Finding a side

Classroom resources

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Students will be able to

• I can find a shorter side of a right-angled triangle.
• I can decide whether to add or subtract the squares.
• I can solve problems with ladders, screens and ramps, and check my answer makes sense.

Timing

Predict and Run 8 min · Investigate 15 min · Modify 12 min · Practise and Make 10 min

Before class

Do the Run board yourself first. Students need the hypotenuse method (previous lesson) and square roots between whole numbers (Between the squares).

The one idea: add the squares to find the hypotenuse, subtract them to find a shorter side. The common mistake is adding for a shorter side (Ken's mistake); the check "the hypotenuse is the longest" catches it.

Running the lesson

1. Predict: students who chose 5.83 added; ask whether the wall can be taller than the ladder.
2. Run: ask what the big square has to be.
3. Investigate: for the screen, share sketches; label the hypotenuse first.
4. Modify: for the shortcut, ask what is being compared.
5. Make: compare 0.5 m in 1.2 m with the 1 in 14 rule.

The hard part (on purpose)

Deciding whether to add or subtract. Ask students to point to the hypotenuse and say whether it is the side they know or the side they want, before any calculation.

When students get stuck

Answers to the checks

1. the ladder: A 5 m ladder leans against a wall. Its foot is 3 m from the wall. How high up the wall does it reach?
   4 m. The ladder is the hypotenuse: 5² − 3² = 25 − 9 = 16, so the height is 4 m. 5.83 m adds the squares, which would make the wall taller than the ladder.
2. the height that fits: The ladder is the hypotenuse, 5 m, and the foot is a = 3 m out. Change the height b until c = 5. Was your prediction right?
   
   b = 4: 9 + 16 = 25. The height's square is 25 − 9 = 16.
3. Add for c, subtract for a side: A right-angled triangle has hypotenuse 13 cm and one shorter side 5 cm. How long is the other side, in cm?
   12
   13² − 5² = 169 − 25 = 144, √144 = 12 cm.
4. Find the mistake: Ken finds a shorter side. Click the line where his working goes wrong.
   line 2
   Line 2: 10 cm is the hypotenuse, so subtract: b² = 100 − 36 = 64, b = 8 cm. His 11.66 cm is longer than the hypotenuse, which is impossible.
5. A screen: A rectangular screen is 48 cm wide and 36 cm high. How long is its diagonal? Write your steps.
   The diagonal is the hypotenuse of a right-angled triangle with sides 48 and 36 (the corners of a rectangle are right angles); $48^2 + 36^2 = 2304 + 1296 = 3600$ (for the hypotenuse, add the squares); $\sqrt{3600} = 60$ cm (take the square root)
   48² + 36² = 3600, √3600 = 60 cm.
6. On the board: a = 6. Change b until the hypotenuse is 10.
   
   b = 8: 100 − 36 = 64 = 8².
7. A ramp: A ramp is 2.6 m long (the sloping side) and goes 2.4 m across the ground. How high does it rise, in m?
   1
   2.6² − 2.4² = 6.76 − 5.76 = 1, √1 = 1 m.
8. Not a whole number: Hypotenuse 7 cm, one side 4 cm. The other side, in cm, to two decimal places?
   5.74
   49 − 16 = 33, √33 ≈ 5.74 cm.
9. A shortcut: Walking round a 30 m by 40 m park is 70 m. How many metres shorter is the path straight across, corner to corner?
   20
   Across: √(900 + 1600) = 50 m, so 70 − 50 = 20 m shorter.
10. Too long: Edge case: Jo finds a shorter side of 13 cm in a triangle whose hypotenuse is 12 cm. What does that tell her?
    She made a mistake: a shorter side can't be longer than the hypotenuse
    The hypotenuse is always the longest side, so she probably added the squares instead of subtracting.
11. A shorter side: Hypotenuse 25 cm, one side 7 cm. The other side, in cm?
    24
    625 − 49 = 576, √576 = 24 cm.
12. A diagonal: A square has sides of 5 cm. How long is its diagonal, in cm, to two decimal places?
    7.07
    25 + 25 = 50, √50 ≈ 7.07 cm.
13. A ladder: A 10 m ladder reaches 8 m up a wall. How far is its foot from the wall, in m?
    6
    100 − 64 = 36, √36 = 6 m.
14. Review: the hypotenuse: Review: the sides at the right angle are 9 cm and 12 cm. Hypotenuse, in cm?
    15
    81 + 144 = 225, √225 = 15 cm.
15. design a ramp: A wheelchair ramp must rise 0.5 m and its sloping surface is 1.3 m long. How far out from the step does it reach along the ground, in m? Write your steps.
    The sloping surface is the hypotenuse: $c = 1.3$, $a = 0.5$ (it is opposite the right angle at the step); $b^2 = 1.3^2 - 0.5^2 = 1.69 - 0.25 = 1.44$ (for a shorter side, subtract the squares); $b = \sqrt{1.44} = 1.2$ m (take the square root)
    1.69 − 0.25 = 1.44, √1.44 = 1.2 m.

