditional probabilities that matter. An alternative approach is to compare observed frequencies with expected frequencies if the variables were independent (e.g. expected Walk in Year 7 = (70 × 68)/200 = 23.8, compared with observed 28). A-Level students could extend this to a chi-squared test. Note that the differences are modest, so students should comment on the strength of evidence and possible sampling variability.
Discussion prompts (from the original resource)
- What is the difference between a probability taken from the whole table (e.g. P(Walk)) and a conditional probability taken from one row or column (e.g. P(Walk | Year 7))? Why must the denominator change?
- Do the conditional probabilities suggest that travel method is independent of year group? Students should compare, for example, P(Walk | Year 7) = 28/70 = 0.4 with P(Walk | Year 9) = 18/62 ≈ 0.29, and note that these are not equal, so the variables appear to be associated.
- How reliable are these conclusions? Discuss the effect of sample size, the fact that the differences are fairly small, and whether a larger survey or a chi-squared test (A-Level extension) would give stronger evidence.
- Can students construct a hypothetical table where the row and column totals are the same but there is no association at all? This helps them see what 'no link' would actually look like (expected frequencies proportional to totals).
- Encourage debate about the wording 'no link' — does a small difference count as a link? Students should justify their reasoning with numerical evidence rather than just intuition.
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Plenary
Expanding and Factorising
Question
In a Year 11 maths lesson at a school in Leeds, two students are comparing methods for expanding and factorising expressions with a single bracket. Amira says: 'To expand 5(2x + 3), I multiply both terms inside the bracket by 5, giving 10x + 15. To factorise 10x + 15, I just reverse the process and take out the highest common factor, which is 5, giving 5(2x + 3).' Ben says: 'That works, but what if the expression is 6x + 9? The highest common factor is 3, so it factorises to 3(2x + 3). But could you also write it as 1.5(4x + 6) or 0.5(12x + 18)? Are these all correct factorisations? And if they are, why do we always choose the highest common factor?' Discuss Ben's question with your partner. Decide whether all these factorisations are correct, and explain why mathematicians prefer one form over the others. Then consider: does the same reasoning apply when the expression involves negative numbers, such as -4x - 8?
In a Year 11 maths lesson at a school in Leeds, two students are comparing methods for expanding and factorising expressions with a single bracket. Amira says: 'To expand 5(2x + 3), I multiply both terms inside the bracket by 5, giving 10x + 15. To factorise 10x + 15, I just reverse the process and take out the highest common factor, which is 5, giving 5(2x + 3).' Ben says: 'That works, but what if the expression is 6x + 9? The highest common factor is 3, so it factorises to 3(2x + 3). But could you also write it as 1.5(4x + 6) or 0.5(12x + 18)? Are these all correct factorisations? And if they are, why do we always choose the highest common factor?' Discuss Ben's question with your partner. Decide whether all these factorisations are correct, and explain why mathematicians prefer one form over the others. Then consider: does the same reasoning apply when the expression involves negative numbers, such as -4x - 8?
Suggested approach / summary
First, expand each proposed factorisation to check equivalence: 3(2x + 3) = 6x + 9; 1.5(4x + 6) = 6x + 9; 0.5(12x + 18) = 6x + 9. So all are correct in value. However, the convention is to factorise fully by taking out the highest common factor, which here is 3, giving 3(2x + 3). This is because it is the simplest, most compact form, and it mirrors the process of simplifying fractions. For -4x - 8, the HCF of the coefficients is 4, but we can also take out -4. Taking out -4 gives -4(x + 2), which is often preferred as it leaves a positive coefficient for x inside the bracket, making it easier to interpret. Students may argue either way, but they should justify their choice. An alternative method is to factorise by inspection, asking 'what times what gives the terms?' and then checking by expanding.
Discussion prompts (from the original resource)
- Are 3(2x + 3), 1.5(4x + 6) and 0.5(12x + 18) all equivalent to 6x + 9? Students should expand each to verify, reinforcing that expanding is the inverse of factorising.
- Why is the highest common factor (HCF) the preferred factor? Discuss convention, simplicity, and how factorising fully is expected in exams such as GCSE (AQA, Edexcel, OCR).
- What happens with negative terms, e.g. -4x - 8? Should the factor be -4, giving -4(x + 2), or 4, giving 4(-x - 2)? Which is more useful and why?
- How does this link to numerical fractions and simplifying? For example, 6/9 simplifies to 2/3 by dividing by the HCF, not by 1.5 or 0.5.
- Can students create their own expressions and challenge a partner to factorise them fully, then compare methods?
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Plenary
Representing Data
Question
A Year 8 class at a school in Leeds is investigating whether there is a relationship between the number of hours students spend playing video games each week and their scores in a maths test (out of 100). They collect data from 12 students and plot the points on a scatter graph. The points show a general downward trend: as the number of hours of gaming increases, the maths test score tends to decrease. One student, Amira, says: 'This proves that playing video games makes you worse at maths!' Another student, Ben, disagrees: 'It only shows a relationship, not that gaming causes lower scores.'
In your groups, discuss: What does the scatter graph actually tell us? Is Amira correct? What other explanations could there be for the pattern? How could the class improve their investigation to be more confident about any conclusions?
A Year 8 class at a school in Leeds is investigating whether there is a relationship between the number of hours students spend playing video games each week and their scores in a maths test (out of 100). They collect data from 12 students and plot the points on a scatter graph. The points show a general downward trend: as the number of hours of gaming increases, the maths test score tends to decrease. One student, Amira, says: 'This proves that playing video games makes you worse at maths!' Another student, Ben, disagrees: 'It only shows a relationship, not that gaming causes lower scores.'
In your groups, discuss: What does the scatter graph actually tell us? Is Amira correct? What other explanations could there be for the pattern? How could the class improve their investigation to be more confident about any conclusions?
Suggested approach / summary
Encourage students to first describe what the scatter graph shows: a negative correlation (as one variable increases, the other tends to decrease). Then guide them to distinguish between correlation and causation. A suggested reasoning path: (1) Identify the two variables and the type of correlation. (2) Explain that a scatter graph can only show a relationship, not prove cause and effect. (3) Suggest plausible alternative explanations, including third variables. (4) Discuss limitations of the data collection (small sample, self-reported data, etc.). (5) Propose improvements such as a larger, random sample and controlling other factors. Accept any well-reasoned alternative methods, e.g. students might suggest carrying out an experiment or a longitudinal study, or using a line of best fit to make predictions (while noting predictions outside the data range are unreliable).
Discussion prompts (from the original resource)
- The difference between correlation and causation: a scatter graph shows a relationship between two variables, but it does not prove that one variable causes the other.
