Why the theorem works | Year 8

01 / LEARN

Understand the idea.

The equation compares square areas

Think of $a^{2}$ , $b^{2}$ and $c^{2}$ as the areas of squares built on the triangle's sides. For a right triangle, the two leg-square areas together equal the hypotenuse-square area.

This is a relationship between squared lengths, not lengths themselves. A $3$ – $4$ – $5$ right triangle has $9 + 16 = 25$ square units on its side squares; it does not claim that $3 + 4 = 5$ .

A rearrangement explains why

In the supplied dissection, four congruent right triangles fit inside a square of side $a + b$ . Their hypotenuses form a central square of side $c$ . Adjacent acute triangle angles sum to $90^{\circ}$ , leaving each central angle $90^{\circ}$ .

The outer square has area ${(a + b)}^{2}$ , and the four triangles have total area $4 \times \frac{a b}{2} = 2 a b$ . Subtraction leaves $c^{2} = {(a + b)}^{2} - 2 a b = a^{2} + b^{2}$ . Because $a$ and $b$ were arbitrary legs, this explains the general theorem.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Compare three squares

Squares on a 3–4–5 right triangle have what areas?

Worked solution

1
Find the areas of the two leg squares.
$3^{2} = 9, 4^{2} = 16$
A square's area equals its side length multiplied by itself.
2
Find the area of the hypotenuse square.
$5^{2} = 25$
The square on the longest side uses that side as its own side length.
3
Compare the areas.
$9 + 16 = 25$
The leg squares have areas 9 and 16 square units, and together equal the 25-square-unit hypotenuse square.

Answer

9, 16 and 25 square units.

EXAMPLE 02

Use the large square

For a = 3 and b = 4, use the dissection to find the central square area.

Worked solution

1
Calculate the outer square's area.
${(3 + 4)}^{2} = 7^{2} = 49$
The outer side consists of one leg of each length.
2
Calculate the four triangles' total area.
$4 \times \frac{3 \times 4}{2} = 24$
Each right triangle has perpendicular base and height 3 and 4, so its area is 6 square units.
3
Subtract to find the central area.
$c^{2} = 49 - 24 = 25$
The outer square is exactly covered by the four triangles and the central square.
4
Find the central square's side.
$c = \sqrt{25} = 5$
The central area is 25 square units, so its positive side length is 5 units.

Answer

25 square units, so c = 5 units.

03 / YOUR TURN

Try it yourself.

Check your understanding

If the leg-square areas are 36 and 64, what is the hypotenuse-square area?

Show answer & reasoning

100 square units.

The hypotenuse-square area is the sum of the leg-square areas: $36 + 64 = 100$ square units. Its side length would be $\sqrt{100} = 10$ units.

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

Squares on two legs have areas 25 and 144 square centimetres. Find the hypotenuse-square area and the hypotenuse length.
3 marks
A square of side 9 cm is drawn on a hypotenuse. One leg-square has area 32 square centimetres. Find the other leg-square area.
2 marks
In the dissection, $a = 5$ and $b = 12$ . Find the total area of the four triangles.
2 marks
For $a = 5$ and $b = 12$ , find the outer square area and central square area.
3 marks
Explain why the 6–8–10 relationship is $36 + 64 = 100$ , not $6 + 8 = 10$ .
2 marks
Expand ${(a + b)}^{2}$ and subtract $2 a b$ . Explain what remains.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Squares on two legs have areas 25 and 144 square centimetres. Find the hypotenuse-square area and the hypotenuse length.
3 marks
Add $25 + 144 = 169$ .
Take the positive square root.
169 square centimetres; 13 cm.
A square of side 9 cm is drawn on a hypotenuse. One leg-square has area 32 square centimetres. Find the other leg-square area.
2 marks
The hypotenuse-square area is $9^{2} = 81$ .
Subtract $81 - 32 = 49$ .
49 square centimetres.
In the dissection, $a = 5$ and $b = 12$ . Find the total area of the four triangles.
2 marks
Each area is $\frac{5 (12)}{2} = 30$ .
Four copies give 120.
120 square units.
For $a = 5$ and $b = 12$ , find the outer square area and central square area.
3 marks
The outer side is 17, so its area is 289.
Subtract the four triangles’ area 120.
289 and 169 square units.
Explain why the 6–8–10 relationship is $36 + 64 = 100$ , not $6 + 8 = 10$ .
2 marks
Those areas are the squared side lengths.
The lengths themselves are not added.
The theorem relates the areas of squares on the sides.
Expand ${(a + b)}^{2}$ and subtract $2 a b$ . Explain what remains.
3 marks
Expansion gives $a^{2} + 2 a b + b^{2}$ .
Removing $2 a b$ leaves the two square terms.
$a^{2} + b^{2}$ .

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics (2026)

Hypotenuse and right-angled triangles; why the theorem works; finding the hypotenuse; finding a shorter side; exact roots versus rounded answers; checking whether a triangle is right-angled; practical length problems.

Official curriculum source ↗

In this lesson

Understand the idea.

The equation compares square areas

A rearrangement explains why

See the method in action.

Compare three squares

Find the areas of the two leg squares.

Find the area of the hypotenuse square.

Compare the areas.

Use the large square

Calculate the outer square's area.

Calculate the four triangles' total area.

Subtract to find the central area.

Find the central square's side.

Try it yourself.

Check your understanding

Keep practising