Raju travels 10 meters towards the north, takes a right turn, and travels 5…

In direction-and-distance problems, track the walker's facing direction after each turn (a right turn rotates the facing 90 degrees clockwise, a left turn rotates it 90 degrees counter-clockwise) and record each leg as a North-South or East-West component. The straight-line distance from the start is the Pythagorean combination of the NET North-South and net East-West displacements, not the sum of the leg lengths.

1. Leg 1: Facing north, Raju walks 10 m north, giving a net displacement of 10 m north (0 m east-west) so far.

2. Turn 1: A right turn from facing north changes his facing to east.

3. Leg 2: Facing east, he walks 5 m east, adding 5 m to the east-west displacement while the north-south displacement stays at 10 m north.

4. Turn 2: A right turn from facing east changes his facing to south.

5. Leg 3: Facing south, he walks 10 m south, which exactly cancels the earlier 10 m north, bringing the net north-south displacement to 0.

6. Net displacement: 0 m north-south and 5 m east, so the straight-line distance from the starting point is 5 m, directly east.

Plotting the path on a grid confirms this: starting at (0, 0), Raju reaches (0, 10) after leg 1, (5, 10) after leg 2, and (5, 0) after leg 3. The distance from (0, 0) to (5, 0) is the square root of (5 squared + 0 squared), which is 5 m, matching the step-by-step result.