Mathematics • Year 9 • Unit 3 • Lesson 1
Pythagoras Review — Mixed Challenge
Pull together every idea from Lesson 1: identifying the hypotenuse, applying $a^2 + b^2 = c^2$, recognising common triples (3-4-5, 5-12-13, 8-15-17, 7-24-25), and noticing multiples. You'll have to choose the right approach, spot a mistake in someone else's working, and tackle an open-ended challenge.
1. Mixed problems — choose the right approach
Each question uses a different idea from Lesson 1. Decide what's being asked before you start writing. Show your working. 3 marks each
1.1 A right-angled triangle has the right angle at vertex $R$. The three sides are $PR = 24$ cm, $QR = 7$ cm, $PQ = 25$ cm. Name the hypotenuse and verify Pythagoras' theorem with the given lengths.
1.2 Test each set of three numbers and state whether it is a Pythagorean triple. Show the squares. (a) 6, 8, 10 (b) 5, 6, 8 (c) 20, 21, 29.
1.3 Without a calculator, decide which set is a Pythagorean triple, justifying with one short sentence: 9, 12, 15 or 9, 13, 15.
1.4 Find the missing side using a Pythagorean triple — no calculator. The two known sides are 5 and 13, and the missing side is the shorter leg. State which side is the hypotenuse and write down the missing leg.
1.5 Show that 6-8-10, 9-12-15 and 12-16-20 are ALL Pythagorean triples by identifying which standard triple each one is a multiple of, and stating the multiplying factor.
1.6 A triangle has sides $a$, $b$ and $c$ with $c$ the longest. State (i) what you would calculate to test whether it is right-angled, and (ii) exactly what result would prove it IS right-angled. Then apply your test to $a = 9$, $b = 40$, $c = 41$ and state the conclusion.
2. Find the mistake
Another student has tried to test whether 5, 12, 13 is a Pythagorean triple. 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 — test 5, 12, 13:
Line 1: Hypotenuse = 13 (the largest of the three numbers).
Line 2: $a^2 + b^2 = 5^2 + 12^2 = 5 + 12 = 17$
Line 3: $c^2 = 13^2 = 169$
Line 4: $17 \ne 169$, so 5, 12, 13 is NOT a Pythagorean triple.
(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 conclusion.
Stuck? Compare Line 2 to the lesson rule for $a^2$ — does $5^2$ mean $5 + 5$ or $5 \times 5$?3. Open-ended challenge — build your own triple
This question has many valid answers — there are infinitely many Pythagorean triples to find. 4 marks
3.1 Find two different Pythagorean triples that BOTH have a hypotenuse less than 30, AND that are NOT just multiples of 3-4-5.
For each triple you find:
(i) Write the three numbers.
(ii) Show the check $a^2 + b^2 = c^2$.
(iii) Say whether your triple is one of the “memorise four” from the lesson, or a multiple of one.
Bonus hint: The lesson lists 5-12-13, 8-15-17 and 7-24-25. Multiples of these (e.g. doubling 5-12-13) also count — but try to find at least one that isn't just a multiple.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Triangle $PQR$
The hypotenuse is $\mathbf{PQ}$ (it sits opposite the right angle at $R$, and is also the longest at 25 cm).
Check: $7^2 + 24^2 = 49 + 576 = 625 = 25^2$ ✓ — Pythagoras' theorem is verified.
1.2 — Test each set
(a) $6^2 + 8^2 = 36 + 64 = 100 = 10^2$ ✓ — $\mathbf{yes}$, a triple (it's $2 \times$(3, 4, 5)).
(b) $5^2 + 6^2 = 25 + 36 = 61$ but $8^2 = 64$. $61 \ne 64$, so $\mathbf{no}$, not a triple.
(c) $20^2 + 21^2 = 400 + 441 = 841 = 29^2$ ✓ — $\mathbf{yes}$, a triple. (This one is not a multiple of any of the memorise-four — Pythagorean triples are a much bigger family than the four common ones.)
1.3 — Without a calculator
$\mathbf{9, 12, 15}$ is the triple. It's just $3 \times$(3, 4, 5), so by the multiples rule it must satisfy Pythagoras' theorem.
9, 13, 15 is not — the middle number doesn't match the pattern, and we can sanity-check that $9^2 + 13^2 = 81 + 169 = 250$ but $15^2 = 225$.
1.4 — Missing shorter leg (5, ?, 13)
Hypotenuse $= \mathbf{13}$ (it's the longest, and we're told the missing side is a leg). This is the 5-12-13 triple, so the missing leg $= \mathbf{12}$.
(Check: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ ✓.)
1.5 — Multiples of standard triples
6-8-10 $= \mathbf{2}\times$(3, 4, 5).
9-12-15 $= \mathbf{3}\times$(3, 4, 5).
12-16-20 $= \mathbf{4}\times$(3, 4, 5).
By the multiples rule, any whole-number multiple of a Pythagorean triple is also a Pythagorean triple.
1.6 — General test, applied to 9-40-41
(i) Calculate $\mathbf{a^2 + b^2}$ and compare with $\mathbf{c^2}$ (the square of the longest side).
(ii) If $a^2 + b^2 = c^2$ (equal), then by the converse of Pythagoras' theorem the triangle IS right-angled.
Apply to 9-40-41: $9^2 + 40^2 = 81 + 1600 = 1681$ and $41^2 = 1681$ ✓. So the triangle $\mathbf{is\ right\text{-}angled}$ and 9, 40, 41 is a Pythagorean triple.
2 — Find the mistake
(a) The mistake is on $\mathbf{Line\ 2}$.
(b) The student wrote $5^2 = 5$ and $12^2 = 12$ — but $5^2 = 5 \times 5 = 25$ and $12^2 = 12 \times 12 = 144$. They have added the numbers instead of squaring them first. This is exactly the trap in the lesson's Common Pitfalls card.
(c) Corrected working:
$a^2 + b^2 = 5^2 + 12^2 = 25 + 144 = 169$.
$c^2 = 13^2 = 169$.
$169 = 169$, so $a^2 + b^2 = c^2$. By the converse of Pythagoras' theorem, $\mathbf{5, 12, 13\ IS\ a\ Pythagorean\ triple}$.
3 — Open-ended challenge (sample solution)
The simplest two from the "memorise four" with hypotenuse less than 30 are 5-12-13 and 8-15-17. Both are not multiples of 3-4-5.
Triple 1: $5, 12, 13$.
Check: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ ✓.
This is one of the memorise-four.
Triple 2: $8, 15, 17$.
Check: $8^2 + 15^2 = 64 + 225 = 289 = 17^2$ ✓.
This is one of the memorise-four.
Other valid choices: $7, 24, 25$ (memorise-four); $20, 21, 29$ (not in lesson, but $400 + 441 = 841 = 29^2$); $10, 24, 26 = 2\times$(5-12-13). Doubled 5-12-13 also counts because its hypotenuse 26 is still under 30 and it isn't a multiple of 3-4-5.
Marking: 2 marks per valid triple (1 for the numbers themselves with hypotenuse $< 30$ and not a 3-4-5 multiple, 1 for clear $a^2 + b^2 = c^2$ working). 4 in total.