13-year-old: Find the minimum third distance from a rectangle corner given 3 m and 5 m

Place corner F at (0,0). Let the rectangle sides be a and b and the diagonal d = √(a²+b²). The three distances from F are a, b and d.

1. Case 1: a = 3, b = 5 → d = √(3²+5²) = √34 ≈ 5.83 m.
2. Case 2: a = 3, d = 5 → 3² + b² = 5² ⇒ 9 + b² = 25 ⇒ b² = 16 ⇒ b = 4 m.
3. Case 3: a = 5, d = 3 is impossible because the diagonal d must be at least as large as each side (3 < 5), so a side cannot exceed the diagonal.

The possible remaining distances are 4 m and √34 m, so the minimum possible remaining distance is 4 m.

Final justification: 4 m is smallest because the only valid assignment that makes 5 the diagonal gives the other side 4 m, and the alternative (both 3 and 5 as sides) gives a larger diagonal √34.