Why the theorem works | Year 8

01 / LEARN

The idea, step by step

The equation compares square areas

Construct a square on each triangle side. In a right triangle, the areas on the two legs add to the area on the hypotenuse. Since a square of side a has area a squared, this gives the familiar equation.

The squares represent areas, not extra side lengths. A statement such as 3 + 4 = 5 is false even though the side lengths 3, 4 and 5 form a right triangle.

A rearrangement explains why

Put four copies of a right triangle with legs a and b inside a square of side a + b. Their other sides bound a central square of side c. Its corners are right angles because the two acute triangle angles together total 90 degrees.

The outer area is ${(a + b)}^{2}$ . The four triangles have total area $2 a b$ . Subtracting them leaves $c^{2} = {(a + b)}^{2} - 2 a b = a^{2} + b^{2}$ . This is a general area argument, not just one numerical example.

A common trap

Adding side lengths instead of their squares.

Instead: The theorem is an area relationship: add squared lengths, then take a square root for a length.

02 / WORKED EXAMPLES

See the thinking

Example 1

Compare three squares

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

Square each side length: $3^{2} = 9$ , $4^{2} = 16$ , $5^{2} = 25$ .
Add the leg-square areas: $9 + 16 = 25$ .
This equals the hypotenuse-square area.

9, 16 and 25 square units.

Example 2

Use the large square

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

The large square has side 7 and area 49.
Each triangle has area $3 (4) / 2 = 6$ ; four have area 24.
The central area is $49 - 24 = 25$ , giving side 5.

25 square units, so c = 5 units.

Check your understanding

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

Show answer

100 square units.

Add the two areas; the hypotenuse length is then 10 units.

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF 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 $5 (12) / 2 = 30$ .
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.
$a^{2} + b^{2}$ .

Checking saved progress…

Continue learning

Finding the hypotenuse

Next lesson

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 ↗

Back to Pythagoras' theorem