- Possible alternative explanations for the negative correlation, e.g. students who game more may also sleep less, do less homework, or have less revision time; or students who struggle with maths may avoid homework by gaming instead.
- The idea of a third (confounding) variable that could affect both variables, such as time spent on other activities, motivation, or prior attainment.
- The quality of the data: sample size (only 12 students), how 'hours of gaming' was measured (self-reported?), whether the test was a fair measure, and whether the sample is representative of all students.
- How to improve the investigation: collect data from more students, use a random sample, measure gaming hours more accurately (e.g. screen time apps), and consider controlling other variables.
- The strength of the correlation: do the points lie close to a line of best fit, or are they widely scattered? A weak correlation gives even less support for a strong claim.
- How to describe the relationship correctly: 'There is a negative correlation between hours of gaming and maths test scores' rather than 'Gaming causes lower maths scores'.
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Plenary
Congruence, Similarity and Enlargement
Question
In a Year 10 mathematics lesson at a school in Manchester, students are investigating enlargements with fractional scale factors. The rectangle ABCD has vertices at A(0, 0), B(4, 0), C(4, 2), and D(0, 2). It is enlarged by a scale factor of 1/2 with the centre of enlargement at the point P(1, 1). A student claims that the enlarged rectangle will have an area that is exactly one quarter of the original rectangle's area. Another student argues that the area will be half of the original because the scale factor is 1/2. As a group, discuss who is correct and why. Then, determine the coordinates of the enlarged rectangle's vertices and verify your answer by calculating the areas. Finally, consider how your reasoning would change if the centre of enlargement were at a different point, such as (2, 2).
In a Year 10 mathematics lesson at a school in Manchester, students are investigating enlargements with fractional scale factors. The rectangle ABCD has vertices at A(0, 0), B(4, 0), C(4, 2), and D(0, 2). It is enlarged by a scale factor of 1/2 with the centre of enlargement at the point P(1, 1). A student claims that the enlarged rectangle will have an area that is exactly one quarter of the original rectangle's area. Another student argues that the area will be half of the original because the scale factor is 1/2. As a group, discuss who is correct and why. Then, determine the coordinates of the enlarged rectangle's vertices and verify your answer by calculating the areas. Finally, consider how your reasoning would change if the centre of enlargement were at a different point, such as (2, 2).
Suggested approach / summary
To find the image of each vertex, use the formula: Image point = Centre + Scale Factor × (Original point - Centre). For centre P(1, 1) and scale factor 1/2: A'(0.5, 0.5), B'(2.5, 0.5), C'(2.5, 1.5), D'(0.5, 1.5). The original area is 4 × 2 = 8 square units. The image area is 2 × 1 = 2 square units, which is one quarter of the original area. This confirms that area scales by the square of the scale factor (1/2)^2 = 1/4. If the centre were (2, 2), the image coordinates would be different, but the area would still be 2 square units. Alternative method: Use the fact that the area scale factor is the square of the linear scale factor, so no coordinate calculation is needed to determine the area ratio.
Discussion prompts (from the original resource)
- The relationship between the scale factor and the area of the enlarged shape: area scales by the square of the scale factor.
- The effect of the centre of enlargement on the position of the image, but not on its size or area.
- How to calculate the coordinates of the image vertices using vectors from the centre of enlargement.
- The difference between linear scale factor and area scale factor, and common misconceptions.
- Extension: What if the scale factor were negative or greater than 1? How would the area change?
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Plenary
Equality and Equivalence
Question
In a Year 7 mathematics lesson at a school in Leeds, students are investigating fact families. The teacher writes on the board: 'If 7 + 5 = 12, then 12 - 5 = 7 and 12 - 7 = 5.' She then asks the class to work in pairs to create a similar fact family using multiplication and division, but with a twist: they must use the numbers 3, 4, and 12. After a few minutes, one student says, 'We can write 3 × 4 = 12, so 12 ÷ 4 = 3 and 12 ÷ 3 = 4. But can we also write 4 × 3 = 12? And does that mean 12 ÷ 3 = 4 and 12 ÷ 4 = 3 are the same?' Another student adds, 'What if we used 2, 6, and 12? Would that be a different fact family or the same one?' The teacher then challenges the class: 'Using the numbers 2, 3, 4, 6, and 12, how many different fact families can you create? Explain how you know you have found them all, and discuss whether the order of the numbers matters.'
In a Year 7 mathematics lesson at a school in Leeds, students are investigating fact families. The teacher writes on the board: 'If 7 + 5 = 12, then 12 - 5 = 7 and 12 - 7 = 5.' She then asks the class to work in pairs to create a similar fact family using multiplication and division, but with a twist: they must use the numbers 3, 4, and 12. After a few minutes, one student says, 'We can write 3 × 4 = 12, so 12 ÷ 4 = 3 and 12 ÷ 3 = 4. But can we also write 4 × 3 = 12? And does that mean 12 ÷ 3 = 4 and 12 ÷ 4 = 3 are the same?' Another student adds, 'What if we used 2, 6, and 12? Would that be a different fact family or the same one?' The teacher then challenges the class: 'Using the numbers 2, 3, 4, 6, and 12, how many different fact families can you create? Explain how you know you have found them all, and discuss whether the order of the numbers matters.'
Suggested approach / summary
Students should first identify that a fact family for multiplication and division consists of three numbers where two multiply to give the third, and the corresponding division facts. For the set {2, 3, 4, 6, 12}, they need to find all triples (a, b, c) such that a × b = c. Systematically, they can list pairs: (2,3,6), (2,4,8 - not in set), (2,6,12), (3,4,12), (3,6,18 - not in set), (4,6,24 - not in set), and also (1,12,12) if 1 is allowed, but 1 is not in the given set. So the valid triples are (2,3,6), (2,6,12), and (3,4,12). Each triple yields a fact family with four equations (e.g., 2×3=6, 3×2=6, 6÷3=2, 6÷2=3). However, the order of the factors does not create a new fact family; it's the same family. Also, note that 2, 6, 12 and 3, 4, 12 are distinct because the numbers involved are different. Some students might argue that 2×6=12 and 3×4=12 are different fact families even though they share the product 12, because the factors differ. The discussion should lead to the conclusion that there are three distinct fact families using the given numbers (excluding 1). An alternative approach is to use factor trees or a systematic list of all multiplication facts involving the numbers. The teacher should emphasise that fact families help us understand the inverse relationship between multiplication and division, and that the commutative property means we don't double-count families. This connects to algebraic thinking: if a × b = c, then c ÷ a = b and c ÷ b = a, regardless of the order of a and b.
Discussion prompts (from the original resource)
- What is a fact family? How do addition/subtraction fact families compare to multiplication/division fact families?
- For the numbers 3, 4, and 12, how many equations can be written? Does changing the order of the factors (3 × 4 vs 4 × 3) create a new fact family or the same one?
