Rectangle minimisation — corner distances (13-year-old) — find the minimum remaining distance given 3 m and 5 m

Cornell-notes style printable solution for a 13-year-old. Lists all possible configurations when two corner distances are 3 m and 5 m, uses Pythagoras to compute the third distance in each case, rejects impossible assignments, and identifies the minimum remaining distance (4 m) with a one-sentence justification.

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Rectangle minimisation — corner distances

Student: 13 years · Objective: Given corner distances 3 m and 5 m, list configurations, compute the third distance, choose the minimum. (Cornell notes)

Cues / Questions
  • What are the three corner distances from one corner?
  • 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?
 Notes / Work (step-by-step)

Draw a rectangle and label one corner F. Let the two sides meeting at F be a and b (these are the side lengths). The distance from F to the opposite corner (the diagonal) is d = √(a2 + b2).

Small sketch to draw:
Draw a rectangle, mark the bottom-left corner F. Label the horizontal side a and the vertical side b. Draw the diagonal from F to the opposite corner and label it d = √(a2 + b2). Put the known distances 3 m and 5 m as two of {a, b, d} in different cases.

All possible sensible assignments of 3 and 5 to {a, b, d}:

  1. Case A: a = 3 m, b = 5 m → compute d.
    d = √(a2 + b2) = √(32 + 52) = √(9 + 25) = √34 ≈ 5.83 m.
  2. Case B: a = 3 m, d = 5 m → compute b using Pythagoras.
    a2 + b2 = d2 → b2 = d2 - a2 = 52 - 32 = 25 - 9 = 16.
    So b = √16 = 4 m (take positive root because length > 0).
  3. Case C: b = 3 m, d = 5 m → symmetric to Case B, so a = 4 m (same algebra).
  4. Impossible case: d = 3 m and one side = 5 m.
    This is impossible because the diagonal d = √(a2 + b2) is always ≥ each side a and b. If d were 3 m but a (or b) were 5 m, that would make a > d, which cannot happen. So any assignment with diagonal = 3 m and a side = 5 m is rejected.

Summary of computed unknown distances

  • Case A (3 & 5 are the two sides): remaining distance (diagonal) = √34 ≈ 5.83 m.
  • Case B (3 is side, 5 is diagonal): remaining side = 4 m.
  • Case C (3 is side, 5 is diagonal, swapped): remaining side = 4 m.
  • Impossible: diagonal = 3 and side = 5 → reject, because diagonal must be ≥ each side.

Which is smallest? The smallest possible remaining distance is 4 m (from Case B or C). The alternative (both sides 3 and 5) gives a diagonal about 5.83 m, which is larger.

Ally-style aside (cheeky and encouraging): You tried on every possibility like different outfits for a courtroom dance — neat labels, neat algebra, and a clean final pick. When you spot a 3-4-5 pattern, sing it out; it saves time!
Final answer (one-sentence justification):

The minimum possible remaining distance is 4 m, because the only feasible assignment that makes one of the given lengths the diagonal (5 m) leads to the other side b = √(52 − 32) = 4 m, while the case where 3 m and 5 m are both sides gives a larger diagonal √34 ≈ 5.83 m.

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