He again turns left and walks for 6 km.
At this point, he turns to his left and walks for 6 km. How far is he from his starting point? (All turns are 90-degree turns only, unless specified.)
This problem involves calculating the final distance from the starting point after a series of movements in different directions.
Rahul starts by walking 6 km North. Let the starting point be O.
Position after Step 1: 6 km North of O.
Rahul turns left (90 degrees) and walks 3 km West.
Current Position: 6 km North and 3 km West of O.
He turns left again (90 degrees) and walks 6 km South.
Analysis: The 6 km South movement cancels out the initial 6 km North movement. The net North-South displacement is now 0 km.
Current Position: 0 km North/South displacement (same latitude as O) and 3 km West of O.
He turns left one more time (90 degrees) and walks 6 km East.
Analysis: He was 3 km West of O. Walking 6 km East means he moves 3 km East to reach the North-South line passing through O, and then another 3 km East.
Net East-West Displacement: -3 km (West) + 6 km (East) = +3 km (East).
The final position has zero North-South displacement and 3 km East displacement from the starting point O.
Therefore, the final distance from the starting point is the net Eastward displacement.
Distance = $\sqrt{(\text{Net East-West Displacement})^2 + (\text{Net North-South Displacement})^2}$
Distance = $\sqrt{(3 \text{ km})^2 + (0 \text{ km})^2}$
Distance = $\sqrt{9 \text{ km}^2}$
Distance = 3 km
Rahul is 3 km away from his starting point.
Point A is $30$ m to the North of point B. Point A is $10$ m to the west of point C. Point C is $20$m to the North of point D. Point E is $20$ m to the East of point D. Point F is $20$ m to the South of point E. What is the shortest distance from the point B to point E.
Rakesh left home and walked $5$km southwards, then turned right and walked $2$km and again turned right and walked $5$ km and finally again turned left and walked $5$ km. The shortest distance between the final position and home is.