Minimum distance F–A in a rectangle (13-year-old) — ACARA v9 aligned — F 3 m from D and 5 m from I

Facts: Four people stand at the four corners of a rectangular room. The corners are named F, I, D, A so that F and I are opposite corners and D and A are opposite corners. We are told: FD = 3 m and FI = 5 m. We must find the minimum possible distance FA (in meters).

Issue: What is the smallest possible value of FA given the facts?

Argument (step-by-step):

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1. Place the rectangle on a coordinate grid for clarity. Put F at the origin: F = (0,0). Because FD = 3 m and FD is a side meeting F, put D at (3,0). Let the other side meeting F have length y, so A = (0,y). Then the opposite corner I is at (3,y).
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2. Compute FI, the diagonal from F to I. Using the distance formula (or Pythagoras),\n FI = sqrt((3-0)^2 + (y-0)^2) = sqrt(3^2 + y^2). We are told FI = 5, so:\n sqrt(9 + y^2) = 5.
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3. Square both sides to solve for y:\n 9 + y^2 = 25 \rarr y^2 = 16 \rarr y = 4 (take the positive length).
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4. Therefore FA, the vertical side from F to A, has length y = 4 m. That is the distance we were asked to find.
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One-sentence rejection of the impossible case (as suggested): A diagonal (hypotenuse) of a rectangle is always longer than either side, so you cannot have a diagonal of 3 m while one of its sides is 5 m.

Conclusion (ruling): The minimum possible distance FA is 4 m.\n
Student answer 4 (with units: 4 m) is correct.

Notes for the teacher: The student enumerated cases, used Pythagoras correctly, labeled units early, and gave a concise justification. The extra one-sentence rejection above — "the hypotenuse is always longer than either leg" — completes the logical guard against the impossible configuration.