Mathematics • Year 9 • Unit 3 • Lesson 3
Shorter Sides — Mixed Challenge
Pull together every idea from Lessons 1-3: choosing between $c^2 = a^2 + b^2$ (missing hypotenuse) and $a^2 = c^2 - b^2$ (missing leg), identifying which side is the hypotenuse, spotting triples to skip the calculator, and sanity-checking your answers. Spot a mistake in someone else's working and tackle an open-ended challenge.
1. Mixed problems — add or subtract?
Each question is missing a different side. Decide first whether to ADD ($c^2 = a^2 + b^2$, missing hypotenuse) or SUBTRACT ($a^2 = c^2 - b^2$, missing leg) before you start writing. Show your working. 3 marks each
1.1 A right triangle has legs $a = 9$ and $b = 12$. Find the hypotenuse — no calculator. State which operation (add or subtract) you used.
1.2 A right triangle has hypotenuse 17 and one leg 15. Find the other leg — no calculator. State which operation you used.
1.3 A right triangle has hypotenuse 25 cm and one leg 24 cm. Find the other leg.
1.4 A right triangle has legs 11 cm and 14 cm. Find the hypotenuse to 2 d.p.
1.5 A right triangle has hypotenuse 26 cm and one leg 10 cm. Find the other leg. (Recognise this as a multiple of a known triple.)
1.6 A right triangle has hypotenuse 10 cm and one leg 8 cm. (i) Find the other leg. (ii) State the perimeter of the triangle. (iii) State the area of the triangle.
2. Find the mistake
Another student has tried to find the missing leg of a right triangle with hypotenuse 13 cm and known leg 12 cm. 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 $a$ when $c = 13$, $b = 12$:
Line 1: Hypotenuse $c = 13$, leg $b = 12$, missing leg $a$.
Line 2: $a^2 = c^2 + b^2 = 13^2 + 12^2$
Line 3: $a^2 = 169 + 144 = 313$
Line 4: So $a = \sqrt{313} \approx 17.69$ cm.
(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? Sanity check: the missing leg MUST be shorter than the hypotenuse 13. But the student got 17.69 — longer than the hypotenuse, which is impossible. So somewhere they used the wrong operation.3. Open-ended challenge — design a "right triangle puzzle"
This question has more than one valid answer. 4 marks
3.1 Design two different right-angled triangles, each with whole-number sides, such that:
• the hypotenuse is between 20 and 30 (inclusive), and
• the two legs are also whole numbers, and
• the two triangles are NOT scaled versions of each other.
For each triangle you design:
(i) State all three sides.
(ii) Verify Pythagoras' theorem with the sides.
(iii) Briefly say whether your triangle is a known triple or a multiple of one.
Hint: The "memorise four" triples are 3-4-5, 5-12-13, 8-15-17 and 7-24-25. Scale them and check which scaled versions have hypotenuse between 20 and 30.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Legs 9 and 12, find hyp
Operation: $\mathbf{ADD}$ (missing hypotenuse). $c^2 = 9^2 + 12^2 = 81 + 144 = 225$, so $c = \mathbf{15}$. (This is $3\times$(3, 4, 5).)
1.2 — Hyp 17, leg 15, find other leg
Operation: $\mathbf{SUBTRACT}$ (missing leg). $a^2 = 17^2 - 15^2 = 289 - 225 = 64$, so $a = \mathbf{8}$. (8-15-17 triple.)
1.3 — Hyp 25, leg 24
$a^2 = 25^2 - 24^2 = 625 - 576 = 49$, so $a = \sqrt{49} = \mathbf{7}$ cm. (7-24-25 triple.)
1.4 — Legs 11 and 14, find hyp
$c^2 = 11^2 + 14^2 = 121 + 196 = 317$, so $c = \sqrt{317} \approx \mathbf{17.80}$ cm.
1.5 — Hyp 26, leg 10
$a^2 = 26^2 - 10^2 = 676 - 100 = 576$, so $a = \sqrt{576} = \mathbf{24}$ cm. (10-24-26 = $2\times$(5-12-13).)
1.6 — Hyp 10, leg 8 — full triangle info
(i) $a^2 = 10^2 - 8^2 = 100 - 64 = 36$, so the other leg $= \mathbf{6}$ cm. (This is $2\times$(3-4-5).)
(ii) Perimeter $= 6 + 8 + 10 = \mathbf{24}$ cm.
(iii) Area $= \tfrac{1}{2} \times 6 \times 8 = \mathbf{24}$ cm$^2$.
(Curiously the perimeter and area are the same number for this triangle — coincidence with this particular scale.)
2 — Find the mistake
(a) The mistake is on $\mathbf{Line\ 2}$.
(b) The student used $+$ instead of $-$. When the missing side is a LEG, the formula is $a^2 = c^2 - b^2$ (SUBTRACT), not $a^2 = c^2 + b^2$. This is exactly the "Adding when you should subtract" trap flagged in the lesson's Common Pitfalls card. The wrong sign gave a missing leg of 17.69 — longer than the hypotenuse 13, which is impossible (a leg must be shorter than the hypotenuse).
(c) Corrected working:
$a^2 = c^2 - b^2 = 13^2 - 12^2 = 169 - 144 = 25$
$a = \sqrt{25} = \mathbf{5}$ cm.
(This is the 5-12-13 triple — the missing leg is 5.)
3 — Open-ended challenge (sample solution)
I'll use 2$\times$(5-12-13) and the standard 7-24-25.
Triangle 1: sides $10, 24, 26$.
Verify: $10^2 + 24^2 = 100 + 576 = 676 = 26^2$ ✓.
This is $2\times$(5-12-13).
Triangle 2: sides $7, 24, 25$.
Verify: $7^2 + 24^2 = 49 + 576 = 625 = 25^2$ ✓.
This is the 7-24-25 memorise-four triple.
These two triangles are NOT scaled versions of each other (their legs and ratios are different).
Other valid choices include: $9, 12, 15$ (3$\times$(3,4,5), hyp 15 — outside range); $20, 21, 29$ (hyp 29, primitive triple); $16, 30, 34 = 2\times$(8,15,17), hyp 34 — outside range. So $\{20, 21, 29\}$ paired with either $\{10, 24, 26\}$ or $\{7, 24, 25\}$ also works.
Marking: 2 marks per valid triangle (1 for whole-number sides with hypotenuse in 20-30 range, 1 for verification working). 4 in total. If both triangles are scaled versions of the same primitive triple, deduct 1 mark.