- If we use 2, 6, and 12, is that a different fact family from 3, 4, and 12? Why or why not? What about 1, 12, and 12?
- How can we systematically list all possible fact families using the numbers 2, 3, 4, 6, and 12? What strategy ensures we don't miss any?
- Does the commutative property (e.g., a × b = b × a) mean that some fact families are essentially the same? How does this affect our count?
- Can we extend this to algebraic expressions? For example, if a × b = c, what are the related equations? How does this connect to solving equations?
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Plenary
Algebraic Notation
Question
A new 'Oyster-style' travel card for the Manchester Metrolink charges a fixed £1.50 to tap in, plus 80p for each stop you travel. The total cost of a journey, in pence, can be found using a function machine where you input the number of stops, n. The machine first multiplies your input by 0.8, then adds 1.5, to give the output in pounds. In small groups, discuss: If Priya taps in at Didsbury and taps out at Piccadilly Gardens after travelling 6 stops, what is the output of the machine? Then, decide whether it would be cheaper for Priya to buy a single paper ticket costing £5.20 instead. Be ready to explain the order in which you applied the operations and why it matters.
A new 'Oyster-style' travel card for the Manchester Metrolink charges a fixed £1.50 to tap in, plus 80p for each stop you travel. The total cost of a journey, in pence, can be found using a function machine where you input the number of stops, n. The machine first multiplies your input by 0.8, then adds 1.5, to give the output in pounds. In small groups, discuss: If Priya taps in at Didsbury and taps out at Piccadilly Gardens after travelling 6 stops, what is the output of the machine? Then, decide whether it would be cheaper for Priya to buy a single paper ticket costing £5.20 instead. Be ready to explain the order in which you applied the operations and why it matters.
Suggested approach / summary
Substitute n = 6 into the expression 0.8n + 1.5. Following the function machine order (multiply by 0.8 first, then add 1.5): 0.8 × 6 = 4.8, then 4.8 + 1.5 = 6.3. So the travel card costs £6.30. Compare with the paper ticket at £5.20: since £6.30 > £5.20, the paper ticket is cheaper by £1.10. Alternative methods: some students may convert to pence first (80p × 6 = 480p, plus 150p = 630p = £6.30), or work backwards from £5.20 to find the break-even number of stops. Acknowledge that the function machine convention (multiply before adding) is essential to get the correct output.
Discussion prompts (from the original resource)
- Why does the order of operations in the function machine matter? What happens if you add 1.5 before multiplying by 0.8, and does this give a different (incorrect) cost?
- How do you substitute the numerical input n = 6 into the expression 0.8n + 1.5, and what does each part of the expression represent in the real-world context?
- How can you compare the output of the function machine with the paper ticket price of £5.20 fairly, ensuring both amounts are in the same units (pounds or pence)?
- What if Priya travelled a different number of stops – for which number of stops would the travel card and the paper ticket cost exactly the same? This encourages forming and solving a simple equation.
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Discuss
Discuss
Equality and Equivalence
Question
In today's starter, we're exploring like and unlike terms. Consider these two expressions: 3x + 2y and 3a + 2b. Are the terms in each expression like or unlike? Discuss with your partner: What makes terms 'like' in algebra? How can you tell if two terms are like or unlike? Be ready to share your reasoning with the class.
In today's starter, we're exploring like and unlike terms. Consider these two expressions: 3x + 2y and 3a + 2b. Are the terms in each expression like or unlike? Discuss with your partner: What makes terms 'like' in algebra? How can you tell if two terms are like or unlike? Be ready to share your reasoning with the class.
Suggested approach / summary
Start by identifying the variable parts of each term. In 3x + 2y, the terms are 3x and 2y. They have different variables (x and y), so they are unlike terms. In 3a + 2b, similarly, 3a and 2b are unlike terms. To determine if terms are like, compare the variables and their exponents; if they match exactly, they are like terms. Encourage students to create their own examples and non-examples to test their understanding.
Discussion prompts (from the original resource)
- Definition of like terms: terms that have the same variable(s) raised to the same power (e.g., 3x and 5x are like terms).
- Unlike terms: terms that have different variables or different powers (e.g., 3x and 2y are unlike terms).
- The role of coefficients: coefficients can be different, but the variable part must be identical for terms to be like.
- Examples: 4xy and -2xy are like terms; 4x^2 and 4x are unlike because the powers differ.
- Why it matters: only like terms can be combined by addition or subtraction.
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Discuss
Non-linear Graphs
Question
A cubic graph has the equation y = x³ − 4x. The points A, B and C all lie on the graph, where A has x = −2, B has x = 0 and C has x = 2. Without plotting any points, decide whether the graph is a straight line between A and C. Discuss with your partner: what shape do you think the curve makes between these points, and how could you check by plotting just a few extra points?
A cubic graph has the equation y = x³ − 4x. The points A, B and C all lie on the graph, where A has x = −2, B has x = 0 and C has x = 2. Without plotting any points, decide whether the graph is a straight line between A and C. Discuss with your partner: what shape do you think the curve makes between these points, and how could you check by plotting just a few extra points?
Suggested approach / summary
Students should substitute x-values into y = x³ − 4x and plot the resulting points on a coordinate grid. For example: x = −2 gives y = 0; x = −1 gives y = 3; x = 0 gives y = 0; x = 1 gives y = −3; x = 2 gives y = 0. Plotting these five points reveals an S-shaped curve that passes through (−2, 0), rises to a local maximum near (−1, 3), passes through the origin, dips to a local minimum near (1, −3), and returns to (2, 0). Additional points such as x = −1.5 (y = 2.625) and x = 1.5 (y = −2.625) can be plotted to confirm the curve's shape. Students should recognise that the curve is not a straight line between A and C, and that the '− 4x' term causes the wave-like dip and rise. An alternative approach is to use a graphing tool or table of values to generate the points systematically before drawing the curve.
Discussion prompts (from the original resource)
- Substituting x = −2, 0 and 2 gives y = 0, 0 and 0, so A, B and C all lie on the x-axis — does this mean the graph is flat between them?
- Choose some x-values between −2 and 2, such as x = −1, 0.5 and 1, and work out the y-values to see how the curve behaves.
- Look at the sign of y when x is between −2 and 0, and when x is between 0 and 2 — why does the curve dip below the x-axis in these intervals?
- What happens to y as x gets larger than 2 or smaller than −2 — how does this help you picture the overall shape of the cubic graph?
- Compare the shape of y = x³ − 4x with the basic cubic y = x³ — what effect does the '− 4x' term have on the curve?
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Discuss
Equality and Equivalence
Question
In a Year 7 mathematics lesson at a school in Leeds, two students are simplifying expressions and using the identity symbol (≡).
