Mathematics • Year 9 • Unit 3 • Lesson 5
Side Labelling — Mixed Challenge
Pull together every idea from Lessons 1-5: identifying the hypotenuse from the right-angle marker, finding sides with $c^2 = a^2 + b^2$ or $a^2 = c^2 - b^2$, recognising Pythagorean triples, AND now labelling sides as hyp/opp/adj relative to a reference angle $\theta$. Spot a labelling mistake and tackle an open-ended challenge.
1. Mixed problems — labelling, Pythagoras, swaps
Each question uses a different combination of ideas from Lessons 1-5. Show your working. 3 marks each
1.1 A right triangle has the right angle at vertex $Q$. $PQ = 7$, $QR = 24$, $PR = 25$. (i) Identify the hypotenuse and state how you know. (ii) With $\theta$ at $P$, identify opp and adj (name and length).
1.2 Same triangle as 1.1 ($PQR$, right angle at $Q$, $PQ = 7$, $QR = 24$, $PR = 25$). With $\theta$ at $R$ instead, identify opp and adj (name and length). State which two sides have swapped names compared with 1.1.
1.3 A right triangle has legs 5 cm (adjacent to $\theta$) and 12 cm (opposite $\theta$). Find the hypotenuse (no calculator).
1.4 A right triangle has hypotenuse 17 cm and opposite $\theta$ of length 15 cm. Find the adjacent (no calculator). State which Pythagorean triple this matches.
1.5 True or false (give a one-sentence justification): "In any right triangle, the hypotenuse is the same side regardless of where $\theta$ is."
1.6 A right triangle has hypotenuse 13 cm and adjacent (relative to $\theta$) of length 5 cm. (i) Find the opposite (no calculator). (ii) Now $\theta$ moves to the OTHER acute angle — state the new opp and new adj.
2. Find the mistake
Another student has tried to label the sides of right triangle $ABC$ ($AB = 3$, $BC = 4$, $AC = 5$, right angle at $B$) with $\theta$ at vertex $C$. Their answer is shown below. Exactly one label is wrong. Spot it, explain why it's wrong, then write out the correct labelling. 3 marks
Student's answer — $\theta$ at $C$:
Hypotenuse = $BC = 4$ (the side touching the right angle).
Opposite ($\theta$) = $AB = 3$ (the side across from $C$, not touching $C$).
Adjacent ($\theta$) = $AC = 5$ (the side next to $\theta$ at $C$ that isn't the hypotenuse).
(a) Which label is wrong?
(b) Explain in one or two sentences why that label is wrong.
(c) Write out the correct labelling (all three sides) with brief reasoning.
Stuck? Hypotenuse rule: it is opposite the right angle (at $B$), so the hyp must be the side NOT touching $B$. Check the student's hyp against that rule.3. Open-ended challenge — design a labelling puzzle
This question has more than one valid answer. 4 marks
3.1 Design a right-angled triangle (with whole-number side lengths) and TWO different positions for $\theta$ (one at each acute vertex), such that:
• the triangle has all three sides as whole numbers (a Pythagorean triple — not necessarily one of the four memorise triples), and
• for one position of $\theta$, the side OPPOSITE is the longer leg; and
• for the other position of $\theta$, the side OPPOSITE is the shorter leg.
For each position of $\theta$:
(i) State which vertex $\theta$ is at.
(ii) Identify hyp, opp and adj (name and length).
(iii) State which two sides have swapped compared with the other position.
Bonus: Verify Pythagoras' theorem holds for your triangle.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Triangle $PQR$, right angle at $Q$, $\theta$ at $P$
(i) Hypotenuse = $\mathbf{PR = 25}$. How I know: it's the side opposite the right angle at $Q$, and it's the longest side (verifying $7^2 + 24^2 = 49 + 576 = 625 = 25^2$ ✓).
