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hscscience Maths Std · Y11
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Module 2 · L3 of 22 ~50 min ⚡ +95 XP available

Pythagoras' Theorem

Know two sides of a right-angled triangle, find the third. Every diagonal, slant height, and distance problem in this module comes back to this one relationship: $a^2 + b^2 = c^2$.

Today's hook, A 5 m ladder leans against a wall with its base 1.5 m from the wall. How high up the wall does it reach, without measuring? There's a formula that makes this trivial.
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Recall, your gut answer first

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
01
Recall, your gut answer first
+5 XP warm-up

A 5 m ladder leans against a wall. You know the ladder is 5 m long and the base is 1.5 m from the wall. Before learning any formula how might you estimate how high up the wall the ladder reaches? What information would you need?

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Come back to this at the end of the lesson.

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The formula you need to own
+5 XP to read

For any right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. If you know two sides, you can always find the third.

Finding the hypotenuse: add the squares, then square root: $c = \sqrt{a^2 + b^2}$.

Finding a shorter side: subtract the squares, then square root: $a = \sqrt{c^2 - b^2}$.

Write the rearranged form before substituting, never rely on memory alone.

In a right triangle the square of the hypotenuse equals the sum of the squares of the other two sides.
$c$ is always opposite the right angle, it is always the longest side
Finding the hypotenuse
ADD the squares of the two known sides, then take the square root. $c = \sqrt{a^2 + b^2}$
Finding a shorter side
SUBTRACT the known shorter side squared from the hypotenuse squared. $a = \sqrt{c^2 - b^2}$
Slant height of a cone
The slant height $\ell$ is the hypotenuse of a triangle formed by radius $r$ and vertical height $h$: $\ell^2 = r^2 + h^2$
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What you'll master
Know

Key facts

  • The relationship $a^2 + b^2 = c^2$ and what each pronumeral represents
  • The two types of Pythagoras problems: finding the hypotenuse vs finding a shorter side
  • How Pythagoras applies to practical and 3D contexts
Understand

Concepts

  • Why identifying the hypotenuse correctly is the critical first step
  • Why finding a shorter side requires rearranging the formula (subtract, not add)
  • How a 3D problem reduces to one or two 2D right-angled triangles
Can do

Skills

  • Correctly identify the hypotenuse in any right-angled triangle regardless of orientation
  • Find any unknown side in a right-angled triangle using Pythagoras' theorem
  • Solve practical problems including slant height of cones and room diagonals
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Key terms
HypotenuseThe longest side of a right-angled triangle, always opposite the right angle.
Right angleA 90° angle, marked with a small square in diagrams. Must be present for Pythagoras to apply.
SurdA square root that cannot be simplified to a whole number (e.g. $\sqrt{5}$, $\sqrt{13}$), leave in this form unless told to give a decimal.
Slant heightThe diagonal height along the face of a cone or pyramid, not the vertical height through the centre.
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The theorem, two operations, one decision

For any right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.

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The theorem, two operations, one decision
core concept

For any right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.

$$a^2 + b^2 = c^2$$

where $c$ is the hypotenuse (the side opposite the right angle) and $a$ and $b$ are the two shorter sides.

This relationship connects all three sides. If you know two sides, you can always find the third.

Two operations, one decision. Finding the hypotenuse: $c = \sqrt{a^2 + b^2}$, square both, add, then square root. Finding a shorter side: $a = \sqrt{c^2 - b^2}$, square both, subtract, then square root. Write the rearranged form before substituting, never rely on memory alone.
Memorise these right-triangle side ratios.

Pythagoras’ theorem: a² + b² = c² where c is the hypotenuse. To find the hypotenuse: c = √(a²+b²). To find a shorter side: a = √(c²−b²). The decision of which operation depends on what is unknown.

Pause, copy Pythagoras' theorem (c² = a² + b²), the rearrangement for the hypotenuse (c = √(a²+b²)), and the rearrangement for a shorter side (a = √(c²−b²)) into your book.

Did you get this? True or false: to find a shorter side of a right-angled triangle, you add $c^2 + b^2$ and take the square root.

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Worked examples · 3 in a row, reveal as you go

Worked examples · 3 in a row, reveal as you go

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · FINDING THE HYPOTENUSE

Find the length of the hypotenuse of a right-angled triangle with shorter sides 6 cm and 8 cm.

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Finding $c$ (hypotenuse) → add $\quad c^2 = a^2 + b^2$
Label: $a = 6$, $b = 8$. We are finding the hypotenuse, so we use the addition form.
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PROBLEM 2 · FINDING A SHORTER SIDE

A right-angled triangle has hypotenuse 15 m and one shorter side of 9 m. Find the other shorter side.

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Finding a shorter side → subtract $\quad a^2 = c^2 - b^2$
$c = 15$ (hypotenuse), $b = 9$ (known shorter side). Write the rearranged form explicitly.
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PROBLEM 3 · SURD ANSWER & CONE SLANT HEIGHT

A cone has a vertical height of 8 cm and a base radius of 6 cm. Find the slant height correct to 2 decimal places.

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Identify the right triangle inside the cone: legs $r = 6$, $h = 8$; hypotenuse = slant height $\ell$
The vertical height, base radius, and slant height form a right-angled triangle inside the cone. Draw this triangle separately.
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Quick check: A right-angled triangle has shorter sides 5 m and 12 m. What is the length of the hypotenuse?

