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.

50–55 min MS-M1 3 MC 3 SA Lesson 3 of 22 Free
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Think First

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?

Type your initial response below — you will revisit this at the end of the lesson.

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Come back to this at the end of the lesson.

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Key Relationships — This Lesson

$a^2 + b^2 = c^2$
$c$ = hypotenuse (side opposite the right angle — always the longest side) $a$, $b$ = the two shorter sides Finding hypotenuse: $c = \sqrt{a^2 + b^2}$  |  Finding shorter side: $a = \sqrt{c^2 - b^2}$
$\ell^2 = r^2 + h^2$
Slant height of a cone — $r$ = base radius, $h$ = vertical height, $\ell$ = slant height The slant height is the hypotenuse of the right triangle inside the cone
Hypotenuse → ADD the squares, then square root   |   Shorter side → SUBTRACT the squares, then square root
PYTHAGORAS' THEOREM a b c (hypotenuse) a² + b² = c² Find hypotenuse: c = √(a² + b²) Find shorter side: a = √(c² − b²) c is always opposite the right angle (90°) — it is always the longest side

Know

  • 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

  • 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

  • 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 Vocabulary

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

Misconceptions to Fix

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Wrong: Capacity and volume are completely different concepts.

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Right: Capacity and volume are related; 1 litre = 1000 cm³ = 1 dm³.

Core Content

01

The Theorem

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 of any right-angled triangle. 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.
COMMON PYTHAGOREAN TRIPLES — MEMORISE THESE 4 3 5 3 3 – 4 – 5 12 5 13 5 – 12 – 13 15 8 17 8 – 15 – 17 Any multiple also works: 6-8-10 9-12-15
02

Identifying the Hypotenuse in Any Orientation

Pythagoras' theorem only works when one angle is exactly 90°. The right angle is marked with a small square — always find it first.

The hypotenuse is always:

  • Opposite the right angle
  • The longest side

These two descriptions always agree. If the side you identified as the hypotenuse is not the longest, you have made an error.

Orientation trap: HSC questions frequently draw triangles in unusual orientations — tilted, flipped, with the right angle at the top. The hypotenuse is still opposite the right angle regardless of how the triangle is drawn. Ignore orientation. Find the right angle square. Label the opposite side as $c$.
03

Pythagoras in Practical Contexts

Pythagoras appears whenever a diagonal, slant, or diagonal distance is involved. The strategy is always: find the hidden right-angled triangle.

ContextThe right triangleFormula
Slant height of a cone Vertical height $h$, base radius $r$, slant height $\ell$ (hypotenuse) $\ell^2 = r^2 + h^2$
Diagonal of a rectangle Length $\ell$, width $w$, diagonal $d$ (hypotenuse) $d^2 = \ell^2 + w^2$
Ladder against a wall Wall height, ground distance, ladder length (hypotenuse) $\text{ladder}^2 = \text{wall}^2 + \text{ground}^2$
Slant height of a pyramid Vertical height, half the base width, slant height (hypotenuse) $\ell^2 = h^2 + (b/2)^2$
3D strategy: Identify a right-angled triangle hidden inside the 3D shape. Draw it separately on your working page. Label the three sides. Apply the theorem to that 2D triangle. This is the professional approach — examiners reward it with clear method marks.
04

Common Mistakes

Mistake 1 — 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.
Mistake 2 — 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.
Mistake 3 — Rounding the square root before the final step
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.

Worked Examples

05
Worked Example 1
Finding the Hypotenuse

Problem

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

Step-by-Step Solution

1
Identify and write the correct form
Finding $c$ (hypotenuse) → add
$c^2 = a^2 + b^2$
Label: $a = 6$, $b = 8$. We are finding the hypotenuse, so we use the addition form.
06
Worked Example 2
Finding a Shorter Side

Problem

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

Step-by-Step Solution

1
Identify and rearrange
Finding a shorter side → subtract
$a^2 = c^2 - b^2$
$c = 15$ (hypotenuse), $b = 9$ (known shorter side). Write the rearranged form explicitly.
07
Worked Example 3
Non-Integer Answer

Problem

Find the length of the hypotenuse of a right-angled triangle with shorter sides 5 cm and 7 cm. Give your answer correct to 2 decimal places.

