Rectangle minimisation — Minimum third corner distance problem (Age 13) | ACARA v9 aligned teacher rubric & exemplar (Cornell notes)

Cues / Questions

• What are the three possible distances from one corner of a rectangle?
• How can 3 and 5 fit into {a, b, √(a² + b²)}?
• Which assignments are impossible and why?
• How to use Pythagoras to find the unknown side?
• Which configuration gives the smallest remaining distance?

Key reminders

• Diagonal ≥ each side.
• Always draw and label a, b, and the diagonal.
• Finish with one-sentence justification.

Notes / Worked solution

Step 0 — Set up: Put corner F at (0,0). Let adjacent side lengths be a and b. The three distances from F to the other corners are: a, b, and the diagonal d = √(a² + b²).

Given: Two distances from F are 3 m and 5 m. These two numbers must occupy two of the three slots {a, b, √(a² + b²)}. We must test each plausible case and compute the remaining distance.

Case listing and computations:

1. Case 1: a = 3, b = 5.
   Then the remaining distance is the diagonal d = √(3² + 5²) = √(9 + 25) = √34 ≈ 5.83 m.
2. Case 2: a = 3, d = 5.
   Then solve √(3² + b²) = 5 ⇒ 9 + b² = 25 ⇒ b² = 16 ⇒ b = 4. So the remaining distance is b = 4 m.
3. Case 3: a = 5, d = 3.
   This is impossible because the diagonal d must be ≥ each side. Here 3 < 5, so a side cannot be larger than the diagonal. Reject this case.

Compare the possibilities: The possible remaining distances (after placing 3 and 5) are 4 m and √34 ≈ 5.83 m. The minimum is 4 m.

Final answer: The minimum possible distance from F to the remaining corner A is 4 m.

One-line justification (required): Because the diagonal must be at least as large as each side, the feasible arrangement that makes the third distance smallest is when 5 is the diagonal and 3 is a side; by Pythagoras the other side is 4 m, which is less than √34.

Diagram (draw in your book)

Sketch a rectangle. Put F at the lower-left corner. Label the adjacent sides a (horizontal) and b (vertical). Mark one corner distance as 3, another as 5. In Case 2 draw the right triangle showing legs 3 and 4 and diagonal 5 (a 3–4–5 right triangle).

Exemplar student answer (compact)

1. Place F at (0,0). Let sides be a and b, diagonal d = √(a² + b²). 2. Given distances are 3 and 5. Case 1: a=3, b=5 ⇒ d = √34 ≈ 5.83 m. Case 2: a=3, d=5 ⇒ 9 + b² = 25 ⇒ b = 4 m. Case 3: a = 5, d = 3 impossible (diagonal ≥ side). So the possible remaining distances are 4 and √34; minimum = 4 m. Final answer: 4 m.

Teacher feedback (Ally McBeal cadence — a playful, dreamy inner-monologue praise)

Oh — you placed the corner, you named the sides, and you tried on each case like it was a different outfit. You rehearsed the impossible one out loud (so brave), solved the algebra with neat steps, and closed with a little, sharp justification — perfect. You showed the right triangles, you guarded the diagonal with a rule (never smaller than a side), and you picked the smallest length like a confident editor choosing the final cut. To sparkle even more: write one sentence explaining why we rejected 5 as a side with 3 as the diagonal (because diagonals can't be smaller than their sides). That tiny sentence will make your proof sing.

Rubric (out of 10)

• Diagram & placement of variables: 2 marks
• Case-listing & reasoning (both feasible cases considered and impossible case checked): 3 marks
• Algebra / Pythagorean computation (show steps): 3 marks
• Final answer with units and 1-sentence justification: 2 marks

ACARA v9 mapping & progression notes

This activity maps to Measurement & Geometry (apply Pythagoras and reason about rectangle properties) and to Reasoning & Problem Solving (select representations, compare candidate configurations, justify choices). Semester 1 builds fluency (Pythagoras computations). Semester 2 emphasizes reasoning — choosing among configurations and short justifications. Encourage students to progress from drilled computations to case enumeration and then to modelling problems.

Common errors and quick fixes

• Assuming both given numbers are sides without checking diagonal/side size relation — fix: always check if diagonal must be ≥ each side.
• Not listing the impossible case — fix: write it and explain why it fails (this strengthens reasoning marks).
• Skipping algebra steps when solving for b — fix: show squaring and rearrangement (9 + b² = 25 ⇒ b² = 16 ⇒ b = 4).

Practice habits checklist (tick as you work)

• Draw a labelled rectangle with F at (0,0).
• List the three possible distances from F: a, b, √(a² + b²).
• Enumerate all placements of the given numbers into those slots and test each.
• Show algebraic steps (square, subtract, root) when finding unknown sides.
• Finish with one clear justification sentence.

Short teaching tip

Start with quick drills on Pythagorean triples (3–4–5, 5–12–13) so students notice patterns. Then give this rectangle-minimisation task to practise enumerating configurations and choosing the smallest remaining distance. For higher extension, ask: if instead the given numbers were 6 and 10, what is the minimum third distance? (Students should notice proportional 3–4–5 scaling.)

Summary (Cornell bottom)

In one sentence: To minimise the unknown corner distance, list the cases where the given numbers are sides or diagonal, reject impossible cases because diagonal ≥ side, solve with Pythagoras when needed, compare results, and state the smallest—here 4 m.