Mathematics • Year 9 • Unit 3 • Lesson 4

Pythagoras Out in the World

Apply Pythagoras' theorem to real-world distance problems: hikers, boats, swimming pools, garden paths, and coordinate-grid maps. Sketch the situation first, identify the two perpendicular legs, then choose ADD (missing hypotenuse) or SUBTRACT (missing leg). Then explain why "6 + 8 = 14 km apart" is wrong.

Apply · Real-World Maths

1. Word problems

Each problem describes a real-world situation. Always sketch, identify the right angle, label the two perpendicular distances as legs, then apply Pythagoras. Show your working — a single final answer with no working only earns half marks.

1.1 — Sailing race. A yacht sails 5 km east, then turns 90° and sails 12 km north to reach a buoy.

(a) Sketch the journey and mark the right angle.
(b) How far is the yacht from its starting point in a straight line? 3 marks

Stuck? East and north are perpendicular — the two legs are 5 km and 12 km. The straight-line gap is the hypotenuse. Spot the 5-12-13 triple.

1.2 — Diagonal corner-to-corner. A garden bed is rectangular, 9 m by 12 m. A diagonal path runs from one corner to the opposite corner.

(a) Find the length of the diagonal path.
(b) If a gardener instead walked along two adjacent sides, how much further would they walk? 3 marks

Stuck? Diagonal = hypotenuse of right triangle with legs 9 and 12 — spot the $3\times$(3-4-5) triple. Walking two sides = 9 + 12 = 21 m.

1.3 — Treasure map distance. On a treasure map, the X is at coordinate $(8, 9)$. The starting point is at $(2, 1)$. The treasure-hunter walks in a straight line from start to X.

(a) Compute $\Delta x$ and $\Delta y$.
(b) Find the straight-line distance from start to X using Pythagoras. 3 marks

Stuck? $\Delta x = 8 - 2$, $\Delta y = 9 - 1$. Then $d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$.

1.4 — Bee-line vs. street grid. Aisha walks 6 blocks east then 8 blocks north along the streets to reach her friend's house. Each block is 100 m.

(a) How far does she walk along the streets?
(b) How far is her friend's house in a straight line (the bee-line)? 3 marks

Stuck on (b)? 6 blocks $\times$ 100 m = 600 m east, 8 blocks $\times$ 100 m = 800 m north. Bee-line distance is the hypotenuse — spot $200\times$(3-4-5).

1.5 — Plane's straight-line speed. A small plane flies 36 km east and 15 km north in 12 minutes.

(a) Find the straight-line distance the plane has actually covered.
(b) Is this distance shorter or longer than the total path it flew (36 + 15 km)? 3 marks

Stuck? Legs 36 and 15 — recognise this as a multiple of 5-12-13? Actually 15-36-? — let's see: 15 and 36 share factor 3, so divide by 3: 5 and 12, hyp 13, scale back: hyp = 39.

2. Explain your thinking

This question is about communication, not just answers. Use full sentences. 4 marks

2.1 A classmate is shown the hiker problem (one walks 6 km N, the other 8 km E). They confidently write "So they are 6 + 8 = 14 km apart." They think their answer is right because adding two distances seems obviously correct.

In your own words, explain (i) what the 14 km actually represents, (ii) why it is NOT the correct straight-line distance, (iii) which lesson rule they have missed, and (iv) the correct straight-line distance.

Stuck? Revisit lesson § "Spot the Trap" — 14 km is the TOTAL distance walked, but each hiker only walked one leg. The straight-line distance is the hypotenuse, not the sum.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Sailing race

(a) Sketch: right triangle with east leg 5 km, north leg 12 km, hypotenuse running from start to the buoy.
(b) $d^2 = 5^2 + 12^2 = 25 + 144 = 169$, so $d = \sqrt{169} = \mathbf{13}$ km. (5-12-13 triple.)

1.2 — Garden diagonal

(a) $d^2 = 9^2 + 12^2 = 81 + 144 = 225$, so $d = \sqrt{225} = \mathbf{15}$ m. ($3\times$(3-4-5).)
(b) Walking two sides $= 9 + 12 = 21$ m. The diagonal is 15 m, so two sides is $21 - 15 = \mathbf{6}$ m further.

1.3 — Treasure map

(a) $\Delta x = 8 - 2 = \mathbf{6}$, $\Delta y = 9 - 1 = \mathbf{8}$.
(b) $d^2 = 6^2 + 8^2 = 36 + 64 = 100$, so $d = \sqrt{100} = \mathbf{10}$ units. ($2\times$(3-4-5).)

1.4 — Bee-line vs. street grid

(a) Streets: $(6 + 8) \times 100 = \mathbf{1400}$ m $= 1.4$ km.
(b) Bee-line: legs 600 m and 800 m, so $d^2 = 600^2 + 800^2 = 360{,}000 + 640{,}000 = 1{,}000{,}000$, $d = \mathbf{1000}$ m $= 1$ km. ($200\times$(3-4-5).) The bee-line saves about 400 m.

1.5 — Plane's straight-line distance

(a) Legs 36 and 15. Both divisible by 3, so this is $3 \times$(5-12-13). $d = 3 \times 13 = \mathbf{39}$ km. (Check: $36^2 + 15^2 = 1296 + 225 = 1521 = 39^2$ ✓.)
(b) Total path flown $= 36 + 15 = 51$ km. Straight-line $= 39$ km. The straight-line distance is $\mathbf{shorter}$ (by 12 km), which makes sense — the bee-line is always the shortest distance between two points.

2.1 — Explain your thinking (sample response)

(i) The 14 km is the total distance walked by the two hikers combined (6 km by one, 8 km by the other) — or by one hiker who walks both legs of the trip. (ii) It is NOT the straight-line distance because the two hikers ended up at different places: one is 6 km north of the campsite, the other is 8 km east. The line connecting them goes diagonally and is shorter than walking around two sides. (iii) The classmate has missed the rule that two perpendicular distances are legs of a right triangle, and the straight-line gap between the endpoints is the hypotenuse — the lesson's Spot the Trap card flags exactly this mistake. (iv) The correct straight-line distance is $\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = \mathbf{10}$ km apart, not 14 km.

Marking: 1 for explaining what 14 km represents; 1 for naming the trap (sum vs. hypotenuse); 1 for correct calculation showing 10 km; 1 for a clear, full-sentence explanation.