Step-by-Step Solution

1
Set up
$c^2 = 5^2 + 7^2 = 25 + 49 = 74$
Finding hypotenuse — add the squares.
08
Worked Example 4
Slant Height of a Cone

Problem

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.

Step-by-Step Solution

1
Identify the right triangle inside the cone
Legs: $r = 6$ cm, $h = 8$ cm
Hypotenuse: slant height $\ell$
The vertical height, base radius, and slant height form a right-angled triangle inside the cone. Draw this triangle separately on your page.
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Pythagoras Fluency and Application

Write the formula, rearrange if needed, substitute, then solve. Show every line.

Section A — Find the Hypotenuse

1 Shorter sides: 3 cm and 4 cm

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2 Shorter sides: 5 m and 12 m

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3 Shorter sides: 7 cm and 9 cm (answer to 2 decimal places)

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4 Shorter sides: 1.5 m and 2 m (answer to 2 decimal places)

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Section B — Find the Shorter Side

5 Hypotenuse 13 cm, one side 5 cm

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6 Hypotenuse 17 m, one side 8 m

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7 Hypotenuse 10 cm, one side 4 cm (answer to 2 decimal places)

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8 Hypotenuse 7.5 m, one side 4.5 m (answer to 2 decimal places)

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Section C — Practical Applications

9 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? Answer to 2 decimal places.

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10 A cone has vertical height 12 cm and base radius 5 cm. Find the slant height. Answer to 2 decimal places.

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11 A rectangular room is 9 m long and 5 m wide. Find the length of the diagonal of the floor to 2 decimal places.

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Show Answers

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Revisit Your Initial Thinking

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

Multiple Choice

1 A right-angled triangle has shorter sides of 9 cm and 12 cm. The length of the hypotenuse is:

A   10.6 cm
B   15 cm
C   21 cm
D   225 cm

? Regarding this topic, 1 A right-angled triangle has shorter sides of 9 cm and 12 cm. The length of the hypotenuse is:

A     10.6 cm
B     15 cm
C     21 cm
D     225 cm
B - Correct!
B — $c^2 = 9^2 + 12^2 = 81 + 144 = 225$. $c = \sqrt{225} = 15$ cm. This is a 9-12-15 triple (the 3-4-5 triple scaled by 3).

2 A right-angled triangle has hypotenuse 20 m and one side of 16 m. The other side is:

A   4 m
B   10 m
C   12 m
D   $\sqrt{514}$ m

? Regarding this topic, 2 A right-angled triangle has hypotenuse 20 m and one side of 16 m. The other side is:

A     4 m
B     10 m
C     12 m
D     $\sqrt{514}$ m
C - Correct!
C — $a^2 = 20^2 - 16^2 = 400 - 256 = 144$. $a = 12$ m. Option A (4) results from subtracting the roots instead of the squares.

3 A cone has vertical height 9 cm and base radius 12 cm. Its slant height is:

A   3 cm
B   7.94 cm
C   15 cm
D   21 cm

? Regarding this topic, 3 A cone has vertical height 9 cm and base radius 12 cm. Its slant height is:

A     3 cm
B     7.94 cm
C     15 cm
D     21 cm
C - Correct!
C — $\ell^2 = 12^2 + 9^2 = 144 + 81 = 225$. $\ell = 15$ cm. Note: $r = 12$ and $h = 9$ — both are the legs, slant height is the hypotenuse.

Short Answer

09

SA 4 2 marks 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.

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10

SA 5 3 marks 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)

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11

SA 6 4 marks 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)

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