R12: Similarity and Trigonometry
Understand that X is inversely proportional to Y is equivalent to X is proportional to 1/Y
📋 Key Concepts
Inverse Proportion: X is inversely proportional to Y means X = k/Y, which is the same as X being proportional to 1/Y.
X ∝ Y means X = kY (direct)
X ∝ 1/Y means X = k/Y (inverse)
Or equivalently: XY = k (constant)
X ∝ 1/Y means X = k/Y (inverse)
Or equivalently: XY = k (constant)
📝 Understanding Inverse Proportion
Key Property: In inverse proportion, as one quantity doubles, the other halves. The product is always constant.
Example 1
X is inversely proportional to Y. When Y = 5, X = 20.
a) Find the equation.
b) Find X when Y = 8.
Solution:
a) X = k/Y
20 = k/5
k = 100
Equation: X = 100/Y
b) X = 100/8 = 12.5
📝 Different Forms
Recognising the forms: X ∝ 1/Y can be written as:
- X = k/Y
- XY = k
- Y = k/X
Example 2
P is inversely proportional to Q. When Q = 3, P = 18.
Find P when Q = 9 using the relationship PQ = k.
Solution:
PQ = k
18 × 3 = 54
When Q = 9: P × 9 = 54
P = 54 ÷ 9 = 6
Notice: Q triples (3→9), so P becomes one-third (18→6)
📝 Real-World Applications
Example 3
The time to complete a task is inversely proportional to the number of workers. If 4 workers take 6 hours, how long for 8 workers?
Solution:
Time × Workers = k
6 × 4 = 24
When Workers = 8: Time × 8 = 24
Time = 24 ÷ 8 = 3 hours
Twice as many workers means half the time.
Example 4
The pressure of a gas is inversely proportional to its volume. When volume is 10 m³, pressure is 50 kPa. Find pressure when volume is 25 m³.
Solution:
Pressure × Volume = k
50 × 10 = 500
When V = 25: P × 25 = 500
P = 20 kPa
📝 Graphs of Inverse Proportion
Shape: The graph of y = k/x is a hyperbola. It gets closer to the axes but never touches them.
Example 5
Sketch y = 12/x for x = 1, 2, 3, 4, 6, 12.
Solution:
| x | 1 | 2 | 3 | 4 | 6 | 12 |
|---|---|---|---|---|---|---|
| y | 12 | 6 | 4 | 3 | 2 | 1 |
The graph is a curve decreasing from left to right.
🧮 Worked Examples — Exam Style
Inverse proportion with powers follows the same recipe as R13: write the model (y = k/xⁿ), find k, then answer.
y is inversely proportional to x. When x = 5, y = 12. Find y when x = 2.
y = k/x → 12 = k/5 → k = 60
y = 60/2 = 30
M1 for k = 60, A1 for 30. Halving x doubles y — sanity-check the direction.
d is inversely proportional to s². When s = 2, d = 45. Find d when s = 3.
d = k/s² → 45 = k/4 → k = 180
d = 180/9 = 20
M1 for k = 180, A1 for 20. Squared inverse: tripling s divides d by NINE — the exam's favourite twist.
Exam wording to expect: "inversely proportional to the square of …", "P ∝ 1/Q".
❓ Practice Questions
Q1: y is inversely proportional to x. When x = 4, y = 15. Find y when x = 10.
Q2: P ∝ 1/Q. When Q = 6, P = 5. Find the constant k.
Q3: The time to paint a fence is inversely proportional to the number of painters. 3 painters take 8 hours. How long for 4 painters?
Q4: a is inversely proportional to b². When b = 2, a = 50. Find a when b = 5.
Q5: Verify: If x doubles and y is inversely proportional to x, then y halves.
Q6: y is inversely proportional to x². When x = 2, y = 12. Find y when x = 6.
Q7: v is inversely proportional to r. When r = 8, v = 9. Find v when r = 12.
