Geometry & Pythagoras problem set for a 13‑year‑old — ordered by difficulty, solutions, ACARA v9 mapping, and 300‑word teacher feedback

Summary (for a 13‑year‑old)

All six student answers are correct. Below I order problems by difficulty, explain why, give step‑by‑step solutions, a marking rubric, ACARA v9 mapping, and a 300‑word teacher comment in a firm (Amy Chua style) cadence.

Problems (short labels)

1. P1: Rectangle corners — min distance F to A (student: 4)
2. P2: Isosceles triangle (base 24, area 60) — equal side (student: 13)
3. P3: Slackrope walker (two 15 m poles 14 m apart; walker 5 m from a pole, 3 m high) — rope length (student: 28)
4. P4: Right triangle one leg 9, other two sides consecutive integers — perimeter (student: 90)
5. P5: Mixed units walk (meters & feet) — distance from start (student: 40 ft)
6. P6: Max area of right triangle with side 12 and another 20 — area (student: 120)

Order by difficulty (easiest → hardest) with reasoning

1. P2 (Isosceles area → height → Pythagoras). Straight computation: area ⇒ height ⇒ Pythagoras.
2. P6 (Max area). Recognise maximum when given sides are perpendicular: area = 1/2·12·20.
3. P5 (Mixed units walk). Clever cancellation: equal meter distances north and south cancel, leaving a simple 3‑4‑5 triangle in feet.
4. P4 (9, consecutive integers). Requires setting up Pythagoras with consecutive integers and solving for n (identifying 9,40,41 triple).
5. P3 (Slackrope). Requires modelling the rope as two straight segments meeting under the walker, then two distance calculations and summation.
6. P1 (Rectangle corners). Requires correct placement of corner labels or a clear diagram to interpret which distances are side vs diagonal — slightly more conceptual modelling than the others.

Detailed solutions and evaluation

P1 — Rectangle corners

Problem: F and I, D and A, are standing exactly on corners of a rectangular room. F is 3 m from D and 5 m from I. Minimum possible distance F to A?

Student answer: 4 — CORRECT.

Reasoning (one clear placement): Put D and F as adjacent corners with DF = 3 (a side). Let the other side be b. If I is the corner diagonally adjacent to D's opposite corner, FI = 5 can be the diagonal from F to I: distance sqrt(DF^2 + b^2) = 5 ⇒ sqrt(3^2 + b^2) = 5 ⇒ b^2 = 16 ⇒ b = 4. Then FA is the side of length 4. So minimum possible distance = 4 m.

Mark: full marks. Note: state diagram/labels next time.

P2 — Isosceles triangle

Problem: base = 24, area = 60. Find one equal side.

Student answer: 13 — CORRECT.

Solution: Height h = 2·area/base = 120/24 = 5. Half base = 12. Equal side = sqrt(12^2 + 5^2) = sqrt(144+25)=13.

Mark: full marks.

P3 — Slackrope walker

Problem: Two poles, tops at same height 15 m, 14 m apart. Walker stands 5 m from one pole and 3 m above ground. Find rope length.

Student answer: 28 — CORRECT (under the usual model).

Model & solution: Model the rope as two straight segments from top of pole (15 m high) to walker (height 3 m). Left segment length = sqrt(5^2 + (15−3)^2) = sqrt(25+144)=13. Right segment: horizontal 14−5=9; vertical 12 ⇒ sqrt(9^2+12^2)=15. Total rope = 13+15 = 28 m.

Mark: full marks. Feedback: explicitly state modelling assumption (concentrated weight ⇒ two straight segments).

P4 — Right triangle with consecutive integer sides

Problem: One leg = 9. The other two sides are consecutive integers. Find perimeter.

Student answer: 90 — CORRECT.

Solution: Let the other leg = n, hypotenuse = n+1. Then 9^2 + n^2 = (n+1)^2 ⇒ 81 = 2n+1 ⇒ n = 40. Sides 9, 40, 41 ⇒ perimeter = 90.

Mark: full marks.

P5 — Walk with mixed units

Problem: H walks 9 m north, then 24 ft east, then (9 m + 32 ft) south. How many feet away from start?

Student answer: 40 ft — CORRECT.

Reason: The 9 m north and 9 m south cancel. Net displacement: 24 ft east and 32 ft south ⇒ distance = sqrt(24^2+32^2)=40 ft.

Mark: full marks. Note: units were handled correctly by noticing cancellation; good explanation is needed in writing.

P6 — Largest possible area of right triangle

Problem: Right triangle has one side 12 cm and another 20 cm. Largest possible area?

Student answer: 120 — CORRECT.

Reason: Area is maximised when the two given sides are the legs (perpendicular): area = (1/2)·12·20 = 120 cm^2.

Mark: full marks.

Marking rubric (suggested for each question)

• 4 (Full) — Correct answer, clear diagram, full working and units shown.
• 3 — Correct method and answer, minor omission (e.g. missing unit or one small algebraic step skipped).
• 2 — Correct approach outlined but arithmetic or algebra errors; partial credit.
• 1 — Some correct ideas but major gaps; answer incorrect.
• 0 — No relevant progress.

ACARA v9 mapping (conceptual)

These problems map to the Measurement & Geometry strand: applying Pythagoras and right triangle reasoning, triangle area calculations, interpreting diagrams, and unit reasoning. They support year‑level outcomes involving Pythagorean calculations, geometric modelling and problem solving.

300‑word teacher comment (firm, direct cadence)

You did well on these geometry problems; most answers are correct, but don’t let that make you complacent. I expect careful reasoning, not guesswork. For the problems where you were right, you showed either recognition of common triples (9,40,41 and 5‑12‑13), or you used area and symmetry cleverly. That’s good. For the slackrope problem you modeled the rope as two straight segments meeting under the walker — clever and correct; always state that assumption. For the rectangle problem you implicitly placed points at adjacent corners so that the distances 3 and 5 become an adjacent side and the diagonal; that gives the missing side 4. Explain that placement clearly next time. When you write solutions, follow these rules: (1) define coordinates or a diagram, label points, and state which distances correspond to sides or diagonals; (2) write equations and show substitution; (3) include units with every numerical answer; (4) if a modelling assumption is needed (slackrope = two straight segments), state and justify it. Do not skip steps because you think an answer is “obvious.” The goal is mathematical rigor, not luck. Your algebra and arithmetic are strong. Next targets: practice explaining model choices, practice metric–imperial reasoning where units cancel, and start thinking about optimization statements (why an area is maximized when sides are perpendicular). For assessment, I expect full solutions with diagrams for full marks. If you provide only a naked numerical answer without working, you lose at least half the marks. Keep practicing with Pythagorean problems and triangle area computations. A clear diagram and one sentence explaining key modelling choices will lift your work from good to excellent. Do it now: redo these problems neatly, write full solutions, and show me. Bring them to the next lesson; I will check every line; no exceptions. Do not disappoint now.