Quick ordering by difficulty (for a 13-year-old)
- Easiest: Question 3 (two poles, shortest rope) — straight Pythagoras, direct calculation.
- Medium: Question 1 (rectangle corners, distances 3 & 5) — reasoning about which distances correspond to sides vs diagonal; Pythagorean insight.
- Hardest: Question 2 (slackrope length) — requires modelling the rope as a parabola and computing an arc length integral (calculus-level). This is beyond typical Year 8–9 routine work.
Worked solutions and evaluation of the student answers
Question 1 (students answer: 4)
Problem summary: Four people stand on the four corners of a rectangle. From corner F, the distances to two other corners are 3 m (to D) and 5 m (to I). 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 three corners are: the two side lengths (call them a and b) and the diagonal sqrt(a^2 + b^2). The set of three numbers {a, b, sqrt(a^2 + b^2)} must include 3 and 5.
Possible assignments:
- Case 1: a = 3 and b = 5. Then the diagonal = sqrt(3^2 + 5^2) = sqrt(34) ≈ 5.83. The three distances are {3, 5, sqrt(34)}; then F–A could be 3, 5, or sqrt(34); the remaining corner distance would be sqrt(34) (~5.83).
- Case 2: a = 3 and diagonal = 5. Then b must satisfy sqrt(3^2 + b^2) = 5 → 9 + b^2 = 25 → b^2 = 16 → b = 4. The three distances are then {3, 4, 5}.
- Case 3: a = 5 and diagonal = 3 is impossible because diagonal ≥ each side.
We want the minimum possible distance from F to A, so we choose the arrangement that makes that remaining distance as small as possible. In Case 2 the remaining distance is 4, which is smaller than sqrt(34) ≈ 5.83. So the minimum possible distance is 4 m.
Evaluation: student answer 4 — correct. Full marks if the student explained why a 3–4–5 right triangle arrangement is needed.
Question 2 (student answer: 28)
Problem summary: A slackrope is tied between two poles, each 15 m high, separated by 14 m horizontally. When a person stands 5 m from one pole (i.e., 5 m along the horizontal from the left pole), the rope there is 3 m above ground. Model the rope as a parabola through the two top points and compute its length.
Reasonable school-level model: Use a parabola y(x) = ax^2 + bx + c with x = 0 at the left pole (top at (0, 15)) and x = 14 at the right pole (top at (14, 15)). Also point (5, 3) lies on the parabola.
Set up equations:
- y(0) = c = 15
- y(14) = 196a + 14b + 15 = 15 → 196a + 14b = 0 → b = -14a
- y(5) = 25a + 5b + 15 = 3
Substitute b = -14a into y(5): 25a + 5(-14a) + 15 = 3 → 25a - 70a = -45a = -12 → a = 12/45 = 4/15. Then b = -14a = -56/15.
So y'(x) = 2ax + b = (8/15)x - 56/15 = (8(x-7))/15. The arc length from x=0 to x=14 is L = ∫_0^{14} sqrt(1 + (y')^2) dx = 2 ∫_0^{7} sqrt(1 + (8/15)^2 t^2) dt, (using symmetry with t = x - 7).
Use the integral formula ∫ sqrt(1 + k^2 t^2) dt = (t/2) sqrt(1 + k^2 t^2) + (1/(2k)) asinh(k t). With k = 8/15 and t from 0 to 7 we get:
L = 7 sqrt(1 + (8/15)^2·49) + (15/8)·asinh(56/15) ≈ 27.055 + 3.803 ≈ 30.86 meters (approx.).
Evaluation: student answer 28 — incorrect. The rope is longer than 28 m under the parabola model; about 30.9 m. Note: the precise physical shape is a catenary, not a parabola; exact catenary arc length would give a slightly different value. Either way, 28 m is too short.
Question 3 (student answer: sqrt 2601)
Problem summary: A 39-ft pole and a 15-ft pole stand 45 ft apart on flat ground. What is the shortest rope connecting their tops?
Shortest straight-line distance between tops: treat the horizontal separation as 45 and vertical difference as 39 - 15 = 24. Distance = sqrt(45^2 + 24^2) = sqrt(2025 + 576) = sqrt(2601) = 51 ft.
Evaluation: student answer "sqrt 2601" — correct; could simplify to 51 ft. Full marks if reasoning is shown.
ACARA v9 mapping (brief)
- Q3: Aligns with Year 8–9 geometry (Pythagoras, distance between points, right triangles).
