Rectangle corner distances — find the minimum remaining distance (Age 13) — given 3 m and 5 m

Enumerate how 3 m and 5 m can be the two known corner distances of a rectangle (they may be two sides or a side and the diagonal). Use Pythagoras to compute the unknown distance in each allowable case, reject impossible assignments (diagonal must be ≥ each side), and conclude the minimum possible remaining distance (4 m). Includes clear algebra and a 3–4–5 recognition.

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Start by labelling the corner F at (0,0). Let the rectangle sides be a and b (both in metres) and the diagonal be d = √(a2 + b2). The three possible distances from F are: a, b, and d = √(a2 + b2).

  1. How can 3 and 5 fit into {a, b, d}?

    There are three placements to check: (i) a = 3, b = 5 (so d is unknown); (ii) a = 3, d = 5 (so b is unknown); (iii) a = 5, d = 3 (so b would be unknown). We must test each.

  2. Case 1: a = 3 m, b = 5 m.

    Then the remaining distance is the diagonal d = √(32 + 52) = √(9 + 25) = √34 ≈ 5.83 m.

  3. Case 2: a = 3 m, d = 5 m.

    Use Pythagoras: a2 + b2 = d2 so 32 + b2 = 52. That is 9 + b2 = 25. Subtract 9: b2 = 16. Take the positive root (lengths are positive): b = 4 m. So the remaining distance is 4 m. (Notice this is the 3–4–5 right triangle shortcut.)

  4. Case 3: a = 5 m, d = 3 m.

    This is impossible because the diagonal d must be at least as long as each side (d ≥ a and d ≥ b). Here d = 3 < a = 5, so a side would be longer than the diagonal, which cannot happen. Reject this case.

Possible remaining distances: from the three cases the allowable remaining distances are 4 m (from Case 2) and √34 ≈ 5.83 m (from Case 1); Case 3 is impossible and rejected.

Minimum: 4 m.

One-sentence justification: The smallest possible remaining distance is 4 m because the only valid configuration that makes one known distance the diagonal is 3 and 5 as a side and diagonal respectively, giving b = √(52 − 32) = 4, and all other valid configurations produce a larger remaining length (the case diagonal = 3 is impossible since a diagonal cannot be shorter than a side).

Teacher rubric (100-word exemplary comment): Exemplary student work: The student placed the corner at (0,0), labelled sides a and b and the diagonal d = √(a2+b2). They enumerated all three placements of 3 and 5 among {a,b,d}, rejected the impossible case because a diagonal cannot be smaller than any side, and used Pythagoras with clear algebra (9+b2=25 ⇒ b2=16 ⇒ b=4). They noted the 3–4–5 right-triangle shortcut, labeled units and right angles, and presented tidy calculations. The final conclusion, that the minimum possible remaining distance is 4 m, is justified succinctly. This earns full marks for clarity and completeness.

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