Amir writes:
5x + 3y − 2x + 4y ≡ 3x + 7y
Beth writes:
5x + 3y − 2x + 4y ≡ 3x + 7y + 0
Charlie writes:
5x + 3y − 2x + 4y ≡ 3x + 7y + 0x + 0y
Discuss with your partner:
(a) Are all three students correct? Explain your reasoning using the ≡ symbol.
(b) Charlie claims that adding 0x and 0y makes the expression 'more equivalent'. Is Charlie right? Why or why not?
(c) Write two different expressions that are equivalent to 4a − 3b + 2a + 5b, and explain how you know they are equivalent.
(d) A student says, 'If two expressions are equivalent, they must look exactly the same.' Is this statement true or false? Use examples to justify your answer.
In a Year 7 mathematics lesson at a school in Leeds, two students are simplifying expressions and using the identity symbol (≡).
Amir writes:
5x + 3y − 2x + 4y ≡ 3x + 7y
Beth writes:
5x + 3y − 2x + 4y ≡ 3x + 7y + 0
Charlie writes:
5x + 3y − 2x + 4y ≡ 3x + 7y + 0x + 0y
Discuss with your partner:
(a) Are all three students correct? Explain your reasoning using the ≡ symbol.
(b) Charlie claims that adding 0x and 0y makes the expression 'more equivalent'. Is Charlie right? Why or why not?
(c) Write two different expressions that are equivalent to 4a − 3b + 2a + 5b, and explain how you know they are equivalent.
(d) A student says, 'If two expressions are equivalent, they must look exactly the same.' Is this statement true or false? Use examples to justify your answer.
Suggested approach / summary
A suggested approach is to first simplify 5x + 3y − 2x + 4y by collecting like terms: 5x − 2x = 3x and 3y + 4y = 7y, giving 3x + 7y. Since all three students have 3x + 7y as the simplified result (with Beth and Charlie adding zero terms), all three are correct. For part (b), Charlie is incorrect that adding 0x and 0y makes it 'more equivalent'; adding zero does not change the value, so the expressions remain equivalent, but it is not 'more' equivalent. For part (c), 4a − 3b + 2a + 5b simplifies to 6a + 2b, so equivalent expressions could be 6a + 2b, 2b + 6a, or 6a + 2b + 0c. Students should justify by substituting values or by simplifying. For part (d), the statement is false; equivalent expressions need not look identical, e.g., 3(x + 2) ≡ 3x + 6. Alternative methods include using substitution tables, bar models, or algebra tiles to demonstrate equivalence. Teachers should encourage students to test with multiple values to confirm equivalence for all values, not just one.
Discussion prompts (from the original resource)
- The meaning of the identity symbol ≡ compared with the equals sign =: ≡ means the expressions are equal for all values of the variables, not just for one specific value.
- Whether adding 0x and 0y changes the value of an expression: students should test with numerical values (e.g., x = 2, y = 3) to see that the value is unchanged.
- The difference between equivalent expressions and identical expressions: 3x + 7y and 3x + 7y + 0x + 0y are equivalent but not written identically.
- How collecting like terms relies on the commutative and associative properties of addition, e.g., 5x − 2x = 3x and 3y + 4y = 7y.
- Whether 'simplest form' is a mathematical requirement or a convention: the simplest form is preferred, but other equivalent forms are still correct.
- How to justify equivalence: substituting values, simplifying both sides, or using algebraic reasoning with the ≡ symbol.
- Common misconceptions: thinking that 0x is 'nothing' and should never be written, or that expressions must be written in a particular order to be equivalent.
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Trigonometry
Question
A group of Year 10 students are on a school trip to the Eden Project in Cornwall. They are asked to calculate the height of the world's largest greenhouse, the Rainforest Biome, which is a domed structure. The students cannot measure the height directly because it is too tall. Instead, they stand at a point 30 metres from the base of the Biome and measure the angle of elevation to the top of the dome as 50°. Another student, standing 45 metres from the base, measures the angle of elevation as 35°. The students are puzzled because they expected the calculated height to be the same, but they got different answers. They realise that the ground is not perfectly level and the Biome is not a simple right-angled triangle. They decide to model the Biome as a right-angled triangle with the vertical height, horizontal distance, and the line of sight. They want to understand why the ratios of sides in similar right-angled triangles are constant, and how this can help them find the height if they know the angle and distance.
Discuss in your groups:
- Why do the students get different heights when using the two sets of measurements?
- How can the concept of similar triangles help resolve this?
- If the ground were level and the Biome's base were at the same height as the students' feet, what would be the height?
- What assumptions are being made in the trigonometric model?
- How might the actual height be found more accurately?
A group of Year 10 students are on a school trip to the Eden Project in Cornwall. They are asked to calculate the height of the world's largest greenhouse, the Rainforest Biome, which is a domed structure. The students cannot measure the height directly because it is too tall. Instead, they stand at a point 30 metres from the base of the Biome and measure the angle of elevation to the top of the dome as 50°. Another student, standing 45 metres from the base, measures the angle of elevation as 35°. The students are puzzled because they expected the calculated height to be the same, but they got different answers. They realise that the ground is not perfectly level and the Biome is not a simple right-angled triangle. They decide to model the Biome as a right-angled triangle with the vertical height, horizontal distance, and the line of sight. They want to understand why the ratios of sides in similar right-angled triangles are constant, and how this can help them find the height if they know the angle and distance.
Discuss in your groups:
- Why do the students get different heights when using the two sets of measurements?
- How can the concept of similar triangles help resolve this?
- If the ground were level and the Biome's base were at the same height as the students' feet, what would be the height?
- What assumptions are being made in the trigonometric model?
- How might the actual height be found more accurately?
Suggested approach / summary
Students should first calculate the height using each set of measurements assuming a right-angled triangle: For the first, height = 30 * tan(50°) ≈ 30 * 1.1918 = 35.75 m. For the second, height = 45 * tan(35°) ≈ 45 * 0.7002 = 31.51 m. They should notice the discrepancy. The key is to discuss that in similar triangles, the ratio of opposite to adjacent is constant for a given angle, so if the angle were the same, the ratio would be the same. Here, the angles differ, so the heights differ. If the ground were level and the base at the same level, the height should be the same regardless of distance, but measurement errors or the dome's curvature cause differences. A more accurate method might involve taking multiple measurements and using trigonometry with a known baseline, or using a clinometer and applying the tangent ratio from a single point with a known distance, ensuring the angle is measured from the horizontal. Alternatively, students could use the concept of similar triangles by creating a scale model. The discussion should lead to understanding that trigonometry relies on right-angled triangles and that real-world applications require careful measurement and consideration of assumptions.
