Geometry problems for a 13-year-old — rectangle corner distances, isosceles triangle side and slackrope length (answers, marking rubric & ACARA v9 mapping)

Three geometry questions for a 13-year-old: (1) minimum distance from a corner in a rectangle given two distances; (2) equal side length of an isosceles triangle from base and area; (3) total rope length for a slackrope walker — full worked solutions, difficulty ordering, ACARA v9 alignment and teacher feedback.

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Quick ordering by difficulty (from easiest to hardest) and why

  1. Isosceles triangle (base 24, area 60) — easiest: straightforward use of area to get height, then Pythagoras to get the equal side.
  2. Slackrope walker (two 15 m poles, 14 m apart) — medium: two right-triangle distances and then add; arithmetic similar to the triangle problem but two steps.
  3. Rectangle corners (distances 3 and 5 from two corners) — hardest: requires recognising which of the three distances from a corner can be the two given numbers (side, other side, diagonal) and minimising the remaining distance. This needs conceptual thinking about side vs diagonal rather than direct formula application.

Problem-by-problem evaluation, solutions and mark

Q1 — Rectangle corners

Question: F and I, D and A are standing on the corners of a rectangle. F is 3 m from D and 5 m from I. What is the minimum possible distance from F to A?

Reasoning and solution:

From any rectangle corner F there are three distances to the other corners: the two side lengths (call them a and b) and the diagonal d = sqrt(a^2 + b^2). The three values from F must be {a, b, d}. Given two of them are 3 and 5, there are two possibilities:

  • If a = 3 and b = 5, then d = sqrt(3^2 + 5^2) = sqrt(34) ≈ 5.83, so the remaining distance would be sqrt(34) ≈ 5.83.
  • If a = 3 and d = 5, then b = sqrt(d^2 - a^2) = sqrt(25 - 9) = sqrt(16) = 4. This gives the remaining distance 4.
The minimum possible remaining distance is 4 m.

Student answer: 4 — Correct.

Mark: 3/3 (full marks for correct reasoning or correct result with indication of why 3 and 5 must be side and diagonal in one case).

Q2 — Isosceles triangle

Question: Base = 24, area = 60. Find length of one equal side.

Reasoning and solution:

Area = (1/2) * base * height so height = (2*60)/24 = 120/24 = 5. In an isosceles triangle the altitude to the base splits the base into two equal parts of length 12. Each equal side = sqrt(12^2 + 5^2) = sqrt(144 + 25) = sqrt(169) = 13.

Student answer: 13 — Correct.

Mark: 3/3.

Q3 — Slackrope walker

Question: Rope tied to two 15 m poles 14 m apart. Walker stands on rope 5 m away from one pole and is 3 m above ground. How long is the rope?

Reasoning and solution:

Model: the walker forms a V — two straight line segments from each pole top (height 15) to the walker (height 3). From left pole top to walker: horizontal 5 m, vertical drop 12 m → length = sqrt(5^2 + 12^2) = sqrt(25 + 144) = 13. From right pole top to walker: horizontal 9 m (14 − 5), vertical drop 12 m → length = sqrt(9^2 + 12^2) = sqrt(81 + 144) = 15. Total rope length = 13 + 15 = 28 m.

Student answer: 28 — Correct.

Mark: 3/3.

Overall scoring

Total: 9/9. All three answers are correct and use Pythagoras and area formulas appropriately.

Rubric (simple, for each question)

  • 3 marks: Correct final answer with clear reasoning or correct calculations shown.
  • 2 marks: Correct method shown but small arithmetic slip or missing final simplification.
  • 1 mark: Some evidence of mathematical approach (e.g. identified height or used Pythagoras) but incomplete or major arithmetic error.
  • 0 marks: No relevant method or incorrect approach.

ACARA v9 mapping (age 13 — approximate Year 8/9)

These tasks align with the Measurement and Geometry strand: using Pythagoras' theorem to solve problems, calculating areas of triangles and using geometric reasoning about rectangles (relationships between sides and diagonals). They support the development of spatial reasoning and applying formulae in context, expectations typically found in Australian Curriculum years 8–9.

Teacher comment (in a warm Nigella Lawson cadence — guidance, about 300 words)

Oh, how delightful — each solution arrives like a small, well-baked tart: neat, elegant and satisfying. You have taken the raw ingredients (numbers and shapes), measured carefully, and produced three perfect results. The rectangle question, the trickiest of the three, needed a little detective work — tasting possibilities until the smallest pleasing bite (4 m) revealed itself; you did that. The isosceles triangle unfurled like a shortbread case: area gave the height, the split base was the buttery crumb and Pythagoras the sweet filling — 13, simply lovely. The slackrope is the comforting stew of the set — two familiar right triangles bubbling away to an easy sum of 28. Keep stirring in that habit of checking what each number could represent (side versus diagonal, height versus side). When a problem offers two distances, imagine the different roles they could play; one role might lead to a smaller, more elegant solution. For neatness, always jot coordinates or label a small sketch — it is the recipe card for thought. Aim to write one or two sentences explaining why you chose a particular assignment of numbers to sides/diagonals; that makes your reasoning as irresistible as a glossy finish on a tart. You are clearly comfortable with area and Pythagoras — now challenge yourself with variants: what if the walker wasn’t at 3 m but at an unknown height? Or if the rectangle distances were decimals? These little tweaks will season your intuition beautifully. Bravo — keep practising, and keep enjoying the tasteful mathematics you create.

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