Farhan travels 4 km in the north east direction and then turns towards the south east to travel another 3 km. What is the straight line distance of Farhan from his starting point now?

Introduction / Context:
This question is a direction and distance problem from reasoning or basic geometry. A person walks along two segments in specific compass directions, and you must find the shortest straight line distance from the starting point to the final point. It combines understanding of directions like north east and south east with right triangle geometry and Pythagoras theorem.

Given Data / Assumptions:
- Farhan first walks 4 km in the north east direction.
- He then turns and walks 3 km in the south east direction.
- Directions are assumed standard: north up, east right on a coordinate plane, with north east at 45 degrees between north and east and south east at 45 degrees between south and east.
- The required answer is the straight line distance between his starting point and his final position.

Concept / Approach:
We can model Farhan's path on a coordinate plane. Let the starting point be the origin. North is the positive y axis and east is the positive x axis. North east at 45 degrees means equal movement in x and y directions. South east means movement in the positive x direction and negative y direction, again at 45 degrees. By resolving each displacement into horizontal and vertical components, we can find the total x and y displacement and then use Pythagoras theorem to find the resultant distance.

Step-by-Step Solution:
Step 1: Represent the first 4 km north east movement. At 45 degrees, horizontal and vertical components are both 4 divided by square root of 2. Step 2: So after the first leg, Farhan is at x = 4 / sqrt(2) and y = 4 / sqrt(2). Step 3: For the second 3 km south east movement, the horizontal component is 3 / sqrt(2) towards east and the vertical component is 3 / sqrt(2) towards south, that is negative y. Step 4: Add the components. Total x displacement is (4 / sqrt(2) + 3 / sqrt(2)) = 7 / sqrt(2). Total y displacement is (4 / sqrt(2) - 3 / sqrt(2)) = 1 / sqrt(2). Step 5: Use Pythagoras theorem. The distance squared from the origin is (7 / sqrt(2))^2 + (1 / sqrt(2))^2 = 49 / 2 + 1 / 2 = 50 / 2 = 25. Step 6: The distance is the square root of 25, which equals 5 km.

Verification / Alternative check:
A quick way to check is to note that the resultant vector components ratio is 7:1, so the displacement roughly lies almost along the east direction with a small north component. A 5 km distance is reasonable since Farhan has walked a total path length of 7 km but changed direction, making the straight line shorter than the path. Any answer greater than 7 km or less than 4 km would be clearly impossible, and 5 km fits the rigorous geometric calculation.

Why Other Options Are Wrong:
4 km would suggest that the second leg cancels most of the first, which is not correct when you compute the components. 6 km and 7 km are longer than the direct path allowed by two line segments of lengths 4 km and 3 km because the maximum straight line distance between start and end could only equal 7 km if he never turned, and would be strictly less in any other configuration. Exact computation shows 5 km is the only correct distance.

Common Pitfalls:
Students often add or subtract the distances directly without resolving directions, for example, incorrectly computing 4 plus 3 or 4 minus 3. Others misinterpret north east and south east as purely north or purely east movements. Drawing a rough diagram and resolving to x and y components greatly reduces mistakes. Remember that when legs of a journey meet at right angles or at symmetric angles, the Pythagoras theorem is very effective.

Final Answer:
Farhan is now at a straight line distance of 5 km from his original starting point.