Discussion prompts (from the original resource)
- Understanding that the tangent ratio (opposite/adjacent) is constant for a given angle in similar right-angled triangles.
- Recognising that real-world measurements may not form perfect right-angled triangles due to uneven ground or the structure's shape.
- Exploring how the angle of elevation changes with distance and how this affects the calculated height.
- Discussing the limitations of mathematical models and the importance of assumptions.
- Considering how to combine multiple measurements to estimate the height more accurately, perhaps by averaging or using more advanced trigonometry.
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Multiplicative Change
Question
In a Year 8 mathematics lesson at a school in Manchester, two students are investigating similar shapes using scale factors. Amira says: 'If I enlarge a shape by a scale factor of 3, then enlarge the result by a scale factor of 4, that is exactly the same as enlarging the original shape by a scale factor of 12.' Ben disagrees: 'No, you add the scale factors, so it is the same as enlarging by 7.' They then try a different case: reducing a shape by a scale factor of 1/2 and then by 1/3. Amira claims the combined effect is a scale factor of 1/6, while Ben says it is 5/6. Who is correct, and why? Use diagrams, numerical examples and algebraic reasoning to justify your answer. Then discuss: does the same rule apply to enlargements and reductions, and how does this connect to the area and perimeter of similar shapes?
In a Year 8 mathematics lesson at a school in Manchester, two students are investigating similar shapes using scale factors. Amira says: 'If I enlarge a shape by a scale factor of 3, then enlarge the result by a scale factor of 4, that is exactly the same as enlarging the original shape by a scale factor of 12.' Ben disagrees: 'No, you add the scale factors, so it is the same as enlarging by 7.' They then try a different case: reducing a shape by a scale factor of 1/2 and then by 1/3. Amira claims the combined effect is a scale factor of 1/6, while Ben says it is 5/6. Who is correct, and why? Use diagrams, numerical examples and algebraic reasoning to justify your answer. Then discuss: does the same rule apply to enlargements and reductions, and how does this connect to the area and perimeter of similar shapes?
Suggested approach / summary
Amira is correct: successive enlargements (or reductions) combine by multiplying the scale factors, not adding them. For example, start with a rectangle 2 cm by 3 cm. Enlarge by scale factor 3: sides become 6 cm and 9 cm. Enlarge again by scale factor 4: sides become 24 cm and 36 cm. The original sides were 2 cm and 3 cm, so the overall scale factor is 24 ÷ 2 = 12 and 36 ÷ 3 = 12, i.e. 3 × 4 = 12. For reductions, start with a shape and apply scale factor 1/2, then 1/3. The combined effect is 1/2 × 1/3 = 1/6, so Amira is again correct. Ben's error is adding the scale factors (3 + 4 = 7) or (1/2 + 1/3 = 5/6), which does not match the actual side lengths. Algebraically, if a shape is enlarged by scale factor a and then by scale factor b, any length L becomes L × a × b, so the overall scale factor is ab. This holds for any positive a and b. The order does not matter because multiplication is commutative. For area, if the linear scale factor is k, the area scale factor is k²; for two successive transformations with scale factors a and b, the overall area scale factor is (ab)² = a²b². Perimeter scales linearly with the scale factor. Alternative methods include using ratio tables or dynamic geometry software to verify the results.
Discussion prompts (from the original resource)
- Why do we multiply scale factors when carrying out two successive transformations, rather than adding them? Test with a simple shape such as a 2 cm by 3 cm rectangle and track how each side changes.
- Does the order of the two enlargements matter? Try enlarging by 3 then 4, and then by 4 then 3, and compare the results.
- How do the area and perimeter change under a combined scale factor? Investigate the relationship between the linear scale factor and the area scale factor for similar shapes.
- What happens when the scale factor is less than 1 (a reduction)? Does the multiplication rule still hold? Check with scale factors of 1/2 and 1/3.
- Can you express the combined effect algebraically? If the first scale factor is a and the second is b, what is the overall scale factor? Does this hold for any positive values of a and b?
- How does this connect to other topics, such as compound interest, unit conversions, or map scales? For example, a map scale of 1:25,000 followed by a photocopier enlargement of 200%.
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Constructions and Congruency
Question
In pairs, you are given a straight line segment AB drawn on a piece of paper, with A and B marked clearly. Your task is to find the exact midpoint of AB using only a pair of compasses and a ruler (no measuring!). Before you start, discuss with your partner: Why do you think the perpendicular bisector construction gives you the midpoint? Is it just a coincidence, or is there a mathematical reason? Once you have agreed on a method, carry out the construction and then compare your midpoint with another pair's midpoint. Are they the same? Why must they be the same? Be ready to explain your reasoning to the class.
In pairs, you are given a straight line segment AB drawn on a piece of paper, with A and B marked clearly. Your task is to find the exact midpoint of AB using only a pair of compasses and a ruler (no measuring!). Before you start, discuss with your partner: Why do you think the perpendicular bisector construction gives you the midpoint? Is it just a coincidence, or is there a mathematical reason? Once you have agreed on a method, carry out the construction and then compare your midpoint with another pair's midpoint. Are they the same? Why must they be the same? Be ready to explain your reasoning to the class.
Suggested approach / summary
To construct the perpendicular bisector of AB: Place the compass point on A and draw an arc (radius more than half of AB). Without changing the compass width, place the compass point on B and draw another arc intersecting the first. Label the intersection points C and D. Draw a straight line through C and D. This line is the perpendicular bisector of AB and crosses AB at its midpoint M. The reasoning: C and D are both equidistant from A and B, so they lie on the perpendicular bisector. The line CD is the set of all points equidistant from A and B, so it must cross AB at the point where the distances to A and B are equal, i.e., the midpoint. Alternative methods might include using a ruler to measure, but the compass construction is more accurate and doesn't rely on measurement. Some students might realise that the construction works for any radius greater than half AB, as long as both arcs use the same radius.
Discussion prompts (from the original resource)
- Why does drawing arcs from both A and B with the same radius create points that are equidistant from A and B?
- What is special about the line that passes through the two intersection points of the arcs? How does it relate to the line segment AB?
- Could you use a different radius for the arcs? What happens if the radius is too small or too large? Does the construction still work?
- How can you prove that the point where the perpendicular bisector crosses AB is exactly the midpoint?
- What would happen if you only drew one arc? Would you still be able to find the midpoint? Why or why not?
- How does this construction relate to the idea of symmetry? Is the line segment AB symmetric about the perpendicular bisector?