Find the hypotenuse or a shorter side using a² + b² = c².
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Common errors · the 3 traps that cost marks

Common errors · the 3 traps that cost marks

Common errors · the 3 traps that cost marks

Trap 01
Adding when finding a shorter side
Writing $a^2 = c^2 + b^2$ instead of $a^2 = c^2 - b^2$ produces an answer larger than the hypotenuse, impossible. Quick check: the answer must be smaller than $c$. If it is not, you added when you should have subtracted.
Trap 02
Misidentifying the hypotenuse
In an unusually oriented triangle, labelling the wrong side as $c$. Fix: ignore orientation entirely. Find the right-angle square. The hypotenuse is directly opposite it, always. Mark it before writing anything.
Trap 03
Rounding the square root too early
Calculating $c^2 = 74$, then writing $c \approx 8.6$, then using 8.6 in further calculations. Keep the exact surd $\sqrt{74}$ through all intermediate steps. Only convert to decimal at the very last step, or store in your calculator's memory.

Fill the gap: A right-angled triangle has hypotenuse $c = 13$ and one side $b = 5$. The other side is $a = \sqrt{13^2 - \square^2} = \sqrt{144} = \square$ — the missing side is and the answer is .

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Quick-fire practice · 11 calculations

Quick-fire practice · 11 calculations

Quick-fire practice · 11 calculations

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Find the hypotenuse: shorter sides 3 cm and 4 cm.

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Find the hypotenuse: shorter sides 5 m and 12 m.

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Find the hypotenuse: shorter sides 7 cm and 9 cm (answer to 2 d.p.).

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Find the hypotenuse: shorter sides 1.5 m and 2 m (answer to 2 d.p.).

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Find the shorter side: hypotenuse 13 cm, one side 5 cm.

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Find the shorter side: hypotenuse 17 m, one side 8 m.

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Find the shorter side: hypotenuse 10 cm, one side 4 cm (answer to 2 d.p.).

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Find the shorter side: hypotenuse 7.5 m, one side 4.5 m (answer to 2 d.p.).

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A 5 m ladder leans against a wall. The base is 2 m from the wall. How high up the wall does the ladder reach (to 2 d.p.)?

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A cone has vertical height 12 cm and base radius 5 cm. Find the slant height (to 2 d.p.).

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A rectangular room is 9 m long and 5 m wide. Find the length of the diagonal of the floor to 2 d.p.

Match it: Match each practical context to the correct Pythagoras formula.

  • Diagonal of rectangle
  • Slant height of cone
  • Ladder against wall
  • ℓ² = r² + h²
  • ladder² = wall² + ground²
  • d² = ℓ² + w²
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Revisit your thinking

Earlier you estimated how high up the wall a 5 m ladder reached with its base 2 m from the wall.

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Revisit your thinking

Earlier you estimated how high up the wall a 5 m ladder reached with its base 2 m from the wall. Let's check:

$h^2 = 5^2 - 2^2 = 25 - 4 = 21$  →  $h = \sqrt{21} \approx 4.58\text{ m}$

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

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Show what you have learned

Multiple choice, then short answer under exam conditions.

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer
ApplyBand 42 marks

Q1. Find the value of $x$ where $x$ is the hypotenuse, and the two shorter sides are 11 cm and 5 cm. Give your answer correct to 2 decimal places. (2 marks)

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ApplyBand 43 marks

Q2. A cone has a vertical height of 10 cm and a base diameter of 12 cm. (a) Write down the radius of the base. (1 mark) (b) Find the slant height of the cone correct to 2 decimal places. (2 marks) (3 marks total)

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AnalyseBand 54 marks

Q3. A builder installs a diagonal brace across a rectangular gate that is 2.4 m wide and 1.8 m tall. (a) Find the length of the diagonal brace correct to 2 decimal places. (2 marks) (b) Timber costs $12.50 per metre. Find the cost of the brace, rounding up to the nearest 10 cm to ensure the timber is long enough. (2 marks) (4 marks total)

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📖 Comprehensive answers (click to reveal)

Drill 1: $c = 5\text{ cm}$  ·  2: $c = 13\text{ m}$  ·  3: $c = \sqrt{130} = 11.40\text{ cm}$  ·  4: $c = \sqrt{6.25} = 2.50\text{ m}$  ·  5: $a = 12\text{ cm}$  ·  6: $a = 15\text{ m}$  ·  7: $a = \sqrt{84} = 9.17\text{ cm}$  ·  8: $a = 6.00\text{ m}$  ·  9: $h = \sqrt{21} = 4.58\text{ m}$  ·  10: $\ell = 13.00\text{ cm}$  ·  11: $d = \sqrt{106} = 10.30\text{ m}$

Q1 (2 marks): $x^2 = 11^2 + 5^2 = 121 + 25 = 146$; $x = \sqrt{146} = \mathbf{12.08\text{ cm}}$ [2].

Q2 (3 marks): (a) $r = 12 \div 2 = \mathbf{6\text{ cm}}$ [1]. (b) $\ell^2 = 6^2 + 10^2 = 36 + 100 = 136$; $\ell = \sqrt{136} = \mathbf{11.66\text{ cm}}$ [2].

Q3 (4 marks): (a) $d^2 = 2.4^2 + 1.8^2 = 5.76 + 3.24 = 9$; $d = \sqrt{9} = \mathbf{3\text{ m}}$ [2]. (b) Brace is exactly 3.0 m, no rounding up needed. Cost $= 3 \times \$12.50 = \mathbf{\$37.50}$ [2].