Mathematics • Year 9 • Unit 3 • Lesson 2
Hypotenuse — Mixed Challenge
Pull together every idea from Lessons 1-2: identifying the hypotenuse, applying $c = \sqrt{a^2 + b^2}$, spotting common triples to skip calculator work, and recognising when rounding is and isn't needed. 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 combination of ideas from Lessons 1-2. Decide whether to spot a triple or to actually compute the square root before you start writing. Show your working. 3 marks each
1.1 Find the hypotenuse of a right triangle with legs 7 and 24 cm — without a calculator. Justify your method in one short sentence.
1.2 Find the hypotenuse of a right triangle with legs 11 cm and 13 cm. Give your answer to 2 d.p.
1.3 A right triangle has legs of length 0.6 m and 0.8 m. Find the hypotenuse exactly (no rounding needed) and state which Pythagorean triple this matches.
1.4 A square has side length 5 cm. Find the length of its diagonal in exact (surd) form AND to 2 d.p.
1.5 The legs of a right triangle are 12 cm and 35 cm. Find the hypotenuse exactly. (Hint: 12-35-37 is a less-common triple — but you can still verify with the calculator.)
1.6 A right triangle has legs of length 30 mm and 40 mm. Find the hypotenuse and give your answer (i) in mm and (ii) in cm. State which triple this matches.
2. Find the mistake
Another student has tried to find the hypotenuse of a right triangle with legs 1.8 m and 3.2 m. 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 — find $c$ when legs are 1.8 m and 3.2 m:
Line 1: $c^2 = 1.8^2 + 3.2^2$
Line 2: $c^2 = 3.24 + 10.24 = 13.48$
Line 3: So $c = 13.48$ m.
Line 4: The ladder is 13.48 m long.
(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 to 2 d.p.
Stuck? Compare Line 3 to the lesson's four-step method. Step 4 is "square root, round" — but the student skipped a key operation. Also: a 13.48 m ladder leaning on a 3.2 m wall is impossible — sanity check.3. Open-ended challenge — design your own right triangle
This question has more than one valid answer. 4 marks
3.1 Design two different right-angled triangles, each with integer (whole number) leg lengths, that have a hypotenuse of EXACTLY $\sqrt{50}$ cm. (Yes, the hypotenuse is allowed to be irrational — only the legs need to be whole numbers.)
For each triangle you design:
(i) State the two leg lengths.
(ii) Show the check $a^2 + b^2 = 50$.
(iii) Briefly say how you found these numbers.
Hint: Find pairs of whole numbers $a$ and $b$ such that $a^2 + b^2 = 50$. There are exactly two essentially different answers (ignoring the trivial $a=0$ case).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Legs 7 and 24, no calculator
This is the 7-24-25 triple from Lesson 1's "memorise four", so $c = \mathbf{25}$ cm. (Verify: $7^2 + 24^2 = 49 + 576 = 625 = 25^2$ ✓.)
1.2 — Legs 11 and 13
$c^2 = 11^2 + 13^2 = 121 + 169 = 290$, so $c = \sqrt{290} \approx \mathbf{17.03}$ cm. (Not a known triple, so calculator is the right tool.)
1.3 — Legs 0.6 m and 0.8 m
$c^2 = 0.6^2 + 0.8^2 = 0.36 + 0.64 = 1.00$, so $c = \sqrt{1} = \mathbf{1}$ m exactly. This matches $0.2 \times$(3, 4, 5), so it's a scaled 3-4-5 triple.
1.4 — Diagonal of a 5 cm square
$d^2 = 5^2 + 5^2 = 25 + 25 = 50$. Exact form: $d = \sqrt{50} = 5\sqrt{2}$ cm. To 2 d.p.: $d \approx \mathbf{7.07}$ cm.
(Any square has diagonal $= $ side $\times \sqrt{2}$.)
1.5 — Legs 12 and 35
$c^2 = 12^2 + 35^2 = 144 + 1225 = 1369$, and $\sqrt{1369} = 37$ exactly (since $37^2 = 1369$). So $c = \mathbf{37}$ cm. This is the 12-35-37 triple.
1.6 — Legs 30 mm and 40 mm
$c^2 = 30^2 + 40^2 = 900 + 1600 = 2500$, so $c = \sqrt{2500} = 50$ mm.
(i) $c = \mathbf{50}$ mm.
(ii) Converting to cm: $50$ mm $= \mathbf{5}$ cm.
This is $10 \times $(3, 4, 5), so it's a multiple of the 3-4-5 triple.
2 — Find the mistake
(a) The mistake is on $\mathbf{Line\ 3}$.
(b) The student forgot to take the square root. They had $c^2 = 13.48$, which means $c = \sqrt{13.48}$, NOT $c = 13.48$. Skipping the square root is exactly the trap flagged in the lesson's Common Pitfalls card. (Sanity check: 13.48 m is absurd for a ladder reaching only 3.2 m up the wall — the answer must be between 3.2 and 5 m.)
(c) Corrected working:
$c^2 = 1.8^2 + 3.2^2 = 3.24 + 10.24 = 13.48$
$c = \sqrt{13.48} \approx \mathbf{3.67}$ m.
The ladder is about 3.67 m long.
3 — Open-ended challenge (sample solution)
We need pairs of whole numbers $a$ and $b$ with $a^2 + b^2 = 50$. Listing squares: 1, 4, 9, 16, 25, 36, 49. Pairs that add to 50:
Triangle 1: $a = 1$, $b = 7$.
Check: $1^2 + 7^2 = 1 + 49 = 50$ ✓.
Method: I tried small squares ($1^2 = 1$) and checked which big square gives a total of 50. $50 - 1 = 49 = 7^2$. ✓
Triangle 2: $a = 5$, $b = 5$.
Check: $5^2 + 5^2 = 25 + 25 = 50$ ✓.
Method: I tried equal legs and found $50 / 2 = 25 = 5^2$. ✓
Those are the only two essentially different answers — Triangle 2 is the diagonal of a 5 cm square (same as question 1.4!), and Triangle 1 is a long thin right triangle.
Marking: 2 marks per valid triangle (1 for the integer leg lengths, 1 for clear working showing $a^2 + b^2 = 50$). 4 in total. Award full marks for the two distinct solutions above; if a student only finds one, award 2 marks.