A and B lie on the same side of a straight line ℓ, at perpendicular distances 3 cm and 5 cm from it. The two feet of those perpendiculars are 6 cm apart. As the point X runs along ℓ, what is the least value that AX + XB can take?

Reflect B in ℓ to get B′, which sits 5 cm on the far side. Since XB = XB′ for every X on ℓ, the smallest value of AX + XB is the straight distance AB′. A and B′ are now on opposite sides, so the vertical gap between them is 3 + 5 = 8 cm while the horizontal gap stays 6 cm, giving AB′ = √(36 + 64) = √100 = 10 cm. The trap 2√10 = √(36 + 4) is the direct distance AB, which no path through ℓ can achieve because A and B are on the same side; 8 cm just adds the two heights; and 14 cm adds 6 to 8 without using Pythagoras.