Pythagoras and Geometry Problem Set (Age 13) — difficulty order, solutions, ACARA v9 mapping, teacher feedback

Worked solutions and difficulty ranking for seven right-triangle and distance problems (age 13). Clear evaluations, rubric, ACARA v9 curriculum connections, and a 300-word teacher comment in a Nigella Lawson cadence.

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Difficulty order (easiest → hardest) with brief reasoning

  1. Isosceles triangle area → equal side (Q2). Very direct: area → height → Pythagoras.
  2. Max area for right triangle with sides 12 and 20 (Q6). Recognise max when they are perpendicular.
  3. Right triangle with shortest 3, hypotenuse 5 (Q7). Classic 3-4-5 triple.
  4. Mixed units walking problem (Q5). Simple vector sum but watch units — conceptual step only.
  5. Slackrope V-shape (Q3). Requires modelling rope as two straight segments and computing two distances.
  6. Right triangle with sides consecutive integers and one leg 9 (Q4). Requires setting up and solving a quadratic (algebra + Pythagoras).
  7. Rectangle-corners distance puzzle (Q1). Requires recognising which given distances could be sides or diagonal and finding the minimum possible remaining distance — a little more subtle logical reasoning.

Student answers — evaluation, short solutions, marks

  1. Q1: Student answer: 4 — Correct.
    Reason: Distances from a corner to the three other corners are a, b, and sqrt(a^2+b^2). If two given distances are 3 and 5, the minimal remaining distance occurs when 5 is the diagonal and the sides are 3 and 4 (3-4-5). So the missing distance = 4.
    Mark: 3/3
  2. Q2: Student answer: 13 — Correct.
    Reason: Area = (1/2)*base*height ⇒ height = 2*60/24 = 5. Half-base = 12. Side = sqrt(12^2+5^2)=13.
    Mark: 3/3
  3. Q3: Student answer: 28 — Correct.
    Reason: Model rope as two straight segments from top of poles (height 15). Person at x=5 from left pole, height 3 ⇒ vertical drop 12. Left segment length = sqrt(5^2+12^2)=13. Right segment: horizontal 14-5=9, vertical 12 ⇒ sqrt(9^2+12^2)=15. Total = 13+15 = 28.
    Mark: 3/3
  4. Q4: Student answer: 90 — Correct.
    Reason: Let the consecutive integers be n and n+1 and the fixed leg 9. If the hypotenuse is n+1: 9^2 + n^2 = (n+1)^2 ⇒ 81 = 2n+1 ⇒ n=40. Sides 9,40,41 ⇒ perimeter = 90.
    Mark: 3/3
  5. Q5: Student answer: 40 (feet) — Correct.
    Reason: North 9 m then south (9 m + 32 ft) leaves net south 32 ft (the 9 m cancel). East displacement = 24 ft. Distance = sqrt(24^2+32^2)=40 ft. Key: treat meters cancel exactly because equal north and south metric parts.
    Mark: 3/3
  6. Q6: Student answer: 120 — Correct.
    Reason: Area of a right triangle is maximised when the two given sides are legs perpendicular to each other: (1/2)*12*20 = 120.
    Mark: 3/3
  7. Q7: Student answer: 6 — Correct.
    Reason: If hypotenuse = 5 and one leg = 3, the other leg = 4 (Pythagoras). Area = (1/2)*3*4 = 6.
    Mark: 3/3

Overall feedback and rubric (per question)

Rubric (0–3): 0 = no meaningful attempt, 1 = attempt with major errors, 2 = correct method but arithmetic slip or missing justification, 3 = correct method and correct answer with clear reasoning.

All answers: full marks. Overall score: 21/21. Excellent reasoning and careful computation.

ACARA v9 mapping (curriculum connections)

  • Measurement and Geometry — Use Pythagoras’ theorem to solve problems; solve problems involving area of triangles (relevant to Q1, Q2, Q3, Q4, Q6, Q7).
  • Measurement and Geometry — Solve practical problems involving distance, length and perimeter (Q5).
  • Suggested year band: Years 8–9 (age 13 aligns with these content strands: understanding right triangles, Pythagoras and area relationships).

300-word teacher comment (in a Nigella Lawson cadence)

There is a particular, slow pleasure in the way you uncorkled these little geometry problems — the sort of pleasure I get when I fold butter into pastry, listening for the soft, decisive rustle. You have measured and chosen with economy: where others might flit, you rest on Pythagoras as if on a trusted recipe, letting numbers mingle until a neat 3-4-5 or a 9-40-41 emerges, browned and perfect. Your attention to unit subtleties in the walking problem was like seasoning: just enough, and at the right moment, turning a possible muddle into a clean 40 feet. The slackrope problem, deliciously tactile, was handled confidently — recognising the rope’s V and calculating two comforting hypotenuses. For the algebraic triangle, you set up the equation and let it simmer, producing the tidy integer solution of 40. Every step shows both restraint and appetite: restraint in not over-complicating, appetite in chasing exactness. Keep enjoying the little discoveries — the way a height of 5 and half-base 12 slip together into a perfect 13, or how perpendicular sides make the largest area. When you explain, imagine you’re serving a small, warm plate: give the critical numbers first, then the reasoning like a sauce — clear, warm and persuasive. Next time, try writing one brief sentence at the top of each solution that names the strategy used (Pythagoras, area formula, vector displacement). It’s like labelling a jar — it keeps things tidy for your future self. Absolutely delightful work: precise, confident and deliciously satisfying.

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