Solution (step-by-step, with units and notation):
Place F at (0,0). Put D at (3,0) so FD = 3 m (one side). Let the other side length be b metres, so the opposite corner I is at (3,b). Then FI is the diagonal: FI = sqrt(3^2 + b^2) = 5 m. Square both sides: 9 + b^2 = 25, so b^2 = 16 and b = 4 m. The remaining corner A is at (0,4), so FA = 4 m. Therefore the distance FA = 4 m.
Notes for Alcumus-style work: When you cant draw, write short justification lines: label points, state values with units (e.g., FD = 3 m), cite the Pythagorean theorem, show algebraic steps, and box the final answer with units (4 m).
She solved it. Fast. She put F at a corner. D at an adjacent corner, 3 m away. I is the opposite corner, 5 m from F. Rectangle. So the other side must be sqrt(5^2−3^2)=4 m. Result: FA = 4 m. Clear. For next year: practice writing each step. Label points. State units: metres. Write theorems used: Pythagoras. Give one-line justifications in drills. Expand to short proofs in exercises. In Intro to Algebra, translate diagrams to coordinates and equations. In Intro to Geometry, write two-column and paragraph proofs. Targets: no omitted units; neat notation for sides a,b; show algebra: b = sqrt(5^2−3^2). Aim: precise, brief, and complete. Weekly proofs: three short ones. Keep speed. Keep clarity. Keep the joy. Schedule: nightly five-minute justification drills, weekly proof write-ups, and monthly review with feedback. Celebrate improvements. Build habit. Parents and tutor note progress; praise precision; nudge for clear units and labels. Daily.