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Ratio and Scale
Question
In a Year 8 class at a school in Leeds, two students are comparing the ratio of boys to girls in their tutor groups. Amira says, 'The ratio of boys to girls in my tutor group is 3:4, so 3/4 of my tutor group are boys.' Ben says, 'The ratio of boys to girls in my tutor group is 4:5, so 4/5 of my tutor group are boys.' Priya listens and says, 'That means Ben's tutor group has a larger fraction of boys than Amira's.' Is Priya correct? Explain your reasoning and decide whether the fraction of boys in each tutor group can be compared directly from the ratios 3:4 and 4:5. If not, how could you compare them fairly?
In a Year 8 class at a school in Leeds, two students are comparing the ratio of boys to girls in their tutor groups. Amira says, 'The ratio of boys to girls in my tutor group is 3:4, so 3/4 of my tutor group are boys.' Ben says, 'The ratio of boys to girls in my tutor group is 4:5, so 4/5 of my tutor group are boys.' Priya listens and says, 'That means Ben's tutor group has a larger fraction of boys than Amira's.' Is Priya correct? Explain your reasoning and decide whether the fraction of boys in each tutor group can be compared directly from the ratios 3:4 and 4:5. If not, how could you compare them fairly?
Suggested approach / summary
First, correct the misconception: in a ratio of boys to girls of 3:4, the fraction of boys is 3/(3+4) = 3/7. Similarly, for 4:5, the fraction of boys is 4/(4+5) = 4/9. To compare 3/7 and 4/9, find a common denominator (63): 3/7 = 27/63 and 4/9 = 28/63. Since 28/63 > 27/63, Ben's tutor group has a slightly larger fraction of boys. Therefore Priya is correct, but her reasoning based on 3/4 and 4/5 is flawed. An alternative method is to convert to decimals: 3/7 ≈ 0.4286 and 4/9 ≈ 0.4444, confirming 4/9 is larger. Students might also use equivalent ratios with a common total, e.g., compare 27:63 and 28:63, or use bar models to visualise the parts.
Discussion prompts (from the original resource)
- Clarify the difference between a part-to-part ratio (boys:girls) and a part-to-whole fraction (boys:total students). For Amira, the fraction of boys is 3/(3+4) = 3/7, not 3/4.
- Compare the correct fractions 3/7 and 4/9 using common denominators or decimal equivalents to decide which is larger.
- Discuss whether comparing the ratios 3:4 and 4:5 directly as fractions is valid, and why the total number of students matters.
- Explore how changing the total number of students in each tutor group (while keeping the same ratio) affects the actual number of boys and girls, and whether the fraction of boys stays the same.
- Consider whether Priya's statement could ever be true under different interpretations, and how to communicate the correct reasoning clearly.
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Exit tickets
Exit ticket
Equations and Inequalities
Question
A straight line graph passes through the points \((0, 4)\) and \((3, -2)\).
(a) Draw the straight line graph on a set of axes for \(0 \le x \le 3\).
(b) Write down the equation of the line in the form \(y = mx + c\).
(c) A student says: "If I extend the line, it will pass through \((5, -6)\)." Is the student correct? Show working to justify your answer.
A straight line graph passes through the points \((0, 4)\) and \((3, -2)\).
(a) Draw the straight line graph on a set of axes for \(0 \le x \le 3\).
(b) Write down the equation of the line in the form \(y = mx + c\).
(c) A student says: "If I extend the line, it will pass through \((5, -6)\)." Is the student correct? Show working to justify your answer.
- Step 1: Plot the two given points: \((0, 4)\) on the y-axis and \((3, -2)\). Join them with a straight line using a ruler, extending it across \(0 \le x \le 3\). Label the line clearly.
- Step 2: Find the gradient \(m\) using \(m = \dfrac{\text{change in } y}{\text{change in } x}\). From \((0,4)\) to \((3,-2)\), the change in \(y\) is \(-2 - 4 = -6\) and the change in \(x\) is \(3 - 0 = 3\). So \(m = \dfrac{-6}{3} = -2\). A common misconception is to mix up the order of subtraction, giving \(+2\); always subtract consistently (second point minus first point).
- Step 3: Find the y-intercept \(c\). The point \((0,4)\) lies on the y-axis, so \(c = 4\). Hence the equation is \(y = -2x + 4\). Check with the other point: when \(x = 3\), \(y = -2(3) + 4 = -2\), which matches \((3,-2)\).
- Step 4: To test the student's claim, substitute \(x = 5\) into \(y = -2x + 4\): \(y = -2(5) + 4 = -10 + 4 = -6\). This gives the point \((5, -6)\).
- Step 5: Since \((5,-6)\) satisfies the equation of the line, the student is correct. The point lies on the extension of the line beyond \(x = 3\). A common error is to assume the line stops at the last plotted point; a straight line continues indefinitely in both directions unless the domain is restricted.
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Forming and Solving Equations
Question
A Year 9 pupil is solving an inequality:
5x + 7 < 3x + 19
They write:
5x + 7 < 3x + 19
2x + 7 < 19
2x < 12
x < 6
Another pupil looks at the answer and says, "That's wrong — the answer should be x > 6 because you have to flip the sign when you move terms across."
(a) Is the second pupil correct? Explain your reasoning.
(b) The first pupil then changes the question to 5x + 7 > 3x + 19. Without solving fully, explain how the answer changes. Give the final inequality.
A Year 9 pupil is solving an inequality:
5x + 7 < 3x + 19
They write:
5x + 7 < 3x + 19
2x + 7 < 19
2x < 12
x < 6
Another pupil looks at the answer and says, "That's wrong — the answer should be x > 6 because you have to flip the sign when you move terms across."
(a) Is the second pupil correct? Explain your reasoning.
(b) The first pupil then changes the question to 5x + 7 > 3x + 19. Without solving fully, explain how the answer changes. Give the final inequality.
- Step 1: Identify what the first pupil did correctly. They subtracted 3x from both sides to get 2x + 7 < 19, then subtracted 7 from both sides to get 2x < 12, then divided both sides by 2 to get x < 6. All three operations are valid because adding or subtracting the same quantity from both sides does not change the direction of an inequality, and dividing by a positive number (2) also keeps the inequality sign the same.
- Step 2: Address the misconception. The second pupil is incorrect. The rule about flipping the inequality sign only applies when you multiply or divide both sides by a negative number. It does not apply when you add or subtract terms across the inequality, even if they involve the variable x. Moving 3x from the right to the left is a subtraction, so the sign stays the same.
- Step 3: Confirm the first pupil's answer is correct: x < 6. For example, test x = 5: 5(5) + 7 = 32 and 3(5) + 19 = 34, and 32 < 34 is true. Test x = 7: 5(7) + 7 = 42 and 3(7) + 19 = 40, and 42 < 40 is false, so the boundary at x = 6 is correct.
- Step 4: For part (b), start from 5x + 7 > 3x + 19. Subtract 3x from both sides: 2x + 7 > 19. Subtract 7 from both sides: 2x > 12. Divide both sides by 2 (positive, so no flip): x > 6.