(ii) opp = $QR = \mathbf{24}$ (across from $P$, not touching $P$); adj = $PQ = \mathbf{7}$ (next to $\theta$ at $P$, but not the hypotenuse).
1.2 — Same triangle, $\theta$ at $R$
opp = $PQ = \mathbf{7}$ (across from $R$, not touching $R$).
adj = $QR = \mathbf{24}$ (next to $\theta$ at $R$, but not the hypotenuse).
Compared with 1.1: opp and adj $\mathbf{swapped}$ (was 24 and 7; now 7 and 24). Hyp = $PR = 25$ unchanged.
1.3 — Legs adj 5, opp 12, find hyp
$c^2 = 5^2 + 12^2 = 25 + 144 = 169$, so $c = \sqrt{169} = \mathbf{13}$ cm. (5-12-13 triple.)
1.4 — Hyp 17, opp 15, find adj
$\text{adj}^2 = 17^2 - 15^2 = 289 - 225 = 64$, so adj $= \sqrt{64} = \mathbf{8}$ cm. This is the $\mathbf{8\text{-}15\text{-}17}$ Pythagorean triple.
1.5 — Hypotenuse fixed under $\theta$-move
$\mathbf{True}$. The hypotenuse is always the side opposite the right angle. Moving $\theta$ between the two acute angles never changes which side is opposite the right angle, so the hyp is fixed.
1.6 — Hyp 13, adj 5, find opp and then swap
(i) $\text{opp}^2 = 13^2 - 5^2 = 169 - 25 = 144$, so opp $= \sqrt{144} = \mathbf{12}$ cm. (5-12-13 triple.)
(ii) After $\theta$ swaps to the other acute angle: new opp = $\mathbf{5}$ cm (was adj); new adj = $\mathbf{12}$ cm (was opp). Hyp $= 13$ cm unchanged.
2 — Find the mistake
(a) The $\mathbf{hypotenuse\ label}$ is wrong (and consequently the adjacent label is too — but the original mistake was on the hypotenuse).
(b) The student wrote hyp = $BC = 4$, but $BC$ touches the right angle at $B$ — so $BC$ is a LEG, not the hypotenuse. The hypotenuse is the side OPPOSITE the right angle: in this triangle that's $AC = 5$.
(c) Correct labelling, $\theta$ at $C$:
hyp = $\mathbf{AC = 5}$ (opposite the right angle at $B$).
opp = $\mathbf{AB = 3}$ (across from $C$ — not touching $C$).
adj = $\mathbf{BC = 4}$ (next to $\theta$ at $C$, but not the hypotenuse).
3 — Open-ended challenge (sample solution)
Use the 3-4-5 triangle: $ABC$, right angle at $B$, $AB = 3$, $BC = 4$, $AC = 5$. Verify Pythagoras: $3^2 + 4^2 = 9 + 16 = 25 = 5^2$ ✓.
Case 1 — $\theta$ at vertex $A$:
hyp = $AC = 5$; opp = $BC = 4$ (the longer leg); adj = $AB = 3$ (the shorter leg).
Case 2 — $\theta$ at vertex $C$:
hyp = $AC = 5$ (unchanged); opp = $AB = 3$ (the shorter leg); adj = $BC = 4$ (the longer leg).
What swapped: opp and adj have exchanged — in Case 1 opp = 4 (longer leg); in Case 2 opp = 3 (shorter leg). Hyp is fixed at 5.
This satisfies all the conditions: whole-number sides (a Pythagorean triple), $\theta$ at each acute vertex in turn, and the OPPOSITE flips from the longer leg to the shorter leg between the two cases.
Other valid choices: any Pythagorean triple works — 5-12-13, 8-15-17, 7-24-25, or any multiple of these (e.g. 6-8-10).
Marking: 1 mark for choosing a valid Pythagorean triple and verifying it; 1 for correct labelling of Case 1; 1 for correct labelling of Case 2; 1 for explicitly identifying that opp and adj have swapped (hyp unchanged). 4 in total.