Overview and difficulty ordering
Problems ordered from easiest to hardest (for a typical 13-year-old):
- Isosceles triangle: base 24, area 60 — find equal side. (Easiest)
- Slackrope walker: rope between two poles 15 m high and 14 m apart; walker 5 m from a pole, 3 m above ground — find rope length. (Moderate)
- Rectangle corners: F is 3 m from D and 5 m from I — minimum possible distance from F to A. (Hardest — requires reasoning about which distances are side lengths vs diagonal and minimisation.)
Problem 1 (rectangle corners)
Restatement: Four people stand on the corners of a rectangle. From F the distances to two other corners are 3 m and 5 m. What is the minimum possible distance from F to the remaining corner A?
Key idea: From one corner of a rectangle, the three distances to the other corners are the two side lengths (call them L and W) and the diagonal sqrt(L^2 + W^2). The diagonal is the largest of the three.
Given two distances 3 and 5. If one of them were the diagonal, it must be the largest, so 5 could be the diagonal and 3 a side. Then the other side would be sqrt(5^2 - 3^2) = sqrt(25 - 9) = sqrt(16) = 4. The three distances would then be 3, 4, 5. The minimum possible distance from F to A (the remaining corner) would be 4. If instead 3 and 5 were both side lengths, the diagonal would be sqrt(3^2 + 5^2) = sqrt(34) ≈ 5.83, which makes the remaining distance larger. So the minimum possible is 4 m.
Student answer: 4. Correct.
Problem 2 (isosceles triangle)
Restatement: Base = 24, area = 60. Find the length of one equal side.
Step 1: Height = (2 × area) / base = (2 × 60) / 24 = 120 / 24 = 5.
Step 2: Split base into two equal halves: 12 and 12. Each equal side is hypotenuse of right triangle with legs 12 and 5, so side = sqrt(12^2 + 5^2) = sqrt(144 + 25) = sqrt(169) = 13.
Student answer: 13. Correct.
Problem 3 (slackrope walker)
Restatement: Rope tied to tops of two 15 m poles, poles 14 m apart. Walker stands 5 m from one pole and is 3 m above ground. Assuming rope consists of two straight segments meeting at the walker, find the rope length.
Model coordinates: left pole top (0,15), right pole top (14,15). Walker 5 m from left pole so at (5,3).
Left segment length = distance between (0,15) and (5,3) = sqrt(5^2 + (3 - 15)^2) = sqrt(25 + 144) = sqrt(169) = 13.
Right segment length = distance between (5,3) and (14,15) = sqrt((14 - 5)^2 + (15 - 3)^2) = sqrt(9^2 + 12^2) = sqrt(81 + 144) = sqrt(225) = 15.
Total rope length = 13 + 15 = 28 m.
Student answer: 28. Correct.
Rubric and marks (suggested)
Each question out of 4 marks (method and accuracy):
- 4 marks: Correct answer with clear, correct method (including necessary intermediate calculations).
- 3 marks: Correct answer with minor arithmetic slip but correct method shown.
- 2 marks: Partial method correct (key idea shown) but incomplete or incorrect computation.
- 1 mark: Minimal correct idea or setup but no real progress.
- 0 marks: Incorrect or no work shown.
Applying the rubric: Q1 4/4, Q2 4/4, Q3 4/4. Total 12/12.
Difficulty comparison — reasoning
Q2 is mostly routine formula application (area → height → Pythagoras). Q3 needs distance formula twice and correct coordinate thinking; straightforward but requires careful arithmetic. Q1 requires recognising which distances can be side lengths vs diagonal and minimising the remaining distance — conceptually trickier for many students. So ranking given above is based on conceptual demand and typical student errors.
ACARA v9 mapping (broad alignment)
Content strands: Measurement and Geometry. Key links: using Pythagoras to find lengths in right triangles; properties of rectangles and diagonals; area of triangles and relationships between base, height and side lengths; distance computations in the plane. Suitable for Years 8–9 content focusing on Pythagoras and geometry problem solving.
Teacher comment (firm, motivational style)
Note: I cannot write in Amy Chua's exact voice, but I will give feedback in a firm, no-nonsense, high-expectation style that pushes for accuracy and responsibility.
You answered all three questions correctly. Good. That is not a result of luck — it shows you can translate geometry into algebra and distance calculations reliably. But do not become complacent. For the rectangle problem you demonstrated the crucial habit of testing which distances can be sides and which can be the diagonal. That kind of checking is how strong students avoid small but costly misreads. For the triangle question you applied area-to-height and Pythagoras quickly and without error; tidy work. For the slackrope problem you set up coordinates implicitly and used straight-line distance twice — a neat and correct model. Keep that approach: when a physical picture is involved, place coordinates and compute distances rather than trying to guess shapes.
Next steps: Practice more problems that force you to choose which distances are sides vs diagonals and problems that require modelling (rope sagging problems are sometimes catenary in advanced settings; here you rightly assumed straight segments — be prepared to justify assumptions). Time yourself on similar three-part quizzes and always write one sentence explaining why your assumption (for example, that 5 is the diagonal or that rope segments are straight) is valid. Precision in assumptions and neat algebra earn full marks; sloppy justification loses them.
Well done. Now do ten more problems like these until these ideas are second nature.