- Step 5: Explain the change: the only difference is the direction of the inequality sign at the start, which is carried through every step because all operations are addition, subtraction, or division by a positive number. The solution set changes from all values less than 6 to all values greater than 6. Final answer: x > 6.
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Exit ticket
Forming and Solving Equations
Question
Aisha is a Year 9 pupil at a school in Leeds. Her teacher writes the equation 4(x − 3) = 2x + 10 on the board and asks the class to solve it. Aisha expands the brackets and writes:
4x − 3 = 2x + 10
(a) Explain the mistake Aisha has made when expanding the brackets.
(b) Solve the equation correctly to find the value of x.
(c) Check your answer by substituting it back into the original equation 4(x − 3) = 2x + 10. Show clearly that both sides are equal.
Aisha is a Year 9 pupil at a school in Leeds. Her teacher writes the equation 4(x − 3) = 2x + 10 on the board and asks the class to solve it. Aisha expands the brackets and writes:
4x − 3 = 2x + 10
(a) Explain the mistake Aisha has made when expanding the brackets.
(b) Solve the equation correctly to find the value of x.
(c) Check your answer by substituting it back into the original equation 4(x − 3) = 2x + 10. Show clearly that both sides are equal.
- Step 1: Part (a): Aisha has only multiplied the first term inside the bracket by 4. She wrote 4x − 3, but she should have multiplied BOTH terms by 4. The correct expansion is 4 × x − 4 × 3 = 4x − 12. This is the classic 'expanding brackets' misconception — the number outside the bracket must multiply every term inside it, not just the first one.
- Step 2: Part (b): Start from the correct expansion: 4(x − 3) = 4x − 12, so the equation becomes 4x − 12 = 2x + 10.
- Step 3: Subtract 2x from both sides to collect the x terms on the left: 4x − 12 − 2x = 2x + 10 − 2x, which gives 2x − 12 = 10.
- Step 4: Add 12 to both sides to isolate the term in x: 2x − 12 + 12 = 10 + 12, which gives 2x = 22.
- Step 5: Divide both sides by 2: x = 11. Remember: whatever you do to one side of the equation, you must do to the other side to keep it balanced.
- Step 6: Part (c): Substitute x = 11 back into the original equation. Left-hand side: 4(11 − 3) = 4 × 8 = 32. Right-hand side: 2 × 11 + 10 = 22 + 10 = 32. Both sides equal 32, so x = 11 is correct.
- Step 7: Key point to remember: a common error is to check using the incorrect expansion (4x − 3) rather than the original bracketed expression. Always substitute into the ORIGINAL equation to check your answer properly.
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Exit ticket
Functions
Question
A function is defined as f(x) = 5 - 2x. A student at a school in Birmingham claims that f(3) = f(-3). Is the student correct? Justify your answer by calculating both values and explaining what this tells you about the function.
A function is defined as f(x) = 5 - 2x. A student at a school in Birmingham claims that f(3) = f(-3). Is the student correct? Justify your answer by calculating both values and explaining what this tells you about the function.
- Step 1: Understand the function notation: f(x) = 5 - 2x means that for any input value x, we substitute it into the expression 5 - 2x to get the output.
- Step 2: Calculate f(3): Substitute x = 3 into the function. f(3) = 5 - 2(3) = 5 - 6 = -1.
- Step 3: Calculate f(-3): Substitute x = -3 into the function. f(-3) = 5 - 2(-3) = 5 + 6 = 11. (Be careful with the negative sign: -2 multiplied by -3 gives +6.)
- Step 4: Compare the two outputs: f(3) = -1 and f(-3) = 11. Since -1 is not equal to 11, the student's claim that f(3) = f(-3) is incorrect.
- Step 5: Explain what this tells you: The function is not symmetric about x = 0. In fact, f(x) is a linear function with a negative gradient, so as x increases, f(x) decreases. This is why f(3) is less than f(-3).
- Step 6: Key misconception to avoid: Do not assume that f(3) and f(-3) are equal just because the input numbers are opposites. Always substitute carefully, especially with negative numbers, and remember that function notation gives a unique output for each input.
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Exit ticket
Three-Dimensional Shapes
Question
A trapezium has parallel sides of length 12 cm and 8 cm, and a perpendicular height of 5 cm. A pupil says: "To find the area, I add the parallel sides and multiply by the height, then divide by 2."
(a) Calculate the area of the trapezium using the pupil's method.
(b) A second pupil claims the area should be 50 cm² because they used the formula ½ × (12 + 8) × 5. Explain whether the second pupil is correct, and identify any mistake in the first pupil's reasoning.
(c) The trapezium is the cross-section of a prism of length 10 cm. Calculate the volume of the prism, stating the units of your answer.
A trapezium has parallel sides of length 12 cm and 8 cm, and a perpendicular height of 5 cm. A pupil says: "To find the area, I add the parallel sides and multiply by the height, then divide by 2."
(a) Calculate the area of the trapezium using the pupil's method.
(b) A second pupil claims the area should be 50 cm² because they used the formula ½ × (12 + 8) × 5. Explain whether the second pupil is correct, and identify any mistake in the first pupil's reasoning.
(c) The trapezium is the cross-section of a prism of length 10 cm. Calculate the volume of the prism, stating the units of your answer.
- Step 1: Recall the formula for the area of a trapezium: Area = ½ × (sum of parallel sides) × perpendicular height. This is the standard formula tested at GCSE (AQA, Edexcel, OCR).
- Step 2: Identify the values: parallel sides a = 12 cm and b = 8 cm, perpendicular height h = 5 cm. Note that the height must be perpendicular to the parallel sides, not the slanted side.
- Step 3: Substitute into the formula: Area = ½ × (12 + 8) × 5 = ½ × 20 × 5 = 10 × 5 = 50 cm².
- Step 4: For part (a), the pupil's method is actually correct: add the parallel sides (12 + 8 = 20), multiply by the height (20 × 5 = 100), then divide by 2 (100 ÷ 2 = 50 cm²). So the pupil's method gives the correct area of 50 cm².
- Step 5: For part (b), the second pupil is correct: ½ × (12 + 8) × 5 = 50 cm². The first pupil's method is also correct, so there is no mistake in their reasoning. A common misconception is to forget to divide by 2, which would give 100 cm², or to use the slanted side instead of the perpendicular height.
- Step 6: For part (c), the volume of a prism = area of cross-section × length. Here, the cross-section is the trapezium with area 50 cm², and the length is 10 cm. So Volume = 50 × 10 = 500 cm³. The units are cubic centimetres (cm³) because it is a volume.
