Pythagoras in Practical Problems
Translate real-world situations into right-triangle diagrams. Apply Pythagoras to coordinate distances, navigation, rectangle diagonals, and worded problems.
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From a starting point, walk 12 metres east, then turn 90° and walk 5 metres south. How far are you from the start, in a straight line? Draw the journey first.
Pythagoras' theorem is the secret weapon for any “straight-line distance” problem — navigation, mapping, ladders, screens, sports fields, gates, framing. The skill is spotting the right triangle hiding in the situation.
Whenever two distances are perpendicular (east+north, vertical+horizontal, width+height), they are the legs of a right triangle. The straight-line distance is the hypotenuse. Draw a quick sketch first — label the legs with the given distances, mark the right angle, then apply $c^2 = a^2 + b^2$.
Know
- Perpendicular distances are legs of a right triangle
- Straight-line distance is the hypotenuse
- Coordinate differences ($\Delta x$, $\Delta y$) act as legs
Understand
- Why drawing a diagram simplifies word problems
- How compass directions create right angles
- Why the diagonal of a rectangle is the hypotenuse of two right triangles
Can Do
- Translate a worded scenario into a right-triangle diagram
- Apply Pythagoras to find distances between points
- Solve ladder, gate, diagonal and navigation problems
Wrong: “The two hikers walked 6 + 8 = 14 km apart.” That's the total distance walked, not the straight-line separation.
Right: Straight-line separation is the hypotenuse: $\sqrt{6^2 + 8^2} = 10$ km.
Wrong: Computing without drawing — missing which sides are perpendicular.
Right: Sketch → label → identify right angle → apply Pythagoras.
To find the distance between two points on a grid, the horizontal difference ($\Delta x$) and the vertical difference ($\Delta y$) are the legs of a right triangle.
For points $A(x_1, y_1)$ and $B(x_2, y_2)$: $\Delta x = x_2 - x_1$, $\Delta y = y_2 - y_1$. The distance is $AB = \sqrt{(\Delta x)^2 + (\Delta y)^2}$. This is just Pythagoras applied to the rise and run between the points.
When someone walks N then E (or any two perpendicular bearings), the legs are the two distances and the hypotenuse is the straight-line gap.
| Journey | Setup |
|---|---|
| 6 km N, 8 km E | $d = \sqrt{6^2+8^2}=10$ km |
| 3 km W, 4 km S | $d = \sqrt{3^2+4^2}=5$ km |
| 12 m E, 5 m N | $d = \sqrt{12^2+5^2}=13$ m |
Watch Me Solve It · 3 examples
- 1SketchTreat N and E as perpendicular legs of a right triangle.North and east are at 90° to each other.
- 2Apply Pythagoras$d^2 = 6^2 + 8^2 = 36 + 64 = 100$
- 3Square root$d = \sqrt{100} = 10$ km
- 1Compute differences$\Delta x = 8-2 = 6$, $\Delta y = 9-1 = 8$
- 2Apply Pythagoras$d^2 = 6^2 + 8^2 = 100$
- 3Root$d = 10$ units
- 1Identify triangleThe diagonal cuts the rectangle into two right triangles with legs 9 and 12.
- 2Compute$d^2 = 9^2 + 12^2 = 81 + 144 = 225$
- 3Root$d = 15$ m
Common Pitfalls
Real-world setup
- Sketch the situation
- Spot two perpendicular distances
- Those are the legs
Coordinates
- $\Delta x = x_2 - x_1$
- $\Delta y = y_2 - y_1$
- $d = \sqrt{(\Delta x)^2+(\Delta y)^2}$
Compass
- N, S, E, W all at 90°
- Two perpendicular legs
- Hyp = straight-line gap
Rectangle diagonal
- $d = \sqrt{L^2 + W^2}$
- Two right triangles form
- Same formula
How are you completing this lesson?
Brain Trainer · 4 problems
Four quick drills to lock in today's skill. Try each, then reveal the answer.
-
1 Walk 9 m east then 12 m north. How far from the start?
$\sqrt{81+144}=\sqrt{225}=15$.15 m -
2 Distance from $(0,0)$ to $(5,12)$.
$\sqrt{25+144}=\sqrt{169}=13$.13 -
3 Diagonal of a 7 m by 24 m field.
$\sqrt{49+576}=\sqrt{625}=25$.25 m -
4 Boat sails 3 km N, then 4 km W. Distance from port.
$\sqrt{9+16}=5$.5 km
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. A rectangular soccer field is 100 m by 60 m. (a) Find the length of the diagonal. (b) A player runs from one corner along the diagonal. By how many metres is this shorter than running 100 m + 60 m along two sides?
Q7. Find the distance between the points $P(-3, 4)$ and $Q(5, -2)$.
Q8. A drone takes off, flies 30 m east, then 40 m north, then 120 m straight up. Find the straight-line distance from the launch point to where it is now (2 d.p.).
Quick Check
1. C — 5-12-13 triple.
2. A — 3-4-5 triple.
3. B — $\sqrt{64+36}=10$ cm.
4. D — 100 m (3-4-5 triple scaled by 20).
5. C — 9-12-15 triple.
Show Your Working Model Answers
Q6 (3 marks): (a) $d=\sqrt{10000+3600}=\sqrt{13600}\approx 116.62$ m [2]. (b) Sides: $100+60=160$ m. Diagonal is shorter by $160-116.62=43.38$ m [1].
Q7 (2 marks): $\Delta x = 8$, $\Delta y = -6$ [1]. $d = \sqrt{64+36} = \sqrt{100} = 10$ units [1].
Q8 (4 marks): Ground distance $=\sqrt{30^2+40^2}=50$ m [1]. The vertical climb of 120 m is perpendicular to the ground distance, forming another right triangle [1]. Total straight-line distance $= \sqrt{50^2+120^2} = \sqrt{2500+14400} = \sqrt{16900} = 130$ m [1]. Notice that 5-12-13 triple appears twice! [1].
Two paths meet
At a junction, road A heads north for 24 km then turns east for 7 km. Road B heads directly from the junction along the straight line that connects the start and end of road A. How much shorter is road B than road A?
Reveal solution
Road A: $24+7=31$ km. Road B: $\sqrt{576+49}=\sqrt{625}=25$ km. Road B is $31-25=6$ km shorter.
Sketch first
Always draw the situation
Spot the legs
Look for perpendicular distances
Distance formula
$d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$
Compass
N, S, E, W all at 90°
Diagonal
Rectangle diagonal $= \sqrt{L^2 + W^2}$
3D too
Pythagoras stacks in three dimensions
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