Solutions

One possible solution for each task. Other correct programs are fine.

one change at a time › On the board

Open task: any program that does what the step asks.

one change at a time › A ramp

Open task: any program that does what the step asks.

one change at a time › Not a whole number

Open task: any program that does what the step asks.

one change at a time › A shortcut

Open task: any program that does what the step asks.

one change at a time › Too long

Open task: any program that does what the step asks.

Stretch and support

Stretch: A 2 m by 1.5 m door: can a 2.6 m wide board be carried through it, tilted along the diagonal?

Support: Use a decision card: 'Know c? Subtract. Want c? Add.' and keep squares to 20² visible.

Coach prompt (optional)

For a student who has done the lesson and is still stuck. They paste it into an AI tool your school allows; it asks questions about this lesson's one idea and doesn't give answers. Students see it on the course recap page only through a link ending in ?coach=1: recap with coach prompts.

You are a patient maths coach for a Year 8 student in Australia. Help me understand ONE idea only, from the lesson "Finding a side".

The idea: Finding unknown side lengths using Pythagoras: I can Students apply Pythagoras' theorem to find the hypotenuse or a shorter side in right-angled triangles, including in practical and coordinate geometry contexts.

What understanding looks like:
- Correctly identifies the hypotenuse in a given right-angled triangle.
- Finds an unknown side correctly using the theorem.
- Applies the theorem to a practical problem context.

Mistakes students often make here:
- A student adds the squares for a shorter side (Ken's mistake).
- A student subtracts the lengths, not the squares.
- A student cannot find the triangle in a word problem.

How to coach me (rules from the Waypoints Kit coach):
- Ask me one short question at a time (2 to 4 sentences) and wait for my answer. Never give me the answer before you have heard my thinking, and don't hide the answer inside your question.
- Start in Discover: ask me to explain the idea in my own words, or to predict what happens in a simple example. I can switch any time: "teach it back" (you play a beginner and I explain), "spot the flaw" (you show me an answer with one common mistake and I find it), or "check" (I explain, and you tell me what is solid and what to revisit).
- If I am stuck or say "just tell me", ask what I think first, then give a small hint or a simpler example. Explain only after I have had a real go.
- If I seem to be missing the idea that comes before this one (Understanding and stating Pythagoras' theorem), check it quickly, then come back.
- Stay in this idea: don't teach later ideas or add extra content. If I ask about something else, say it is for another lesson and bring me back.
- After you help, ask me one reflective question, such as how I know my answer is right or what I would check.
- Praise my effort and method, not my ability.
- You can be wrong: if I rely on something you said, remind me to check it against my notes or the lesson. If you don't know, say so.
- Use simple words.
- Stop when I can explain the idea in my own words and get one new example right. Then tell me to go back to the lesson "Finding a side".

Exit ticket

A right-angled triangle has hypotenuse 17 cm and one side 8 cm. Find the other side.

289 − 64 = 225, √225 = 15 cm.

What each step is for

• the ladder: Committing to a height for the 5 m ladder with its foot 3 m out shows whether they subtract the squares or add them as for a hypotenuse.
• the height that fits: Changing b until the hypotenuse is 5 shows that the missing square is the big square minus the known one.
• add or subtract?: For a shorter side subtract the squares; Ken adds, and gets a side longer than the hypotenuse. A real problem starts by spotting the right angle and the hypotenuse; students own the working and pick the key reason.
• one change at a time: One change at a time: a side on the board, a ramp with decimals, a side that is not a whole number, a shortcut saving, then the edge case of a shorter side longer than the hypotenuse.
• any side: A shorter side, a diagonal, a ladder and a review, mixed; adding for a shorter side is the planted slip.
• design a ramp: Designing a ramp from its rise and sloping length is the subtract case in a real setting, with decimals and a sense check.

Where it leads

Next: Right, acute or obtuse? (node 3): using the theorem backwards to test a triangle.

Where it leads: distance between two points, trigonometry (Year 10), and 3D diagonals of boxes.