- Step 7: Key concept: The perpendicular height is essential; using the slanted side would give an incorrect area. Also, always check units: area in cm², volume in cm³. This question targets White Rose Maths Year 9, Autumn, Block 4, Step 6: Find Area of 2D Shapes, and extends to volume of a prism.
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Exit ticket
Trigonometry
Question
A vertical flagpole stands on horizontal ground. A student, Priya, measures the length of the flagpole's shadow and the distance from the base of the flagpole to a point on the ground. She records the following information:
• The shadow of the flagpole is 8.4 metres long.
• The angle of elevation of the top of the flagpole from the tip of the shadow is 52°.
Priya wants to calculate the height of the flagpole. She writes:
"I will use the sine ratio because I know the hypotenuse and I want the opposite side."
(a) Explain why Priya's choice of method is incorrect.
(b) Calculate the correct height of the flagpole, giving your answer to 1 decimal place.
(c) Describe a different measurement Priya could take, along with one angle, that would allow her to use the sine ratio to find the height of the flagpole. Explain your reasoning.
A vertical flagpole stands on horizontal ground. A student, Priya, measures the length of the flagpole's shadow and the distance from the base of the flagpole to a point on the ground. She records the following information:
• The shadow of the flagpole is 8.4 metres long.
• The angle of elevation of the top of the flagpole from the tip of the shadow is 52°.
Priya wants to calculate the height of the flagpole. She writes:
"I will use the sine ratio because I know the hypotenuse and I want the opposite side."
(a) Explain why Priya's choice of method is incorrect.
(b) Calculate the correct height of the flagpole, giving your answer to 1 decimal place.
(c) Describe a different measurement Priya could take, along with one angle, that would allow her to use the sine ratio to find the height of the flagpole. Explain your reasoning.
- Step 1: Understand the scenario: The flagpole is vertical and the ground is horizontal, so the flagpole, its shadow and the line from the top of the flagpole to the tip of the shadow form a right-angled triangle. The right angle is at the base of the flagpole.
- Step 2: Identify the known sides and angles: The shadow (8.4 m) is the side adjacent to the 52° angle, because the angle is measured at the tip of the shadow between the ground and the line of sight to the top of the flagpole. The height of the flagpole is the side opposite the 52° angle. The hypotenuse is the line from the top of the flagpole to the tip of the shadow, which is not given.
- Step 3: Part (a): Priya is incorrect because the sine ratio requires the hypotenuse and the opposite side. Here, the hypotenuse is unknown and the adjacent side is known. She should use the tangent ratio, which links the opposite and adjacent sides: tan(θ) = opposite ÷ adjacent.
- Step 4: Part (b): Set up the tangent equation: tan(52°) = height ÷ 8.4. Rearrange to find the height: height = 8.4 × tan(52°).
- Step 5: Calculate: tan(52°) ≈ 1.2799 (to 4 decimal places). So height = 8.4 × 1.2799 ≈ 10.751 metres. Rounding to 1 decimal place gives 10.8 metres.
- Step 6: Part (c): To use the sine ratio, Priya needs the hypotenuse and the opposite side (the height), or the hypotenuse and the angle. For example, she could measure the distance from the tip of the shadow to the top of the flagpole (the hypotenuse) and the angle of elevation from the tip of the shadow. Then sin(52°) = height ÷ hypotenuse, so height = hypotenuse × sin(52°). Alternatively, she could stand at a known distance from the base and measure the angle of elevation to the top; if she measures the hypotenuse directly (e.g., using a tape from her eye to the top), she could use sine. The key is that the sine ratio requires the hypotenuse and the opposite side.
- Step 7: Common misconception to highlight: Students often confuse which side is adjacent and which is opposite, especially when the angle is not at the base of the pole. Always identify the sides relative to the given angle. Also, remember that the shadow length is the adjacent side, not the hypotenuse.
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Exit ticket
Changing the Subject
Question
The formula for the surface area of a closed cylinder is
S = 2πr² + 2πrh
where r is the radius and h is the height. A Year 11 student at a school in Leeds is given a cylinder with surface area S = 250 cm² and height h = 8 cm. Rearrange the formula to make r the subject, then use it to find the radius of this cylinder. Give your answer correct to 2 decimal places.
The formula for the surface area of a closed cylinder is
S = 2πr² + 2πrh
where r is the radius and h is the height. A Year 11 student at a school in Leeds is given a cylinder with surface area S = 250 cm² and height h = 8 cm. Rearrange the formula to make r the subject, then use it to find the radius of this cylinder. Give your answer correct to 2 decimal places.
- Step 1: Identify the target variable: we need r on its own. Notice that r appears in two terms — 2πr² and 2πrh — so this is a more complex rearrangement than a simple one-step formula.
- Step 2: Subtract 2πrh from both sides to collect the r² term on its own: S − 2πrh = 2πr². (A common misconception is to divide by 2π first, which leaves r² + rh on one side and does not isolate r.)
- Step 3: Divide both sides by 2π to isolate r²: (S − 2πrh) ÷ (2π) = r². At this point r² is the subject, not r.
- Step 4: Take the square root of both sides. Since r is a length, we take the positive root only: r = √[(S − 2πrh) ÷ (2π)]. This is the formula with r as the subject.
- Step 5: Substitute the values S = 250 and h = 8 into the rearranged formula. First work out 2πrh = 2 × π × r × 8, but since r is unknown we keep it symbolic: 2π × 8 = 16π ≈ 50.265. So S − 2πrh = 250 − 50.265 = 199.735.
- Step 6: Now divide by 2π: 199.735 ÷ (2π) = 199.735 ÷ 6.283 ≈ 31.789. So r² ≈ 31.789.
- Step 7: Take the square root: r ≈ √31.789 ≈ 5.638. Rounding to 2 decimal places gives r ≈ 5.64 cm.
- Step 8: Check: substitute r = 5.64 and h = 8 back into the original formula. 2π(5.64)² + 2π(5.64)(8) ≈ 199.8 + 283.5 ≈ 483.3, which is not 250 — this reveals that the substitution in Step 5 was done incorrectly. The correct substitution gives r² = (250 − 2πrh) ÷ 2π, but since r appears on both sides, we must solve the quadratic 2πr² + 16πr − 250 = 0 instead. Using the quadratic formula: r = [−16π + √((16π)² + 4 × 2π × 250)] ÷ (2 × 2π) = [−50.265 + √(2526.6 + 6283.2)] ÷ 12.566 = [−50.265 + √8809.8] ÷ 12.566 = [−50.265 + 93.86] ÷ 12.566 ≈ 3.47 cm.
- Step 9: Final answer: r ≈ 3.47 cm (to 2 decimal places). The key learning point is that when the subject appears in more than one term, simply rearranging by inverse operations is not enough — you must form and solve a quadratic equation, or use an iterative method. Always check your answer by substitution.
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