📐 Distance Calculation & Pythagoras
Calculate shortest distances using Pythagorean theorem, trace multi-step paths, find net displacement, and solve distance-based direction problems from banking exams
Distance Calculation & Pythagoras
Most direction problems ask not just "which direction?" but also "how far?" This lesson teaches you to calculate the shortest distance (straight-line distance) from the starting point using the net displacement method and the Pythagorean theorem.
Net Displacement Method
After tracing a path, calculate:
- Net vertical displacement = Total North - Total South
- Net horizontal displacement = Total East - Total West
If one is zero → the person is directly in a cardinal direction. If both are non-zero → use the Pythagorean theorem for shortest distance, and the person is in an intermediate direction (NE, NW, SE, SW).
The Pythagorean Theorem
Shortest Distance = sqrt(vertical² + horizontal²)
When to use:
- When the net path forms a right-angled triangle
- When both vertical and horizontal displacements are non-zero
Solved Example 1: Multi-Point Path
Q: A man starts walking from point A in the North and walks 5m to reach Point B. Then he turns to the right and walks for 2m to reach point C. Now he takes a left turn and goes 5m to reach point D. After that he turns to his left and walks for 14m left and reaches point E. How far and in which direction is point E with respect to point B?
Pro Content Locked
Upgrade to Pro to access this lesson and all other premium content.
₹99 charged monthly · Cancel anytime
- All Agriculture & Banking Courses
- AI Lesson Questions (100/day)
- AI Doubt Solver (50/day)
- Glows & Grows Feedback (30/day)
- AI Section Quiz (20/day)
- 22-Language Translation (100/day)
- Recall Questions (20/day)
- AI Quiz (15/day)
- AI Quiz Paper Analysis (100/day)
- AI Step-by-Step Explanations (100/day)
- Spaced Repetition Recall (FSRS)
- AI Tutor
- Immersive Text Questions
- Audio Lessons — Hindi & English
- Mock Tests & Previous Year Papers
- Summary & Mind Maps
- XP, Levels, Leaderboard & Badges
- Generate New Classrooms
- Voice AI Teacher (AgriDots Live)
- AI Revision Assistant
- Knowledge Gap Analysis
- Interactive Revision (LangGraph)
🔒 Secure via Razorpay · Cancel anytime · No hidden fees
Distance Calculation & Pythagoras
Most direction problems ask not just "which direction?" but also "how far?" This lesson teaches you to calculate the shortest distance (straight-line distance) from the starting point using the net displacement method and the Pythagorean theorem.
Net Displacement Method
After tracing a path, calculate:
- Net vertical displacement = Total North - Total South
- Net horizontal displacement = Total East - Total West
If one is zero → the person is directly in a cardinal direction. If both are non-zero → use the Pythagorean theorem for shortest distance, and the person is in an intermediate direction (NE, NW, SE, SW).
The Pythagorean Theorem
Shortest Distance = sqrt(vertical² + horizontal²)
When to use:
- When the net path forms a right-angled triangle
- When both vertical and horizontal displacements are non-zero
Solved Example 1: Multi-Point Path
Q: A man starts walking from point A in the North and walks 5m to reach Point B. Then he turns to the right and walks for 2m to reach point C. Now he takes a left turn and goes 5m to reach point D. After that he turns to his left and walks for 14m left and reaches point E. How far and in which direction is point E with respect to point B?
Solution — trace step by step:
- A → B: 5m North
- B → C: Right turn (now facing East), 2m East
- C → D: Left turn (now facing North), 5m North
- D → E: Left turn (now facing West), 14m West
Diagram:
E ←————14m————— D
↑
5m (North)
|
B ——→ C
↑ (2m East)
5m
|
A (start)
E relative to B:
- Vertical: 5m North (B to C to D is 5m North from C, but C is at same level as B... wait)
- B → C is East (same North level as B)
- C → D is 5m North
- So D is 5m North and 2m East of B
- D → E is 14m West
- E is 5m North and (2 - 14) = 12m West of B
- Horizontal: 2m East - 14m West = 12m West of B
Net: 5m North and 12m West of B
Shortest distance = sqrt(5² + 12²) = sqrt(25 + 144) = sqrt(169) = 13m
Direction: North-West of B (both North and West components)
Answer: (c) 13m, North-West
Solved Example 2: Complex Path with Pythagoras
Q: While running in a park, John ran 83m towards South. He then took a left turn and ran 71m until he took a right turn to run 23m. He paused for a while and then 43m towards his left until he took another left turn to run 106m. How far is he from the starting point?
Solution:
- Move 1: 83m South
- Move 2: Left turn from South = East. 71m East
- Move 3: Right turn from East = South. 23m South
- Move 4: Left turn from South = East. 43m East (wait, "43m towards his left" — facing South, left = East). Hmm, re-read: "then 43m towards his left" — this means he turned left. From South (after move 3), left = East. 43m East.
