ACARA v9 — Year 8 (13‑year‑old): Measurement & Geometry — rectangular room distances (student: 4 m)

Task recap (short): Four people stand on the four corners of a rectangle labeled F, I, D and A. From F we measure two known distances: F to D = 3 m and F to I = 5 m. What is the minimum possible distance from F to A (in metres)?

Key idea (suitable for a 13‑year‑old): Put F at one corner of a rectangle. The other three corners relative to F must be:

• a point a metres along one side (call that side length a),
• a point b metres along the other side (side length b),
• and the diagonally opposite corner a distance sqrt(a² + b²) from F (the rectangle diagonal).

So the three distances from F to the other corners are exactly: a, b and √(a² + b²). We are told two of these distances are 3 and 5 (metres). The third distance is what we must find — and we want its minimum possible value.

1. Consider the possibilities for how 3 and 5 match {a, b, √(a²+b²)}.
2. If 3 and 5 were both side lengths (a = 3 and b = 5), then the diagonal would be √(3² + 5²) = √34 ≈ 5.83, so the remaining distance would be ≈5.83 (bigger than 5).
3. If one of 3 or 5 is a side and the other is the diagonal, you can get a smaller remaining side. For example, suppose a = 3 (a side) and √(a² + b²) = 5 (the diagonal). Then:
4. Calculate the other side b from 5² = 3² + b² → 25 = 9 + b² → b² = 16 → b = 4.

So in that case the three distances from F are 3, 4 and 5 metres. The remaining distance (the minimum possible) is 4 m.

Why this is the minimum: If 3 and 5 are both sides, the third distance (the diagonal) is √34 > 5. If 3 is the diagonal and 5 is a side, you'd get b = √(3² - 5²) which is impossible (negative under the root). So the only valid assignment that gives a distance smaller than 5 is treating 5 as the diagonal and 3 as a side, which produces the other side 4. Therefore the minimum possible distance from F to A is 4 metres.

Suggested diagram to draw (simple):

   I (w,0)
    ┌───────────┐
    │           │
    │           │
F (0,0)       A (w,h)
    │           │
    └───────────┘
   D (0,h)

Label distances from F: F→I = w, F→D = h, F→A = √(w²+h²).
Use a = 3 and diagonal = 5 to get the other side = 4.

Student performance notes (ACARA v9 aligned):

• Correct final answer: 4 — demonstrates understanding of the Pythagorean relationship for rectangle corners.
• Verbal justification: Student later explained the reasoning correctly (showed conceptual understanding and could connect values to side/diagonal roles).
• Missing elements: Student submitted no written working, omitted units (metres), and did not include a diagram. These are important for clear communication and full evidence of reasoning.

Next steps / Teacher guidance:

• Ask the student to rework the question in writing: draw the rectangle, label sides, and show the calculation (3² + ?² = 5² → ? = 4). Mark the unit m on the final answer.
• Give one or two similar problems (swap numbers so the diagonal is larger) to practise identifying which values are legs vs diagonal and to reinforce using Pythagoras correctly.
• Assessment focus: require a diagram and units for full credit. Verbal reasoning is strong — convert that into clear written steps next time.

Suggested feedback to the student (short): Good job — your answer 4 is correct and your verbal reasoning shows you understand why (Pythagoras). Next time, please draw the rectangle, show the working steps, and write the units “m” with your final answer so your reasoning is fully clear on paper.

Links to ACARA v9 themes: Measurement & Geometry — using Pythagoras, representing problems with diagrams, and communicating solutions clearly are expected skills at this stage of secondary schooling.