---
marp: true
theme: default
html: true
---

# Finding a side

Area, volume, circles and Pythagoras · Pythagoras

<!-- Speaker note: Students use Pythagoras to find the hypotenuse or a shorter side (subtracting the squares), with non-whole answers to two decimal places and a sense check, in contexts such as ladders, screens and ramps (AC9M8M06, node 2). About 45 minutes (Predict and Run 8 min · Investigate 15 min · Modify 12 min · Practise and Make 10 min). -->

---

## Today

**We are learning to** pythagoras finds any side. For the hypotenuse, add the squares: c² = a² + b². For a shorter side, subtract from the hypotenuse's square: b² = c² − a². Then take the square root, and check the hypotenuse is still the longest.

- 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.

<!-- Speaker note: 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. -->

---

## Why it matters

> A 5 m ladder leans against a wall with its foot 3 m out. How high up the wall does it reach? Screens, ramps, roofs and ladders all hide a right-angled triangle.

<!-- Speaker note: Pythagoras' theorem links all three sides, so it can find any one of them: - **The hypotenuse:** add the squares. $c^2 = a^2 + b^2$. - **A shorter side:** the hypotenuse's square minus the other side's square. $b^2 = c^2 - a^2$. Then take the square root. The answer often lies between two whole numbers: give it to two decimal places, and check that the hypotenuse is still the longest side. -->

---

<span style="background:#dbeafe; color:#1a1a1a; border:1.5px solid #3b82f6; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">💬 Discussion</span>

## Predict

**A 5 m ladder leans against a wall. Its foot is 3 m from the wall. How high up the wall does it reach?**

- 8 m
- 4 m
- About 5.83 m
- 2 m

<!-- Speaker note: What happens: 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. Take two or three predictions before anyone runs it. -->

<!-- Speaker note: Teacher: students who chose 5.83 added; ask whether the wall can be taller than the ladder. -->

---

<span style="background:#fef3c7; color:#1a1a1a; border:1.5px solid #f59e0b; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✍️ Activity</span>

## Try it

{{embed: https://studio.learningwaypoints.com/tutorials/_playground-board/index.html?id=deck%2Fmaths%2Fmeasurement%2Ffinding-a-side%2Fstep-2&code=widget%3A+pythag%0Aa%3A+3%0Aslider+b%3A+1%2C+8%2C+1%0Aunits%3A+m%0A&fresh=1}}

[Open this step in the tutorial](https://studio.learningwaypoints.com/tutorials/maths/measurement/finding-a-side/?full=step-2#step-2)

<!-- Speaker note: Watch the biggest square as b changes: you need it to be 25. -->

<!-- Speaker note: Teacher: ask what the big square has to be. -->

<style scoped>pre, marp-pre { font-size: 0.5em; line-height: 1.25; } section { font-size: 25px; }</style>

---

<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## What happened?

* 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.

<!-- Speaker note: Reveal after running. Ask who predicted it, and what made them think so. -->

---

<span style="background:#dbeafe; color:#1a1a1a; border:1.5px solid #3b82f6; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">💬 Discussion</span>

## Add for c, subtract for a side

Here are both kinds.

Now try one: hypotenuse 13 cm, one side 5 cm.

$$c^2 = 6^2 + 8^2 = 100,\ c = 10$$

1) $b^2 = 10^2 - 6^2 = 100 - 36 = 64,\ b = 8$ — Missing a shorter side: subtract from the hypotenuse's square.
2) $b^2 = 7^2 - 4^2 = 33,\ b = \sqrt{33} \approx 5.74$ — Not a perfect square: between 5 and 6 (25 and 36), about 5.74.
3) $5.74 < 7$ — Check: the hypotenuse must still be the longest.