✅ Answers
- y = 6
- k = 30
- 6 hours
- a = 8 (a = 200/b²)
- Original: y = k/x. New: y' = k/(2x) = y/2 ✓
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. k = xy = 4 × 15 = 60 → y = 60/10 = 6
Q2. P = k/Q → 5 = k/6 → k = 30
Q3. Total work = 3 × 8 = 24 painter-hours → 24 ÷ 4 = 6 hours
Q4. k = ab² = 2² × 50 = 200 → a = 200/25 = 8
Q5. y = k/x → y′ = k/(2x) = (k/x)/2 = y/2 → halves ✓
Q6. k = x²y = 4 × 12 = 48 → y = 48/36 = 4/3
Q7. k = rv = 8 × 9 = 72 → v = 72/12 = 6
📌 Key Facts to Memorise
- Inverse: y = k/x · inverse square: y = k/x² · find k from ONE data pair
- Inverse intuition: as x grows, y SHRINKS proportionally
- x doubles → y halves; x triples → y thirds; x doubles in y = k/x² → y quarters
- k = the product x × y (for y = k/x)
- Graph of y = k/x: hyperbola, two branches, asymptotes on the axes
🗺️ In the Exam
Inverse proportion: 3–5 marks on Papers 2/3. The x² version separates grade 5 from grade 6 at Foundation/Higher boundary.
Links to: direct proportion (R7/R10), graphs of proportion (R8), substitution (A2).
🧭 Board Focus — OCR
Spec J560. Three papers of 100 marks each: Paper 1 non-calculator, Papers 2 and 3 calculator, at Foundation/Higher tiers. OCR marks problem-solving explicitly, and its papers tend to front-load multi-step worded contexts — practise extracting the mathematics from prose before calculating.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- Inverse proportion: one increases as other decreases
- Use XY = k to find the constant quickly
- If x doubles, y halves (and vice versa)
- Graph is a curve, not a straight line
- Check: product of values should be constant
🧠 Problem-Solving Strategies
Problem-Solving
Similar shapes have equal angles and sides in the same ratio (scale factor). For length scale factor k, the area scale factor is k² and the volume scale factor is k³. Use SOH CAH TOA to find missing sides and angles in right-angled triangles. Always identify the hypotenuse, opposite and adjacent sides relative to the given angle.
Multi-Step Problem
Two similar cones have heights 6 cm and 15 cm. The smaller cone has volume 80 cm³. Find the volume of the larger cone.
Solution: Linear scale factor = 15/6 = 2.5. Volume scale factor = 2.5³ = 15.625. Volume of larger cone = 80 × 15.625 = 1250 cm³.
⚠️ Common Errors
Watch Out!
1. Wrong: Using the linear scale factor for area: if lengths double, area doubles too Correct: If lengths double (scale factor 2), area multiplies by 2² = 4 and volume by 2³ = 8.
2. Wrong: Mixing up opposite and adjacent sides when using trigonometry Correct: Label sides relative to the angle: opposite is across from the angle, adjacent is next to it (not the hypotenuse).
3. Wrong: Using the area scale factor to find a length Correct: To find a length from an area ratio, take the square root. If area ratio = 9, length factor = √9 = 3.
✍️ 6-Mark Exam Question
Extended Answer
6 marks: Two similar pyramids have base areas of 20 cm² and 45 cm². The height of the smaller pyramid is 8 cm. (a) Find the height of the larger pyramid. (b) The smaller pyramid has volume 64 cm³. Find the volume of the larger pyramid. (c) Show that the volume scale factor equals the linear scale factor cubed.
(a) Area ratio = 45/20 = 2.25. Linear scale factor = √2.25 = 1.5. Height = 8 × 1.5 = 12 cm.
(b) Volume scale factor = 1.5³ = 3.375. Volume = 64 × 3.375 = 216 cm³.
(c) Linear SF = 1.5. Linear SF³ = 1.5³ = 3.375. Volume ratio = 216/64 = 3.375. They are equal ✓
Mark scheme: M1 area ratio, M1 square root for linear SF, A1 height = 12cm, M1 volume SF, A1 216cm³, A1 verification shown
📊 AO3: Reason & Interpret
Reasoning and Interpretation
A model car is built at 1:24 scale. The real car is 4.2 m long.
(a) How long is the model car in cm?
(b) The model has a surface area of 350 cm². Estimate the real car's surface area in m².
(c) A student says "If the model weighs 0.8 kg, the real car weighs 0.8 × 24 = 19.2 kg." Why is this wrong?
Answers: (a) 4.2 m ÷ 24 = 0.175 m = 17.5 cm. (b) Area SF = 24² = 576. Real area = 350 × 576 = 201,600 cm² = 20.16 m². (c) Weight depends on volume, not length. Volume SF = 24³ = 13,824. Real weight ≈ 0.8 × 13,824 = 11,059 kg ≈ 11 tonnes. The student used the linear SF instead of the volume SF.
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