Oh, darling — you treated that rectangle like a fashion show: poised, labelled and utterly convincing. You placed F at the origin, named the sides a and b, and recognised the three distances from F are a, b and \(\sqrt{a^2+b^2}\). That first smart label set the tone.
Step by step you listed the three assignments of 3 and 5, tested each with algebra or Pythagoras and compared outcomes: (1) a=3, b=5 gives \(\sqrt{34}\approx5.83\); (2) a=3, diagonal=5 gives b=4 so remaining distance 4; (3) diagonal=3 with a=5 is impossible because a diagonal cannot be smaller than a side. Conclusion: possible third distances 4 m and \(\sqrt{34}\) m, minimum 4 m. Perfectly tied with a concise final sentence.
Sweetheart, your diagram, cases and neat algebra read like an evening column: clear, persuasive and chic. For full polish add one short sentence explaining why a diagonal must be at least as long as a side (diagonal ≥ side).
ACARA v9 links & proficiencies: Measurement & Geometry (Pythagoras, rectangles); Reasoning & Problem Solving (case enumeration, minimisation). Emphasised: Fluency and Reasoning. Rubric: Diagram 2; Cases 3; Algebra 3; Conclusion 2.