<!-- Speaker note: Model this live on the board before revealing anything: write each line and say why, pausing to ask the class what comes next. Then click to reveal the lines and compare. Script. Start: write c^2 = 6^2 + 8^2 = 100,c = 10 and say "Missing the hypotenuse: add the squares." Line 1: write b^2 = 10^2 - 6^2 = 100 - 36 = 64,b = 8 and say "Missing a shorter side: subtract from the hypotenuse's square." Line 2: write b^2 = 7^2 - 4^2 = 33,b = 33 5.74 and say "Not a perfect square: between 5 and 6 (25 and 36), about 5.74." Line 3: write 5.74 < 7 and say "Check: the hypotenuse must still be the longest." -->

<!-- Speaker note: Teacher: for the screen, share sketches; label the hypotenuse first. -->

<style scoped>pre, marp-pre { font-size: 0.5em; line-height: 1.25; } section { font-size: 26px; }</style>

---

<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## 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?**

* Answer: 12

<!-- Speaker note: 13² − 5² = 169 − 25 = 144, √144 = 12 cm. -->

---

<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Find the mistake

**Ken finds a shorter side. Click the line where his working goes wrong.**

```
 1  A right-angled triangle has hypotenuse 10 cm and one shorter side 6 cm. Find the other side.
 2  b² = 10² + 6²
 3  = 100 + 36 = 136
 4  b = √136 ≈ 11.66 cm
```

* Answer: line 2

<!-- Speaker note: 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. -->

---

<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## A screen

**A rectangular screen is 48 cm wide and 36 cm high. How long is its diagonal? Write your steps.**

* Answer: 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)

<!-- Speaker note: 48² + 36² = 3600, √3600 = 60 cm. -->

---

<span style="background:#fef3c7; color:#1a1a1a; border:1.5px solid #f59e0b; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✍️ Activity</span>

## One change at a time



- **On the board:** Each task changes one thing.
- **A ramp:** 
- **Not a whole number:** 
- **A shortcut:** 
- **Too long:** 

[Open this step](https://studio.learningwaypoints.com/tutorials/maths/measurement/finding-a-side/?full=step-4#step-4)

<!-- Speaker note: What this step is for: 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. Hints, in order: The big square must be 100. Take away 36. For students who need more support: Use a decision card: 'Know c? Subtract. Want c? Add.' and keep squares to 20² visible. For early finishers: A 2 m by 1.5 m door: can a 2.6 m wide board be carried through it, tilted along the diagonal? 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: A student adds the squares for a shorter side (Ken's mistake). Ask: Is your answer shorter than the hypotenuse? Fix: Subtract: hypotenuse squared minus the known side squared. / A student subtracts the lengths, not the squares. Ask: What does the theorem say about the squares? Fix: Square both, subtract, then root. / A student cannot find the triangle in a word problem. Ask: Where is the right angle? Which length is opposite it? Fix: Sketch it, mark the right angle, label c first. -->

<!-- Speaker note: Teacher: for the shortcut, ask what is being compared. -->

---

<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## A shorter side

**Hypotenuse 25 cm, one side 7 cm. The other side, in cm?**

* Answer: 24

<!-- Speaker note: 625 − 49 = 576, √576 = 24 cm. -->

---

<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## A diagonal

**A square has sides of 5 cm. How long is its diagonal, in cm, to two decimal places?**

* Answer: 7.07

<!-- Speaker note: 25 + 25 = 50, √50 ≈ 7.07 cm. -->

---

<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## A ladder

**A 10 m ladder reaches 8 m up a wall. How far is its foot from the wall, in m?**

* Answer: 6

<!-- Speaker note: 100 − 64 = 36, √36 = 6 m. -->

---

<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Review: the hypotenuse

**Review: the sides at the right angle are 9 cm and 12 cm. Hypotenuse, in cm?**

* Answer: 15

<!-- Speaker note: 81 + 144 = 225, √225 = 15 cm. -->

---

<span style="background:#fef3c7; color:#1a1a1a; border:1.5px solid #f59e0b; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✍️ Activity</span>

## Make: design a ramp

Then: real ramps should rise no more than 1 m for every 14 m along. Is this ramp too steep?

<!-- Speaker note: What this step is for: Designing a ramp from its rise and sloping length is the subtract case in a real setting, with decimals and a sense check. Hints, in order: The sloping length is the hypotenuse. / 1.3² − 0.5² = 1.69 − 0.25. -->

<!-- Speaker note: Teacher: compare 0.5 m in 1.2 m with the 1 in 14 rule. -->

---

<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## 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.

<!-- Speaker note: Students answer on paper or a mini whiteboard before the reveal. Anyone who cannot answer needs the main idea again next lesson. Where this 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. -->