Rectangle corner distances: find minimum remaining distance given 3 m and 5 m — Age 13

75-word step-by-step summary for a 13-year-old: list the three distances (a, b, diagonal), test placements of 3 and 5, use Pythagoras to solve 9+b^2=25⇒b=4, reject impossible diagonal-smaller-than-side case, conclude minimum remaining distance is 4 m.

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Place corner F with sides a, b and diagonal d = √(a^2+b^2). The three distances from F are a, b and d. Cases: (1) a=3, b=5 ⇒ d=√(9+25)=√34≈5.83 m. (2) a=3, d=5 ⇒ 9+b^2=25 ⇒ b^2=16 ⇒ b=4 m. (3) a=5, d=3 impossible because diagonal ≥ each side. Possible remaining distances: 4 m or √34, so minimum = 4 m. Justification: the diagonal cannot be shorter than a side, so discard the impossible case and minimum remaining distance is 4 m.

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