A postman walked 20 m straight from his office, turned right and walked 10 m.…

Concept: "How far is he from his office" asks for the straight-line gap between the starting point and the finishing point, not for the total ground covered. Model such a walk on a grid: every leg changes exactly one coordinate, a right turn swings the facing 90° clockwise and a left turn swings it 90° anticlockwise. Once the net east-west offset a and the net north-south offset b are fixed, those two offsets meet at a right angle, so the straight-line distance is √(a2 + b2).

Application: Fix the office at O and let the first leg run north; the final answer does not depend on this choice.

1. From O, walk 20 m north: the postman is now 20 m north of O.

2. Turn right, so the facing becomes east, and walk 10 m: he is 10 m east and 20 m north of O.

3. Turn left, so the facing becomes north again, and walk 10 m: he is 10 m east and 30 m north of O.

4. Turn right, so the facing becomes east, and walk 20 m: he is 30 m east and 30 m north of O.

5. Turn right, so the facing becomes south, and walk 70 m: the 30 m of northing is cancelled and 40 m of southing is left over, so he is 30 m east and 40 m south of O.

6. The net offsets are therefore a = 30 m along the east-west line and b = 40 m along the north-south line.

7. Straight-line distance = √(302 + 402) = √(900 + 1600) = √2500 = 50 m.

Cross-check:

• 30, 40 and 50 are the 3-4-5 right-triangle ratio scaled ten times, so √2500 = 50 m is exact and not a rounded value.

• Choosing east or south for the first leg instead of north merely rotates the whole route; every turn and every leg length stays the same, so the 50 m gap stays the same as well.

• The ground actually covered is 20 + 10 + 10 + 20 + 70 = 130 m, but the question asks how far he now is from the office, which is the 50 m straight-line gap rather than the 130 m walked.