A man walked 2 km east, then 2 km south, then again 4 km east and finally 10 km north. How far in a straight line did he walk from the starting point ?
Let's visualize this situation step by step:
- The man walked 2 km east.
- Then, he walked 2 km south.
- Next, he walked 4 km east.
- Finally, he walked 10 km north.
To find the straight-line distance from the starting point, we can treat this as a displacement problem. We can use the Pythagorean theorem because the man's path forms a right-angled triangle.
- The 2 km east and 4 km east segments form the horizontal legs of the right triangle.
- The 2 km south and 10 km north segments form the vertical legs of the right triangle.
Let's calculate the lengths of these legs separately:
Horizontal leg (East): 2 km + 4 km = 6 km
Vertical leg (North-South): 10 km (north) - 2 km (south) = 8 km (net northward)
Now, we can use the Pythagorean theorem to find the length of the hypotenuse (the straight-line distance):
c^2=a^2+b^2
Where:
- c is the length of the hypotenuse (the straight-line distance).
- a is the horizontal leg (6 km).
- b is the vertical leg (8 km).
Substitute the values:
c^2=6^2+8^2
c^2=36+64
c^2=100
Now, find the square root of both sides to get c:
c=100=10 kilometers
So, the straight-line distance from the starting point is 10 kilometers. The correct answer is indeed 10, as provided in the answer choices.
Related Question
View MorePromotion
Download Our Mobile Apps