- Q1: Fits Year 8–9 reasoning with Pythagorean theorem and properties of rectangles; problem solving and spatial reasoning.
- Q2: Involves modelling with quadratic functions (Year 9–10) but computing arc length requires calculus (beyond standard Year 9). You can still link it to Victorian/ACARA outcomes on modelling, using functions and interpreting graphs, but the final integral is outside typical Year 9 requirements.
Grading rubric (useful for a teacher)
| Criterion | Excellent (A) | Satisfactory (C) | Developing (E) |
|---|---|---|---|
| Answer correctness | Correct final answer. | Correct final answer but missing some steps. | Incorrect final answer but some correct ideas. |
| Reasoning / method | Clear, logical step-by-step justification (diagram if helpful). | Some reasoning given, minor gaps. | Little or no correct reasoning shown. |
| Presentation | Neat, labels/units included, simplified results. | Readable but missing units or simplification. | Hard to follow, no units. |
For each question, give full marks for correct result with clear reasoning; 0.5 marks for correct result with no explanation; partial marks for correct intermediate steps.
Short feedback for the student
- Q1: Well done — 4 is correct. Remember that from one corner you can get side lengths and the diagonal; a 3–4–5 right triangle is the key.
- Q2: Your answer 28 is too small. This problem needs modelling the sag as a parabola and then computing the arc length — that integral gives about 30.9 m. Don’t be discouraged; this problem is harder than typical school geometry problems.
- Q3: Correct (sqrt 2601 = 51). Next time simplify the square root to give the neat number 51.
700-word Sailor Moon–style summary (for fun and memory)
In the name of moonlit reasoning, a young geometry guardian stands at the edge of a rectangular room, hands on hips, hair shining like the sweep of a compass. A problem sparkles before her: four friends on corners, numbers whispering—3 and 5—like charms in a sailor’s pocket. She closes her eyes and thinks of right triangles, of ancient Pythagoras singing the secret tune: square the legs, add them up, and the diagonal will answer. "From one corner," she says, "three distances appear: the two sides and the diagonal. If two of them are 3 and 5, then perhaps the diagonal is 5 and the other side must be 4." The 3–4–5 triangle glows like a moon crystal and the smallest distance—4—becomes the chosen one, sweet and correct. Then the slackrope beckons, swaying like a ribbon in twilight. Two poles stand proud, fifteen meters each, fourteen meters apart, and a walker balances five from a pole at a low, brave height of three. "This is not a simple right-angle spell," she hums. "The rope bows, a gentle parabola of the heart." She writes a quadratic, fitting three points like three moons on a night sky, and hears the derivative whisper the slope that changes with each step. To measure the rope’s length is to measure a curve—an arc that asks for a deeper magic: an integral. She weaves the formula, splits the span at the center, and then with careful calculation finds the rope is longer than 28, shining out near thirty point nine meters. The lesson is tender: some problems wear the cloak of algebra, some demand the mysterious calculus of arcs. Finally, two poles of different heights stand like rival princesses upon the earth—one thirty-nine, one fifteen—forty-five apart in the plain. The guardian draws a straight line between their crowns; it is the shortest path, bold and true. She measures the horizontal and vertical differences, squares and adds, and the result is as clear as starlight: fifty-one feet. "A straightforward Pythagorean truth," she smiles, and the number leaps into place like a victorious team. She compares the three trials: the simplest was the direct distance between two tops, a friendly right triangle that needs nothing more than the Pythagorean spell. A bit trickier was the mystery of corners—recognizing which numbers are sides and which are diagonals, letting the 3–4–5 secret impart the minimum. The hardest was the slackrope, a curve that must be modelled and measured; it humbles her, for not all geometry is flat and neat. In this moonlit lesson the guardian learns to sort problems by the tools they ask for: straight-line Pythagorean power, triangle insight, or the extended magic of functions and arc length. She learns that a correct answer is honored most when accompanied by a clear path of reason. She promises, with sailor’s courage, to always sketch a diagram, label the lengths, and ask: which numbers are sides, which are diagonals, and which require a deeper curve. Thus, armed with simple algebra, keen logic, and a respect for more advanced calculus where needed, she solves and shines—an honest student and a geometry guardian under the gentle glow of the moon.
If you want, I can: 1) draw diagrams for each problem, 2) show the full algebraic steps for the Q2 integral with intermediate numeric values, or 3) give practice problems that progress from easiest (Q3-type) to hardest (Q2-type).