Mathematics • Year 9 • Unit 3 • Lesson 4
Practical Pythagoras — Mixed Challenge
Pull together every idea from Lessons 1-4: identifying right triangles hidden in real-world problems, choosing between $c^2 = a^2 + b^2$ and $a^2 = c^2 - b^2$, computing coordinate distances, recognising triples to skip work, and sanity-checking your setup. Spot a mistake in someone else's working and tackle an open-ended challenge.
1. Mixed problems — sketch first, then choose
Each question is a real-world Pythagoras problem. Sketch first, identify the hypotenuse, decide whether you need ADD or SUBTRACT, and only then start computing. Show your working. 3 marks each
1.1 Find the distance from $A(1, 1)$ to $B(7, 9)$. Give your answer in surd (exact) form AND to 2 d.p.
1.2 A square has side 8 cm. Find the length of its diagonal exactly (surd form) and to 2 d.p.
1.3 A 5 m ladder leans on a wall. The base is 1.4 m from the wall. How high up the wall does the ladder reach, to 2 d.p.?
1.4 Two friends start at the same point. Tash walks 11 km due east; Sam walks 60 km due north. How far apart are they? (Hint: 11-60-61 is a Pythagorean triple.)
1.5 A rectangular soccer pitch is 90 m long and 45 m wide. Find the length of the diagonal to 2 d.p. Then state how much further it is than running the long side (90 m).
1.6 A drone is flying at altitude 30 m above ground level. Looking down, it spots a target 40 m horizontally away. What is the straight-line distance from the drone to the target? (Use a known triple.)
2. Find the mistake
Another student has tried to find the distance from $A(-2, 3)$ to $B(4, -5)$. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — distance $AB$ from $(-2, 3)$ to $(4, -5)$:
Line 1: $\Delta x = 4 - (-2) = 6$
Line 2: $\Delta y = -5 - 3 = -8$
Line 3: $d^2 = (\Delta x)^2 + (\Delta y)^2 = 6^2 + (-8) = 36 - 8 = 28$
Line 4: $d = \sqrt{28} \approx 5.29$ units.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? Look at Line 3 — they squared $\Delta x$ correctly, but did they square $\Delta y$? Remember $(-8)^2 = +64$, not $-8$.3. Open-ended challenge — design a journey
This question has more than one valid answer. 4 marks
3.1 Design two different journeys, each made up of TWO perpendicular straight-line legs (e.g. east then north, or north then west), such that:
• each leg is a whole-number distance in km between 1 and 30,
• the straight-line distance between start and end is also a whole-number of km, and
• the two journeys are NOT scaled versions of each other.
For each journey:
(i) State the two leg distances and directions.
(ii) Compute and check the straight-line distance using Pythagoras.
(iii) Briefly say which Pythagorean triple your journey matches.
Hint: Look at the four memorise triples (3-4-5, 5-12-13, 8-15-17, 7-24-25) and their multiples — anything with all sides under 30 km will work.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Distance $A(1,1)$ to $B(7,9)$
$\Delta x = 7 - 1 = 6$, $\Delta y = 9 - 1 = 8$. $d^2 = 6^2 + 8^2 = 36 + 64 = 100$, so $d = \sqrt{100} = \mathbf{10}$ units exactly.
(Surd form: $\sqrt{100}$. To 2 d.p.: $10.00$.) ($2\times$(3-4-5).)
1.2 — Diagonal of 8 cm square
$d^2 = 8^2 + 8^2 = 64 + 64 = 128$. Exact: $d = \sqrt{128} = 8\sqrt{2}$ cm. To 2 d.p.: $d \approx \mathbf{11.31}$ cm.
1.3 — Ladder 5 m, base 1.4 m
The ladder is the hypotenuse, so we SUBTRACT: $a^2 = 5^2 - 1.4^2 = 25 - 1.96 = 23.04$, so $a = \sqrt{23.04} = \mathbf{4.80}$ m (exact, since $23.04 = 4.8^2$).
1.4 — Tash and Sam
$d^2 = 11^2 + 60^2 = 121 + 3600 = 3721 = 61^2$, so $d = \mathbf{61}$ km apart. (11-60-61 is a primitive Pythagorean triple — less famous, but still clean.)
1.5 — Soccer pitch diagonal
$d^2 = 90^2 + 45^2 = 8100 + 2025 = 10{,}125$, so $d = \sqrt{10{,}125} \approx \mathbf{100.62}$ m.
Long side is 90 m, so the diagonal is $100.62 - 90 = \mathbf{10.62}$ m further.
1.6 — Drone line-of-sight
Vertical leg 30, horizontal leg 40. $d^2 = 30^2 + 40^2 = 900 + 1600 = 2500$, so $d = \sqrt{2500} = \mathbf{50}$ m. ($10\times$(3-4-5).)
2 — Find the mistake
(a) The mistake is on $\mathbf{Line\ 3}$.
(b) The student forgot to SQUARE $\Delta y$. They wrote "$+ (-8)$" instead of "$+ (-8)^2$". The Pythagorean formula squares each difference, so $(-8)^2 = +64$, NOT $-8$. (Squaring removes the negative sign — sign doesn't matter for distance.)
(c) Corrected working:
$\Delta x = 6$, $\Delta y = -8$
$d^2 = (\Delta x)^2 + (\Delta y)^2 = 6^2 + (-8)^2 = 36 + 64 = 100$
$d = \sqrt{100} = \mathbf{10}$ units.
(This is the $2\times$(3-4-5) triple again.)
3 — Open-ended challenge (sample solution)
Journey 1: 5 km east then 12 km north.
Check: $d^2 = 5^2 + 12^2 = 25 + 144 = 169 = 13^2$, so $d = \mathbf{13}$ km ✓.
Triple: 5-12-13.
Journey 2: 8 km north then 15 km east.
Check: $d^2 = 8^2 + 15^2 = 64 + 225 = 289 = 17^2$, so $d = \mathbf{17}$ km ✓.
Triple: 8-15-17.
These two journeys are NOT scaled versions of each other — different triples.
Other valid pairs include: 3 km E + 4 km N (hyp 5); 6 km E + 8 km N (hyp 10) — but these two ARE scaled versions of each other, so they wouldn't count together. The 7-24-25 triple is also a clean choice (7 km E + 24 km N, hyp 25).
Marking: 2 marks per valid journey (1 for whole-number leg distances 1-30 km, 1 for verification working giving whole-number hypotenuse). 4 in total. If both journeys are scaled versions of the same triple, deduct 1 mark.