Wait, let me re-read: "He paused for a while and then 43m towards his left until he took another left turn to run 106m."
- After move 3, facing South. "43m towards his left" = turns left = East, walks 43m East
- "took another left turn to run 106m" = left from East = North, 106m North
Net displacement:
- North-South: 83m S + 23m S - 106m N = 106m S - 106m N = 0
- East-West: 71m E + 43m E = 114m E
Wait: 83 + 23 = 106 South, and 106 North → net vertical = 0
Shortest distance = 71 + 43 = 114m East
Hmm, but the options are 100, 112, 114, 113. Let me re-check.
Actually, re-reading: "then 43m towards his left" — after pausing, he was still facing South from move 3. His left = East. So 43m East. Then left turn from East = North. 106m North.
N-S: 83 + 23 = 106 South. 106 North. Net = 0. E-W: 71 + 43 = 114 East.
Answer: (3) 114m
Solved Example 3: Large Path with Multiple Turns
Q: Mohan ran 340 steps towards North and then ran 90 steps after turning to his left. He then ran 130 steps towards his right. He then turned towards his right and ran 90 steps. From there, he turned towards his right and ran another 240 steps. How far and in which direction is Mohan with respect to the starting point?
Solution:
- Move 1: 340 steps North
- Move 2: Left from North = West. 90 steps West
- Move 3: Right from West = North. 130 steps North
- Move 4: Right from North = East. 90 steps East
- Move 5: Right from East = South. 240 steps South
Net displacement:
- N-S: (340 + 130) North - 240 South = 470 - 240 = 230 steps North
- E-W: 90 West - 90 East = 0
Answer: (a) 230 steps, North
Solved Example 4: Total Distance Covered (Not Shortest)
Q: Raj is facing his house, 50 meters away, facing North. He walks 15m straight before meeting Ravi. Raj takes a left turn (West), walks 8m, then takes a left turn (South) and walks 6m to reach Ravi's house. How much total distance has Raj covered from his starting position to reach his own house (going back the same path to Ravi's house, then straight to his house)?
Solution:
- Raj's path to Ravi's house: 15m N + 8m W + 6m S = 29m total
- Going back same path: 29m (retracing)
- From starting position, Raj's house is 50m North
- But after retracing, Raj is back at start → walks 50m to his house
Wait — re-read: "he goes back by the same path that he used to reach Ravi's house, and then walks on the straight path which leads to his house"
- Path to Ravi's house: 15 + 8 + 6 = 29m
- Retrace back to start: 6 + 8 + 15 = 29m
- Then straight to his house: 50m
Hmm but that gives 29 + 29 + 50 = 108m which isn't an option.
Actually, re-reading more carefully: "from his starting position to reach his house" — meaning total path from start. He goes to Ravi's house (29m), comes back same path (29m), then goes to his own house (50m) = 29 + 29 + 50 = 108? But options are 78, 50, 64, 93.
Let me reconsider: After meeting Ravi (15m from start), they walk TOGETHER to Ravi's house. So from the meeting point: 8m W + 6m S = 14m more. Total to Ravi's house from start = 15 + 8 + 6 = 29m.
Return same path from Ravi's house to meeting point: 6 + 8 = 14m. Then from meeting point (which is 15m from start, 35m from house): walks straight to his house = 35m.
Total = 29 + 14 + 35 = 78m.
Answer: (1) 78
Common Pythagorean Triplets
Memorize these — they appear frequently:
| a | b | c (hypotenuse) |
|---|---|---|
| 3 | 4 | 5 |
| 5 | 12 | 13 |
| 6 | 8 | 10 |
| 7 | 24 | 25 |
| 8 | 15 | 17 |
| 9 | 12 | 15 |
| 9 | 40 | 41 |
| 12 | 16 | 20 |
Also remember multiples: 3-4-5 → 6-8-10 → 9-12-15 → 12-16-20
Two-Person Path Problems
Q: Abhi and Asha start cycle race from point A. They both start in East direction. After cycling for 7m, Abhi continues straight while Asha takes a left turn. They both cycle for 6m before turning right and left directions respectively. (1) Asha cycles for 8m and takes a right turn. She cycles for 5m before turning right again. (2) Abhi cycles for 4m and takes a left turn. He cycles for 6m before turning to left again. If both stop, how much respective distance they have to travel to meet each other on their current paths?
Strategy for two-person problems:
- Trace each person's path separately on the same diagram
- Identify their final positions and facing directions
- Find where their extended paths intersect
- Calculate distances from each person to the intersection point
This type requires careful diagram work with both paths plotted.
Tips for Speed
- Draw first, think later — always start with a compass and trace
- Cancel opposite movements — North-South and East-West cancel out
- Recognize Pythagorean triplets — saves calculation time
- Label every turn — write the new facing direction at each turn
- Double-check the question — "from starting point" vs "from point B" vs "from his house"