Question

Samesh walks 12 km North, turns right and walks 4 km, then turns right and walks 9 km. How far and in what direction is he from his starting point?

• A. 10 km North
• B. 10 km North-East
• C. 5 km North
• D. 5 km North-East
• E. 6 km North

Hint:

Direction problems with distances often form a right-angled triangle. Use the Pythagorean theorem ($a^2 + b^2 = c^2$) to find the displacement.

Answer 1

The Correct Option is D

Step 1: Analyse options.
- Samesh moves 12 km North. - He turns right and moves 4 km East. - He turns right again and moves 9 km South. - Net vertical distance: $12\text{ (North)} - 9\text{ (South)} = 3\text{ km North}$. - Net horizontal distance: $4\text{ km East}$. - Shortest distance: $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ km}$.
Step 2: Conclusion.
He is 5 km away in the North-East direction. Final Answer: (d) 5 km North-East