A man walks 2 km towards East and then he turns to South and walks 6 km. Again, he turns to East and walks 4 km, after this he turns to North and walks 14 km. How far is he from his starting point?

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Accepted Answer

The man's movement can be broken into steps:  
1. He moves **2 km East**, then **6 km South**, followed by **4 km East**, and finally **14 km North**.

2. **Net movement in the East-West direction:**  
   - Total distance East = $$ 2 + 4 = 6 \, 	ext{km} $$.  
   - Net distance in the East direction = $$ 6 \, 	ext{km} $$.

3. **Net movement in the North-South direction:**  
   - Total distance North = $$ 14 \, 	ext{km} $$.  
   - Total distance South = $$ 6 \, 	ext{km} $$.  
   - Net distance in the North direction = $$ 14 - 6 = 8 \, 	ext{km} $$.

4. The man's final position forms a right triangle with the starting point. Using the Pythagoras theorem:  
   $$ 
   H^2 = (	ext{East-West distance})^2 + (	ext{North-South distance})^2 
   $$  
   $$ 
   H^2 = (6)^2 + (8)^2 
   $$  
   $$ 
   H^2 = 36 + 64 = 100 
   $$  
   $$ 
   H = \sqrt{100} = 10 \, 	ext{km}.
   $$

Thus, the man is **10 km** away from his starting point.

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